-------------------------------------------------------------------------- Welcome to Lecture 4 of Notes on Financial Mathematics, by Scot Adams and Fernando Reitich. We begin with a -------------------------------------------------------------------------- correction to someting I said in the last lecture. I said that -------------------------------------------------------------------------- forwards have initial value equal to zero, and this *is* true of -------------------------------------------------------------------------- *most* forwards. By -------------------------------------------------------------------------- initial value of an asset, we mean its value at the time when the contract is signed. Since our sellers don't seek any profit, a forward with no initial value has the property that -------------------------------------------------------------------------- no money changes hands at time zero. Technically, a forward need not have this property, and -------------------------------------------------------------------------- a forward that is *not* of initial value zero -------------------------------------------------------------------------- is called an off-market forward. -------------------------------------------------------------------------- That is, what we've been calling -------------------------------------------------------------------------- an obligation derivative -------------------------------------------------------------------------- is more typically called an off-market forward. -------------------------------------------------------------------------- Also, as we mentioned in Lecture 3, what we've been calling -------------------------------------------------------------------------- the receivable dollars of a forward -------------------------------------------------------------------------- is typically called its price, or forward price. By the -------------------------------------------------------------------------- price of an asset, one often means its initial value. However, that's *not* true of forwards, and those who work with them are, of course, used to this, and don't get confused. It does leave a hole in the terminology, though, and you may wonder how to refer to -------------------------------------------------------------------------- the initial value of a forward, and the answer is that -------------------------------------------------------------------------- it's typically called, not its price, but -------------------------------------------------------------------------- its mark-to-market value. Again, this amount -------------------------------------------------------------------------- is usually zero, and, when it's not, the forward is called an -------------------------------------------------------------------------- off-market forward. Okay. We now begin a review of concepts from the Calculus that we'll need in this lecture. -------------------------------------------------------------------------- Let f be the squaring function. In these lectures, we'll draw a strong distinction between functions, on the one hand, and expressions of x, on the other. Here, -------------------------------------------------------------------------- f is a function, while -------------------------------------------------------------------------- x squared is an expression of x, as is -------------------------------------------------------------------------- f(x). For any function, there's a corresponding expression of x, and, also, a corresponding expression of t. For the squaring function, the corresponding expression of t is -------------------------------------------------------------------------- t squared. Of course, given a function, one has an expression for any possible independent variable, not just x and t. While expressions are important, they do rely on a somewhat arbitrary choice of independent variable, and so some people prefer to avoid them, and to focus on the underlying function. In this example, the function -------------------------------------------------------------------------- f is the squaring function. Some people are uncomfortable simply leaving an -------------------------------------------------------------------------- open space where the variable would go, and prefer to -------------------------------------------------------------------------- insert a dot. To evaluate -------------------------------------------------------------------------- f at 3, the standard notation is to surround the number 3 by -------------------------------------------------------------------------- parentheses, to the left of the function. Here we get three squared, or -------------------------------------------------------------------------- 9. The -------------------------------------------------------------------------- left hand side is read "f of 3". For expressions of t, we'll evaluate at t equals 3 by -------------------------------------------------------------------------- surrounding the expression by brackets, and putting t equals 3 as a subscript. Again, we get 3 squared or -------------------------------------------------------------------------- 9. For expressions of x, -------------------------------------------------------------------------- we do something similar. This notation extends to other possible independent variables. Sometimes we want to calculate how much a function or expression changes as the input moves from one number to another. So, for example, -------------------------------------------------------------------------- this denotes the change in x squared, as x changes from 3 to 4. It's equal to 4 squared minus 3 squared, or -------------------------------------------------------------------------- 16 minus 9, or 7. For expressions of t, -------------------------------------------------------------------------- we do something similar. For functions, we'll indicate the same concept with -------------------------------------------------------------------------- a vertical bar followed by the two numbers in the superscript and subscript positions. Let's move on to differentiation. -------------------------------------------------------------------------- Here's the same expression of x and -------------------------------------------------------------------------- the same expression of t. Here's -------------------------------------------------------------------------- the same function, -------------------------------------------------------------------------- sometimes denoted dot squared. To differentiate an expression of x with respect to x, we -------------------------------------------------------------------------- precede the expression by -------------------------------------------------------------------------- d over dx, read simply "dee dee x". In this case, the derivative is -------------------------------------------------------------------------- 2 x For expressions of t, -------------------------------------------------------------------------- we use dee dee t, which denotes differentiation with respect to t. Here, we get -------------------------------------------------------------------------- 2 t. For functions, -------------------------------------------------------------------------- we follow the function by a prime. Thus the derivative of squaring is -------------------------------------------------------------------------- doubling, sometimes denoted 2 times -------------------------------------------------------------------------- dot. Now take this doubling function, and form the corresponding expression of x, namely -------------------------------------------------------------------------- f'(x). This gives -------------------------------------------------------------------------- 2x, -------------------------------------------------------------------------- which also appears up here. -------------------------------------------------------------------------- Same thing for t. Next, we describe some of our conventions. We -------------------------------------------------------------------------- allow for a variety of notations, in particular -------------------------------------------------------------------------- differentiation with respect to x as dee dee x, -------------------------------------------------------------------------- differentiation with respect to t as dee dee t, and -------------------------------------------------------------------------- differentiation as prime. In *our* lectures, we do -------------------------------------------------------------------------- *not* allow certain things as well. For example, we don't allow -------------------------------------------------------------------------- dee dee x on a function, only on an expression of x. We don't allow -------------------------------------------------------------------------- prime to denote dee dee t -------------------------------------------------------------------------- or dee dee u. Any of -------------------------------------------------------------------------- these is bad form. Finally, we have some notational conventions that aren't perfectly clean, but *are* useful in some circumstances, simply because they shorten the amount that has to be read to parse a complicated formula. We describe these as notations that are -------------------------------------------------------------------------- usually avoided. That is to say, when we do use these notations, we'll try to be as clear as possible about what they mean. In this category is the use of -------------------------------------------------------------------------- a dot for dee dee t. One surrounds the expression by brackets or parentheses, and puts a -------------------------------------------------------------------------- dot in the superscript. Similarly, we may sometimes use -------------------------------------------------------------------------- prime as an abbreviation for dee dee x. We said above that -------------------------------------------------------------------------- prime for dee dee u is verboten, but, in fact, we should qualify that to say that we -------------------------------------------------------------------------- might sometimes use prime as an abbreviation for dee dee u, but *if* we choose to do such a thing, it's particularly important to say something to highlight the exception. Same thing for other independent variables, with the exception of "t". -------------------------------------------------------------------------- A small dot will be reserved for dee dee t, and so there's no need ever to use prime for dee dee t. Also, we -------------------------------------------------------------------------- don't allow dot to go on a function. Let's -------------------------------------------------------------------------- clear some space, and go through some examples. Recall that -------------------------------------------------------------------------- dee dee x of sin x is cos x. Thus, if we set -------------------------------------------------------------------------- u equal to sin x, -------------------------------------------------------------------------- then we get -------------------------------------------------------------------------- dee dee x of u, and when we take dee dee x of a single variable, we allow ourselves to shorten the notation, -------------------------------------------------------------------------- moving the variable into the numerator. We thus get -------------------------------------------------------------------------- dee u dee x. Note that the dependent variable -------------------------------------------------------------------------- $u$ here is *not* a function. It's an -------------------------------------------------------------------------- expression of $x$, namely sine x. The underlying function is -------------------------------------------------------------------------- sine, with derivative cosine. It causes a small degree of heartburn, but we *do* sometimes allow ourselves to use -------------------------------------------------------------------------- prime for dee dee x, in which case sin x prime is cos x. Since u is sin x, we get -------------------------------------------------------------------------- u prime here. When we write something like -------------------------------------------------------------------------- this, we'd typically elucidate verbally that prime is being used to mean dee dee x. -------------------------------------------------------------------------- Dee dee t of t to the fifth is five t to the fourth, so if we set -------------------------------------------------------------------------- v equal to t to the fifth, -------------------------------------------------------------------------- then we get dee dee t of v, or -------------------------------------------------------------------------- dee v dee t. Using the dot notation, we can -------------------------------------------------------------------------- shorten this a bit. Since v is equal to t to the fifth, we get -------------------------------------------------------------------------- v dot. Note that, when applying dot to a single variable, the dot is typically placed -------------------------------------------------------------------------- directly above the letter, *not* as a superscript. -------------------------------------------------------------------------- Here we see the sine function, and its -------------------------------------------------------------------------- avatars as expressions of x and t. -------------------------------------------------------------------------- Here are the derivatives. Note again that -------------------------------------------------------------------------- dee dee x of f of x is -------------------------------------------------------------------------- f prime of x, and that -------------------------------------------------------------------------- dee dee t of f of t is -------------------------------------------------------------------------- f prime of t. -------------------------------------------------------------------------- Here's cosine, in three forms. -------------------------------------------------------------------------- Here we see the derivatives. Note that, while the derivative of sine is cosine, the derivative of cosine is the -------------------------------------------------------------------------- negative of sine. Once again, the derivatives of the expressions are obtained from the -------------------------------------------------------------------------- derivative of the underlying function, -------------------------------------------------------------------------- reincarnated as expressions. -------------------------------------------------------------------------- Here's the logarithm naturalis, or natural logarithm, or log base e, or, simply, log. We display it in the usual -------------------------------------------------------------------------- three forms. -------------------------------------------------------------------------- Here are the derivatives. In particular, we get that f prime is -------------------------------------------------------------------------- one over dot. Once again, the derivatives of the expressions are obtained from the -------------------------------------------------------------------------- derivative of the underlying function, -------------------------------------------------------------------------- reincarnated as expressions. Is this getting old? Let's state the general fact, once and for all. Say you take any function -------------------------------------------------------------------------- $f$, then -------------------------------------------------------------------------- change it to an expression of x, and then -------------------------------------------------------------------------- differentiate with respect to x. If you now take -------------------------------------------------------------------------- the derivative of f and -------------------------------------------------------------------------- change *that* to an expression of x, then you always -------------------------------------------------------------------------- get the same result, just by definition of -------------------------------------------------------------------------- dee dee x and prime. We have a -------------------------------------------------------------------------- similar fact for the independent variable t, or for *any* independent variable. Expressions and functions need not be continuous. For example, -------------------------------------------------------------------------- let the expression u of x be as given here. Note that we've designated -------------------------------------------------------------------------- $u$ as the *de*pendent variable; computation of $u$ depends on -------------------------------------------------------------------------- $x$. So then $x$ is the independent variable. -------------------------------------------------------------------------- Here's the graph of u against x. We say that -------------------------------------------------------------------------- u is caglad in x, where caglad is an abbreviation of the French expression -------------------------------------------------------------------------- "continue a gauche, limite a droite", which, translated from the French, becomes "continuous from the left, with limits from the right". Indeed, even at x equals 2, one sees -------------------------------------------------------------------------- continuity from the left, and the existence of a -------------------------------------------------------------------------- limit from the right. -------------------------------------------------------------------------- The corresponding expression of -------------------------------------------------------------------------- t will be denoted -------------------------------------------------------------------------- v on this slide. -------------------------------------------------------------------------- Here's the graph of v against t. -------------------------------------------------------------------------- Note that v is caglad in t. -------------------------------------------------------------------------- The corresponding function will be denoted f on this slide. -------------------------------------------------------------------------- Here's the graph of f. -------------------------------------------------------------------------- Note that f is caglad. Now we begin -------------------------------------------------------------------------- the Product Rule. Suppose we take the product of two expressions of x, -------------------------------------------------------------------------- f(x) times -------------------------------------------------------------------------- g(x), and say we want to compute -------------------------------------------------------------------------- the derivative of the product. I should say "derivative *with respect to x*", but, on this slide, all derivatives will be with respect to x, so I'll omit saying it. Here, we call -------------------------------------------------------------------------- f(x) the first part of the product, and -------------------------------------------------------------------------- g(x) the second part of the product. As a result, the Product Rule is sometimes called -------------------------------------------------------------------------- differentiation by parts. According to the Product Rule, we take -------------------------------------------------------------------------- the first part, -------------------------------------------------------------------------- which is f(x), -------------------------------------------------------------------------- times the derivative of the second part, -------------------------------------------------------------------------- which we'll fill in later. Next comes -------------------------------------------------------------------------- plus the derivative of the first part, -------------------------------------------------------------------------- which we'll fill in later, -------------------------------------------------------------------------- times the second part, -------------------------------------------------------------------------- which is g(x). Next, we take the derivative of -------------------------------------------------------------------------- f(x), and -------------------------------------------------------------------------- put it where the arrow points, -------------------------------------------------------------------------- like so. Finally, we take the derivative of -------------------------------------------------------------------------- g(x), and -------------------------------------------------------------------------- put it where the arrow points, -------------------------------------------------------------------------- like so. -------------------------------------------------------------------------- This completes the formula. There's a similar formula for -------------------------------------------------------------------------- expressions of t, and, as you can imagine, a formula for any independent variable. There's also a formula for functions, in which -------------------------------------------------------------------------- the derivative of fg is equal to -------------------------------------------------------------------------- the first part times -------------------------------------------------------------------------- the derivative of the second -------------------------------------------------------------------------- plus the derivative of the first part -------------------------------------------------------------------------- times the second. Let's do an example. Let's compute the -------------------------------------------------------------------------- derivative, with respect to u, of -------------------------------------------------------------------------- u squared e to the u. Throughout this slide, derivative will mean "derivative with respect to u". -------------------------------------------------------------------------- Here's the first part. -------------------------------------------------------------------------- Here's the second. -------------------------------------------------------------------------- Here's the first part. -------------------------------------------------------------------------- Here we'll put the derivative of the second. -------------------------------------------------------------------------- Next comes plus. -------------------------------------------------------------------------- Here we'll put the derivative of the first part. -------------------------------------------------------------------------- Here's the second. The derivative of -------------------------------------------------------------------------- u squared goes -------------------------------------------------------------------------- where the arrow points, -------------------------------------------------------------------------- like so. The derivative of -------------------------------------------------------------------------- e to the u goes -------------------------------------------------------------------------- where the arrow points, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- finishing the problem. There are actually many versions the Product Rule. Let's do the -------------------------------------------------------------------------- same problem -------------------------------------------------------------------------- another way. Let's take -------------------------------------------------------------------------- the derivative of the first part -------------------------------------------------------------------------- times the second -------------------------------------------------------------------------- plus the first part -------------------------------------------------------------------------- times the derivative of the second. -------------------------------------------------------------------------- We now fill in this derivative, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- and this derivative, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- finishing the problem. Now we move on to the -------------------------------------------------------------------------- Quotient Rule, which we -------------------------------------------------------------------------- display using functions. We'll refer to -------------------------------------------------------------------------- the numerator, f, as "high" and -------------------------------------------------------------------------- the denominator, g, as "low". Then -------------------------------------------------------------------------- gf prime becomes -------------------------------------------------------------------------- "low dee high", where -------------------------------------------------------------------------- "dee" is short for "derivative". -------------------------------------------------------------------------- The minus sign is identified by the word -------------------------------------------------------------------------- "less". -------------------------------------------------------------------------- fg prime is -------------------------------------------------------------------------- "high dee low". Moving to the -------------------------------------------------------------------------- denominator is conveyed by the word -------------------------------------------------------------------------- "underneath". Finally, -------------------------------------------------------------------------- g squared is rendered as -------------------------------------------------------------------------- "low squared". So our mnemonic for remembering the Quotient Rule becomes -------------------------------------------------------------------------- "low dee high less high dee low, and underneath, low squared'll go". This comes from a cute little tune in a wonderful book on the Calculus published many moons ago by Bonic et al. We now leave it as an -------------------------------------------------------------------------- exercise to write out the corresponding Quotient Rule formulas for -------------------------------------------------------------------------- expressions of x and -------------------------------------------------------------------------- expressions of t. Of course, there's one such formula for any possible independent variable. -------------------------------------------------------------------------- Here's a sample problem. In this slide, all differentiation is with respect to t. -------------------------------------------------------------------------- High is sin t and -------------------------------------------------------------------------- low is cosine t. -------------------------------------------------------------------------- "Low dee high" is -------------------------------------------------------------------------- cos t followed by some empty space where we'll eventually insert "dee high". -------------------------------------------------------------------------- "Less" is a minus sign. -------------------------------------------------------------------------- "High dee low" is -------------------------------------------------------------------------- sin t followed by some empty space where we'll eventually insert "dee low". -------------------------------------------------------------------------- "And underneath" tells us to -------------------------------------------------------------------------- divide. -------------------------------------------------------------------------- "Low squared'll go" tells us to put -------------------------------------------------------------------------- the square of cosine t in the down under. To complete the problem, we take the derivative of -------------------------------------------------------------------------- cosine t, -------------------------------------------------------------------------- and put it here, -------------------------------------------------------------------------- like so, and then take the derivative of -------------------------------------------------------------------------- sine t, -------------------------------------------------------------------------- and put it here, -------------------------------------------------------------------------- like so. Now let's remember our trig. -------------------------------------------------------------------------- Sine t divided by -------------------------------------------------------------------------- cosine t is -------------------------------------------------------------------------- tangent t, so the work we've just done gives us -------------------------------------------------------------------------- dee dee t of tangent t. In -------------------------------------------------------------------------- this numerator, the -------------------------------------------------------------------------- two minus signs cancel, and we have -------------------------------------------------------------------------- cosine squared t plus sine squared t. We leave -------------------------------------------------------------------------- the denominator (pause...............) -------------------------------------------------------------------------- unchanged. According to Pythagorus, -------------------------------------------------------------------------- cosine squared plus sine squared is one, so we get -------------------------------------------------------------------------- one over cosine squared t. Tapping into the recesses of our long term memory, we recover that the reciprocal of cosine is secant, and so the final answer can be rendered as -------------------------------------------------------------------------- secant squared t. Restating this in terms of functions, we see that -------------------------------------------------------------------------- the derivative of tangent is secant squared. The basic rules of differentiation are Product, Quotient and -------------------------------------------------------------------------- Chain, these three, but the greatest of these is the Chain Rule, which tells us how to differentiate an expression of the form -------------------------------------------------------------------------- f of g of x. This breaks up as: an expression, -------------------------------------------------------------------------- g of x, plugged into a function, -------------------------------------------------------------------------- f. Chain Rule says that to find -------------------------------------------------------------------------- dee dee x of this, we -------------------------------------------------------------------------- take the derivative of the function, -------------------------------------------------------------------------- like so, and then -------------------------------------------------------------------------- plug in the expression, -------------------------------------------------------------------------- like so, and then -------------------------------------------------------------------------- multiply by the derivative of the expression -------------------------------------------------------------------------- like so. Of course, -------------------------------------------------------------------------- $x$ can be replaced by any other independent variable, as usual. As an example, let's do -------------------------------------------------------------------------- log of sine of x, so the expression is -------------------------------------------------------------------------- sine x and the function is -------------------------------------------------------------------------- log. To find -------------------------------------------------------------------------- dee dee x of this, we -------------------------------------------------------------------------- take the derivative of the function, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- plug in the expression, -------------------------------------------------------------------------- like so, and then we -------------------------------------------------------------------------- multiply by the derivative of the expression, -------------------------------------------------------------------------- like so. In this case, we end up with -------------------------------------------------------------------------- dee dee x of sine x, which is -------------------------------------------------------------------------- cosine x, and that finishes the problem, unless you want to do some trig gyrations, and note that -------------------------------------------------------------------------- cosine x over sine x is -------------------------------------------------------------------------- cotangent x. Next, let's do -------------------------------------------------------------------------- log of tangent of x to the seventh. The expression is -------------------------------------------------------------------------- tangent x to the seventh, and the function is -------------------------------------------------------------------------- log. We take the derivative of -------------------------------------------------------------------------- the function -------------------------------------------------------------------------- like so, then plug in -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so. It remains to ------------------------------------------------------------------------- multiply by the derivative of -------------------------------------------------------------------------- the expression. This expression is, itself, -------------------------------------------------------------------------- an expression plugged into -------------------------------------------------------------------------- a function, so we recycle back through the Chain Rule. That is, we take the derivative of -------------------------------------------------------------------------- the function, -------------------------------------------------------------------------- like so, then plug in -------------------------------------------------------------------------- the expresion, -------------------------------------------------------------------------- like so, and then multiply by the derivative of -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so, and that finishes the problem, except for the usual desultory trig manipulations, which we omit. Let's put all these rules together, and do a -------------------------------------------------------------------------- practice differentiation problem. If you take -------------------------------------------------------------------------- this expression, get out your calculator and plug a number, like -------------------------------------------------------------------------- 3, into it, you'll go through many steps to get the answer, but the last thing you'll do is -------------------------------------------------------------------------- this division. That last step in evaluation is the first thing you handle in differentiation. That is, we begin with the Quotient Rule. We get -------------------------------------------------------------------------- low -------------------------------------------------------------------------- dee high -------------------------------------------------------------------------- less high -------------------------------------------------------------------------- dee low -------------------------------------------------------------------------- and underneath -------------------------------------------------------------------------- low squared'll go. Note that this -------------------------------------------------------------------------- seven was increased to -------------------------------------------------------------------------- fourteen when we took low squared. Now we take the derivative of -------------------------------------------------------------------------- this, and put it -------------------------------------------------------------------------- here. We use the Product Rule, one version of which asks for -------------------------------------------------------------------------- the derivative of the first part times -------------------------------------------------------------------------- the second -------------------------------------------------------------------------- plus the first part times -------------------------------------------------------------------------- the derivative of the second. The derivative of -------------------------------------------------------------------------- this goes -------------------------------------------------------------------------- here, and the derivative of -------------------------------------------------------------------------- this goes -------------------------------------------------------------------------- here. Really, I should be saying "the derivative with respect to x" to be completely correct, but, well, life is short. We move on, and endeavor to compute the derivative of -------------------------------------------------------------------------- this expression, and put it -------------------------------------------------------------------------- here. This expression can be written -------------------------------------------------------------------------- like this. All we've done is to move -------------------------------------------------------------------------- this seven from the inside -------------------------------------------------------------------------- to the outside, making it clear that the expression is itself -------------------------------------------------------------------------- an expression plugged into -------------------------------------------------------------------------- a function. We call on the Chain Rule. We take -------------------------------------------------------------------------- the derivative of the function -------------------------------------------------------------------------- and plug in the expression. It remains to multiply by the dervative of the expression, but let's put our work so far -------------------------------------------------------------------------- into the answer, taking care to move -------------------------------------------------------------------------- this six from the outside -------------------------------------------------------------------------- to the inside, giving a more professional looking result, with fewer parentheses. We now return to the problem of multiplying by the derivative of -------------------------------------------------------------------------- this expression. This expression is again -------------------------------------------------------------------------- an expression plugged into -------------------------------------------------------------------------- a function, and we call on the Chain Rule once more. We take the derivative of -------------------------------------------------------------------------- the function -------------------------------------------------------------------------- like so, plug in -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so, and then multiply by the derivative of -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- and this finishes the problem. (pause...............) Next, take any expression, -------------------------------------------------------------------------- f(x), then take its -------------------------------------------------------------------------- logarithm, and then -------------------------------------------------------------------------- differentiate the result. This'll give what's called the -------------------------------------------------------------------------- logarithmic derivative of f(x). Again, technically, we should say "the logarithmic derivative **with respect to x**, but we give ourselves some room for imprecision, for, otherwise, the complications of grammar will interfere with the flow and parsability of the exposition. That is to say, we ain't got time, and it'd just confuse things anyway. We have here -------------------------------------------------------------------------- an expression inside -------------------------------------------------------------------------- a function, so we take -------------------------------------------------------------------------- the derivative of the function -------------------------------------------------------------------------- plug in the expression -------------------------------------------------------------------------- and multiply by the derivative of the expression. Next, let's multiply both sides -------------------------------------------------------------------------- by f(x). We -------------------------------------------------------------------------- eliminate the crossed out stuff, move -------------------------------------------------------------------------- the RHS to the left -------------------------------------------------------------------------- and the LHS to the right, and the result is the -------------------------------------------------------------------------- Principle of Logarithmic Differentiation, which compares the -------------------------------------------------------------------------- derivative of f(x) to -------------------------------------------------------------------------- the logarithmic derivative of f(x). Again, to get the logarithmic derivative of -------------------------------------------------------------------------- f(x), you take the -------------------------------------------------------------------------- log and then the -------------------------------------------------------------------------- derivative. In words, the -------------------------------------------------------------------------- Principle of Logarithmic Differentiation asserts that -------------------------------------------------------------------------- to compute the derivative of an expression, for example, -------------------------------------------------------------------------- the derivative of f(x), we should -------------------------------------------------------------------------- multiply the expression, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- by its logarithmic derivative, -------------------------------------------------------------------------- like so. To see this in action, let's compute -------------------------------------------------------------------------- dee dee x of x to the x. We multiply -------------------------------------------------------------------------- the expression by -------------------------------------------------------------------------- the logarithmic (pause...............) -------------------------------------------------------------------------- derivative (pause...............) -------------------------------------------------------------------------- of the expression. In many cases, the -------------------------------------------------------------------------- logarithm simplifies the expression. For example, -------------------------------------------------------------------------- here, log of x to the x -------------------------------------------------------------------------- is x log x, and the Product Rule gives -------------------------------------------------------------------------- the derivative of the first part -------------------------------------------------------------------------- times the second -------------------------------------------------------------------------- plus the first part -------------------------------------------------------------------------- times the derivative of the second. Multiplication by -------------------------------------------------------------------------- one does nothing and -------------------------------------------------------------------------- x times one over x -------------------------------------------------------------------------- is one. -------------------------------------------------------------------------- Addition is commutative, and -------------------------------------------------------------------------- we're done. Let's move on ..... to -------------------------------------------------------------------------- Integration. To say that an expression, -------------------------------------------------------------------------- capital F(x), is an antiderivative of another expression, little f(x), -------------------------------------------------------------------------- with respect to x, -------------------------------------------------------------------------- means that dee dee x of capital F(x) is little f(x). That is, an antiderivative is the opposite of a derivative. -------------------------------------------------------------------------- For example, an antiderivative of sine x is minus cosine x, because -------------------------------------------------------------------------- sine x is dee dee x of minus cosine x. Once you have one antiderivative, the rest differ from the one by constants, so, for example, minus cosine x -------------------------------------------------------------------------- plus five is another antiderivative of sine x, and *every* antiderivative of sine x is of that form: minus cosine x plus a constant. The -------------------------------------------------------------------------- indefinite integral of f(x) -------------------------------------------------------------------------- with respect to x, by definition, -------------------------------------------------------------------------- is the set of all antiderivatives of f(x) with respect to x. -------------------------------------------------------------------------- For example, the indefinite integral of sine x with respect to x is -------------------------------------------------------------------------- the set of all expressions of the form minus cosine x plus C, where C ranges over all constants. Note that this set is the same as -------------------------------------------------------------------------- the set of all expressions of the form minus cosine x plus 2C, where C ranges over all constants. Standard notational practice in the Calculus, dating back to the pleistocene era, dictates that the set braces are to be -------------------------------------------------------------------------- dropped, creating odd looking equations that indicate that -------------------------------------------------------------------------- 2C is the same as -------------------------------------------------------------------------- C. Well, in fact, the set of doubled constants *is* just the set of constants, so -------------------------------------------------------------------------- this equality *is* true, but one has to bear in mind the missing set braces to avoid confusion. One of the most important facts from the Calculus is that antiderivatives don't just connect to *in*definite integration, but also to definite integration. If you can find -------------------------------------------------------------------------- an antiderivative of an expression little f(x), then its -------------------------------------------------------------------------- definite integral from a to b is equal to -------------------------------------------------------------------------- the change in that antiderivative between a and b. So, *if* the antiderivative is -------------------------------------------------------------------------- capital F(x), then we get -------------------------------------------------------------------------- capital F of b minus capital F of a. Again, remember that, for technical correctness, we should tack on the words -------------------------------------------------------------------------- "with respect to x" repeatedly, but we -------------------------------------------------------------------------- don't, because we're too cool by half. This result is called the -------------------------------------------------------------------------- Fundamental Theorem of Calculus. Actually, we'll present *two* forms of the Fundamental Theorem, and this is only -------------------------------------------------------------------------- the *First* Form. In an expression like -------------------------------------------------------------------------- this, if we change "x" to, say -------------------------------------------------------------------------- "t", then nothing really changes, because one just evaluates t -------------------------------------------------------------------------- at b and -------------------------------------------------------------------------- at a, and subtracts, so one still gets -------------------------------------------------------------------------- capital F of b minus capital F of a. While -------------------------------------------------------------------------- this *is* technically correct, it's odd to mix the independent variables "x" and "t" on a single line like this, and so one would likely then -------------------------------------------------------------------------- in the integral, change "x" to -------------------------------------------------------------------------- "t", as well. This has no effect on the *value* of the integral. For the same reason, it would be typical, -------------------------------------------------------------------------- here, to change "x" to -------------------------------------------------------------------------- "t", as well. The antiderivative is now -------------------------------------------------------------------------- with respect to t, which goes -------------------------------------------------------------------------- unsaid. One might also choose to change -------------------------------------------------------------------------- this to function notation -------------------------------------------------------------------------- like so, and this statement is correct as well. Next, let's replace each -------------------------------------------------------------------------- "b" with the variable -------------------------------------------------------------------------- "x". We move -------------------------------------------------------------------------- the LHS up, and then -------------------------------------------------------------------------- the RHS, so -------------------------------------------------------------------------- these two expressions of x are equal. We take -------------------------------------------------------------------------- dee dee x of both sides. Remember that, whatever -------------------------------------------------------------------------- capital F is, and whatever the constant -------------------------------------------------------------------------- a is, -------------------------------------------------------------------------- capital F of a is just a constant, and so -------------------------------------------------------------------------- differentiates to zero. -------------------------------------------------------------------------- Capital F is an antiderivative of little f, so -------------------------------------------------------------------------- dee dee x of capital F of x is -------------------------------------------------------------------------- little f of x. We now -------------------------------------------------------------------------- move left and right hand sides down, which yields -------------------------------------------------------------------------- Version A of our *Second* Form of the Fundamental Theorem of Calculus. *This* form tells us that the derivative of -------------------------------------------------------------------------- this expression is -------------------------------------------------------------------------- f of x. Put another way, if we seek an antiderivative of -------------------------------------------------------------------------- f(x), and if we know how to do definite integration, then, -------------------------------------------------------------------------- for any $a$, -------------------------------------------------------------------------- this is an antiderivative. We state this as -------------------------------------------------------------------------- Version B: An antiderivative of f(x) is -------------------------------------------------------------------------- the integral from $a$ to $x$ of f(t) dee t, where -------------------------------------------------------------------------- $a$ is arbitrary, and could even be minus infinity. The -------------------------------------------------------------------------- First Form said that, if you can find an -------------------------------------------------------------------------- antiderivative, then you can -------------------------------------------------------------------------- do definite integrals. You simply evaluate the change in the antiderivative. The Second Form says: If you can do -------------------------------------------------------------------------- definite integrals, then you can -------------------------------------------------------------------------- find antiderivatives. You simply integrate from some $a$ to $x$. Of course, once you've found one antiderivative, you've found them all, simply by adding all possible constants. We therefore have -------------------------------------------------------------------------- Version C of the Second Form, which says that to get the -------------------------------------------------------------------------- indefinite integral of f(x), we can simply take -------------------------------------------------------------------------- this antiderivative, -------------------------------------------------------------------------- like so, and -------------------------------------------------------------------------- add all possible constants. Next topic is -------------------------------------------------------------------------- Integration by Substitution. Suppose we're examining -------------------------------------------------------------------------- an indefinite integral, and decide to replace every x by some -------------------------------------------------------------------------- expression of u, say psi of u. This replacement is called subsitution, and we say that we are substituting psi of u for x. If we make this substition in -------------------------------------------------------------------------- f of x, -------------------------------------------------------------------------- we get f of psi of u, and you might wonder whether -------------------------------------------------------------------------- this integral is the same as the original. In fact, -------------------------------------------------------------------------- they're not. The problem is that, if we're going to -------------------------------------------------------------------------- substitute psi of u for x, then we should calculate that -------------------------------------------------------------------------- dee x dee u is psi prime of u, and so -------------------------------------------------------------------------- dee x is psi prime of u dee u, so this -------------------------------------------------------------------------- dee x, on substitution, becomes -------------------------------------------------------------------------- psi prime of u dee u. Now the two -------------------------------------------------------------------------- integrals *are* equal. We admonish you -------------------------------------------------------------------------- not to forget this psi prime of u, without which the formula would simply be wrong. -------------------------------------------------------------------------- This is the basic formula for Integration by Substition, when one substitutes, -------------------------------------------------------------------------- an expression of u -------------------------------------------------------------------------- for x. Of course, for any two variables, we have an Integration by Substitution formula; you could, for example, replace "x" and "u" by "z" and "t" throughout this slide, if you want. We can also, if we wish, interchange "x" and "u" throughout this slide, -------------------------------------------------------------------------- like so. In Calculus courses, victims are often confronted with an indefinite integration problem that's in -------------------------------------------------------------------------- this form, and they're expected to figure out that a clever de-substitution, -------------------------------------------------------------------------- like this, can simplify the integral. One decides to de-substitute -------------------------------------------------------------------------- psi of x to u, and then one calculates -------------------------------------------------------------------------- this, and then de-substitutes -------------------------------------------------------------------------- psi prime of x dee x to dee u, obtaining -------------------------------------------------------------------------- this integral of u. Assuming -------------------------------------------------------------------------- this new integral is one we can do, then, after it's done, we can substitute -------------------------------------------------------------------------- psi of x back in for u, which then gives the answer to the -------------------------------------------------------------------------- original problem. For example, -------------------------------------------------------------------------- here's an indefinite integral in x. Being brilliant beyond belief, we immediately recognize that we should de-substitute -------------------------------------------------------------------------- u for sine x, which appears -------------------------------------------------------------------------- here. Then -------------------------------------------------------------------------- dee u is cosine x dee x, which appears -------------------------------------------------------------------------- here. De-substitution yields -------------------------------------------------------------------------- this simpler integral of u which, being brilliant beyond belief, we know how to do. We get -------------------------------------------------------------------------- the set of expressions of the form e to the u plus a constant. We now substitute -------------------------------------------------------------------------- this for -------------------------------------------------------------------------- u, and get -------------------------------------------------------------------------- the set of expressions of the form e to the sine x plus a constant, which is the answer to -------------------------------------------------------------------------- the original problem. There's also an Integration by Substitution formula for -------------------------------------------------------------------------- definite integrals. The limits of integration on the LHS are -------------------------------------------------------------------------- a and b, so x is ranging from a to b. On the RHS, we have u ranging -------------------------------------------------------------------------- from psi of a to psi of b, and this makes perfect sense, because -------------------------------------------------------------------------- $u$ is psi of $x$. Incidentally, for this to work, we do -------------------------------------------------------------------------- need some hypotheses: For example, we need that -------------------------------------------------------------------------- psi is continuous on the closed interval from a to b, and -------------------------------------------------------------------------- differentiable on the open interval from a to b. In these lectures, we'll sometimes omit tame hypotheses like -------------------------------------------------------------------------- these, understanding that, in the situations we encounter, the hypotheses will typically hold. Let's do an example with psi of x -------------------------------------------------------------------------- equal to x plus 5. We leave -------------------------------------------------------------------------- f to be a general function. We calculate that -------------------------------------------------------------------------- psi prime of x is now the derivative of x + 5, or -------------------------------------------------------------------------- 1, so -------------------------------------------------------------------------- psi prime of x dee x is -------------------------------------------------------------------------- dee x. Again, -------------------------------------------------------------------------- psi of x dee x is -------------------------------------------------------------------------- dee x. -------------------------------------------------------------------------- Here, we can change psi of x to -------------------------------------------------------------------------- x+5. -------------------------------------------------------------------------- Here we can change psi of b and psi of a to -------------------------------------------------------------------------- b+5 and a+5. We -------------------------------------------------------------------------- clear some room and, on the right hand side, -------------------------------------------------------------------------- change u -------------------------------------------------------------------------- to x. We move the -------------------------------------------------------------------------- RHS to the left, and the -------------------------------------------------------------------------- LHS to the right, obtaining -------------------------------------------------------------------------- this equation, which tells us that, if we're examining an -------------------------------------------------------------------------- integral in x, and, if we replace -------------------------------------------------------------------------- $x$ by -------------------------------------------------------------------------- x+5, then -------------------------------------------------------------------------- dx -------------------------------------------------------------------------- is unchanged, but -------------------------------------------------------------------------- the limits of integration -------------------------------------------------------------------------- are decreased by five, compensating for the -------------------------------------------------------------------------- addition of five to the variable x. Note that, if we plug -------------------------------------------------------------------------- a plus 5 in for -------------------------------------------------------------------------- x in -------------------------------------------------------------------------- f(x), we get f of a plus 5. If we plug -------------------------------------------------------------------------- a in for -------------------------------------------------------------------------- x (pause...............) in -------------------------------------------------------------------------- f(x+5), we again get f of a plus 5, the same thing. Similarly, if we plug -------------------------------------------------------------------------- b plus 5 in for -------------------------------------------------------------------------- x here, we get the same thing as if we plug -------------------------------------------------------------------------- b in for -------------------------------------------------------------------------- x here. Let's now return to -------------------------------------------------------------------------- *in*definite integrals, for a moment. There are those who don't like -------------------------------------------------------------------------- these two formulas, and who prefer that -------------------------------------------------------------------------- psi prime of x be thought of, not as dee u dee x, but as -------------------------------------------------------------------------- a one-by-one matrix, whose only entry is dee u dee x. "Why would anyone be so silly?" I hear you think. Well, it turns out that this new matrix definition of psi prime lends itself well to a multivariable generalization, as we'll see in a moment. To distinguish between the usual derivative, and this odd one-by-one -------------------------------------------------------------------------- matrix form of the derivative, some people refer to the latter as the -------------------------------------------------------------------------- *differential* of psi (instead of *derivative*), and use -------------------------------------------------------------------------- capital D psi to denote it. We'll -------------------------------------------------------------------------- stick with psi prime in this lecture, and ask you to figure out, from context, whether psi prime is in matrix form, or not. If it *is* in matrix form, then there's a problem in that it's not really accurate to put a -------------------------------------------------------------------------- matrix, even a one-by-one matrix, here, in the middle of formula where everything else is just numbers. Incidentally, there are many different number systems around: real, complex, mod p, p-adic, and others. So the meaning of the term "number" varies with context. Recognizing the imprecision of the word "number", the hoity-toity professional mathematician typically says "scalar" to refer to a number. Of course, the imprecision is then simply transferred to another word, but, well, that's just how it is. In the present situation, we have a -------------------------------------------------------------------------- matrix, albeit a one-by-one matrix, inserted into the middle of an otherwise scalar equation, so it doesn't quite fit. The only entry in this matrix is -------------------------------------------------------------------------- dee u dee x, and, one way to refer to that entry is to take -------------------------------------------------------------------------- the determinant of the matrix, because the determinant of a one-by-one matrix *is* the unique entry inside. So, for those who want psi prime of x to be a one-by-one -------------------------------------------------------------------------- matrix, we can fix up the substitution equation, by -------------------------------------------------------------------------- putting in a determinant. Let's -------------------------------------------------------------------------- clear some room and -------------------------------------------------------------------------- change back to definite integrals, -------------------------------------------------------------------------- from $a$ to $b$ and from -------------------------------------------------------------------------- psi of a to psi of b. Let's assume that -------------------------------------------------------------------------- $a$ is less than $b$, and let's let -------------------------------------------------------------------------- D denote the open interval from a to b. Then we can take -------------------------------------------------------------------------- the integral from a to b, and -------------------------------------------------------------------------- replace it by -------------------------------------------------------------------------- the integral over D. Now let -------------------------------------------------------------------------- E denote the image of D under psi. If we assume that dee u dee x is -------------------------------------------------------------------------- positive on D, then psi is increasing on D. If we also assume that -------------------------------------------------------------------------- psi is continuous at the endpoints a and b of D, then we can conclude that -------------------------------------------------------------------------- psi of a is less than psi of b, and, also, that -------------------------------------------------------------------------- E is the open interval from psi of a to psi of b. Then we can take -------------------------------------------------------------------------- the integral from psi of a to psi of b, and -------------------------------------------------------------------------- replace it by -------------------------------------------------------------------------- the integral over E. We needed to assume -------------------------------------------------------------------------- continuity of psi at the endpoints of D in order to be sure of -------------------------------------------------------------------------- these two statements. It turns out that -------------------------------------------------------------------------- this substitution formula doesn't require -------------------------------------------------------------------------- continuity at $a$ and $b$, so we can -------------------------------------------------------------------------- drop these three statements, and this -------------------------------------------------------------------------- substitution formula continues to be true. We do need that the determinant of psi prime is -------------------------------------------------------------------------- positive on D, for, otherwise, this substitution formula could be off by a sign. This concludes *one*-variable Integration by Substitution. Let's now move on to multivariable Integration by Substitution. -------------------------------------------------------------------------- Let D be any open subset of the plane. Remember that a set is open if it omits all of its boundary points. Every element of D is a point in the plane, and so has -------------------------------------------------------------------------- two coordinates. Our change of variables function will continue to be called -------------------------------------------------------------------------- psi, but, instead of taking scalars to scalars, *this* psi now takes points in -------------------------------------------------------------------------- D and maps them to points in -------------------------------------------------------------------------- the plane. We'll assume that our psi is -------------------------------------------------------------------------- continuously differentiable, and we'll describe what that means later. In the last slide, psi had one scalar input and one scalar output. This new psi takes in -------------------------------------------------------------------------- two scalar inputs, and has -------------------------------------------------------------------------- two scalar outputs. Let -------------------------------------------------------------------------- E denote the set of output points. That is, one takes every point in D puts it through psi, and then plots all the output points. So E is a -------------------------------------------------------------------------- subset of the plane. Suppose we -------------------------------------------------------------------------- integrate an expression f(u,v) over the set E, with respect to u and v. Let's take that expression -------------------------------------------------------------------------- f(u,v), and substitute -------------------------------------------------------------------------- psi(x,y) for (u,v). If we now -------------------------------------------------------------------------- integrate this expression of x and y over D, then, in keeping with the one-variable substitution formula, we'd -------------------------------------------------------------------------- *not* expect the two integrals in front of us to be equal, and, indeed, they're not. We should again -------------------------------------------------------------------------- remember psi prime, though there's some question as to how to define it, when psi has -------------------------------------------------------------------------- two inputs and two outputs. To address this issue, note that, because psi has -------------------------------------------------------------------------- two outputs, it follows that we can write psi as a pair of expessions -------------------------------------------------------------------------- ( alpha(x,y), beta(x,y) ). Here, -------------------------------------------------------------------------- alpha and beta are functions from D to the scalars. So alpha and beta both have -------------------------------------------------------------------------- two scalar inputs and -------------------------------------------------------------------------- one scalar output. This makes for four different possible derivatives, namely: -------------------------------------------------------------------------- dee dee x of alpha, -------------------------------------------------------------------------- dee dee y of alpha, -------------------------------------------------------------------------- dee dee x of beta, and -------------------------------------------------------------------------- dee dee y of beta. Correctly, we wrote -------------------------------------------------------------------------- alpha of x y and -------------------------------------------------------------------------- beta of x y, but, in speaking, we allow ourselves the sloppiness of dropping "x,y" and simply saying "alpha" and "beta". So what's meant by -------------------------------------------------------------------------- psi prime, given that we have -------------------------------------------------------------------------- four possible results? The answer is: We take all of them, and -------------------------------------------------------------------------- assemble them into a two-by-two matrix. This matrix is often referred to as the differential of psi, and is sometimes denoted -------------------------------------------------------------------------- capital D psi, but, in this lecture, -------------------------------------------------------------------------- we'll continue to use psi prime. Incidentally, we said earlier that psi is -------------------------------------------------------------------------- continuously differentiable, and that simply means that all -------------------------------------------------------------------------- four of the expressions in this matrix are continuous at each point (x,y) in D. As before, -------------------------------------------------------------------------- we don't get equality here, because we have a -------------------------------------------------------------------------- matrix in the middle of a scalar equation. As before, we fix this by -------------------------------------------------------------------------- taking the determinant of psi prime. Mathematicians have worked hard to prove -------------------------------------------------------------------------- equality here, *assuming* -------------------------------------------------------------------------- the determinant of psi prime is positive on D, without which the formula could be off by a sign. We'll just take this formula as a fact, but, for someone who's interested in how it's proved, there are, by now, many books on Multivariable Calculus. In this formula, we've substituted -------------------------------------------------------------------------- for ( u , v ), the expression psi of ( x , y ). We must remember that, pursuant to that substitution, one also substitutes, for -------------------------------------------------------------------------- dee u dee v, the determinant of psi prime times dee x dee y. The particular variables are unimportant, and we can interchange ( x , y ) with ( u , v ) throughout this slide, -------------------------------------------------------------------------- like so. If we prefer ( r , theta ) to ( u , v), we can also make that change, -------------------------------------------------------------------------- like so. Let's now do an example. Let D be -------------------------------------------------------------------------- the open set given here. We'll graph it in a moment. Let alpha and beta be -------------------------------------------------------------------------- as given here; remember that -------------------------------------------------------------------------- psi is ( alpha , beta ). You may recognize -------------------------------------------------------------------------- this as the polar coordinates change of variables. To take -------------------------------------------------------------------------- dee dee r of alpha, we think of theta as fixed, and differentiate -------------------------------------------------------------------------- alpha with respect to r, -------------------------------------------------------------------------- obtaining cosine theta. Similarly, -------------------------------------------------------------------------- dee dee *theta* of alpha is minus r sine theta. Differentiating -------------------------------------------------------------------------- *beta* with resepect to r, -------------------------------------------------------------------------- we get sine theta, and differentiating beta with respect to *theta*, -------------------------------------------------------------------------- we get r cosine theta. We've now computed four derivatives: -------------------------------------------------------------------------- This, -------------------------------------------------------------------------- this, -------------------------------------------------------------------------- this and -------------------------------------------------------------------------- this. -------------------------------------------------------------------------- Psi prime is just the two-by-two matrix whose entries are those same four expressions of r and theta. To compute the -------------------------------------------------------------------------- determinant of psi prime, we take -------------------------------------------------------------------------- this times this, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- minus this times this, -------------------------------------------------------------------------- like so. -------------------------------------------------------------------------- These two minus signs cancel, and we get -------------------------------------------------------------------------- r cosine squared theta plus r sine squared theta, and we can -------------------------------------------------------------------------- factor out r. By Pythagorus, -------------------------------------------------------------------------- cosine squared plus sine squared is one, -------------------------------------------------------------------------- and we get that det of psi prime is r, which is -------------------------------------------------------------------------- positive whenever (r,theta) is in -------------------------------------------------------------------------- $D$, because, in that case, r is -------------------------------------------------------------------------- here, in the positive reals. Remember that -------------------------------------------------------------------------- psi is ( alpha , beta ) and remember -------------------------------------------------------------------------- our substitution formula. We've now computed that -------------------------------------------------------------------------- det of psi prime *is* positive and, in fact, -------------------------------------------------------------------------- is $r$, so we can replace -------------------------------------------------------------------------- det of psi prime -------------------------------------------------------------------------- by $r$. (pause...............) -------------------------------------------------------------------------- Psi is alpha beta, -------------------------------------------------------------------------- alpha is r cosine theta and -------------------------------------------------------------------------- beta is r sine theta, so -------------------------------------------------------------------------- psi is ( r cosine theta , r sine theta ) . We can therefore change -------------------------------------------------------------------------- psi here to -------------------------------------------------------------------------- ( r cosine theta , r sine theta ). -------------------------------------------------------------------------- Let's make some room. We'll also adjust the text a bit more -------------------------------------------------------------------------- tightly. Next, we'll -------------------------------------------------------------------------- graph D. We identify -------------------------------------------------------------------------- 2 pi on the vertical axis, and D is then -------------------------------------------------------------------------- this strip, which extends -------------------------------------------------------------------------- infinitely far to the right. The boundary consists of -------------------------------------------------------------------------- this ray on the bottom, -------------------------------------------------------------------------- this ray on the top -------------------------------------------------------------------------- and this vertical line segment. Remember that D is open, so it contains none of its boundary points. Next, let's -------------------------------------------------------------------------- identify a typical point (r,theta) in D, and let's find its image -------------------------------------------------------------------------- under psi. -------------------------------------------------------------------------- Here's the definition of psi, namely ( r cosine theta , r sine theta ). If we plot -------------------------------------------------------------------------- the ray that makes angle -------------------------------------------------------------------------- theta radians with the -------------------------------------------------------------------------- positive horizontal axis, and then -------------------------------------------------------------------------- travel a distance r from the origin, then, by trigonometry, the point we reach is -------------------------------------------------------------------------- ( r cosine theta , r sine theta ) . Next, keep theta fixed, but let -------------------------------------------------------------------------- $r$ range from 0 to infinity. In the coordinate plane on the left, we get -------------------------------------------------------------------------- this ray. The image of these points under psi is then -------------------------------------------------------------------------- this ray on the right. Note that, -------------------------------------------------------------------------- since r is never 0, we don't get -------------------------------------------------------------------------- the origin. (pause...............) Now let -------------------------------------------------------------------------- theta range from 0 to 2 pi. On the left, we get -------------------------------------------------------------------------- all of D, which we remember doesn't include the -------------------------------------------------------------------------- top or bottom boundary points. In other words, theta is never -------------------------------------------------------------------------- 0, and theta is never -------------------------------------------------------------------------- 2 pi. On the right coordinate plane, -------------------------------------------------------------------------- this angle sweeps from 0 to 2 pi radians, that is from 0 to 360 degrees. Theta is never actually *equal* to 0 or 2 pi, so we never get the -------------------------------------------------------------------------- positive horizontal axis. Also, again, D doesn't include -------------------------------------------------------------------------- this vertical segment, so, on the right, we never get -------------------------------------------------------------------------- the origin. On the other hand, on the right, we do get all other points in the plane, so we get -------------------------------------------------------------------------- all of R^2 except the positive horizontal axis and the origin. The -------------------------------------------------------------------------- excluded points are those -------------------------------------------------------------------------- whose first coordinate is nonnegative, and whose -------------------------------------------------------------------------- second coordinate is zero. Recall that -------------------------------------------------------------------------- E is the image of psi, so we've just calculated that E is -------------------------------------------------------------------------- all of R^2 -------------------------------------------------------------------------- except that set. We use this -------------------------------------------------------------------------- backslash notation for set-theoretic subtraction. -------------------------------------------------------------------------- This set that's being subtracted is one-dimensional, and so it has no area. As such, it makes no difference in -------------------------------------------------------------------------- this integral at the bottom, and we may -------------------------------------------------------------------------- replace E by R^2 without changing the result. A double integral over R^2 is the same as -------------------------------------------------------------------------- two integrals from minus infinity to infinity. On the -------------------------------------------------------------------------- other integral, (r,theta) is ranging over -------------------------------------------------------------------------- $D$, so r ranges -------------------------------------------------------------------------- from 0 to infinty and theta ranges -------------------------------------------------------------------------- from 0 to 2 pi, so the -------------------------------------------------------------------------- double integral over D can be -------------------------------------------------------------------------- replaced by a -------------------------------------------------------------------------- dee r integral from 0 to infinty and -------------------------------------------------------------------------- a dee theta integral from 0 to 2 pi. Remember that we substituted -------------------------------------------------------------------------- ( r cosine theta , r sine theta ) for ( x , y ), which is the same as replacing -------------------------------------------------------------------------- x by r cosine theta and y by r sine theta. Pursuant to that, we substituted -------------------------------------------------------------------------- r dee r dee theta for dee x dee y, and the -------------------------------------------------------------------------- r appearing there is just the determinant of psi prime, and it should *not* be forgotten. -------------------------------------------------------------------------- We record this differential substitution. So, to do a polar coordinates change of variables, we need to remember the -------------------------------------------------------------------------- three formulas appearing here. Actually, there's a fourth worth recording. If we square -------------------------------------------------------------------------- these two formulas, and add the results, then, on the left hand side, we get -------------------------------------------------------------------------- x squared plus y squared. On the right, we get r squared cosine squared theta plus r squared sine squared theta, but, by Pythagorus, -------------------------------------------------------------------------- that's just r squared. Our next review topic is -------------------------------------------------------------------------- Integration by Parts. We start with a -------------------------------------------------------------------------- product of two expressions of x, f(x) and g(x), and we wish to -------------------------------------------------------------------------- integrate this product -------------------------------------------------------------------------- with respect to x. To do Integration by Parts, we should identify one of the parts, say -------------------------------------------------------------------------- g(x), and then find an -------------------------------------------------------------------------- antiderivative of it, say -------------------------------------------------------------------------- capital G of x. -------------------------------------------------------------------------- This up arrow indicates antidifferentiation, and so -------------------------------------------------------------------------- the derivative of capital G is little g. In general, we'll take the view that differentiation is relatively easy compared to antidifferentiation, and the heavy lifting of antidifferentiation is represented by an -------------------------------------------------------------------------- up-arrow, whereas differentiation, which is easier, will be represented by a -------------------------------------------------------------------------- down-arrow. To do Integration by Parts, we should now take the other part, -------------------------------------------------------------------------- f(x), and -------------------------------------------------------------------------- differentiate it with respect to x -------------------------------------------------------------------------- obtaining f'(x), which is the same as saying that -------------------------------------------------------------------------- the derivative of f is f prime. Note that -------------------------------------------------------------------------- $f$ is above $f'$. Also, -------------------------------------------------------------------------- capital G is above little g. We indicate this by putting in -------------------------------------------------------------------------- the words "up". That which is not "up" is -------------------------------------------------------------------------- "down". We started with -------------------------------------------------------------------------- f and g, so we label them -------------------------------------------------------------------------- "old". We *then* found -------------------------------------------------------------------------- capital G and f prime, so we label *them* -------------------------------------------------------------------------- "new". With all this labeling, the buzz phrase for Integration by Parts is -------------------------------------------------------------------------- up up, which is the same as -------------------------------------------------------------------------- f capital G -------------------------------------------------------------------------- minus the integral of -------------------------------------------------------------------------- new new, which is the same as -------------------------------------------------------------------------- f prime capital G, and -------------------------------------------------------------------------- that gives the Integration by Parts formula. Again, the phrase to remember is -------------------------------------------------------------------------- up up -------------------------------------------------------------------------- minus the integral of -------------------------------------------------------------------------- new new. Let's do an example. Let's integrate -------------------------------------------------------------------------- x cosine x dee x. An antiderivative of -------------------------------------------------------------------------- cosine x is -------------------------------------------------------------------------- sine x, because -------------------------------------------------------------------------- dee dee x of sine x is cosine x. The derivative of -------------------------------------------------------------------------- x is -------------------------------------------------------------------------- 1, because -------------------------------------------------------------------------- dee dee x of x is 1. A comment before continuing. When we implement Integration by Parts, we end up with the integral of -------------------------------------------------------------------------- new new, and a key question is whether *that* integral is going to be any easier than than the original integral of -------------------------------------------------------------------------- old old. In the present problem, antidifferentiation moved us from -------------------------------------------------------------------------- cosine (pause...............) to -------------------------------------------------------------------------- sine. From the perspective of integration, the two most basic trig functions, sine and cosine, are at the same level of complexity. We highlight that by noting that sine x is -------------------------------------------------------------------------- still just a trig function, and no more complicated than cosine x. For the other factor, we went from -------------------------------------------------------------------------- x to -------------------------------------------------------------------------- 1, and this moved us from a polynomial of degree 1 to a polynomial of degree 0, and *that* represents a distinct *de*crease in complexity. We highlight *that* by noting that 1 is a -------------------------------------------------------------------------- polynomial of lower degree. Then -------------------------------------------------------------------------- new new will be less complicated to integrate than -------------------------------------------------------------------------- old old, so Integration by Parts is working. Implementing Integration by Parts, we now take -------------------------------------------------------------------------- up up, -------------------------------------------------------------------------- like so, and -------------------------------------------------------------------------- subtract the integral of -------------------------------------------------------------------------- new new, -------------------------------------------------------------------------- like so. Multiplication by -------------------------------------------------------------------------- 1 does nothing, so -------------------------------------------------------------------------- we drop it, and -------------------------------------------------------------------------- this is the result of Integration by Parts. An antiderivative of -------------------------------------------------------------------------- sine x is -------------------------------------------------------------------------- minus cosine x. To get *all* antiderivatives, we -------------------------------------------------------------------------- add "C", an arbitrary constant. -------------------------------------------------------------------------- This minus sign cancels -------------------------------------------------------------------------- the other one, and we get -------------------------------------------------------------------------- plus cosine x. You may think that -------------------------------------------------------------------------- this minus sign should also change -------------------------------------------------------------------------- this "plus C" to "minus C", but the *set* of all negated constants is the same as the *set* of all constants, so it would make -------------------------------------------------------------------------- no difference. -------------------------------------------------------------------------- This completes the problem. (Pause) -------------------------------------------------------------------------- The remainder of this lecture will have five parts. -------------------------------------------------------------------------- We'll talk about limits of powers of functions after renormalization, which is one of the two basic ingredients in the Central Limit Theorem. The other ingredient is the Fourier Transform. We'll also discuss some -------------------------------------------------------------------------- other limits. Then we'll cover some basic skills needed to do -------------------------------------------------------------------------- financial mathematics integrals. Then we'll get back to the -------------------------------------------------------------------------- pricing problem that we introduced at the end of Lecture 3. We'll -------------------------------------------------------------------------- reduce the pricing problem to an integral. This reduction'll involve those two basic ingredients in the Central Limit Theorem, though we won't actually state that theorem until the end of this lecture. The integral that we'll end up needing to calculate will be of a type that we practiced when doing -------------------------------------------------------------------------- financial mathematics integrals, and we'll use the skills that we developed at that time -------------------------------------------------------------------------- to calculate the intgral, thereby solving the -------------------------------------------------------------------------- pricing problem. Finally, we'll review our work, and extract from it, a version of -------------------------------------------------------------------------- the Central Limit Theorem, which we call the 50-50 Central Limit Theorem. -------------------------------------------------------------------------- Let's take a break! -------------------------------------------------------------------------- Welcome to the Second Act of Lecture 4 of Notes on Financial Mathematics, by Scot Adams and Fernando Reitich. In this act of this lecture, we focus on some important limits. Consider, for example, -------------------------------------------------------------------------- this expression of n. To get some sense of the limit as n approaches infinity, we plug in -------------------------------------------------------------------------- n equals one million, giving -------------------------------------------------------------------------- this. -------------------------------------------------------------------------- These cancel, and the rest almost does, so the answer is -------------------------------------------------------------------------- close to 1. In fact, it's not hard to show that, given -------------------------------------------------------------------------- two polynomials of the same degree, then the -------------------------------------------------------------------------- limit at infinity is the quotient of their leading coefficients. -------------------------------------------------------------------------- This limit is one, as well, by the same logic. We now proceed to a basic finance -------------------------------------------------------------------------- Problem: Suppose we invest one dollar in the bank for one year. -------------------------------------------------------------------------- How many dollars will be there at the end of the year? Clearly this depends on the interest rate, so let's -------------------------------------------------------------------------- assume that the annual *nominal* interest rate is some number r, but let's *also* assume that the account is -------------------------------------------------------------------------- convertible $n$ times per year, which means that one $n$th of the nominal rate is given out each $n$th of a year. So, to -------------------------------------------------------------------------- solve this, we imagine that we -------------------------------------------------------------------------- start with our 1 dollar, and then add on interest of -------------------------------------------------------------------------- r over n with each conversion. That is, we -------------------------------------------------------------------------- multiply by 1 plus r over n, but we do this -------------------------------------------------------------------------- n times per year for -------------------------------------------------------------------------- one year, which means we do this multiplication -------------------------------------------------------------------------- n times. This gives an -------------------------------------------------------------------------- answer of 1 plus r over n raised to the $n$th power. Now we pose another -------------------------------------------------------------------------- Question: What if we move to -------------------------------------------------------------------------- continuous compounding, which means that n is made to approach infinity. That is, in the -------------------------------------------------------------------------- preceding answer, we take the -------------------------------------------------------------------------- limit as n approaches infinity, and ask -------------------------------------------------------------------------- to compute that limit. To solve this problem, our approach will be to expand the -------------------------------------------------------------------------- $n$th power, using the binomial formula. The binomial formula says that, to compute -------------------------------------------------------------------------- $a$ plus $b$ to the $n$, multiply together -------------------------------------------------------------------------- $n$ "a"s, then -------------------------------------------------------------------------- change one to a "b", then -------------------------------------------------------------------------- another, and then -------------------------------------------------------------------------- continue on to -------------------------------------------------------------------------- $n$ "b"s. Then insert the binomial coefficients of -------------------------------------------------------------------------- n choose 1 -------------------------------------------------------------------------- n choose 2 and so on, and then -------------------------------------------------------------------------- add up the results. On the right hand side, -------------------------------------------------------------------------- we drop the last term, and ask you simply to remember that it's there. In the end, we'll let $n$ approach infinity, so this finite sum will, in the limit, turn into an infinite series. Our goal is to expand 1 plus r over n to the $n$, and so we replace -------------------------------------------------------------------------- a by 1, -------------------------------------------------------------------------- like so, and, in a moment, we'll replace b by r over n. -------------------------------------------------------------------------- This 1 to the n -------------------------------------------------------------------------- is equal to 1. -------------------------------------------------------------------------- These are 1 as well, and if we multiply 1 by -------------------------------------------------------------------------- b, b squared and so on, -------------------------------------------------------------------------- we get b, b squared and so on. The rest we leave -------------------------------------------------------------------------- untouched. Now we replace -------------------------------------------------------------------------- b by r over n, -------------------------------------------------------------------------- like so. We cancel -------------------------------------------------------------------------- these "n"s, and, -------------------------------------------------------------------------- here, we distribute squaring over division to get -------------------------------------------------------------------------- r squared over n squared, giving -------------------------------------------------------------------------- this. We -------------------------------------------------------------------------- interchange 2 factorial with n squared, giving -------------------------------------------------------------------------- this. The rest we leave -------------------------------------------------------------------------- untouched. (pause...............) -------------------------------------------------------------------------- Here we have a quotient of equal degree polynomials in $n$, and we noted earlier that -------------------------------------------------------------------------- it tends to 1, as n tends to infinity. -------------------------------------------------------------------------- These don't change as $n$ changes, and so we see that -------------------------------------------------------------------------- the limit is a -------------------------------------------------------------------------- series that begins like this. If you calculate a few more terms, you can convince yourself that this is the Taylor expansion at 0 of -------------------------------------------------------------------------- e to the r. e to the r is also called the -------------------------------------------------------------------------- exponential of r. Let's record the -------------------------------------------------------------------------- same limit a little less colorfully up above. To summarize, -------------------------------------------------------------------------- if $r$ is used for the nominal interest rate, then each dollar, invested with continuous compounding, after one year, grows to -------------------------------------------------------------------------- e to the r dollars, answering our question. Then -------------------------------------------------------------------------- e to the r is the risk-free factor, meaning the factor by which we multiply to go from the size of a bank account at the start of a year to its size at the end. Again, this is all assuming that we're working -------------------------------------------------------------------------- under continuous compounding. Let's -------------------------------------------------------------------------- clear some room. This limit is true for all -------------------------------------------------------------------------- $r$, so we can replace $r$ by -------------------------------------------------------------------------- minus x squared. This -------------------------------------------------------------------------- minus sign can be -------------------------------------------------------------------------- brought to the outside, -------------------------------------------------------------------------- as can squaring, though note that -------------------------------------------------------------------------- $n$ changes to its square root inside the square. Doing -------------------------------------------------------------------------- this rewriting, we end up with -------------------------------------------------------------------------- this, and the limit is still -------------------------------------------------------------------------- $e$ to the minus $x$ squared, for any $x$. Note that we divided $x$ -------------------------------------------------------------------------- by root n, in other words, by the square root of -------------------------------------------------------------------------- the exponent. We can visualize -------------------------------------------------------------------------- this limit using graphs of functions, as follows. -------------------------------------------------------------------------- Here's the graph of -------------------------------------------------------------------------- 1 minus x squared, a parabola. Let's examine what happens as we take powers of 1 minus x squared. -------------------------------------------------------------------------- Here's the fifth power. -------------------------------------------------------------------------- Here's the hundredth. -------------------------------------------------------------------------- Here's the ten thousandth. As the exponent grows, the graph tends toward a discontinuous function that spikes up to -------------------------------------------------------------------------- 1 at 0, but is equal to -------------------------------------------------------------------------- 0 everywhere else between minus 1 and 1. We now avoid loss of continuity by doing -------------------------------------------------------------------------- some stretching. We call this stretching step -------------------------------------------------------------------------- "renormalization", without which the graphs won't tend to a continuous limit. In algebraic terms, renormalization -------------------------------------------------------------------------- replaces x by x over some large constant. Our limit calculation from before would indicate that the correct constant to use is the square root of -------------------------------------------------------------------------- the exponent, the square root of 10 thousand. -------------------------------------------------------------------------- Here's the result of that renormalization. Now ... look closely at the graph in front of you, and watch for changes, as we -------------------------------------------------------------------------- switch to e to the minus x squared. The graph didn't show any observable change, because 10,000 is so large, and the -------------------------------------------------------------------------- $n$th power of 1 minus x squared, after -------------------------------------------------------------------------- renormalization, tends to e to the minus x squared. "Renormalization", in *this* lecture, will always mean replacing x by x over the square root of -------------------------------------------------------------------------- the exponent. Now define -------------------------------------------------------------------------- f(x) to be 1-x^2. The $n$th power is -------------------------------------------------------------------------- this. We replace x by -------------------------------------------------------------------------- x over root n. This is the renormalized $n$th power of f(x), and is equal to -------------------------------------------------------------------------- this, and so the -------------------------------------------------------------------------- limit as n tends to infinity -------------------------------------------------------------------------- is e to the minus x squared, for every x. Now suppose we change the definition of f(x) to -------------------------------------------------------------------------- this. All we've done is to tack on -------------------------------------------------------------------------- a cubic term, x cubed. We'll see, in a *moment*, that, for this new f(x), -------------------------------------------------------------------------- the limit of the renormalized $n$th power -------------------------------------------------------------------------- is again e to the minus x squared. So the -------------------------------------------------------------------------- answer was unaffected by the introduction of -------------------------------------------------------------------------- this cubic term. On the other hand, suppose we change the -------------------------------------------------------------------------- quadratic term from minus x squared to -------------------------------------------------------------------------- minus x squared *over 5*. We'll see, in a *moment*, that, for this new f(x), -------------------------------------------------------------------------- the limit of the renormalized $n$th power -------------------------------------------------------------------------- is e to the minus x squared *over 5*. Next, we redefine -------------------------------------------------------------------------- f(x) once again, tacking on -------------------------------------------------------------------------- this cubic term, x cubed over 8. We'll see, in a *moment*, that, for this new f(x), -------------------------------------------------------------------------- the limit of the renormalized $n$th power -------------------------------------------------------------------------- is again e to the minus x squared over 5. So the -------------------------------------------------------------------------- answer was unaffected by the introduction of -------------------------------------------------------------------------- this cubic term. Let's now tack on a -------------------------------------------------------------------------- quartic term; the coefficient's unimportant, and we, at random, picked -------------------------------------------------------------------------- minus pi. Any guesses as to the limit of the renormalized $n$th power now? We'll see, in a *moment*, that it's still -------------------------------------------------------------------------- e to the minus x squared over 5, so the quartic term made no difference in the answer. We'd like to capture all these various limits into a single statement. The concept that underlies them all is that of a MacLaurin approximation, which some listeners may know as a Taylor approximation about 0. To say that an expression -------------------------------------------------------------------------- f(x) has 2nd order MacLaurin approximation -------------------------------------------------------------------------- 1 minus x squared over 2 -------------------------------------------------------------------------- means that, if we define -------------------------------------------------------------------------- q(x) to *be* 1 minus x squared over 2, then -------------------------------------------------------------------------- f and q agree at zero, -------------------------------------------------------------------------- as do their first derivatives -------------------------------------------------------------------------- and their second derivatives. -------------------------------------------------------------------------- Second order means that we only go up to the second derivative. Since we have a specific function -------------------------------------------------------------------------- q, we can calculate -------------------------------------------------------------------------- q'(x) and -------------------------------------------------------------------------- q''(x), and then -------------------------------------------------------------------------- evaluate q at 0, and then -------------------------------------------------------------------------- q' at 0, and then -------------------------------------------------------------------------- q'' at 0. Because -------------------------------------------------------------------------- q(0) is 1, -------------------------------------------------------------------------- this says that -------------------------------------------------------------------------- f(0) is 1. Because -------------------------------------------------------------------------- q'(0) is 0, -------------------------------------------------------------------------- this says that -------------------------------------------------------------------------- f'(0) is 0. Because -------------------------------------------------------------------------- q''(0) is minus 1, -------------------------------------------------------------------------- this says that -------------------------------------------------------------------------- f''(0) is minus 1. So to say that -------------------------------------------------------------------------- f(x) has 2nd order MacLaurin approximation -------------------------------------------------------------------------- 1 minus x squared over 2, is just to say that -------------------------------------------------------------------------- f(0) is 1, that f'(0) is 0 and that f''(0) is minus 1. Let's run through a few -------------------------------------------------------------------------- examples of expressions with that second order MacLaurin approximation. Easiest is -------------------------------------------------------------------------- 1 minus x squared over 2 itself. We can -------------------------------------------------------------------------- then, if we want, tack on a cubic, and then -------------------------------------------------------------------------- a quartic, and then -------------------------------------------------------------------------- a quintic. Another example is -------------------------------------------------------------------------- 2 plus x minus e to the x. For this f(x), check out -------------------------------------------------------------------------- these equations, by differentiating and evaluating. For our use later in *this* lecture, the most important example is -------------------------------------------------------------------------- cosine x. Again, for this last f(x), check out the truth of -------------------------------------------------------------------------- these equations, by differentiating and evaluating. Let's pause to look at graphs of the renormalized powers of -------------------------------------------------------------------------- cosine x. We start with -------------------------------------------------------------------------- cosine x itself, then move on to the -------------------------------------------------------------------------- renormalized 10th power, then to the -------------------------------------------------------------------------- renormalized 1,000th power, then to the -------------------------------------------------------------------------- renormalized 10,000th power. Look closely at the graph in front of you, and watch for changes as we -------------------------------------------------------------------------- switch to e to the minus x squared over 2. Note the lack of any observable change to the graph. Okay. -------------------------------------------------------------------------- Let's assume we have *some* f(x) with 2nd order MacLaurin approximation -------------------------------------------------------------------------- 1 minus x squared over 2. f(x) could be, for example, cosine x. Let's -------------------------------------------------------------------------- assume also that f'' is continuous at 0. We'll -------------------------------------------------------------------------- prove that the limit of the renormalized $n$th power of f(x) is -------------------------------------------------------------------------- e to the minus x squared over 2. We'll consider later what happens as we change -------------------------------------------------------------------------- the quadratic to, say minus x squared, or x squared over 5. For concreteness, we'll only -------------------------------------------------------------------------- verify this equation for x equals 3; that is, we'll verify -------------------------------------------------------------------------- this statement, but the argument we'll give will work for *any* value of x. The first step is to simplify the problem by taking the logarithm of both sides. On the -------------------------------------------------------------------------- left hand side, after the limit, we have -------------------------------------------------------------------------- this expression, and, if we take its -------------------------------------------------------------------------- logarithm, we can use the properties of log to bring -------------------------------------------------------------------------- this exponent down front, -------------------------------------------------------------------------- like so. On the -------------------------------------------------------------------------- right hand side, we take the -------------------------------------------------------------------------- logarithm. Remembering that -------------------------------------------------------------------------- log and exp are inverses, we see that log of e to the whatever is just -------------------------------------------------------------------------- whatever, and, here, this "whatever" is -------------------------------------------------------------------------- minus 9 over 2. So it suffices to prove -------------------------------------------------------------------------- this. In fact, if we can prove -------------------------------------------------------------------------- this limit, then, exponentiating, we get -------------------------------------------------------------------------- this limit. So we can now focus on proving -------------------------------------------------------------------------- this. Next step is to substitute -------------------------------------------------------------------------- x for 1 over root n. -------------------------------------------------------------------------- Then 3x is 3 over root n. Moreover, -------------------------------------------------------------------------- x squared is 1 over n, which, on reciprocating, gives that -------------------------------------------------------------------------- 1 over x squared is $n$. Substituting, we replace -------------------------------------------------------------------------- $n$ by -------------------------------------------------------------------------- 1 over x squared, and -------------------------------------------------------------------------- 3 over root n by -------------------------------------------------------------------------- 3 x, which yields -------------------------------------------------------------------------- this. As -------------------------------------------------------------------------- $n$ tends to infinity, -------------------------------------------------------------------------- x tends to zero, so the -------------------------------------------------------------------------- limit as n tends to infinity gets replaced by -------------------------------------------------------------------------- the limit as x tends to zero, and it now suffices to prove -------------------------------------------------------------------------- this. In fact, if we can prove -------------------------------------------------------------------------- this limit, then, working backward, -------------------------------------------------------------------------- this substitution implies -------------------------------------------------------------------------- this limit. So we can now focus on proving -------------------------------------------------------------------------- this. (pause...............) -------------------------------------------------------------------------- This expression can be written as a quotient, namely -------------------------------------------------------------------------- log of f of 3 x over x squared. Remember that we're assuming -------------------------------------------------------------------------- that f(0) is 1, so, as x tends to zero, -------------------------------------------------------------------------- f(3x) tends to 1. So, formally, we get -------------------------------------------------------------------------- this limit. However -------------------------------------------------------------------------- log of 1 is zero, so this gives the indeterminate form -------------------------------------------------------------------------- zero over zero, for which the standard next step is to apply -------------------------------------------------------------------------- l'Hopital's Rule, which tells us to take the derivative of -------------------------------------------------------------------------- this, and divide it by the derivative of -------------------------------------------------------------------------- this. We therefore need to compute -------------------------------------------------------------------------- these two derivatives. The one on the bottom is easy; it's just -------------------------------------------------------------------------- 2 x. For the one above, we have -------------------------------------------------------------------------- an expression plugged into -------------------------------------------------------------------------- a function, so we use the Chain Rule, and take the derivative of -------------------------------------------------------------------------- the function, -------------------------------------------------------------------------- like so and plug in -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so. It remains to multiply by the derivative of -------------------------------------------------------------------------- the expression. Now, this expression is itself an -------------------------------------------------------------------------- expression plugged into a -------------------------------------------------------------------------- function, so we use the Chain Rule again, and take the derivative of -------------------------------------------------------------------------- the function, -------------------------------------------------------------------------- like so, plug in -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so, and multiply by the derivative of -------------------------------------------------------------------------- the expression, -------------------------------------------------------------------------- like so. Following l'Hopital, we take the quotient of the derivatives of -------------------------------------------------------------------------- these two expressions, and we now -------------------------------------------------------------------------- want to prove this. In -------------------------------------------------------------------------- this expression, we can put -------------------------------------------------------------------------- this in the numerator, -------------------------------------------------------------------------- like so, and then put -------------------------------------------------------------------------- this in the denominator, -------------------------------------------------------------------------- like so. Because -------------------------------------------------------------------------- f'(0) is 0, -------------------------------------------------------------------------- the numerator tends to zero as $x$ tends to 0. Because of -------------------------------------------------------------------------- this x in the denominator, -------------------------------------------------------------------------- the denominator tends to zero, as well. So, formally, we get the indeterminate form -------------------------------------------------------------------------- zero over zero, and again appeal to -------------------------------------------------------------------------- l'Hopital's Rule, which says to take the derivative of -------------------------------------------------------------------------- the numerator divided by the derivative of -------------------------------------------------------------------------- the denominator. The result is that we -------------------------------------------------------------------------- want to prove this. We leave it as an exercise for you to check these derivatives, using the Chain and Product Rules. Note that the Quotient Rule never makes an appearance, because we never differentiate a quotient; instead, following l'Hopital, we take a quotient of derivatives. You can verify that -------------------------------------------------------------------------- *this* limit is no longer an indeterminate form, so we *can't* apply l'Hopital again, which is a good thing, since I'm sick of it anyway. Let's just try to calculate *this* limit directly. Since -------------------------------------------------------------------------- f(0) is 1, -------------------------------------------------------------------------- f(3x) tends to one, and multiplication by 1 doesn't do anything. Since -------------------------------------------------------------------------- x tends to zero, -------------------------------------------------------------------------- this whole term in the denominator tends to zero. Since -------------------------------------------------------------------------- f''(0) is minus 1, -------------------------------------------------------------------------- this factor in the numerator tends to minus 1, and with -------------------------------------------------------------------------- these two 3s, the numerator tends to -------------------------------------------------------------------------- minus 9. On the other hand, the denominator tends to -------------------------------------------------------------------------- 2, and so the limit is, indeed, -------------------------------------------------------------------------- minus 9 over 2, finishing the proof. (pause...............) So -------------------------------------------------------------------------- this now verified for x equals 3, and a similar argument works for any value of x. We invite the listener to review the proof, and to identify the point or points at which -------------------------------------------------------------------------- this assumption was used. Let's -------------------------------------------------------------------------- clear some room, and then -------------------------------------------------------------------------- move the text up to make *more* room. We state our findings as a -------------------------------------------------------------------------- theorem, which asserts that -------------------------------------------------------------------------- if f(x) has 2nd order MacLaurin approximation -------------------------------------------------------------------------- 1 minus x squared over 2 and if -------------------------------------------------------------------------- f'' is continuous at 0, -------------------------------------------------------------------------- then the limit of the renormalized $n$th power of f(x) is -------------------------------------------------------------------------- e to the minus x squared over 2, for all x. -------------------------------------------------------------------------- Here's another theorem, the only difference being that -------------------------------------------------------------------------- minus x squared over 2 is replaced by -------------------------------------------------------------------------- minus 2 (x squared). The theorem above can be proved in the same way as the one below, and, in fact, we can capture both of these two theorems, and more, in -------------------------------------------------------------------------- this result, in which we've used -------------------------------------------------------------------------- minus k x squared, and in which -------------------------------------------------------------------------- k can be *any* real number. Again, the proof goes through in exactly the same way: take logarithms, substitute x equals 1 over root n and apply l'Hopital's Rule twice. This completes our analysis of limits of renormalized $n$th powers. We now move on to another collection of limits. We start -------------------------------------------------------------------------- with the graph of e to the minus x squared, sometimes called -------------------------------------------------------------------------- the bell curve. Note -------------------------------------------------------------------------- here, how this curve -------------------------------------------------------------------------- hugs the x axis to the right, and, -------------------------------------------------------------------------- here, how it -------------------------------------------------------------------------- hugs the x axis to the left. We point out that this "hugging" is very tight, and one says that the bell curve has -------------------------------------------------------------------------- "thin tails". Let's contrast this to the graph of -------------------------------------------------------------------------- 1 over 1 plus x squared. Note that this graph also hugs the x-axis, but much more slowly, and one often hears that *this* graph has -------------------------------------------------------------------------- "fat tails". Fatness is measured -------------------------------------------------------------------------- vertically, but no one ever says "tall tails" in *this* context. In this lecture, we'll focus on -------------------------------------------------------------------------- the bell curve and we'll now argue again that it -------------------------------------------------------------------------- hugs the x-axis, but we'll argue *formally*, in terms of limits. To begin, note that -------------------------------------------------------------------------- if we plug x equals a thousand into -------------------------------------------------------------------------- minus x squared, we get -------------------------------------------------------------------------- minus a million, which is a -------------------------------------------------------------------------- very negative number. If instead of minus x squared, we exponentiate, -------------------------------------------------------------------------- and use (pause...............) -------------------------------------------------------------------------- e to the minus x squared, and again, -------------------------------------------------------------------------- evaluate at a thousand then we get -------------------------------------------------------------------------- e to the minus one million, which is -------------------------------------------------------------------------- 1 over e to the one million. That's the reciprocal of a very large number, and so is -------------------------------------------------------------------------- very close to zero. Indeed, by an epsilon-delta style of argument, one can prove that -------------------------------------------------------------------------- the limit is zero. This limit simply says that -------------------------------------------------------------------------- the bell curve hugs the x-axis to the right. If we -------------------------------------------------------------------------- plug x equals *minus* a thousand into minus x squared, we still get -------------------------------------------------------------------------- minus a million, because squaring a -------------------------------------------------------------------------- negative number is the *same* as squaring the -------------------------------------------------------------------------- corresponding positive number. Again, minus a million -------------------------------------------------------------------------- is very negative. We therefore find that -------------------------------------------------------------------------- e to the minus x squared evaluated at -------------------------------------------------------------------------- *minus* a thousand is also -------------------------------------------------------------------------- very close to zero, and -------------------------------------------------------------------------- the limit as x approaches -------------------------------------------------------------------------- *minus* infinity -------------------------------------------------------------------------- is zero. That is, -------------------------------------------------------------------------- the bell curve hugs the x-axis to the *left*. We now start this analysis over again, but with another polynomial of degree 2, namely -------------------------------------------------------------------------- minus x squared plus 5 x minus 4. Plugging in a large number -------------------------------------------------------------------------- like 1000, we get -------------------------------------------------------------------------- this, and the -------------------------------------------------------------------------- negative million dominates -------------------------------------------------------------------------- the other two terms, so the result is -------------------------------------------------------------------------- very negative. -------------------------------------------------------------------------- Exponentiating, we get something -------------------------------------------------------------------------- close to zero. In the -------------------------------------------------------------------------- limit, as x approaches infinity, we get -------------------------------------------------------------------------- zero. Taking the -------------------------------------------------------------------------- limit as x approaches -------------------------------------------------------------------------- minus infinity, we again get -------------------------------------------------------------------------- zero, because, once again, -------------------------------------------------------------------------- quadratic term in the exponent is minus x squared, and it dominates -------------------------------------------------------------------------- the other two terms, because they're of lower degree. There's nothing special about the numbers -------------------------------------------------------------------------- 5 and minus 4, and we observe, -------------------------------------------------------------------------- for any real numbers a and b, -------------------------------------------------------------------------- that the limits at -------------------------------------------------------------------------- plus *or* minus infinity of -------------------------------------------------------------------------- the exponential of -------------------------------------------------------------------------- minus x squared plus a x plus b -------------------------------------------------------------------------- is zero. Replacing -------------------------------------------------------------------------- minus x squared by -------------------------------------------------------------------------- minus x squared over 2, we see the same result. Again, the -------------------------------------------------------------------------- quadratic term in the exponent dominates -------------------------------------------------------------------------- the other two terms. In point of fact, -------------------------------------------------------------------------- for any positive c, we can use -------------------------------------------------------------------------- minus c x squared -------------------------------------------------------------------------- and get the same result. Next, let -------------------------------------------------------------------------- i denote the square root of minus 1. We go back to the study of -------------------------------------------------------------------------- the exponential of minus x squared plus 5 x minus 4, but we now think of x as a complex number, -------------------------------------------------------------------------- say s+ti, where s and t are real. It's traditional to use $z$ s a complex variable, not $x$, but, in this lecture, we'll flout that tradition. Our next goal is to -------------------------------------------------------------------------- fix t, and let s tend toward plus or minus infinity, and analyze what happens. Eventually, we'll see that, even in this complex numbers setting, the limits are zero. For an -------------------------------------------------------------------------- example, let's fix t equals minus 3. We can't show you the graph of -------------------------------------------------------------------------- this expression, which maps complex numbers to complex numbers, since that would require four real dimensions. However, we can discuss what's going on in the domain, which is the set of complex numbers. If -------------------------------------------------------------------------- this is the complex plane, then -------------------------------------------------------------------------- the origin is zero. Let -------------------------------------------------------------------------- this point be minus 3 i. Then -------------------------------------------------------------------------- this is the set of points of the form s plus ti, where, remember, -------------------------------------------------------------------------- t is minus 3, but s is allowed to vary. We want to analyze the behavior of -------------------------------------------------------------------------- our exponential, as -------------------------------------------------------------------------- s approaches infinity, and as -------------------------------------------------------------------------- s approaches minus infinity. So, in -------------------------------------------------------------------------- this expression, we replace -------------------------------------------------------------------------- t by minus 3 giving -------------------------------------------------------------------------- this. Now -------------------------------------------------------------------------- we copy minus 4 -------------------------------------------------------------------------- down here. Next, -------------------------------------------------------------------------- plus 5 x, evaluated at -------------------------------------------------------------------------- s minus 3 i gives -------------------------------------------------------------------------- 5 times s minus 3 i. Finally, -------------------------------------------------------------------------- minus x squared, evaluated at -------------------------------------------------------------------------- s minus 3 i, expands out to -------------------------------------------------------------------------- s squared minus 6 s i *minus* 9. Next, we collect together real parts, like -------------------------------------------------------------------------- minus s squared, (pause...............) -------------------------------------------------------------------------- minus minus 9, (pause...............) -------------------------------------------------------------------------- 5 s (pause...............) -------------------------------------------------------------------------- and minus 4, (pause...............) giving -------------------------------------------------------------------------- this. Next, we collect together imaginary parts. These are -------------------------------------------------------------------------- minus minus 6 s (pause...............) ------------------------------------------------------------------------- and 5 times minus 3, giving RECERR: I said 5 times minus i ------------------------------------------------------------------------- 6 s minus 15. Putting together real and imaginary parts, we get -------------------------------------------------------------------------- this. (pause...............) -------------------------------------------------------------------------- 9 minus 4 is 5. (pause...............) -------------------------------------------------------------------------- Exponentiating, we get the product of the exponentials of -------------------------------------------------------------------------- this and -------------------------------------------------------------------------- this, -------------------------------------------------------------------------- like so. Our goal is to show that this tends to 0, as s tends to plus or minus infinity. A complex expression tends to zero if and only if its -------------------------------------------------------------------------- absolute value does. The absolute value of a product is the -------------------------------------------------------------------------- product of the absolute values. Next, -------------------------------------------------------------------------- for any r, e to the r i is -------------------------------------------------------------------------- cosine r plus i sine r. -------------------------------------------------------------------------- Taking absolute values, and assuming that $r$ is real, we compute the absolute value as -------------------------------------------------------------------------- the square root of the sum of the squares of the real and imaginary parts. Pythagorus tells us that this -------------------------------------------------------------------------- is 1. In the case where r is -------------------------------------------------------------------------- 6 s minus 15, we now see that -------------------------------------------------------------------------- this absolute value is 1, and so can be -------------------------------------------------------------------------- dropped. Next, -------------------------------------------------------------------------- for any real number r, e to the r is positive, so -------------------------------------------------------------------------- its absolute value -------------------------------------------------------------------------- is just itself. Applying this to the case where r is -------------------------------------------------------------------------- minus s squared plus 5 s plus 5, we now see that we can -------------------------------------------------------------------------- drop these absolute value bars. We've already analyzed expressions like -------------------------------------------------------------------------- this, and found that the -------------------------------------------------------------------------- quadratic in the exponent dominates -------------------------------------------------------------------------- the other two terms, and so this expression -------------------------------------------------------------------------- tends to zero as s tends to plus or minus infinity. Writing this in a different way, we see that -------------------------------------------------------------------------- this expression (pause...............) -------------------------------------------------------------------------- has limit (pause...............) at -------------------------------------------------------------------------- plus or minus infinity (pause...............) -------------------------------------------------------------------------- equal to zero. A complex number is small if and only if -------------------------------------------------------------------------- its absolute value is small, so the fact that the absolute value is tending to zero tells us that -------------------------------------------------------------------------- the limit without absolute value bars is zero, as well. We conclude that, -------------------------------------------------------------------------- on the complex plane, -------------------------------------------------------------------------- this expression, along -------------------------------------------------------------------------- this line, tends to zero as -------------------------------------------------------------------------- the real part of $x$ tends to infinity or -------------------------------------------------------------------------- minus infinity. There's really nothing special about -------------------------------------------------------------------------- this particular degree two polynomial. Quite generally, if we form -------------------------------------------------------------------------- this expression, -------------------------------------------------------------------------- for *any* complex numbers a and b, and -------------------------------------------------------------------------- any *positive* real number c, and if we fix -------------------------------------------------------------------------- any real number t, then -------------------------------------------------------------------------- the limit as s tends to plus or minus infinity is -------------------------------------------------------------------------- zero. Later in this lecture, we'll find ourselves taking a close interest in -------------------------------------------------------------------------- this particular expression of x, never mind why now. Note that if you add the -------------------------------------------------------------------------- two exponents and do some algebra, you'll end up with a quardratic where -------------------------------------------------------------------------- minus x squared over 2 is the -------------------------------------------------------------------------- leading term. If we set -------------------------------------------------------------------------- c to 1 over 2, then the leading term in the exponent down here is -------------------------------------------------------------------------- minus x squared over 2, also. So the sum of -------------------------------------------------------------------------- these exponents is equal to -------------------------------------------------------------------------- this, *provided* that we set -------------------------------------------------------------------------- c equal to 1 over 2, and also make good choices for -------------------------------------------------------------------------- $a$ and $b$. We leave it to you to figure out what $a$ and $b$ should be. We now -------------------------------------------------------------------------- evaluate at s plus t i, where -------------------------------------------------------------------------- t is a fixed real number, and where -------------------------------------------------------------------------- s tends to plus or minus infinity. Then -------------------------------------------------------------------------- this shows that the two limits above -------------------------------------------------------------------------- are both zero. This is true for *any* real number t, but, later in this lecture, -------------------------------------------------------------------------- we'll be particularly interested in values of t that are between minus 5 and zero, never mind why now. If -------------------------------------------------------------------------- this is the complex plane, -------------------------------------------------------------------------- with origin zero, and if -------------------------------------------------------------------------- this is minus 5 i, then -------------------------------------------------------------------------- here's a line of s+ti, with t about midway between minus 5 and 0. -------------------------------------------------------------------------- Here's another, with t closer to minus 5, and -------------------------------------------------------------------------- here's a third, with t closer to zero. We conclude that, along all three of these lines, the -------------------------------------------------------------------------- expression in question tends to zero -------------------------------------------------------------------------- as the real part of $x$ tends to infinity, or -------------------------------------------------------------------------- minus infinity. This is true not just for -------------------------------------------------------------------------- these three lines, but for *any* horizontal line, and, in particular, of any horizontal line with imaginary part between minus 5 and 0. Let's -------------------------------------------------------------------------- clear some room, take -------------------------------------------------------------------------- this expression and -------------------------------------------------------------------------- copy it twice. Next, fix some positive number -------------------------------------------------------------------------- $c$ and integrate the expression -------------------------------------------------------------------------- from c to c minus 5i, -------------------------------------------------------------------------- like so. If we let -------------------------------------------------------------------------- $c$ tend to infinity, then the line segment -------------------------------------------------------------------------- moves to the right and the -------------------------------------------------------------------------- integrand tends to zero uniformly, so the integral -------------------------------------------------------------------------- thens to zero as well. Similarly, if we -------------------------------------------------------------------------- integrate the expression -------------------------------------------------------------------------- from minus c minus 5i to minus c, -------------------------------------------------------------------------- like so, and let -------------------------------------------------------------------------- $c$ tend to infinity, then -------------------------------------------------------------------------- this line segment moves to the left, and the -------------------------------------------------------------------------- integrand tends to zero uniformly, so the integral -------------------------------------------------------------------------- tends to zero. This concludes our work on limits, and what -------------------------------------------------------------------------- remains are -------------------------------------------------------------------------- finanical math integrals, -------------------------------------------------------------------------- reduction of the pricing problem to an integral, -------------------------------------------------------------------------- calculation of that integral and -------------------------------------------------------------------------- a statement of the 50-50 Central Limit Theorem. -------------------------------------------------------------------------- Time to kick back! -------------------------------------------------------------------------- Welcome to the Third Act of Lecture 4 of Notes on Financial Mathematics, by Scot Adams and Fernando Reitich. In this act, we focus on some important integrals. The first calculates the -------------------------------------------------------------------------- total area under e to the minus x squared over 2, and we'll -------------------------------------------------------------------------- prove that that integral -------------------------------------------------------------------------- is root 2 pi. Let's define -------------------------------------------------------------------------- $I$ to be the integral. We then wish to show that $I$ is -------------------------------------------------------------------------- root 2 pi. Because the -------------------------------------------------------------------------- integrand is positive, -------------------------------------------------------------------------- the integral must be positive. If we could show that -------------------------------------------------------------------------- $I$ squared were 2 pi, then $I$ would be plus or minus root 2 pi, RECERR: left out the word "root" but, because -------------------------------------------------------------------------- $I$ is positive, we'd get the -------------------------------------------------------------------------- positive square root, and be done. So -------------------------------------------------------------------------- this is what we *want* to prove. Next, take -------------------------------------------------------------------------- this expression of x and y and note that, by properties of the exponential function, this is -------------------------------------------------------------------------- the product of two exponentials. -------------------------------------------------------------------------- We integrate over the plane. -------------------------------------------------------------------------- This factor does *not* depend on x, and may therefore be moved outside of the -------------------------------------------------------------------------- integral with respect to x, -------------------------------------------------------------------------- like so. -------------------------------------------------------------------------- This integral is exactly -------------------------------------------------------------------------- this, which is -------------------------------------------------------------------------- $I$. (pause...............) -------------------------------------------------------------------------- We leave the rest alone. Whatever $I$ is, it's a constant, and so can be -------------------------------------------------------------------------- brought outside the integral. -------------------------------------------------------------------------- We leave the rest alone. -------------------------------------------------------------------------- This is equal to -------------------------------------------------------------------------- this; in fact, the only difference is the variable of integration. Above, it's -------------------------------------------------------------------------- x; below, it's -------------------------------------------------------------------------- y. Then -------------------------------------------------------------------------- both of these are equal to -------------------------------------------------------------------------- $I$, and we end up with -------------------------------------------------------------------------- $I$ squared. (pause...............) We -------------------------------------------------------------------------- drop the intermediate steps and -------------------------------------------------------------------------- move $I$ squared up, to make room. Next, we change variables to polar coordinates. Remember that -------------------------------------------------------------------------- "x squared plus y squared is r squared" is one of the four key equations in that change of variables, and so -------------------------------------------------------------------------- "x squared plus y squared is r squared", on substitution, becomes -------------------------------------------------------------------------- r squared. The differential substitution is the equation -------------------------------------------------------------------------- dee x dee y equals r dee r dee theta, and remember that -------------------------------------------------------------------------- r is the determinant of psi prime, and should *not* be forgotten. We therefore replace -------------------------------------------------------------------------- dee x dee y by -------------------------------------------------------------------------- r dee r dee theta. The integrals change to -------------------------------------------------------------------------- these. Remember that we let r range -------------------------------------------------------------------------- from zero to infinity, and we let theta range -------------------------------------------------------------------------- from zero to 2 pi, and this covers all but the origin and positive horizontal axis, and these don't matter, because they're of zero area. Next, we substitute -------------------------------------------------------------------------- s equals r squared over 2, so that -------------------------------------------------------------------------- r squared over 2 becomes -------------------------------------------------------------------------- e to the minus s. The differential substitution now is the equation -------------------------------------------------------------------------- dee s equals r dee r, so -------------------------------------------------------------------------- r dee r becomes -------------------------------------------------------------------------- dee s. -------------------------------------------------------------------------- We leave the rest alone. Note that, as r ranges -------------------------------------------------------------------------- from zero to infinity, -------------------------------------------------------------------------- r squared over 2 also ranges -------------------------------------------------------------------------- from zero to infinity. To do -------------------------------------------------------------------------- this integral, we identify an -------------------------------------------------------------------------- antiderivative, and -------------------------------------------------------------------------- evaluate between 0 and infinity. Evaluating at -------------------------------------------------------------------------- infinity, we get -------------------------------------------------------------------------- minus zero. Evaluating at -------------------------------------------------------------------------- zero, we get -------------------------------------------------------------------------- minus 1. Then the -------------------------------------------------------------------------- difference is -------------------------------------------------------------------------- 1. So we end up integrating -------------------------------------------------------------------------- 1 from 0 to 2 pi, and this gives -------------------------------------------------------------------------- 2 pi. So -------------------------------------------------------------------------- $I$ squared *is* -------------------------------------------------------------------------- 2 pi, and -------------------------------------------------------------------------- that's what we wanted. -------------------------------------------------------------------------- Quod erat demonstratum. We now move on to -------------------------------------------------------------------------- a slightly more complicated integral. If it weren't for -------------------------------------------------------------------------- this 5 x, we could just pull the constant -------------------------------------------------------------------------- e to the 8 out of the integral, and we'd be back to integrating -------------------------------------------------------------------------- e to the minus x squared over 2, and we just found *that* integral to be root 2 pi. So the linear term -------------------------------------------------------------------------- 5 x is the troublemaker here. The secret to -------------------------------------------------------------------------- this kind of integral is to recognize that the -------------------------------------------------------------------------- full exponent is a polynomial of degree two, namely 5 x plus 8 minus x squared over 2. By completing the square, we can eliminate the linear term of -------------------------------------------------------------------------- 5 x, which is exactly the troublemaker term. Completing the square isn't hard, but it's especially easy when the quadratic in the exponent is -------------------------------------------------------------------------- minus x squared over 2. In *that* situation, just remember to -------------------------------------------------------------------------- replace x by x plus the linear coefficient. In *this* problem, -------------------------------------------------------------------------- the linear term is 5 x, so the linear coefficient is therefore 5. So, we -------------------------------------------------------------------------- replace x by x plus 5. Then -------------------------------------------------------------------------- 5 x becomes -------------------------------------------------------------------------- 5 times ( x plus 5 ) , and -------------------------------------------------------------------------- x squared becomes -------------------------------------------------------------------------- x plus 5 squared. The differential -------------------------------------------------------------------------- dee x (pause...............) -------------------------------------------------------------------------- doesn't change. To compensate for adding 5 to x, we subtract 5 from the -------------------------------------------------------------------------- limits of integration, but, as they're plus and minus infinity, this -------------------------------------------------------------------------- has no effect, as we'll explain on the next slide. We now expand -------------------------------------------------------------------------- 5 times x plus 5 to -------------------------------------------------------------------------- 5x plus 25. We now expand -------------------------------------------------------------------------- x plus 5 squared to -------------------------------------------------------------------------- x squared plus 10 x plus 25. -------------------------------------------------------------------------- We leave the rest alone. The exponential of a sum is the product of the exponentials, so -------------------------------------------------------------------------- this is equal to -------------------------------------------------------------------------- this, while -------------------------------------------------------------------------- here, we also get three factors: -------------------------------------------------------------------------- e to the minus x squared over 2, -------------------------------------------------------------------------- e to the minus 10 x over 2 or minus 5 x and -------------------------------------------------------------------------- e to the minus 25 over 2. The acid test of whether we really did the right -------------------------------------------------------------------------- replacement is whether the linear terms in the exponent cancel, and -------------------------------------------------------------------------- voila, they do. The product of -------------------------------------------------------------------------- these two is -------------------------------------------------------------------------- e to the 25 over 2, and -------------------------------------------------------------------------- this is (pause...............) -------------------------------------------------------------------------- e to the 8. -------------------------------------------------------------------------- This (pause...............) -------------------------------------------------------------------------- we leave inside (pause...............) -------------------------------------------------------------------------- the integral. Remember that -------------------------------------------------------------------------- this integral is -------------------------------------------------------------------------- root 2 pi, and -------------------------------------------------------------------------- we leave the rest alone, -------------------------------------------------------------------------- completing the problem. Let's -------------------------------------------------------------------------- back up a bit to the point where we'd just decided to -------------------------------------------------------------------------- replace x by x plus 5. -------------------------------------------------------------------------- Here's a number line. Our integral is from -------------------------------------------------------------------------- minus infinity to -------------------------------------------------------------------------- infinity, which covers every real number -------------------------------------------------------------------------- from left to right. If you plug, say, -------------------------------------------------------------------------- x equals 0 in the integrand above, you get exactly the same answer as you get if you -------------------------------------------------------------------------- plug x equals minus 5 in the one below. So, in moving from the top integrand to the bottom, in terms of evaluation, -------------------------------------------------------------------------- 0 moves to minus 5. In fact, to compensate for -------------------------------------------------------------------------- adding 5 to x, any evaluation has to move -------------------------------------------------------------------------- 5 units in the negative direction. In the top integral, we integrated -------------------------------------------------------------------------- from minus infinity to infinity. Moving all of those numbers -------------------------------------------------------------------------- 5 units to the left, we still integrate -------------------------------------------------------------------------- from minus infinity to infinity, so, in the bottom integral, the limits of integration are still -------------------------------------------------------------------------- minus infinity and -------------------------------------------------------------------------- infinity. Intuitively, we assert that -------------------------------------------------------------------------- infinity minus 5 is infinity, so -------------------------------------------------------------------------- this (pause...............) -------------------------------------------------------------------------- is unchanged. Similarly, speaking again intuitively, -------------------------------------------------------------------------- subtracting 5 does nothing to minus infinity, and so -------------------------------------------------------------------------- this (pause...............) -------------------------------------------------------------------------- is unchanged. -------------------------------------------------------------------------- Here, again, is the problem and answer. Of course, the numbers -------------------------------------------------------------------------- 5 and 8 were hardly crucial. We could rework the problem with other numbers, but, instead, let's generalize, and -------------------------------------------------------------------------- replace 5 by a and 8 by b. Then -------------------------------------------------------------------------- e to the 5x + 8 changes to -------------------------------------------------------------------------- e to the ax + b, and we -------------------------------------------------------------------------- leave the rest of the integral alone. If you go through all the computations again, you'll find that -------------------------------------------------------------------------- e to the 25 over 2 changes to -------------------------------------------------------------------------- e to the $a$ squared over 2, and that -------------------------------------------------------------------------- e to the 8 changes to -------------------------------------------------------------------------- e to the b, and that -------------------------------------------------------------------------- root 2 pi is unchanged. -------------------------------------------------------------------------- *This* formula is valid -------------------------------------------------------------------------- for all real numbers a and b. Next, we specialize to the case where -------------------------------------------------------------------------- b is zero. Then -------------------------------------------------------------------------- this goes away, and -------------------------------------------------------------------------- this goes away, as well, since e to the zero is 1. Also, if we set b equal to zero, then we should no longer say -------------------------------------------------------------------------- for all b. We end up with -------------------------------------------------------------------------- this statement. (pause...............) -------------------------------------------------------------------------- Here we have an expression of a, and it's complex analytic in a. -------------------------------------------------------------------------- This is also complex analytic in a. It's a general fact that, if two complex analytic expressions agree at -------------------------------------------------------------------------- every real number, then they agree at -------------------------------------------------------------------------- every complex number. Remembering that -------------------------------------------------------------------------- i is the square root of minus one, let's -------------------------------------------------------------------------- replace a by minus i t, whereupon -------------------------------------------------------------------------- e to the a x changes to -------------------------------------------------------------------------- e to the minus i t x. Since -------------------------------------------------------------------------- a is minus i t, -------------------------------------------------------------------------- a squared is -------------------------------------------------------------------------- *plus* i squared t squared, which is i squared t squared, which is -------------------------------------------------------------------------- minus t squared. Then -------------------------------------------------------------------------- a squared over 2 changes to -------------------------------------------------------------------------- minus t squared over 2. -------------------------------------------------------------------------- Nothing else changes. -------------------------------------------------------------------------- This formula is true for any complex number t, but we'll only need that it's true -------------------------------------------------------------------------- for real numbers t. -------------------------------------------------------------------------- Here's our next problem. By properties of exponentials -------------------------------------------------------------------------- this expands to -------------------------------------------------------------------------- this, and -------------------------------------------------------------------------- the rest we leave alone. We -------------------------------------------------------------------------- pull the constant e to the 8 out of the integration, -------------------------------------------------------------------------- and leave the rest alone. To compute -------------------------------------------------------------------------- this integral, we set -------------------------------------------------------------------------- t equals minus 5, so that -------------------------------------------------------------------------- this becomes -------------------------------------------------------------------------- e to the plus 5 i x and -------------------------------------------------------------------------- these two integrals are the same Plugging t equals minus 5 -------------------------------------------------------------------------- over here on the right, -------------------------------------------------------------------------- the right hand side becomes -------------------------------------------------------------------------- this. (pause...............) -------------------------------------------------------------------------- Minus 5 squared -------------------------------------------------------------------------- 25. Remembering -------------------------------------------------------------------------- e to the 8, we get -------------------------------------------------------------------------- this and -------------------------------------------------------------------------- that's our final answer. We next do the -------------------------------------------------------------------------- same problem, by another method. Since the -------------------------------------------------------------------------- linear term in the exponent has coefficient 5 i, we -------------------------------------------------------------------------- replace x by x plus 5 i. Then -------------------------------------------------------------------------- $x$ (pause...............) changes to -------------------------------------------------------------------------- x plus 5i, (pause...............) and -------------------------------------------------------------------------- x squared (pause...............) changes to -------------------------------------------------------------------------- (x plus 5i) squared. (pause...............) -------------------------------------------------------------------------- The rest stays the same. Expanding -------------------------------------------------------------------------- this, we get -------------------------------------------------------------------------- 5ix *minus* 25. Expanding -------------------------------------------------------------------------- this, we get -------------------------------------------------------------------------- x squared plus 10 i x *minus* 25. -------------------------------------------------------------------------- The rest stays the same. Expanding -------------------------------------------------------------------------- this, (pause...............) -------------------------------------------------------------------------- we get three factors. Expanding -------------------------------------------------------------------------- this, (pause...............) we *also* get three factors: -------------------------------------------------------------------------- e to the minus x squared over 2, (pause...............) -------------------------------------------------------------------------- e to the minus 10 i x over 2, or minus 5 i x (pause...............) and -------------------------------------------------------------------------- e to the minus minus 25 over 2, or plus 25 over . (pause...............) -------------------------------------------------------------------------- These cancel, -------------------------------------------------------------------------- these give -------------------------------------------------------------------------- e to the minus 25 over 2, and -------------------------------------------------------------------------- e to the 8 also -------------------------------------------------------------------------- comes outside the integral. This, by contrast -------------------------------------------------------------------------- remains inside the integral. The differential dee x -------------------------------------------------------------------------- is unchanged. -------------------------------------------------------------------------- This, we recall, is -------------------------------------------------------------------------- root 2 pi, and -------------------------------------------------------------------------- the rest is unchanged. -------------------------------------------------------------------------- This is the -------------------------------------------------------------------------- same answer as before, but by other means. Let's -------------------------------------------------------------------------- back up a bit to the point where we'd just decided to -------------------------------------------------------------------------- replace x by x plus 5i. -------------------------------------------------------------------------- Here's a complex plane. The integral on top is -------------------------------------------------------------------------- from minus infinity -------------------------------------------------------------------------- to infinity, which covers every real number -------------------------------------------------------------------------- from left to right. Plugging -------------------------------------------------------------------------- 0 into the top integrand is the same as plugging -------------------------------------------------------------------------- minus 5 i into the bottom one. So, in terms of evaluation, -------------------------------------------------------------------------- zero moves to minus 5i. If we do this, not just for zero, but for -------------------------------------------------------------------------- every real number, we see that the bottom integral *should* be over -------------------------------------------------------------------------- *this* line, and so we really should -------------------------------------------------------------------------- subtract 5 i from the limits of integration to compensate for -------------------------------------------------------------------------- adding 5 i to x. These -------------------------------------------------------------------------- two lines are different, so we should be careful *neither* to equate -------------------------------------------------------------------------- infinity minus 5 i with infinity *nor* to equate -------------------------------------------------------------------------- minus infinity minus 5 i with minus infinity. So (pause...............) even though, in the last slide, we got -------------------------------------------------------------------------- the right answer, did we make a mistake getting there? It turns out that we didn't, as we'll show in just a moment. In just a moment, we'll study the analytic properties of -------------------------------------------------------------------------- the integrand on -------------------------------------------------------------------------- this strip which extends -------------------------------------------------------------------------- infinitely far to the left and right. We'll show that the -------------------------------------------------------------------------- left to right integral along the bottom of the strip is equal to the -------------------------------------------------------------------------- left to right integral along the top. -------------------------------------------------------------------------- Here's the integral along the bottom, and, in just a moment, we'll show that it's -------------------------------------------------------------------------- equal to the integral along the top, where -------------------------------------------------------------------------- the integrand stays the same throughout. So, even though equality fails -------------------------------------------------------------------------- here, our -------------------------------------------------------------------------- two integrals, in the end, agree, and so we made no mistake, as long as we can prove -------------------------------------------------------------------------- this equality, and here's how we do it: Take any positive number -------------------------------------------------------------------------- c, and integrate -------------------------------------------------------------------------- this expression (pause...............) -------------------------------------------------------------------------- from minus c to c, then -------------------------------------------------------------------------- to c minus 5 i, then -------------------------------------------------------------------------- to minus c minus 5 i, then -------------------------------------------------------------------------- back to minus c. The -------------------------------------------------------------------------- integrand is analytic for x in -------------------------------------------------------------------------- this rectangular region, so Cauchy's Theorem from Complex Analysis tells us that the sum of -------------------------------------------------------------------------- these four integrals is equal to zero. That is, if we compute -------------------------------------------------------------------------- these four integrals of -------------------------------------------------------------------------- this integrand -------------------------------------------------------------------------- with respect to x, -------------------------------------------------------------------------- and add, -------------------------------------------------------------------------- then we get zero. Note that we don't have enough room in the -------------------------------------------------------------------------- top line to fit four copies of -------------------------------------------------------------------------- the integrand dee x, so we omit all that, just to save space. If we let -------------------------------------------------------------------------- c tend to infinity, then -------------------------------------------------------------------------- the right edge moves -------------------------------------------------------------------------- this way. In the preceding act of this lecture, we showed that -------------------------------------------------------------------------- this integrand tends to zero uniformly, as we move to the right along any horizontal line. Using that, we saw that -------------------------------------------------------------------------- the right edge integral, which appears -------------------------------------------------------------------------- here, tends to zero. Similarly, as c -------------------------------------------------------------------------- tends to infinity, -------------------------------------------------------------------------- the left edge moves -------------------------------------------------------------------------- this way, and -------------------------------------------------------------------------- this expression again tends to zero uniformly, and so we can prove that -------------------------------------------------------------------------- this integral tends to zero as well. So, letting c tend to infinity, -------------------------------------------------------------------------- both of these tend to zero, and, in the limit, we get -------------------------------------------------------------------------- this. Interchanging -------------------------------------------------------------------------- these limits of integration, and changing -------------------------------------------------------------------------- this plus to a minus, we get -------------------------------------------------------------------------- this, so -------------------------------------------------------------------------- this is equal to -------------------------------------------------------------------------- this, which says that the -------------------------------------------------------------------------- left to right integral along the bottom of the strip is equal to the -------------------------------------------------------------------------- left to right integral along the top. In other words, -------------------------------------------------------------------------- this *is* true. Okay. To do the next integral, we first need a -------------------------------------------------------------------------- definition. The -------------------------------------------------------------------------- error function, denoted -------------------------------------------------------------------------- capital Phi, is given by -------------------------------------------------------------------------- this equation. As finanical mathematicians, we consider this to be one of our basic or primitive functions, much as a surveyor might consider trig functions to be basic. It's a fact that -------------------------------------------------------------------------- this error function can't be expressed in terms of exponential, trig and polynomial functions, even using composition and inverses. It really is a brand new function. The good news is that, nowadays, many calculators and computer applications have this error function in their library. So you can think of -------------------------------------------------------------------------- Phi as being just one more button on the calculator. As an exercise, find a calculator with such a button, and compute -------------------------------------------------------------------------- Phi of minus 2. If we now -------------------------------------------------------------------------- replace t by minus t, and -------------------------------------------------------------------------- dee t by minus 1 times dee t, we get -------------------------------------------------------------------------- this. Note that -------------------------------------------------------------------------- the limits of integration were negated. We now -------------------------------------------------------------------------- interchange them, and, at the same time, -------------------------------------------------------------------------- eliminate this minus 1, giving -------------------------------------------------------------------------- this. If we multiply -------------------------------------------------------------------------- *this* by root 2 pi, we get -------------------------------------------------------------------------- this. If we multiply -------------------------------------------------------------------------- this by root 2 pi, we get -------------------------------------------------------------------------- this, so these are -------------------------------------------------------------------------- equal. There's nothing special about the number -------------------------------------------------------------------------- 2. The same formula holds with 2 replaced by any number -------------------------------------------------------------------------- x. Note that, here -------------------------------------------------------------------------- x is a *lower* limit of integration, and we find -------------------------------------------------------------------------- *minus* x inside Phi, whereas, here, -------------------------------------------------------------------------- x is an *upper* limit, and we find -------------------------------------------------------------------------- x itself inside Phi, not minus x. In either case, we should -------------------------------------------------------------------------- always remember, and never forget "root 2 pi." Okay. -------------------------------------------------------------------------- Here's our next problem. It's similar to problems we've tackled before, except for the finite -------------------------------------------------------------------------- lower limit of integration. Following the process described before, we identify the -------------------------------------------------------------------------- linear term in the exponent. The linear coefficient is 4, and so we -------------------------------------------------------------------------- replace x by x plus 4. -------------------------------------------------------------------------- 4 x (pause...............) becomes -------------------------------------------------------------------------- 4 times x plus 4, (pause...............) -------------------------------------------------------------------------- x squared (pause...............) becomes -------------------------------------------------------------------------- x plus 4 squared, (pause...............) -------------------------------------------------------------------------- dx is unchanged, and -------------------------------------------------------------------------- on the integral, we compensate for -------------------------------------------------------------------------- adding 4 to x by subtracting 4 from the limits of integration. Speaking intuitively, -------------------------------------------------------------------------- infinity minus 4 is -------------------------------------------------------------------------- infinity. Speaking precisely, -------------------------------------------------------------------------- 6 minus 4 is -------------------------------------------------------------------------- 2. (pause...............) -------------------------------------------------------------------------- 4 times (x plus 4) (pause...............) expands to -------------------------------------------------------------------------- 4x plus 16, (pause...............) -------------------------------------------------------------------------- (x plus 4) squared (pause...............) expands to -------------------------------------------------------------------------- x squared plus 8 x plus 16, (pause...............) and -------------------------------------------------------------------------- the rest is unchanged. -------------------------------------------------------------------------- This (pause...............) expands to -------------------------------------------------------------------------- this product. (pause...............) -------------------------------------------------------------------------- This also expands to a product of three factors: -------------------------------------------------------------------------- e to the minus x squared over 2 (pause...............), -------------------------------------------------------------------------- e to the minus 8 x over 2, or minus 4 x (pause...............), and -------------------------------------------------------------------------- e to the minus 16 over 2. Now for the magic of completing the square: -------------------------------------------------------------------------- The exponentiated linear terms cancel, leaving us with only -------------------------------------------------------------------------- constants and -------------------------------------------------------------------------- e to the minus x squared over 2. We pull the constants outside the integral. -------------------------------------------------------------------------- These two give -------------------------------------------------------------------------- e to the 16 over 2, and -------------------------------------------------------------------------- e to the 9 -------------------------------------------------------------------------- comes out unchanged. -------------------------------------------------------------------------- Also, e to the minus squared over 2 -------------------------------------------------------------------------- remains inside the integral. The differential dee x -------------------------------------------------------------------------- is unchanged. -------------------------------------------------------------------------- This can be computed with the error function. It's equal to -------------------------------------------------------------------------- this. Note that, because -------------------------------------------------------------------------- 2 is a *lower* limit of integration, it appears with a -------------------------------------------------------------------------- minus sign. Also, we -------------------------------------------------------------------------- remembered and didn't forget "root 2 pi". -------------------------------------------------------------------------- These factors -------------------------------------------------------------------------- remain, and -------------------------------------------------------------------------- here's our answer. If you have a calculator that can compute the -------------------------------------------------------------------------- error function Phi, then you can compute the answer in decimal form, but we leave it -------------------------------------------------------------------------- as shown. Here's an -------------------------------------------------------------------------- exercise: Compute -------------------------------------------------------------------------- this integral. We provide a -------------------------------------------------------------------------- hint. If you -------------------------------------------------------------------------- compute this and -------------------------------------------------------------------------- this, and then -------------------------------------------------------------------------- add the two results together, then that'll solve the problem. To compute -------------------------------------------------------------------------- this integral, since the linear term in the exponent is -------------------------------------------------------------------------- 3 x, you should replace x by x plus 3. To compute -------------------------------------------------------------------------- *this* integral, bring the constant -------------------------------------------------------------------------- minus 2 out of the integral. Note that, because -------------------------------------------------------------------------- minus 7 is a lower limit, it needs to be negated and your answer will involve Phi of *plus* 7. Also: remember and don't forget "root 2 pi". To summarize, in order to compute -------------------------------------------------------------------------- this kind of integral, we should distribute -------------------------------------------------------------------------- this exponentiated quadratic and -------------------------------------------------------------------------- this integral over -------------------------------------------------------------------------- this subtraction, and then use the skills we developed earlier to finish the problem. -------------------------------------------------------------------------- Here's our next problem. If it weren't for -------------------------------------------------------------------------- this positive part, this problem would be similar to the problem given in the last slide. To do *this* integral, it helps to figure out where -------------------------------------------------------------------------- this expression is positive, negative and zero. -------------------------------------------------------------------------- This expression increases as x increases, and there's a *number*, -------------------------------------------------------------------------- call it $a$, at which it's -------------------------------------------------------------------------- zero. Below that $a$, -------------------------------------------------------------------------- this is negative; above, it's positive. We solve for the unknown $a$ by, first -------------------------------------------------------------------------- adding 2, then -------------------------------------------------------------------------- taking logarithms, then -------------------------------------------------------------------------- subtracting 9. For values of $x$ below this number $a$, -------------------------------------------------------------------------- this expression is negative, and the -------------------------------------------------------------------------- positive part of a negative number is zero. We can therefore -------------------------------------------------------------------------- eliminate all integration below -------------------------------------------------------------------------- this number $a$, -------------------------------------------------------------------------- like so. For values of $x$ above $a$, -------------------------------------------------------------------------- the positive part doesn't do anything, because -------------------------------------------------------------------------- this expression is positive, and the positive part of a positive number is itself. We therefore -------------------------------------------------------------------------- drop the positive part, and end up with a problem very similar to the problem we saw in the last slide. Following what we learned there, the next step is to distribute -------------------------------------------------------------------------- this exponentiated quadratic across -------------------------------------------------------------------------- this subtraction, -------------------------------------------------------------------------- like so. Note that -------------------------------------------------------------------------- the exponentiated quadratic now appears in two places. Next step is to distribute -------------------------------------------------------------------------- this integral across -------------------------------------------------------------------------- this subtraction, -------------------------------------------------------------------------- like so. Note that -------------------------------------------------------------------------- the integral now appears in two places. We also snuck -------------------------------------------------------------------------- this constant outside of the integral. We've run out of room, but, by the wonders of PowerPoint animation, -------------------------------------------------------------------------- we're saved. -------------------------------------------------------------------------- This integral -------------------------------------------------------------------------- is this. Note that, as -------------------------------------------------------------------------- $a$ is a *lower* limit of integration, -------------------------------------------------------------------------- we get a minus sign. Also, -------------------------------------------------------------------------- we remembered and didn't forget "root 2 pi". -------------------------------------------------------------------------- The rest is unchanged. -------------------------------------------------------------------------- x is 1 times x, and so the linear coefficient in the exponent is -------------------------------------------------------------------------- 1. So we change -------------------------------------------------------------------------- x to x plus 1. On the -------------------------------------------------------------------------- limits of integration, we compensated by -------------------------------------------------------------------------- subtracting 1. Expanding, -------------------------------------------------------------------------- we get this. An important observation is that the exponentiated linear terms cancel, and we didn't even write them down. Specifically, -------------------------------------------------------------------------- This x cancels against the minus x that comes from expanding -------------------------------------------------------------------------- this square. The square gives 2 x, but we also have a -------------------------------------------------------------------------- minus sign, and we -------------------------------------------------------------------------- divide by 2, giving minus x. Next, we -------------------------------------------------------------------------- pull constants outside the integral. We remember -------------------------------------------------------------------------- this term, -------------------------------------------------------------------------- memorize $a$ -------------------------------------------------------------------------- partition the slide -------------------------------------------------------------------------- and pull the term down unchanged. We move what's at the bottom -------------------------------------------------------------------------- up a bit to make room. We -------------------------------------------------------------------------- remember $a$. -------------------------------------------------------------------------- This integral is -------------------------------------------------------------------------- this. The -------------------------------------------------------------------------- lower limit of integration -------------------------------------------------------------------------- was negated, and we -------------------------------------------------------------------------- remembered and didn't forget "root 2 pi". -------------------------------------------------------------------------- The rest is unchanged. We run to our calculators, -------------------------------------------------------------------------- compute $a$, and plug the result in -------------------------------------------------------------------------- here. Our calculators have a button that can do -------------------------------------------------------------------------- the error function, and, with a bit of work, -------------------------------------------------------------------------- we get to this numerical answer. So ... is time finally for a beer??? No, wait ... -------------------------------------------------------------------------- here's our next problem, and as k varies from -------------------------------------------------------------------------- zero to 1 to 2 and so on, it's really an infinite sequence of problems. If Sisyphus can do it, so can we! We begin with -------------------------------------------------------------------------- k equals 0. To -------------------------------------------------------------------------- solve this, we remember -------------------------------------------------------------------------- Version C of the Second Form of the Fundamental Theorem of Calculus. We set -------------------------------------------------------------------------- f(x) equal to -------------------------------------------------------------------------- e to the minus x squared over 2. -------------------------------------------------------------------------- Then f(t) is -------------------------------------------------------------------------- e to the minus t squared over 2, and we -------------------------------------------------------------------------- leave the rest alone. -------------------------------------------------------------------------- We clear some room. This works for -------------------------------------------------------------------------- any $a$, even -------------------------------------------------------------------------- minus infinity. Now remember the definition of -------------------------------------------------------------------------- the error function Phi. We -------------------------------------------------------------------------- multiply by root 2 pi and -------------------------------------------------------------------------- add an arbitrary constant C, whereupon we get -------------------------------------------------------------------------- equality here and -------------------------------------------------------------------------- this is the answer -------------------------------------------------------------------------- to our question. Don't be put off by the presence of the -------------------------------------------------------------------------- error function in the answer. To us, this is no different from seeing an answer to an integration problem that involves sine or cosine or logarithm. The error function is just one of our primitive fucntions. So k equals 0 is done. Now we move on to -------------------------------------------------------------------------- k equals 1. This is handled by making the de-substitution -------------------------------------------------------------------------- u for x squared over 2. Then -------------------------------------------------------------------------- this becomes -------------------------------------------------------------------------- e to the minus u. The differential substitution is -------------------------------------------------------------------------- dee u equals x dee x, so -------------------------------------------------------------------------- x dee x becomes -------------------------------------------------------------------------- dee u. -------------------------------------------------------------------------- Integrating e to the minus u with respect to u, -------------------------------------------------------------------------- we get this, and substituting -------------------------------------------------------------------------- x squared over 2 back in, we get -------------------------------------------------------------------------- this answer to the k equals 1 problem. For k equals 1, the -------------------------------------------------------------------------- $x$ in the problem changed to a -------------------------------------------------------------------------- minus sign in the answer. We record, for posterity, that, among all of -------------------------------------------------------------------------- these antiderivatives, the -------------------------------------------------------------------------- simplest is found by setting C to 0, which gives -------------------------------------------------------------------------- minus e to the minus x squared over 2. Take note of -------------------------------------------------------------------------- this minus sign. Now we move on to -------------------------------------------------------------------------- k equals 2. We want to compute -------------------------------------------------------------------------- this integral, and the first step is to take -------------------------------------------------------------------------- x squared, and write it as -------------------------------------------------------------------------- x times x, and -------------------------------------------------------------------------- leave everything else alone. We now apply Integration by Parts, -------------------------------------------------------------------------- differentiating one factor. We need to antidifferentiate -------------------------------------------------------------------------- the other. This antidifferentiation problem is just the -------------------------------------------------------------------------- k equals 1 case from the last slide, and the simplest antiderivative is -------------------------------------------------------------------------- minus e to the minus x squared over 2. We take note of this -------------------------------------------------------------------------- minus sign. Each -------------------------------------------------------------------------- new factor is simpler to integrate than the corresponding -------------------------------------------------------------------------- old factor, so Integration by Parts is working *very* well. We follow our little buzz phrase, and take -------------------------------------------------------------------------- up up, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- minus the integral of -------------------------------------------------------------------------- new new, -------------------------------------------------------------------------- like so. We've run out of room, so we take the -------------------------------------------------------------------------- stuff at the bottom of the screen, and -------------------------------------------------------------------------- move it up a bit. The -------------------------------------------------------------------------- two minus signs cancel, giving -------------------------------------------------------------------------- this, and -------------------------------------------------------------------------- this is just the -------------------------------------------------------------------------- k equal 0 case from a few slides ago, and, if you look back, you'll find -------------------------------------------------------------------------- this. The rest remains -------------------------------------------------------------------------- unchanged, and -------------------------------------------------------------------------- this is our k equals 2 answer. Now we move on to -------------------------------------------------------------------------- k equals 3. We want to compute -------------------------------------------------------------------------- this integral, and the first step is to take -------------------------------------------------------------------------- x cubed, and write it as -------------------------------------------------------------------------- x squared times x, and -------------------------------------------------------------------------- leave everything else alone. We now apply Integration by Parts, -------------------------------------------------------------------------- differentiating one factor. We need to -------------------------------------------------------------------------- antidifferentiate the other. This antidifferentiation problem is just the k equals 1 case from a previous slide, and recall the -------------------------------------------------------------------------- simplest antiderivative. We take note of -------------------------------------------------------------------------- this minus sign. Note that the two -------------------------------------------------------------------------- new factors are both simpler to integrate than the -------------------------------------------------------------------------- old factors, so Integration by Parts is working *very* well. We follow our little buzz phrase, and take -------------------------------------------------------------------------- up up, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- minus the integral of -------------------------------------------------------------------------- new new, -------------------------------------------------------------------------- like so. We've run out of room, so we take the stuff at the bottom of the screen, and -------------------------------------------------------------------------- move it up a bit. The -------------------------------------------------------------------------- two minus signs cancel, giving -------------------------------------------------------------------------- this. Note that we also -------------------------------------------------------------------------- snuck this constant out of the integral. -------------------------------------------------------------------------- This is just the -------------------------------------------------------------------------- k equals 1 case from a few slides ago, and it's equal to -------------------------------------------------------------------------- this. We take note of -------------------------------------------------------------------------- this minus sign. We multiply by -------------------------------------------------------------------------- 2, (pause...............) -------------------------------------------------------------------------- like so, and the rest remains -------------------------------------------------------------------------- unchanged. We're adding on -------------------------------------------------------------------------- 2 times every constant, which is no different from adding on every constant, so, the factor of 2 -------------------------------------------------------------------------- is superfluous, and -------------------------------------------------------------------------- this is our k equals 3 answer. Now we move on to -------------------------------------------------------------------------- the general case, where we take -------------------------------------------------------------------------- x to the k, and write it as -------------------------------------------------------------------------- (x to the k minus 1) times x and -------------------------------------------------------------------------- leave everything else unchanged. We -------------------------------------------------------------------------- differentiate and -------------------------------------------------------------------------- antidifferentiate, and then take -------------------------------------------------------------------------- up up, -------------------------------------------------------------------------- like so, -------------------------------------------------------------------------- minus the integral of -------------------------------------------------------------------------- new new, -------------------------------------------------------------------------- like so. The two -------------------------------------------------------------------------- minuses -------------------------------------------------------------------------- cancel and, at the same time, we -------------------------------------------------------------------------- sneak the constant out of the integral. Now -------------------------------------------------------------------------- this integral is the same as -------------------------------------------------------------------------- this one, except that we have -------------------------------------------------------------------------- k minus 2 here, and -------------------------------------------------------------------------- k here. So, if we were to want to do, say, the -------------------------------------------------------------------------- k equals 4 problem, we could get to the answer, as long as we'd already done the -------------------------------------------------------------------------- k equals 2 problem, which we *have*. In other words, we've provided a recursive way of answering all of -------------------------------------------------------------------------- these problems. To get any answer, one has to work one's way up to it. We leave it as an -------------------------------------------------------------------------- exercise to work out the k equals -------------------------------------------------------------------------- 4 problem, using the recursive formula below. Okay. In the -------------------------------------------------------------------------- remainder of this lecture, we'll -------------------------------------------------------------------------- recall our pricing problem from the last lecture. Using limits of renormalized powers, and using also Fourier transforms, we'll -------------------------------------------------------------------------- reduce the problem to the calculation of an integral, then use our newly acquired integration skills to -------------------------------------------------------------------------- calculate the integral, solving the -------------------------------------------------------------------------- pricing problem. Finally, we'll -------------------------------------------------------------------------- state a version of the -------------------------------------------------------------------------- Central Limit Theorem. -------------------------------------------------------------------------- Time for another break. -------------------------------------------------------------------------- Welcome to the Fourth Act of Lecture 4 of Notes on Financial Mathematics, by Scot Adams and Fernando Reitich. Remember -------------------------------------------------------------------------- Kyle who wants to buy a -------------------------------------------------------------------------- call option on -------------------------------------------------------------------------- 5,000 shares of ABC at a strike of -------------------------------------------------------------------------- 5,000 dollars with a term of -------------------------------------------------------------------------- 30 days. -------------------------------------------------------------------------- Gail sells this option. We -------------------------------------------------------------------------- assume a spot price of $1 per share. We also assume that -------------------------------------------------------------------------- the stock ticks up or down a small amount each second, with uptick and downtick factors -------------------------------------------------------------------------- as given. We also assume -------------------------------------------------------------------------- this one-second risk-free factor. Our -------------------------------------------------------------------------- goal is to find the right, or -------------------------------------------------------------------------- arbitrage-free price. As usual, we'll begin by working out the -------------------------------------------------------------------------- payoff function, which we'll denote by -------------------------------------------------------------------------- f(S), where S is the final share price. *If* Kyle exercises, then Gail will provide him with -------------------------------------------------------------------------- 5,000 shares for -------------------------------------------------------------------------- 5,000 dollars. At share price S, the 5,000 shares cost Gail -------------------------------------------------------------------------- 5,000 S dollars, but that cost is offset by the -------------------------------------------------------------------------- 5,000 dollar strike price that Kyle pays her to exercise. Of course, it's reasonable to assume Kyle'll only exercise if the net amount he'll receive -------------------------------------------------------------------------- is positive. After all, who would choose to lose money, given the *option* not to? As usual, once we have a payoff function, we have an -------------------------------------------------------------------------- exercise: Graph it. The various numbers we're handling here are quite long, and so we'll give each a name. For example, there's the -------------------------------------------------------------------------- number of seconds in 30 days, which calculates to -------------------------------------------------------------------------- more than 2 and a half million. We'll call this number -------------------------------------------------------------------------- N. Next, there's the -------------------------------------------------------------------------- uptick factor, which we'll call -------------------------------------------------------------------------- u, and the -------------------------------------------------------------------------- downtick factor, which we'll call -------------------------------------------------------------------------- d, and the -------------------------------------------------------------------------- risk-free factor, which we'll call -------------------------------------------------------------------------- rho. We'll denote the one-second interest rate by -------------------------------------------------------------------------- iota, which is just -------------------------------------------------------------------------- the risk-free rate minus 1. We record that -------------------------------------------------------------------------- 1 plus iota is rho, that is, 1 plus the interest rate is the risk-free factor. Okay. We next work out the risk-neutral uptick and downtick probabilities. -------------------------------------------------------------------------- On a number line, we plot the -------------------------------------------------------------------------- uptick and downtick factors, and the -------------------------------------------------------------------------- risk-free factor. For pedagogical purposes, I chose these numbers so that -------------------------------------------------------------------------- rho is *exactly* halfway between $d$ and $u$, which means that -------------------------------------------------------------------------- the risk-neutral uptick and downtick probabilities are both -------------------------------------------------------------------------- 50 percent. In the -------------------------------------------------------------------------- 50-50 world, we can calculate expected returns of -------------------------------------------------------------------------- stock (pause...............) and -------------------------------------------------------------------------- bank. First, for the -------------------------------------------------------------------------- bank, the (pause...............) -------------------------------------------------------------------------- expected value is -------------------------------------------------------------------------- 50 percent of rho L plus -------------------------------------------------------------------------- 50 percent of rho L, which is (pause...............) -------------------------------------------------------------------------- 100 percent of rho L. A loan of -------------------------------------------------------------------------- $L$ will grow to -------------------------------------------------------------------------- rho L, after one second. The increase is -------------------------------------------------------------------------- rho L minus L, or -------------------------------------------------------------------------- (rho minus 1) times L, or -------------------------------------------------------------------------- iota times L, so the -------------------------------------------------------------------------- expected return is by a factor of -------------------------------------------------------------------------- iota. For the (pause...............) -------------------------------------------------------------------------- stock the -------------------------------------------------------------------------- expected value is -------------------------------------------------------------------------- 50% of uS plus -------------------------------------------------------------------------- 50% of dS. Because -------------------------------------------------------------------------- rho is exactly halfway between -------------------------------------------------------------------------- d and u, we get an -------------------------------------------------------------------------- expected value of rho S. So -------------------------------------------------------------------------- S dollars invested in the stock for one second can be *expected*, on average, in the risk-neutral world, to increase to -------------------------------------------------------------------------- rho S, so the expected amount of increase here is -------------------------------------------------------------------------- rho S minus S, or -------------------------------------------------------------------------- (rho minus 1) times S, or -------------------------------------------------------------------------- iota times S, which gives an -------------------------------------------------------------------------- expected return of by a factor of -------------------------------------------------------------------------- iota. The risk neutral world, in this problem, is a -------------------------------------------------------------------------- 50-50 world, and what's special about these probabilities is *not* that we necessarily believe that they're the real-world probabilities. What's special in this imaginary risk-neutral world is that -------------------------------------------------------------------------- these two expected returns are equal, so that any self-financed portfolio of -------------------------------------------------------------------------- stock and bank will have -------------------------------------------------------------------------- an expected return of iota each second. In other words, the expected value of any self-financed portfolio will grow by a factor of -------------------------------------------------------------------------- rho each second. Okay. Because of the size of the number N of subperiods, it's infeasible to use templates, as we did in the past, but we make a bit of an attempt anyway. We'll render, in black, the evolution of the underlying -------------------------------------------------------------------------- ABC share price, which starts out at a spot price of -------------------------------------------------------------------------- 1 dollar per share, and then evolves by the uptick or downtick factor to -------------------------------------------------------------------------- $u$ or $d$, over the first second. Over the *second* second, it again changes by a factor of $u$ or $d$ and -------------------------------------------------------------------------- these are the possible share prices after two seconds. We invite the interested listener to -------------------------------------------------------------------------- review the values of $u$ and $d$ and to compute -------------------------------------------------------------------------- $u$ squared, $ud$ and $d$ squared. At the end of the third second, the price may hit one of -------------------------------------------------------------------------- four possible values. This -------------------------------------------------------------------------- continues on for N steps, and N is so large that there's -------------------------------------------------------------------------- NO WAY we can show you all of them, but, if we -------------------------------------------------------------------------- continue to the ending share prices, we have many possibilities ranging from -------------------------------------------------------------------------- u to the N, representing N upticks, down to -------------------------------------------------------------------------- d to the N, representing N downticks. Another two possible ending share prices are -------------------------------------------------------------------------- u to the (N minus 1) d, -------------------------------------------------------------------------- u to the (N minus 2) d squared. Of course we -------------------------------------------------------------------------- can't list them all. We now move from the price of the underlying ABC, rendered in black, to the -------------------------------------------------------------------------- contingent claim, which we'll render in red. Remember that the contingent claim is the amount that Gail needs, in her hedge, to meet her obligation to Kyle, at the end of the 30 day term. That is, it's the ending value of the derivative. The connection between the underlying and the derivative is the payoff function -------------------------------------------------------------------------- f. That is, to get the contingent claim, we plug the -------------------------------------------------------------------------- ending underlying prices into the payoff function, f, -------------------------------------------------------------------------- like so. We want to price the derivative, so we want to move backward in time, to get from the contingent claim back to the initial price of the option. The inefficient way to do this is to solve huge numbers of equations in huge numbers of unknowns. The clever and efficient way to get the result is to move to -------------------------------------------------------------------------- the risk-neutral world, where any self-financed portfolio has an expected return of iota per second, and so its expected value grows by a factor of rho each second. The -------------------------------------------------------------------------- price, P, of the option is just the amount Gail needs in order to set up the hedge, so it's -------------------------------------------------------------------------- the initial value of the hedge. Gail, never steals from the hedge, and never puts money into it. She simply adjusts it. That is, the hedge is self-financing. In the risk-neutral world, the expected value grows by a factor of rho each second, so, since there are N seconds in 30 days, it grows by a factor of -------------------------------------------------------------------------- rho to the N over the 30 day term. So -------------------------------------------------------------------------- rho to the N times P, -------------------------------------------------------------------------- is the expected final value of the hedge. The collection of possible final values of the hedge *is* the -------------------------------------------------------------------------- contingent claim, so we seek to compute -------------------------------------------------------------------------- the expected contingent claim, working in the -------------------------------------------------------------------------- risk-neutral world. To compute this expected value, we should find the probability of -------------------------------------------------------------------------- N upticks and 0 downticks, and multiply it by -------------------------------------------------------------------------- f of u to the N. We should then find the probability of -------------------------------------------------------------------------- N-1 upticks and 1 downtick, and multiply *it* by -------------------------------------------------------------------------- f of u to the (N minus 1) d. We should then find the probability of -------------------------------------------------------------------------- N-2 upticks and 2 downticks, and multiply *it* by -------------------------------------------------------------------------- f of u to the (N minus 2) d squared. We should then -------------------------------------------------------------------------- continue on, until we finally find the probability of -------------------------------------------------------------------------- 0 upticks and N downticks, and multiply *it* by -------------------------------------------------------------------------- f of d to the N. Adding all those results gives the -------------------------------------------------------------------------- expected contingent claim, from which we can get the -------------------------------------------------------------------------- price of the option. Remembering that "Coin-Flippers got Price", we seek to relate this huge pricing computation to a -------------------------------------------------------------------------- coin-flipping game. Since the risk-neutral world is, here, a 50-50 world, we -------------------------------------------------------------------------- flip a fair (or 50-50) coin, and we do it N times. The payoff rule for this coin-flipping game is as follows: If, in those N flips, we see -------------------------------------------------------------------------- H heads and T tails, then we -------------------------------------------------------------------------- pay out f of u to the H d to the T. Moreover, this payoff is done -------------------------------------------------------------------------- 30 days from now. We've set up this coin-flipping game so that its -------------------------------------------------------------------------- expected payout, which we call E, is exactly the same as the -------------------------------------------------------------------------- expected contingent claim, which, in turn, is equal to -------------------------------------------------------------------------- rho to the N P, and so we have -------------------------------------------------------------------------- rho to the N P here, as well. Our goal is to price this option, so we want to compute the price -------------------------------------------------------------------------- P, and now we see that it's equal to -------------------------------------------------------------------------- rho to the minus N times E. Remember that rho is -------------------------------------------------------------------------- 1 plus iota. Since -------------------------------------------------------------------------- 1 plus iota to the minus N is the 30 day discount factor on the cost of money, we see that the option price is, as usual -------------------------------------------------------------------------- the discounted expected payout for the coin-flipping game. All this is well and good, but we're still faced with the seemingly daunting problem of computing -------------------------------------------------------------------------- this expected payout, and there's the rub. Fortunately, we'll see that we have a nice little fact called the Central Limit Theorem that'll bring us home. We can highlight all the main ideas in the Central Limit Theorem by doing an -------------------------------------------------------------------------- easier problem than the pricing problem. We'll get back to pricing in a moment, but let's cut our teeth on -------------------------------------------------------------------------- computing the *probability* that -------------------------------------------------------------------------- the number of heads minus the number of tails is between -------------------------------------------------------------------------- minus root N and -------------------------------------------------------------------------- plus root N. Our plan is that, once we gain some experience with -------------------------------------------------------------------------- probability calculation, we can move on to calculating some easy expected values and then, gradually, work our way up to the expected value, E, of that coin-flipping payout from the last slide. Okay. Let -------------------------------------------------------------------------- X be H minus T over root N. We want to -------------------------------------------------------------------------- compute the probability that -------------------------------------------------------------------------- this is true, but, if we divide -------------------------------------------------------------------------- this inequality by root N, then we get -------------------------------------------------------------------------- this simpler looking inequality. So we still have our -------------------------------------------------------------------------- easier problem, but it's been slightly restated. Next, let -------------------------------------------------------------------------- H_1 be the number of heads after the first flip, so H_1 may be 0 or 1 depending on whether the first flip comes up tails or heads. Let -------------------------------------------------------------------------- H_2 be the number of heads after the second flip, so H_2 may be 0 or 1 or 2. -------------------------------------------------------------------------- We continue on to -------------------------------------------------------------------------- H_N which is the number of heads after the Nth flip, and it could be any integer from 0 to N. H_N is, in fact, the same as -------------------------------------------------------------------------- H, the number of heads after the N flips of the coin. So, -------------------------------------------------------------------------- for every integer, j, from 1 to N, we've let -------------------------------------------------------------------------- H_j be the number of heads after the jth flip, and we noted that -------------------------------------------------------------------------- H is H_N. We also define -------------------------------------------------------------------------- T_j be the number of tails after the jth flip, -------------------------------------------------------------------------- so T is T_N. Let's also define -------------------------------------------------------------------------- D_j to be the difference between H_j and T_j. Now, -------------------------------------------------------------------------- X is (H minus T over root N). It's therefore -------------------------------------------------------------------------- (H_N minus T_N over root N), which is -------------------------------------------------------------------------- D_N over root N. Our -------------------------------------------------------------------------- easier problem involves -------------------------------------------------------------------------- X, but we propose to work our way up to X. In the next few slides, we'll first study -------------------------------------------------------------------------- D_1, (pause...............) then -------------------------------------------------------------------------- D_2. Then we'll boldly jump to -------------------------------------------------------------------------- D_N, and, finally -------------------------------------------------------------------------- D_N over root N, which is -------------------------------------------------------------------------- X. Actually, to understand what happens when we divide by a constant like -------------------------------------------------------------------------- root N, we'll throw in -------------------------------------------------------------------------- D_1 divided by a constant, -------------------------------------------------------------------------- say, by 7. Okay. We begin with the first difference, -------------------------------------------------------------------------- D_1, which is -------------------------------------------------------------------------- the difference between H_1 and T_1. After the first flip, we have either -------------------------------------------------------------------------- 1 head and 0 tails, or -------------------------------------------------------------------------- 0 heads and 1 tail. Moreover, there's a -------------------------------------------------------------------------- 50-50 chance of either. Note that we use an -------------------------------------------------------------------------- up-arrow for heads and a -------------------------------------------------------------------------- down-arrow for tails. We plot the possible results on -------------------------------------------------------------------------- a number line, and, half of the time, D_1 ends up -------------------------------------------------------------------------- 1, whereas the other half, it ends up -------------------------------------------------------------------------- minus 1. We record the ending probabilities -------------------------------------------------------------------------- here. In -------------------------------------------------------------------------- this box, we see data that, together, are called -------------------------------------------------------------------------- the distribution of D_1. One sometimes says -------------------------------------------------------------------------- probability distribution, (pause...............) or -------------------------------------------------------------------------- measure, (pause...............) or -------------------------------------------------------------------------- probability measure, but we'll say -------------------------------------------------------------------------- "distribution" in these lectures. -------------------------------------------------------------------------- D_1 is referred to as a -------------------------------------------------------------------------- random variable, because its value varies in a random way. One of our goals, for the next lecture, is to give a firm mathematical definition of a random variable, but, for now, we content ourselves with the following intuitive description: It's -------------------------------------------------------------------------- a variable whose value is determined by random events, like a coin flip. Incidentally, note that -------------------------------------------------------------------------- T_1 minus H_1 has *exactly* the same *distribution* as -------------------------------------------------------------------------- H_1 minus T_1. Both come up -------------------------------------------------------------------------- one half of the time and -------------------------------------------------------------------------- minus one the other half. So -------------------------------------------------------------------------- two different random variables can have the same distribution. Next, we simplify the graphic for H_1 minus T_1 -------------------------------------------------------------------------- like this. From the -------------------------------------------------------------------------- distribution, we can proceed to what's called its -------------------------------------------------------------------------- generating function, or, sometimes, its -------------------------------------------------------------------------- "moment generating function". First, we choose a variable, and, in this lecture, we'll use the variable -------------------------------------------------------------------------- "z". For us, a generating function technically -------------------------------------------------------------------------- *won't* be a function, but, rather, an -------------------------------------------------------------------------- expression of z. However, if we start saying "generating expression" in public, we'll get funny looks, so we stick with the traditional term -------------------------------------------------------------------------- "generating function". To compute it, we place -------------------------------------------------------------------------- z to the 1 next to 1, and -------------------------------------------------------------------------- z to the minus 1 next to minus 1. We record that -------------------------------------------------------------------------- z to the 1 is z. By definition, to get the generating function of -------------------------------------------------------------------------- this distribution, we take -------------------------------------------------------------------------- this times -------------------------------------------------------------------------- this (pause...............) plus -------------------------------------------------------------------------- this times -------------------------------------------------------------------------- this, (pause...............) which gives -------------------------------------------------------------------------- this expression. From the generating function of a distribution, we can proceed to its -------------------------------------------------------------------------- Fourier transform. First, remember that -------------------------------------------------------------------------- i denotes the square root of minus one. We now choose another variable, and, in this lecture, we'll use -------------------------------------------------------------------------- "t". Our Fourier transforms will be expressions of t. Another common variable to use in Fourier transforms is the Greek letter -------------------------------------------------------------------------- "xi", but, in these lectures, we'll stick to -------------------------------------------------------------------------- "t". Remember that, in *this* context, "t" has nothing to do with time. To go from the generating function to the Fourier transform, we -------------------------------------------------------------------------- replace z by e to the minus i t, -------------------------------------------------------------------------- like so. This may all seem a bit desultory, and you may be wondering: Why do all this? Well, first off, we'll see in a few slides that this cookbook definition of Fourier transform is very useful for calculating some difficult probabilities and expected values. Still, cookbook definitions are always a little unsatisfying. Unfortunately, it's beyond the scope of these lectures to go into great detail about motivation, but maybe it'd be good to hint that there's an -------------------------------------------------------------------------- infinite-dimensional version of the Spectral Theorem from Linear Algebra, and all these manipulations are motivated by an infinite-dimensional generalization of simultaneous diagonalization of commuting matrices. In the case of -------------------------------------------------------------------------- generating functions and Fourier transforms, we are, in some sense, simultaneously diagonalizing all the translation operators on the L two space of the real numbers. If you should want to learn more about the ideas that underlie these cookbook definitions, the place to start would be a Functional Analysis course that includes the -------------------------------------------------------------------------- Infinite-dimensional Spectral Theorem. In *this* lecture, there's no need to worry about *any* of that. Suffice it to say that we'll see, in a few slides, that these cookbook definitions are *very* useful. First, though, let's do some algebraic simplification. Remember -------------------------------------------------------------------------- this formula, which is easily verified by power series expansion. Replacing t by minus t, we get -------------------------------------------------------------------------- this equation. Remember that cosine of minus t is -------------------------------------------------------------------------- cosine of t whereas sine of minus t is -------------------------------------------------------------------------- the negative of sine of t. Now take -------------------------------------------------------------------------- point five times the first equation, -------------------------------------------------------------------------- and add it (pause...............) to -------------------------------------------------------------------------- point five times the second. On the -------------------------------------------------------------------------- left hand side, we get -------------------------------------------------------------------------- this, and, on the -------------------------------------------------------------------------- right hand side, -------------------------------------------------------------------------- these cancel, and we get -------------------------------------------------------------------------- cosine t. So the Fourier transform of the -------------------------------------------------------------------------- distribution of D_1 *is* -------------------------------------------------------------------------- cosine t. (pause...............) Another important concept is the -------------------------------------------------------------------------- inverse Fourier transform. For example, if you tell me you're thinking of a distribution whose Fourier transform is -------------------------------------------------------------------------- cosine t, and if I'm smart enough to know that cosine t is equal to -------------------------------------------------------------------------- this, then I can work my way back to -------------------------------------------------------------------------- the generating function, from which I can see that there are probabilities of -------------------------------------------------------------------------- point 5 at 1 and point 5 at minus 1, thereby recovering, from the -------------------------------------------------------------------------- Fourier transform, -------------------------------------------------------------------------- the distribution from which it came. Of course, I can't tell whether it came from -------------------------------------------------------------------------- H_1 minus T_1, or from -------------------------------------------------------------------------- T_1 minus H_1, or, for that matter, from any other random variable with the same -------------------------------------------------------------------------- distribution as D_1. So we can't work back from the -------------------------------------------------------------------------- Fourier transform all the way to the random variable; we can only get back to its distribution. Next, let's analyze what happens to the Fourier transform if we divide a random variable by a constant. For example, -------------------------------------------------------------------------- what about D_1 (over 7)? What happens to the Fourier transform? To answer this, on the next slide, we'll replace -------------------------------------------------------------------------- D_1 by (D_1) (over 7), and -------------------------------------------------------------------------- eliminate this text. D_1 (over 7) has possible values not -------------------------------------------------------------------------- 1 and minus 1, but one seventh and minus one seventh, so we'll divide these two numbers, -------------------------------------------------------------------------- 1 and minus 1, by 7. Therefore we'll also divide -------------------------------------------------------------------------- this 1 and this minus 1 by 7. It'll be convenient, -------------------------------------------------------------------------- in these two equations, to -------------------------------------------------------------------------- replace t by t over 7. Finally, we'll -------------------------------------------------------------------------- eliminate this text, to make room to work out the -------------------------------------------------------------------------- generating function and Fourier transform of (pause...............) -------------------------------------------------------------------------- (D_1) (over 7). To get the generating function, we take -------------------------------------------------------------------------- this times -------------------------------------------------------------------------- this (pause...............) plus -------------------------------------------------------------------------- this times -------------------------------------------------------------------------- this, (pause...............) -------------------------------------------------------------------------- like so. (pause...............) -------------------------------------------------------------------------- Replacing z by e to the minus i t, we get -------------------------------------------------------------------------- the Fourier transform. Averaging -------------------------------------------------------------------------- these two equations, we see that that Fourier transform is -------------------------------------------------------------------------- cosine t over 7. So, to answer the question -------------------------------------------------------------------------- "What about D_1 over 7", we now see that, in terms of the Fourier transform, the answer is -------------------------------------------------------------------------- replace t by t over 7. This, in fact, works quite generally: If you know the Fourier transform of the distribution of a random variable, and you then divide by a constant to get a new random variable, then the new Fourier transform is obtained from the old one by replacing t by t over the constant. (pause...............) We now move on to -------------------------------------------------------------------------- D_2. For D_2, one has -------------------------------------------------------------------------- two coin flips, and each flip has a -------------------------------------------------------------------------- fifty-fifty chance of heads or tails. We put in -------------------------------------------------------------------------- a number line. In the case of -------------------------------------------------------------------------- 2 heads and 0 tails, D_2 is -------------------------------------------------------------------------- 2. In the case of -------------------------------------------------------------------------- 1 head and 1 tail, D_2 is -------------------------------------------------------------------------- 0. In the case of -------------------------------------------------------------------------- 0 heads and 2 tails, D_2 is -------------------------------------------------------------------------- minus 2. The probability of -------------------------------------------------------------------------- heads-heads is (pause...............) -------------------------------------------------------------------------- point 2 5. The probability of -------------------------------------------------------------------------- heads-tails is (pause...............) -------------------------------------------------------------------------- point 2 5 and of -------------------------------------------------------------------------- tails-heads is (pause...............) -------------------------------------------------------------------------- point 2 5, as well. Then the total probability that D_2 is 0 is -------------------------------------------------------------------------- point 5. Finally, the probability of -------------------------------------------------------------------------- tails-tails is (pause...............) -------------------------------------------------------------------------- point 2 5. We therefore represent the distribution of D_2 graphically -------------------------------------------------------------------------- like this. To find the -------------------------------------------------------------------------- generating function, we place -------------------------------------------------------------------------- z to the 2 next to 2, -------------------------------------------------------------------------- z to the 0 next to 0 and -------------------------------------------------------------------------- z to the minus 2 next to minus 2. We record that -------------------------------------------------------------------------- z to the 0 is 1. The generating function is then -------------------------------------------------------------------------- this times this (pause...............) -------------------------------------------------------------------------- plus this times this (pause...............) -------------------------------------------------------------------------- plus this times this. Finally, we get to a key point: *This* generating function is -------------------------------------------------------------------------- a perfect square. In fact, it's the square of (pause...............) -------------------------------------------------------------------------- the generating function of the distribution of D_*1*. (pause...............) To get the -------------------------------------------------------------------------- Fourier transform, we -------------------------------------------------------------------------- replace z by e to the minus i t, and we already know that, when we do that -------------------------------------------------------------------------- here, we get cosine t, so the result for D_*2* will be -------------------------------------------------------------------------- the square of cosine t, or, -------------------------------------------------------------------------- cosine squared t. Okay. We now boldly jump to -------------------------------------------------------------------------- D_N. There is -------------------------------------------------------------------------- no way I can show you its distribution, since that'd involve plotting over 2 and a half million numbers and probabilities on a number line. Looking at the -------------------------------------------------------------------------- generating function, there's -------------------------------------------------------------------------- no way I can show it to you in expanded form, with over 2 and a half million terms. However, because of the magic of generating functions, we *can* see that the answer is -------------------------------------------------------------------------- the Nth power of -------------------------------------------------------------------------- the generating function of the distribution of D_*1*. The -------------------------------------------------------------------------- Fourier transform is, similarly, -------------------------------------------------------------------------- the Nth power of -------------------------------------------------------------------------- the Fourier transform of the distribution of D_*1*. It's -------------------------------------------------------------------------- cosine to the N of t. We're actually interested in the random variable -------------------------------------------------------------------------- X, which, remember, is -------------------------------------------------------------------------- D_N over root N. To find *its* Fourier transform, we -------------------------------------------------------------------------- replace t by t over root N -------------------------------------------------------------------------- here. Let's -------------------------------------------------------------------------- clear some room, and -------------------------------------------------------------------------- make the change. Again, D_N over root N is just -------------------------------------------------------------------------- X. Let's -------------------------------------------------------------------------- clear some room, and -------------------------------------------------------------------------- reformat. Here's the point: Even though it might seem very daunting to understand -------------------------------------------------------------------------- X in terms of its -------------------------------------------------------------------------- *distribution*, the trick of abandoning distributions in favor of -------------------------------------------------------------------------- their Fourier transforms *vastly* simplifies things. This may seem very clever, and you may wonder what mad scientist or mathematician came up with -------------------------------------------------------------------------- generating functions and Fourier transforms. Well, it only *appears* clever. In fact, as we suggested earlier, from the right perspective, mathematicians have simply followed their noses to develop -------------------------------------------------------------------------- Fourier analysis. They first understood that diagonal matrices are more easily manipulated than general matrices, and, applying this thinking to translation operators on L two of R, they were led, ineluctably, to develop Infinite-Dimensional -------------------------------------------------------------------------- Spectral Theory, and, in the process, the basics of -------------------------------------------------------------------------- Fourier analysis. Incidentally, this perspective, while right, (pause...............) is probably ahistorical. In any case, setting aside the question of how clever or historically accurate all of this is, you may also wonder what *value* there is in knowing the -------------------------------------------------------------------------- Fourier transform of the distribution of X. How will *that* help us to solve our -------------------------------------------------------------------------- problem, which was to -------------------------------------------------------------------------- compute the probability that -------------------------------------------------------------------------- X is between minus 1 and 1? To answer this, first note that, -------------------------------------------------------------------------- this Fourier transform is the renormalized Nth power of cosine. Because N is so large, that's very -------------------------------------------------------------------------- close to the limit of the renormalized powers of cosine, and we know how to do that kind of limit. In this case, you may remember that we got -------------------------------------------------------------------------- e to the minus t squared over 2. This is *so* important that, for practice, let's pause to -------------------------------------------------------------------------- verify this for t equals 3. Replacing t by 3, we -------------------------------------------------------------------------- want to show this, and we -------------------------------------------------------------------------- move it to the top to make room. We begin by taking logarithms. On the left, we have -------------------------------------------------------------------------- this. (pause...............) -------------------------------------------------------------------------- Copying it, and, taking its -------------------------------------------------------------------------- logarithm, we can move -------------------------------------------------------------------------- this n down front, -------------------------------------------------------------------------- like so. On the -------------------------------------------------------------------------- right, when we take its -------------------------------------------------------------------------- logarithm, we can use that log and exp are inverses, so we get -------------------------------------------------------------------------- minus 3 squared over 2, or -------------------------------------------------------------------------- minus 9 over 2. It therefore suffices to prove -------------------------------------------------------------------------- this. Next step is to substitute -------------------------------------------------------------------------- x for 1 over root n. Then -------------------------------------------------------------------------- 3 x is 3 over root n. Also, -------------------------------------------------------------------------- x squared is 1 over n and, reciprocating, we get that -------------------------------------------------------------------------- 1 over x squared is n. Substituting, we replace -------------------------------------------------------------------------- n (pause...............) by -------------------------------------------------------------------------- 1 over x squared and -------------------------------------------------------------------------- 3 over root n by -------------------------------------------------------------------------- 3 x, (pause...............) -------------------------------------------------------------------------- like so. As -------------------------------------------------------------------------- n approaches infinity, -------------------------------------------------------------------------- x approaches zero, so we get -------------------------------------------------------------------------- the limit as x approaches zero, and we now -------------------------------------------------------------------------- want to show that *this* limit is -------------------------------------------------------------------------- minus 9 over 2. Algebra changes -------------------------------------------------------------------------- this (pause...............) to -------------------------------------------------------------------------- this, and we leave it to you to see that this limit is a -------------------------------------------------------------------------- zero over zero indeterminate form, which just begs for an application of -------------------------------------------------------------------------- l'Hopital's Rule. We must calculate -------------------------------------------------------------------------- dee dee x of the numerator and -------------------------------------------------------------------------- dee dee x of the denominator. -------------------------------------------------------------------------- The denominator's easy. In the numerator, we have an -------------------------------------------------------------------------- expression inside a -------------------------------------------------------------------------- function, so we take the derivative of the -------------------------------------------------------------------------- function, (pause...............) -------------------------------------------------------------------------- like so, and plug in the -------------------------------------------------------------------------- expression, (pause...............) -------------------------------------------------------------------------- like so. It remains to multiply by the dervative of the -------------------------------------------------------------------------- expression, but this expression is itself an -------------------------------------------------------------------------- expression inside a -------------------------------------------------------------------------- function, so we take the derivative of the -------------------------------------------------------------------------- function, (pause...............) -------------------------------------------------------------------------- like so, and plug in the -------------------------------------------------------------------------- expression, (pause...............) -------------------------------------------------------------------------- like so. Finally, we multiply by the derivative of the -------------------------------------------------------------------------- expression, (pause...............) -------------------------------------------------------------------------- like so. L'Hopital's Rule now tells us to divide the derivative of -------------------------------------------------------------------------- this by the derivative of -------------------------------------------------------------------------- this, (pause...............) -------------------------------------------------------------------------- like so. -------------------------------------------------------------------------- This simplifies. In the numerator, we put -------------------------------------------------------------------------- this, (pause...............) -------------------------------------------------------------------------- like so. In the denominator, we put -------------------------------------------------------------------------- this, (pause...............) -------------------------------------------------------------------------- like so. We again get a -------------------------------------------------------------------------- zero over zero indeterminate form, and, with the tenacity of the truly obsessed, we again apply -------------------------------------------------------------------------- l'Hopital's Rule, which tells us to take the derivative of -------------------------------------------------------------------------- this, and divide it by the derivative of -------------------------------------------------------------------------- this. We leave it to you to work through the computations, and get -------------------------------------------------------------------------- this. In those computations, your allies are the Product and Chain Rules, and powerful allies they are, yes. We finally arrive at a -------------------------------------------------------------------------- limit that is *not* an indeterminate form, so our l'Hopital's Rule days are, thankfully, over. As x tends to zero, -------------------------------------------------------------------------- this tends to 1, as does -------------------------------------------------------------------------- this. Because -------------------------------------------------------------------------- x is approaching zero, -------------------------------------------------------------------------- this term tends to zero. We arrive at -------------------------------------------------------------------------- minus three times three divided by -------------------------------------------------------------------------- two, and that's -------------------------------------------------------------------------- minus nine over two, so we get what we wanted, and -------------------------------------------------------------------------- this is now verified for t equals 3. Its verification for any other value of t is similar. This now leads us to the -------------------------------------------------------------------------- Key idea in *our* presentation of the Central Limit Theorem. The Fourier transform of the distribution of -------------------------------------------------------------------------- X (pause...............) is -------------------------------------------------------------------------- this, so the inverse Fourier transform of -------------------------------------------------------------------------- this is the distribution of -------------------------------------------------------------------------- X. If we could somehow take the inverse Fourier transform of -------------------------------------------------------------------------- e to the minus t squared over 2, we'd get some distribution. If we could then find a random variable with that distribution, then that random variable should be, distributionally, very close to X. Let's write out that thought. -------------------------------------------------------------------------- Let X tilde be some random variable that we someday, somehow manage to find, and suppose its distribution has Fourier transform -------------------------------------------------------------------------- e to the minus t squared over 2. -------------------------------------------------------------------------- Here's the Fourier transform of the distribution of -------------------------------------------------------------------------- X. Then, because X and -------------------------------------------------------------------------- X tilde have distributions with -------------------------------------------------------------------------- close Fourier transforms, if there's any justice in the world, -------------------------------------------------------------------------- X and X tilde should be close in the sense that their distributions are close. In fact, the theory of random variables is developed to the point where this notion of closeness is a completely rigorous concept called "closeness -------------------------------------------------------------------------- in distribution", or, sometimes, -------------------------------------------------------------------------- "weak-*" closeness. In this lecture, we content ourselves with -------------------------------------------------------------------------- intuition, but be aware that there is rigorous mathematics that underlies it. To learn more, you'd want graduate level courses on probability theory and functional analysis. Remember that our immediate goal was -------------------------------------------------------------------------- this and, if we can find X tilde, and if our -------------------------------------------------------------------------- approximations are good, then, after replacing X -------------------------------------------------------------------------- by X tilde, we should get the same answer, up to several decimal places of accuracy. Now, how are we to find this X tilde? Well, in a moment, we'll simply show you a random variable, traditionally called, not X tilde, but rather -------------------------------------------------------------------------- Z whose distribution has -------------------------------------------------------------------------- this as its Fourier transform, so we can just use Z for X tilde. However, that leaves open the question of how to find Z if you're not told up front what it is. In other words, how does the inverse Fourier transform work? This is also too advanced for this lecture, but you should note that Fourier Theory is quite well-developed, and, if you want to see more, you'll want to take a course on Fourier analysis. We want to describe the distribution of this random variable -------------------------------------------------------------------------- Z. First, though, let me mention that, up to now, every random variable we've studied has had only a finite number of possible values. For example, -------------------------------------------------------------------------- D_2 could have exactly three possible values: -------------------------------------------------------------------------- 2, 0 or minus 2, depending on whether we have two heads, a head and a tail, or two tails, in the first two coin flips. Even -------------------------------------------------------------------------- D_N, while much more complicated, has only -------------------------------------------------------------------------- finitely many possible values. The theory of random variables, which we'll describe in the next lecture, is robust enough to account for random variables with infinitely many possible values. As we'll soon see, this -------------------------------------------------------------------------- Z has an infinite range of possible values. In fact, for *each* number -------------------------------------------------------------------------- x on the real number line, we'll put an infinitesimal amount of probability at x. Specifically, we'll put -------------------------------------------------------------------------- e to the minus x squared over 2 dee x at x. Note that -------------------------------------------------------------------------- dee x is an infinitesimal, so the point x is *not* given a positive probability. Moreover, we -------------------------------------------------------------------------- do this for *all* real numbers x. Before we go on, the astute listener might have noticed a small -------------------------------------------------------------------------- mistake. You see, if we take -------------------------------------------------------------------------- the probability at x, and -------------------------------------------------------------------------- add up over -------------------------------------------------------------------------- *all* x, by integrating -------------------------------------------------------------------------- from minus infinity to infinity, you may remember that this integral computes to -------------------------------------------------------------------------- root 2 pi. We want for the sum of the probabilities in a probability distribution to be 1, *not* root 2 pi, so -------------------------------------------------------------------------- we divide by root 2 pi, and the mistake -------------------------------------------------------------------------- goes away. When we get to a more rigorous development of random variables, we'll describe Z more precisely, but, for now, we simply ask you to accept that a random variable with -------------------------------------------------------------------------- this intuitively described distribution exists, and to watch how we do computations with this Z. For example, let's -------------------------------------------------------------------------- compute the probability that Z is equal to 7. To -------------------------------------------------------------------------- solve this, we take the -------------------------------------------------------------------------- probability at x and -------------------------------------------------------------------------- add up for all x between 7 and 7. Of course, this integral -------------------------------------------------------------------------- is zero because the -------------------------------------------------------------------------- upper and lower limits of integration are the same. There's *no* positive probability that Z will be equal to any one *particular* number, like 7. By contrast, let's -------------------------------------------------------------------------- compute the probability that Z is between 2 and 3. As before, we -------------------------------------------------------------------------- take the probability at x, but now we -------------------------------------------------------------------------- add up, for all x between 2 and 3. Remembering that -------------------------------------------------------------------------- Phi of x is an antiderivative, we use the Fundamental Theorem of Calculus, and -------------------------------------------------------------------------- evaluate between 2 and 3, which gives -------------------------------------------------------------------------- Phi of 3 minus Phi of 2. Running to our Phi-enabled calculator, we see that, to four decimals, this is -------------------------------------------------------------------------- point 0 2 1 4. In other words, the probability that Z is between 2 and 3 is -------------------------------------------------------------------------- 2.14 percent, to two decimals. Next, let's find the -------------------------------------------------------------------------- generating function of the distribution of Z. For each x, we put -------------------------------------------------------------------------- z to the x next to x. We then multiply -------------------------------------------------------------------------- this (pause...............) -------------------------------------------------------------------------- by this, (pause...............) and -------------------------------------------------------------------------- add up over -------------------------------------------------------------------------- all values of x from minus infinity to infinity. We leave this integral as an -------------------------------------------------------------------------- exercise, and move on to the -------------------------------------------------------------------------- Fourier transform. We replace -------------------------------------------------------------------------- z (pause...............) by -------------------------------------------------------------------------- e to the minus i t. -------------------------------------------------------------------------- This is an integral of a type that we've worked out in the past. It's equal to -------------------------------------------------------------------------- e to the minus t squared over 2. This is *so* important that, for practice, let's pause to -------------------------------------------------------------------------- verify this for t equals 3. Replacing t by 3, we -------------------------------------------------------------------------- want to show this. We start -------------------------------------------------------------------------- at the right hand side, and copy it -------------------------------------------------------------------------- here. We pull a -------------------------------------------------------------------------- constant out of the integral. Because the -------------------------------------------------------------------------- linear term in the exponent has coefficient minus 3 i, we follow our muse, and -------------------------------------------------------------------------- replace x by x minus 3 i, which yields -------------------------------------------------------------------------- this. Remember that, by Cauchy's Theorem, we do *not* need to subtract minus 3 i from -------------------------------------------------------------------------- these limits of integration. -------------------------------------------------------------------------- This expands (pause...............) -------------------------------------------------------------------------- like this, and to expand -------------------------------------------------------------------------- this, we start with -------------------------------------------------------------------------- the square of x minus 3 i, which is -------------------------------------------------------------------------- x squared *minus* 6 i x *minus* 9. Multiplying that by -------------------------------------------------------------------------- minus 1 over 2, we get -------------------------------------------------------------------------- *minus* x squared over 2 *plus* 3 i x *plus* 9 over 2, so the expansion of -------------------------------------------------------------------------- this exponent is done. By properties of exponentials, we get -------------------------------------------------------------------------- e to the *minus* x squared over 2 times e to the *plus* 3 i x times e to the *plus* 9 over 2, (pause...............) -------------------------------------------------------------------------- and the rest stays the same. By the magic of completing the square, -------------------------------------------------------------------------- these two factors cancel. -------------------------------------------------------------------------- These two constants multiply together to give -------------------------------------------------------------------------- e to the minus 9 over 2, -------------------------------------------------------------------------- and the rest stays the same. -------------------------------------------------------------------------- This is equal to 1, so we get -------------------------------------------------------------------------- e to the minus 9 over 2, which *is* -------------------------------------------------------------------------- e to the minus 3 squared over 2, which is -------------------------------------------------------------------------- the right hand side, and we've proved what we wanted to prove. (pause...............) So -------------------------------------------------------------------------- this is now verified for t equals 3, and a similar argument works for any value of t. Recall again the -------------------------------------------------------------------------- key idea, which involves finding some random variable whose distribution has Fourier transform -------------------------------------------------------------------------- e to the minus t squared over 2. Such a variable will be -------------------------------------------------------------------------- close to X in the sense that their distributions are close. Now we see that the -------------------------------------------------------------------------- Fourier transform of the distribution of -------------------------------------------------------------------------- Z *is* (pause...............) -------------------------------------------------------------------------- e to the minus t squared over 2, so -------------------------------------------------------------------------- Z is close to X in distribution, which we also record -------------------------------------------------------------------------- here. Our -------------------------------------------------------------------------- easier problem is to compute the probability that X is between minus 1 and 1. Since the distributions of Z and X are close, this should be -------------------------------------------------------------------------- about the same as the probability that -------------------------------------------------------------------------- *Z* is between minus 1 and 1. We take -------------------------------------------------------------------------- the probability at (little) x, -------------------------------------------------------------------------- like so, and -------------------------------------------------------------------------- add up, for all x between minus 1 and 1. This gives an -------------------------------------------------------------------------- approximate solution to our -------------------------------------------------------------------------- easier problem. An antiderivative is -------------------------------------------------------------------------- Phi of x, and we -------------------------------------------------------------------------- evaluate between minus 1 and 1. Our handy-dandy Phi-calculating calculator now tells us that, to two decimals, the answer is -------------------------------------------------------------------------- 68.27 percent. This is an approximate solution to the -------------------------------------------------------------------------- easier problem. Remember that N is over two and a half million. *Because* N is so large, we expect -------------------------------------------------------------------------- this answer to be accurate to several decimals. There are, in fact, ways of checking on accuracy, by bounding the error in our approximation. If you're interested in learning about how to do that kind of error-analysis, one uses a result called the -------------------------------------------------------------------------- Berry-Esseen Theorem. As usual, we omit such advanced topics in these lectures, for lack of time. We now work our way up from -------------------------------------------------------------------------- this easier (pause...............) -------------------------------------------------------------------------- probability problem to the original pricing problem, which was an -------------------------------------------------------------------------- expected value problem. Remember that our -------------------------------------------------------------------------- goal was to compute the expected value of -------------------------------------------------------------------------- f of u to the H d to the T, where -------------------------------------------------------------------------- H is the number of heads, after N flips of a fair coin, and -------------------------------------------------------------------------- T is the number of tails. Also, remember that the payoff function -------------------------------------------------------------------------- f(S) is this. This calculation is, perhaps, still too hard for us, so we again simplify, and come up with a -------------------------------------------------------------------------- new easier problem: We'll compute the expected value of a -------------------------------------------------------------------------- less complicated random variable, namely f(D_2). Remember the -------------------------------------------------------------------------- distribution of D_2. There's a -------------------------------------------------------------------------- 25 percent chance that D_2 is -------------------------------------------------------------------------- 2, in which case f(D_2) is -------------------------------------------------------------------------- f(2). There's a -------------------------------------------------------------------------- 50 percent chance that D_2 is -------------------------------------------------------------------------- 0, in which case f(D_2) is -------------------------------------------------------------------------- f(0). Finally, there's a -------------------------------------------------------------------------- 25 percent chance that D_2 is -------------------------------------------------------------------------- minus 2, in which case f(D_2) is -------------------------------------------------------------------------- f of minus 2. To get the -------------------------------------------------------------------------- expected value of f(D_2), we multiply -------------------------------------------------------------------------- this (pause...............) times -------------------------------------------------------------------------- this, (pause...............) then -------------------------------------------------------------------------- this (pause...............) times -------------------------------------------------------------------------- this, (pause...............) then -------------------------------------------------------------------------- this (pause...............) times -------------------------------------------------------------------------- this, (pause...............) and *then* -------------------------------------------------------------------------- add up. We leave it as an exercise for you to plug -------------------------------------------------------------------------- 2, 0 and minus 2 into -------------------------------------------------------------------------- f, and to compute that the expected value -------------------------------------------------------------------------- is 1 thousand 2 hundred 50 (1,250). (pause...............) We -------------------------------------------------------------------------- clear some room, and note that there's nothing special about f. The same computation would work for any function -------------------------------------------------------------------------- g. For example, we can -------------------------------------------------------------------------- define g(S) to be, say, -------------------------------------------------------------------------- 5 (e to the S) plus S squared. We leave it as an -------------------------------------------------------------------------- exercise to compute the -------------------------------------------------------------------------- expected value of *g* of D_2, with -------------------------------------------------------------------------- this definition of g. Next, let's -------------------------------------------------------------------------- clear some room, -------------------------------------------------------------------------- change back to f and move on to another problem, namely, to -------------------------------------------------------------------------- compute the expected value of -------------------------------------------------------------------------- f(*Z*). Remember the -------------------------------------------------------------------------- distribution of Z. There's -------------------------------------------------------------------------- this infinitesimal probability that Z is -------------------------------------------------------------------------- x, in which case f(Z) is -------------------------------------------------------------------------- f(x). To get the -------------------------------------------------------------------------- expected value of f(Z), we multiply -------------------------------------------------------------------------- this (pause...............) times -------------------------------------------------------------------------- this, and (pause...............) *then* -------------------------------------------------------------------------- add up. Here, adding up means integrating over -------------------------------------------------------------------------- all real numbers x from minus infinity to infinity. We can -------------------------------------------------------------------------- pull the constant out of the integral. In -------------------------------------------------------------------------- this equation, we can change -------------------------------------------------------------------------- S (pause...............) to -------------------------------------------------------------------------- x. We can then plug -------------------------------------------------------------------------- f(x) into the integral, and use our newfound integration skills to find the answer, but I'm too lazy myself, so I leave this work as another -------------------------------------------------------------------------- exercise. Now let's move from -------------------------------------------------------------------------- f(Z) (pause...............) to -------------------------------------------------------------------------- f(*X*). Remembering that -------------------------------------------------------------------------- Z is close to X, we see that our -------------------------------------------------------------------------- last solution is a good -------------------------------------------------------------------------- approximate solution to the problem of finding the -------------------------------------------------------------------------- expected value of f(*X*). I continue my lazy ways, and leave that integral as an -------------------------------------------------------------------------- exercise. There's nothing special about the function -------------------------------------------------------------------------- f, so we drop it both -------------------------------------------------------------------------- here (pause...............) and -------------------------------------------------------------------------- here. The same logic works for any *reasonable* function -------------------------------------------------------------------------- g, and, in the last act of this lecture, the meaning of the word "reasonable" will be made more precise. Now, if -------------------------------------------------------------------------- this (pause...............) were -------------------------------------------------------------------------- equal to g of capital X, for some reasonable g, we'd be in a position to achieve our -------------------------------------------------------------------------- goal, to a high degree of accuracy. Unfortunately, -------------------------------------------------------------------------- this expression involves H and T, not X. However, there is a connection between H, T and X, namely -------------------------------------------------------------------------- the definition of X as H minus T over root N. Remember that -------------------------------------------------------------------------- N is the number of coin flips, which is the same as the number of heads plus the number of tails. Then -------------------------------------------------------------------------- H plus T is N. If we multiply -------------------------------------------------------------------------- X by root N, we get -------------------------------------------------------------------------- X root N. If we multiply -------------------------------------------------------------------------- this by root N, we get -------------------------------------------------------------------------- H minus T, so these are -------------------------------------------------------------------------- equal. Adding -------------------------------------------------------------------------- these two equations, -------------------------------------------------------------------------- these cancel, and we get -------------------------------------------------------------------------- this. We copy -------------------------------------------------------------------------- this equation, (pause...............) -------------------------------------------------------------------------- over here. We negate -------------------------------------------------------------------------- this equation, (pause...............) -------------------------------------------------------------------------- over here. Adding (pause...............) -------------------------------------------------------------------------- these two equations, -------------------------------------------------------------------------- these cancel, and we get -------------------------------------------------------------------------- this. We now take -------------------------------------------------------------------------- the equation on the left, and -------------------------------------------------------------------------- divide it by 2. We now take -------------------------------------------------------------------------- the equation on the right, and -------------------------------------------------------------------------- divide *it* by 2. We've now written -------------------------------------------------------------------------- H and T in terms of X. Remember that -------------------------------------------------------------------------- N is a specific number, a bit over 2 and a half million. We can plug these expressions for H and T in -------------------------------------------------------------------------- here, to turn this -------------------------------------------------------------------------- expression of H and T into -------------------------------------------------------------------------- an expression of X, for which we have an -------------------------------------------------------------------------- approximate solution, assuming $g$ isn't unreasonable. Okay -- let's go through this in small steps. First, -------------------------------------------------------------------------- u to the H is -------------------------------------------------------------------------- this, (pause...............) and -------------------------------------------------------------------------- d to the T is -------------------------------------------------------------------------- this. To get -------------------------------------------------------------------------- u to the H d to the T, we multiply -------------------------------------------------------------------------- u to the H by -------------------------------------------------------------------------- d to the T. -------------------------------------------------------------------------- This product is equal to -------------------------------------------------------------------------- u d to the N over 2 -------------------------------------------------------------------------- This product is equal to -------------------------------------------------------------------------- this, where, because of the -------------------------------------------------------------------------- minus sign, we -------------------------------------------------------------------------- divide. Next, we define -------------------------------------------------------------------------- C to *be* u d to the N over 2. Remember that -------------------------------------------------------------------------- u, d and N are all known, and we'll eventually use their values calculate -------------------------------------------------------------------------- C. This definition allows us to replace -------------------------------------------------------------------------- this (pause...............) by -------------------------------------------------------------------------- C. Next, define -------------------------------------------------------------------------- k by this equation. Again, -------------------------------------------------------------------------- u, d and N are all known, and we'll eventually use their values calculate -------------------------------------------------------------------------- k. (pause...............) -------------------------------------------------------------------------- Exponentiating k, and remembering that exp and -------------------------------------------------------------------------- log are inverses, we get -------------------------------------------------------------------------- u over d to the (root N) over 2. If we raise -------------------------------------------------------------------------- this to the -------------------------------------------------------------------------- capital X power, we get -------------------------------------------------------------------------- exactly this. So -------------------------------------------------------------------------- this underlined expression is -------------------------------------------------------------------------- e to the k raised to the -------------------------------------------------------------------------- capital X power. We're interested in -------------------------------------------------------------------------- this expression, which we put -------------------------------------------------------------------------- here. This is f of -------------------------------------------------------------------------- u to the H d to the T, which is f of -------------------------------------------------------------------------- C [e to the k (capital X)], which we put -------------------------------------------------------------------------- here. We now define a specific function g by -------------------------------------------------------------------------- this equation. This g is "reasonable", in a sense that we'll make precise in the last act of this lecture. Anyway, replacing -------------------------------------------------------------------------- little x by capital X, we see that -------------------------------------------------------------------------- f of C [e to the k (capital X)] is equal to -------------------------------------------------------------------------- g of *capital* X. Then -------------------------------------------------------------------------- these two are equal, which we can re-state -------------------------------------------------------------------------- here. At this point, we no longer have a -------------------------------------------------------------------------- new easier problem, but, rather, -------------------------------------------------------------------------- a restatement of our goal. So our goal now becomes to calculate -------------------------------------------------------------------------- this approximate expected value. We move the -------------------------------------------------------------------------- definition of g(x) -------------------------------------------------------------------------- over here. In the payoff function, -------------------------------------------------------------------------- factor out 5,000. Changing -------------------------------------------------------------------------- S to -------------------------------------------------------------------------- C e to the k x, we find that g(x) is given by -------------------------------------------------------------------------- this expression. Again, in the last act of this lecture, we'll address the question of whether this g(x) is -------------------------------------------------------------------------- reasonable. For now, though, lte's focus on the -------------------------------------------------------------------------- approximate solution. We can now -------------------------------------------------------------------------- write out g(x), and -------------------------------------------------------------------------- leave the rest unchanged. We recall -------------------------------------------------------------------------- our definitions of C and k, as well as the values of -------------------------------------------------------------------------- u and d, and of -------------------------------------------------------------------------- N. (pause...............) Plugging -------------------------------------------------------------------------- u, d and N into -------------------------------------------------------------------------- the formula for C, we find that -------------------------------------------------------------------------- C (pause...............) is equal to -------------------------------------------------------------------------- this. (pause...............) Plugging -------------------------------------------------------------------------- u, d and N into -------------------------------------------------------------------------- the formula for k, we find that -------------------------------------------------------------------------- k (pause...............) is equal to -------------------------------------------------------------------------- this. We leave the computation of -------------------------------------------------------------------------- this approximate expected value for the next act, but we've already developed all the necessary integration skills. So, (pause...............) it -------------------------------------------------------------------------- remains to finish (pause...............) the -------------------------------------------------------------------------- pricing problem, which amounts to -------------------------------------------------------------------------- calculating the integral from the last slide. A reasonable complaint is that, while we've now exposed all the ideas in the Central Limit Theorem, we haven't actually stated it. We leave *that* for another -------------------------------------------------------------------------- act in this many-act lecture. Now, though, -------------------------------------------------------------------------- let's take a break. -------------------------------------------------------------------------- Welcome to the Fifth Act of Lecture 4 of Notes on Financial Mathematics, by Scot Adams and Fernando Reitich. -------------------------------------------------------------------------- Here's the approximate expected value problem that we left unsolved at the end of the last act. Remember that -------------------------------------------------------------------------- C and k are known constants. Let's -------------------------------------------------------------------------- move the problem up to the top of the page, -------------------------------------------------------------------------- bring the constant out of the integral, and -------------------------------------------------------------------------- leave the rest alone. -------------------------------------------------------------------------- This expression increases as x increases, and we need to figure where it's zero. We therefore substitute -------------------------------------------------------------------------- $a$ for $x$, and set the resulting expression of $a$ -------------------------------------------------------------------------- to zero. We solve for $a$ by -------------------------------------------------------------------------- adding 1, (pause...............) -------------------------------------------------------------------------- dividing by C, (pause...............) -------------------------------------------------------------------------- taking logarithms, (pause...............) -------------------------------------------------------------------------- simplifying and (pause...............) -------------------------------------------------------------------------- dividing by k. Remember that -------------------------------------------------------------------------- C and k are -------------------------------------------------------------------------- known numbers, and we'll eventually use their values to calculate -------------------------------------------------------------------------- $a$. (pause...............) The -------------------------------------------------------------------------- positive part of a negative number is zero, so the -------------------------------------------------------------------------- integrand here is zero, for all x less than a, so we may as well -------------------------------------------------------------------------- integrate from *$a$* to infinity. In that range, -------------------------------------------------------------------------- this expression is positive, and the -------------------------------------------------------------------------- positive part of a positive number is itself, so -------------------------------------------------------------------------- this "plus" has no effect, and we -------------------------------------------------------------------------- eliminate it. We -------------------------------------------------------------------------- leave everything else alone. Next step is to distribute -------------------------------------------------------------------------- the exponentiated quadratic and -------------------------------------------------------------------------- the integral over -------------------------------------------------------------------------- this subtraction, giving (pause...............) -------------------------------------------------------------------------- this. Note that -------------------------------------------------------------------------- the exponentiated quadratic has been distributed, as has -------------------------------------------------------------------------- the integral. Also, we snuck the -------------------------------------------------------------------------- constant C outside the integral. We move this -------------------------------------------------------------------------- up to the top to make more room. -------------------------------------------------------------------------- This integral is equal to -------------------------------------------------------------------------- this, and note that, because -------------------------------------------------------------------------- $a$ is in the *lower* limit of the integral, it gets -------------------------------------------------------------------------- negated. Also, we -------------------------------------------------------------------------- remembered and didn't forget "root 2 pi". For the -------------------------------------------------------------------------- other integral, we identify the -------------------------------------------------------------------------- linear term in the exponent, and we see that the coefficient is k, so we -------------------------------------------------------------------------- replace x by x plus k, which has -------------------------------------------------------------------------- no effect on dee x, and only a small effect on -------------------------------------------------------------------------- the limits of integration. Specifically, to compensate for adding k to x, we subtract k from -------------------------------------------------------------------------- *a*, (pause...............) giving -------------------------------------------------------------------------- a minus k. Expanding -------------------------------------------------------------------------- this, we get -------------------------------------------------------------------------- e to the k squared, -------------------------------------------------------------------------- e to the minus x squared over 2, and -------------------------------------------------------------------------- e to the minus k squared over 2. If you were watching carefully, you may have noticed that we skipped the exponentiated linear terms -------------------------------------------------------------------------- e to the k x and -------------------------------------------------------------------------- e to the minus 2 k x over 2, or minus k x. Skipping them was okay because -------------------------------------------------------------------------- they cancel, which is really the whole point of adding $k$ to $x$. To compute -------------------------------------------------------------------------- this integral, we can pull -------------------------------------------------------------------------- these constants to the -------------------------------------------------------------------------- outside of the integral, but -------------------------------------------------------------------------- this depends on x, and so -------------------------------------------------------------------------- must remain inside. -------------------------------------------------------------------------- This integral is equal to -------------------------------------------------------------------------- this, and, as usual, because -------------------------------------------------------------------------- $a$ minus $k$ is in the *lower* limit, we -------------------------------------------------------------------------- negate it. Also, we -------------------------------------------------------------------------- remembered and didn't forget "root 2 pi". We -------------------------------------------------------------------------- clear some room, and move the result up, and then -------------------------------------------------------------------------- clear some more room, and move the result up again. -------------------------------------------------------------------------- These cancel and, on the outside of -------------------------------------------------------------------------- the brackets, we have -------------------------------------------------------------------------- 5,000. On the inside, we have -------------------------------------------------------------------------- C times (pause...............) -------------------------------------------------------------------------- this (pause...............) -------------------------------------------------------------------------- minus this. Remember that we earlier computed -------------------------------------------------------------------------- k and C, and we now plug them into the -------------------------------------------------------------------------- expression for $a$, and get -------------------------------------------------------------------------- this. Using these numbers, -------------------------------------------------------------------------- C e to the k squared over 2 is equal to -------------------------------------------------------------------------- this, (pause...............) -------------------------------------------------------------------------- k minus a is equal to -------------------------------------------------------------------------- this (pause...............) and -------------------------------------------------------------------------- minus a is equal to -------------------------------------------------------------------------- this. Finally, a calculator that can compute the -------------------------------------------------------------------------- error function yields our long-sought -------------------------------------------------------------------------- answer. (pause...............) Okay. Now that we have the -------------------------------------------------------------------------- answer, does anyone remember the question? You have to go all the way back to our -------------------------------------------------------------------------- coin-flipping game in Act 4, where our goal was to find -------------------------------------------------------------------------- the expected value E. All this work was directed toward that end, and we now have, -------------------------------------------------------------------------- up to a good approximation, -------------------------------------------------------------------------- E. Again, the approximation is good to several decimals, because of the accuracy of the Central Limit Theorem, when -------------------------------------------------------------------------- N is over 2 and a half million. A bound on the error can be found through Berry-Esseen and tail estimates, but that's outside the scope of these lectures. We really want, not E, but the -------------------------------------------------------------------------- price, P, of Kyle's option, so, now that we have -------------------------------------------------------------------------- E, we need -------------------------------------------------------------------------- rho to the minus N. We look up the value of -------------------------------------------------------------------------- rho, and compute -------------------------------------------------------------------------- rho to the minus N. The price, -------------------------------------------------------------------------- P, of the option is just rho to the minus N times E, which is -------------------------------------------------------------------------- approximately this, and it should be accurate to several decimals. So Gail should charge Kyle -------------------------------------------------------------------------- 384 dollars and 87 cents for this option. (pause...............) Okay. -------------------------------------------------------------------------- Let's go back to our assumptions. No bank'll ever offer a -------------------------------------------------------------------------- per second rate. For this option, the term was 30 days, and one might get, from the bank, a statement of their -------------------------------------------------------------------------- 30-day risk-free factor. Of course, if we know -------------------------------------------------------------------------- this number, then, by taking its -------------------------------------------------------------------------- Nth root, we can work our way back to -------------------------------------------------------------------------- the one-second risk-free factor. In any real-life problem, -------------------------------------------------------------------------- this assumption would probably be stated as -------------------------------------------------------------------------- a 30-day risk-free factor, and we'd have to derive, from that, the -------------------------------------------------------------------------- per second risk-free factor. Similarly, it's unlikely that any market analyst would give Gail a -------------------------------------------------------------------------- per second volatility assumption the way it's shown here, and we want, eventually, to replace this assumption with a more reasonably stated one, but that requires some preliminary work. Remember that -------------------------------------------------------------------------- these numbers were called u and d, the one-second uptick and downtick factors. If -------------------------------------------------------------------------- $s$ is the ABC share price at the start of some second, then, -------------------------------------------------------------------------- at the end of that second, according to our model, it'll go either -------------------------------------------------------------------------- up to s u or -------------------------------------------------------------------------- down to s d, and we put in -------------------------------------------------------------------------- a multiplication sign to stress the fact that -------------------------------------------------------------------------- u and d are multiplicative factors. Financial mathematicians typically study not $s$, but rather -------------------------------------------------------------------------- log s, which'll go either -------------------------------------------------------------------------- up to log s plus log u or -------------------------------------------------------------------------- down to log s plus log d, and we put in an -------------------------------------------------------------------------- addition sign to stress that -------------------------------------------------------------------------- these are (pause...............) -------------------------------------------------------------------------- additive terms, *not* multiplicative factors. The point is that taking logarithms causes -------------------------------------------------------------------------- multiplication to become -------------------------------------------------------------------------- addition, and addition is easier. Let's imagine that our real world is a -------------------------------------------------------------------------- 50-50 world, which is a little suspicious because, for this problem, the risk-neutral world was also 50-50. Still, let's suppose. In that case, Gail's market analyst, who studies the real world, and not some imaginary one, will notice that -------------------------------------------------------------------------- log s will increase, on average, by the average of -------------------------------------------------------------------------- these two numbers, log u and log d, each second. -------------------------------------------------------------------------- Here's the average of log u and log d. To get the average log price change over *30 days*, instead of one second, -------------------------------------------------------------------------- multiply by N, the number of seconds in 30 days. Plugging in -------------------------------------------------------------------------- these two numbers for -------------------------------------------------------------------------- $u$ and $d$, the expected change computes to -------------------------------------------------------------------------- this number. Since it's positive, the -------------------------------------------------------------------------- log price, and therefore the -------------------------------------------------------------------------- price, is trending up. In any one second, the price might go -------------------------------------------------------------------------- down by a factor of $d$, but the overall drift is upward, assuming upticks and downticks happen with -------------------------------------------------------------------------- 50-50 probabilities. We record the 30-day expected log price change -------------------------------------------------------------------------- as an assumption. This kind of assumption is usually called a -------------------------------------------------------------------------- drift assumption, and we might call -------------------------------------------------------------------------- this (pause...............) the -------------------------------------------------------------------------- 50-50 drift equation, stressing the fact that it uses -------------------------------------------------------------------------- 50-50 probabilities. Incidentally, what we've been calling a -------------------------------------------------------------------------- volatility assumption is really a combined volatility *and drift* assumption, so our terminology has been somewhat sloppy. To get at -------------------------------------------------------------------------- volatility *alone*, we *might* take the difference of -------------------------------------------------------------------------- u and d, the way we did back in Lecture 3, but a more sophisticated view is that we should measure how far apart -------------------------------------------------------------------------- log u and log d are, and so, change -------------------------------------------------------------------------- this plus sign (pause...............) -------------------------------------------------------------------------- to a minus sign. Also, for reasons that'll be explained in the next lecture, we'll multiply by some odd-looking factors, namely -------------------------------------------------------------------------- the square root of N and -------------------------------------------------------------------------- the geometric mean of 50% and 50%. Plugging in $u$ and $d$, we get -------------------------------------------------------------------------- this number. -------------------------------------------------------------------------- Let's clear some room. The -------------------------------------------------------------------------- number we just calculated is called -------------------------------------------------------------------------- the 30-day price volatility. Again, it was calculated using -------------------------------------------------------------------------- 50-50 probabilities and so, we might want to keep track of that, and call -------------------------------------------------------------------------- this formula the -------------------------------------------------------------------------- 50-50 volatility equation. Traditional notation is -------------------------------------------------------------------------- sigma for the volatility and -------------------------------------------------------------------------- mu for the drift. It's also traditional to use -------------------------------------------------------------------------- e to the r for the risk-free factor, so that -------------------------------------------------------------------------- $r$ is the nominal interest rate in continuous compounding. In this lecture, we use the word "drift" to mean -------------------------------------------------------------------------- expected log price change. Let's examine the -------------------------------------------------------------------------- 50-50 drift and volatility equations. Again, note the -------------------------------------------------------------------------- odd-looking factors in the volatility equation, which'll be explained in the next lecture, when we talk about standard deviation and independence. We'll call these two equations the -------------------------------------------------------------------------- 50-50 equations. -------------------------------------------------------------------------- This number is N, the number of 1-second subperiods in the 30 day term. To hightlight that, we'll sometimes call these the 50-50 -------------------------------------------------------------------------- *N-subperiod* equations, and we may sometimes call each one individually, an -------------------------------------------------------------------------- N-subperiod equation. Here's the point: Since we somehow knew -------------------------------------------------------------------------- the one-second uptick and downtick factors, we were able to -------------------------------------------------------------------------- plug them in, and compute -------------------------------------------------------------------------- the 30-day drift and volatility, but bear in mind that it would be more typical to reverse this: Gail's market analyst would tell us -------------------------------------------------------------------------- the 30-day drift and volatility, and we'd solve for -------------------------------------------------------------------------- the one-second uptick and downtick factors, -------------------------------------------------------------------------- obtaining these numbers. Okay. Remember that -------------------------------------------------------------------------- N is a large number, the number of seconds in 30 days. The 30-day -------------------------------------------------------------------------- risk-free factor was calculated from the one-second risk-free factor, -------------------------------------------------------------------------- rho, but, in real life, we'd reverse that: Gail's banker would tell us that -------------------------------------------------------------------------- this is the 30-day risk free factor, then we'd take its Nth root, and get -------------------------------------------------------------------------- the one-second risk-free factor. Similarly, we calculated the -------------------------------------------------------------------------- 30-day drift and volatility from -------------------------------------------------------------------------- the one-second uptick and downtick factors, but, in real life, we'd reverse that: Gail's market analyst would tell us that -------------------------------------------------------------------------- these are the 30-day drift and volatility for ABC stock, and, by solving the 50-50 N-subperiod equations, we'd find the -------------------------------------------------------------------------- one-second uptick and downtick factors. By an *amazing* coincidence, -------------------------------------------------------------------------- rho turns out to be *exactly* halfway between -------------------------------------------------------------------------- u and d. Ordinarily, that wouldn't happen, but, in this situation, the *risk-neutral* uptick and downtick probabilities are *exactly* 50-50, same as the real-world uptick and downtick probabilities. Most of this lecture has been devoted to using ideas from the 50-50 Central Limit Theorem to show, via risk-neutral pricing, that the -------------------------------------------------------------------------- option price is approximately -------------------------------------------------------------------------- 384 dollars and 87 cents. We now propose a series of challenging -------------------------------------------------------------------------- exercises. First, we propose that you through the same computations, but with -------------------------------------------------------------------------- N changed to 10, so Gail is planning to adjust her portfolio 10 times in 30 days, not once per second. Thus a subperiod of the 30 day term is now 3 days, not one second, and there are 10 subperiods. We leave it as an exercise for you to take the 10th root of -------------------------------------------------------------------------- this, and get a new value of -------------------------------------------------------------------------- rho. Using -------------------------------------------------------------------------- these numbers, and solving the 50-50 *10*-subperiod equations, you can find the new values of the -------------------------------------------------------------------------- uptick and downtick factors. -------------------------------------------------------------------------- Here are the 50-50 N-subperiod equations, from back when N was -------------------------------------------------------------------------- this number. We find the number N -------------------------------------------------------------------------- here. We also find it -------------------------------------------------------------------------- in the odd-looking factors, -------------------------------------------------------------------------- under the square root. We change N to 10 -------------------------------------------------------------------------- like so, and you should now solve these two equations for -------------------------------------------------------------------------- u and d and then -------------------------------------------------------------------------- plug them in here. So, we're now assuming that, in each subperiod, the real-world probabilities of uptick and downtick are 50-50 and, under that assumption, we choose our -------------------------------------------------------------------------- uptick and downtick factors to conform to the known -------------------------------------------------------------------------- 30-day drift and volatility estimated by Gail's market analyst. Now, drift and volatility are based on empircal data, probably assuming that the future'll be something like the past. By contrast, the assumption of a -------------------------------------------------------------------------- 50-50 chance of uptick-downtick once every three days is our -------------------------------------------------------------------------- *model* for the evolution of the price of the underlying stock, and a model just an assumption. This model is called the -------------------------------------------------------------------------- binomial model, and the -------------------------------------------------------------------------- "bi" in "binomial" signifies that the price of the underlying has two choices for how it changes, in each subperiod, given by multiplying by -------------------------------------------------------------------------- $u$ or $d$. To be more specific, we'll call this model the 50-50 -------------------------------------------------------------------------- 10-subperiod binomial model. One of the major themes of financial mathematics is the analysis of how the derivative price changes as we change our model for the underlying price. Okay. When N was -------------------------------------------------------------------------- this number, it turned out that -------------------------------------------------------------------------- rho was *exactly* halfway between -------------------------------------------------------------------------- $u$ and $d$. In this new 10-subperiod model, the new -------------------------------------------------------------------------- rho you calculate will *not* turn out to be exactly halfway between the new -------------------------------------------------------------------------- $u$ and $d$. So, this time around, the risk-neutral world will *not* be a 50-50 world, even though, according to our model, the -------------------------------------------------------------------------- real world *is*. It'll be difficult for you to proceed to an -------------------------------------------------------------------------- approximate price, using risk-neutral pricing, because the new risk-neutral probabilities aren't 50-50, and we've so far only exposed ideas from the 50-50 Central Limit Theorem. In a future lecture, we'll talk about other Central Limit Theorems, and it'll become feasible to go through all the work in this lecture, and calculate an -------------------------------------------------------------------------- approximate price. If you enjoy a challenge, and maybe already know something of the Central Limit Theorem, perhaps you want to try to do that now. Anyway, to compute the approximate price, we need first to compute -------------------------------------------------------------------------- these three values, rho, $u$ and $d$. These are inputs to our -------------------------------------------------------------------------- model. These inputs are often called model parameters, and the process of going from -------------------------------------------------------------------------- real-world data to the -------------------------------------------------------------------------- values of the parameters is called -------------------------------------------------------------------------- calibration of the model. Models must be calibrated, or they're useless, so calibration is a major topic for financial analysts. Here, calibration is relatively easy, given that Gail's market analyst supplies the -------------------------------------------------------------------------- 30-day drift and volatility, and her banker supplies the -------------------------------------------------------------------------- 30-day risk-free factor. Next, we change the number of subperiods, -------------------------------------------------------------------------- N, to 100, -------------------------------------------------------------------------- changing the model. Here -------------------------------------------------------------------------- N is now 100. You should now -------------------------------------------------------------------------- recalibrate (pause...............) and -------------------------------------------------------------------------- calculate a new approximate price. Next we change -------------------------------------------------------------------------- N to 1,000, -------------------------------------------------------------------------- and go through the same thing again. Next, we change -------------------------------------------------------------------------- N to 10,000, -------------------------------------------------------------------------- and go through the same thing again. A reasonable question to ask is what happens to the price as -------------------------------------------------------------------------- N tends to infinity, that is, as Gail considers adjusting her hedge more and more often. The limiting price is sometimes called the -------------------------------------------------------------------------- continuous-hedging price, or Black-Scholes price. When we let N tend to infinity, we'll see in a later lecture that our -------------------------------------------------------------------------- 50-50 $N$-subperiod binomial model tends toward what's called the -------------------------------------------------------------------------- Black-Scholes model. This is a model that is driven by the Central Limit Theorem, and which therefore predicts -------------------------------------------------------------------------- thin tails in price distributions. Nowadays, it's commonly accepted that thin-tails are actually relatively rare, and, in fact, fat-tailed distributions make the finance world go round. So we want, eventually, to move beyond Black-Scholes, but it pays to understand *well* that which we seek to transcend, so we'll spend a good deal of time on Black-Scholes. The price of the option, in the Black-Scholes model, is exactly the price that's given by -------------------------------------------------------------------------- the Black-Scholes Option Pricing Formula, and we'll show you that formula on the next slide, though we won't derive it until a later lecture. On the next slide, we'll see that that formula -------------------------------------------------------------------------- gives a dollar price of 384 point 866 434. Note how close -------------------------------------------------------------------------- these two numbers are, and that's a reflection of the fact that, with such a large value of -------------------------------------------------------------------------- N, Gail has almost achieved continuous hedging by hedging once per second. Okay. Let's take a look at that Black-Scholes Option Pricing Formula. We first recall -------------------------------------------------------------------------- mu, sigma and e to the r. These are the 30-day drift, volatility and risk-free factor, respectively. Remember that Kyle's option had a strike price of 5,000 dollars, and -------------------------------------------------------------------------- K is a traditional notation for the dollar strike price. We -------------------------------------------------------------------------- divide K by the 30-day risk-free factor to get the present value of the strike price. We denote that present value by -------------------------------------------------------------------------- K', and it computes to -------------------------------------------------------------------------- this number. In Kyle's option, at the spot price of $1 per share, the initial cost of the promised 5,000 shares is -------------------------------------------------------------------------- 5,000 dollars. This initial dollar price of the promised underlying stock is typically denoted by -------------------------------------------------------------------------- S_0. By the way, notice that, in Kyle's option, -------------------------------------------------------------------------- S_0 and K are equal, and to indicate that, one says that Gail is selling -------------------------------------------------------------------------- "at the money". Two important quantities the Black-Scholes formula are -------------------------------------------------------------------------- d plus and d minus, and they're defined here. The -------------------------------------------------------------------------- Black-Scholes price is then -------------------------------------------------------------------------- S_0 Phi of d plus minus K prime Phi of d minus. Plugging in the numbers above, we get -------------------------------------------------------------------------- this, which then gives -------------------------------------------------------------------------- these values for d plus and d minus. We grab our Phi-calculating calculator, and compute -------------------------------------------------------------------------- Phi of d plus and Phi of d minus. Plugging in, we get -------------------------------------------------------------------------- this, which calculates to -------------------------------------------------------------------------- 384 point 866 434. (pause...............) Okay. Let's -------------------------------------------------------------------------- clear the calculations and examine the -------------------------------------------------------------------------- formula more closely. We've made no attempt to derive this -------------------------------------------------------------------------- Black-Scholes Option Pricing Formula. That'll come later. For now, we simply examine it. One interesting thing that we should notice right now about this formula, is that -------------------------------------------------------------------------- the drift mu isn't used, and Gail's market analyst need -------------------------------------------------------------------------- not bother estimating it. Strangely, even though, for each N, we need mu as a parameter in the N-subperiod binomial model, when we let N approach infinity, and we move to continuous hedging, mu becomes irrelevant. It may also seem strange that, in calculating the price of an option, we don't care whether the underlying is drifting up or down, nor do we care how quickly. We only care about the -------------------------------------------------------------------------- volatility, along with the -------------------------------------------------------------------------- risk-free factor, the -------------------------------------------------------------------------- spot price of the promised underlying and the -------------------------------------------------------------------------- strike price. Of these, -------------------------------------------------------------------------- four Black-Scholes parameters, -------------------------------------------------------------------------- volatility is the *only* one whose value is difficult to know. In this lecture, we've pretended that there are clever market analysts who somehow just know how to estimate it. Perhaps there are, but, in fact, what often happens in practice is that -------------------------------------------------------------------------- volatility is unclear, but, people who are looking to buy a certain option see, on their computer screens, that a -------------------------------------------------------------------------- price on that option is offered by a seller, with no explanation of how the price was calculated. Looking at that price, these buyers then wonder what volatility would have produced it -------------------------------------------------------------------------- in the Black-Scholes formula. In the vernacular of the quantitative analyst, -------------------------------------------------------------------------- solving for sigma is called "backing out" the volatility from the price. *That* volatility is called the -------------------------------------------------------------------------- "implied volatility" of the option. We'll see, in a later lecture, that the -------------------------------------------------------------------------- Black-Scholes price increases as the volatility increases, so there's at most one volatility for any given price. Implied volatilities are used by options traders in the way that home-buyers use interest rates. To wit: If you're comparing several possible home purchase agreements, there may be different conditions and terms on each one, making them hard to compare. However the interest rate gives you one basic dimensionless number to which you'll likely pay close attention. In today's home finance market, if you see an interest rate of, say, 25%, you'll likely *not* take that home financing agreement. So it is with options and implied volatility. A buyer comparing an array of options may have difficulty making the comparison, because each contract has its own peculiarities. It helps a great deal to have a single number for each that shows how costly it is. In today's options market, if you see an implied volatility of, say, 45%, you'll likely *not* buy the option. Okay. You might wonder if -------------------------------------------------------------------------- this formula ought not be called the -------------------------------------------------------------------------- *50-50* Black-Scholes Option Pricing formula, since the price it gives is a limit of *50-50* binomial prices. To explain why -------------------------------------------------------------------------- *not*, let's assign a few more -------------------------------------------------------------------------- exercises. You'll now work on a limit of -------------------------------------------------------------------------- 65-35 binomial prices. We'll start at -------------------------------------------------------------------------- 10 subperiods. Taking the 10th root of -------------------------------------------------------------------------- this number (pause...............), -------------------------------------------------------------------------- you get rho. Solving the 65-35 10-subperiod equations with -------------------------------------------------------------------------- these numbers for 30-day drift and volatility, -------------------------------------------------------------------------- you get u and d. -------------------------------------------------------------------------- Here are the *50-50* 10-subperiod equations. We find the probabilities -------------------------------------------------------------------------- here, and also in the -------------------------------------------------------------------------- odd-looking factors -------------------------------------------------------------------------- down here. We change them to 65-35 -------------------------------------------------------------------------- like so, and you should now solve these two equations -------------------------------------------------------------------------- for u and d and then (pause) -------------------------------------------------------------------------- write them in here. This completes calibration. Next, find the risk-neutral probabilities, and do risk-neutral pricing. With the help of the Central Limit Theorem, find an -------------------------------------------------------------------------- approximate price. Then set -------------------------------------------------------------------------- N to 100 and -------------------------------------------------------------------------- go through the process again. Then set -------------------------------------------------------------------------- N to 1,000 and -------------------------------------------------------------------------- go through the process again. Then set -------------------------------------------------------------------------- N to 10,000 and -------------------------------------------------------------------------- go through the process again. Then set -------------------------------------------------------------------------- N to 1 million and -------------------------------------------------------------------------- go through the process again. Letting -------------------------------------------------------------------------- N tend to inifinity, we'll show, in a later lecture, that these 65-35 -------------------------------------------------------------------------- prices converge, as N tends to infinity, to -------------------------------------------------------------------------- the *same* Black-Scholes price that we saw before. So, if you were to do the exercises on this slide correctly, then -------------------------------------------------------------------------- these two numbers would turn out to be very close. In fact, when I ran through the computations, I got -------------------------------------------------------------------------- exactly the same number, to 6 decimals. There's nothing special about -------------------------------------------------------------------------- 65-35 or 50-50. In fact, for *any* choice of real-world uptick-downtick probabilities, the -------------------------------------------------------------------------- the N-subperiod binomial model tends to the -------------------------------------------------------------------------- Black-Scholes model. This centrality and universality of Black-Scholes is part of what makes it such an attractive starting point, before moving on to more sophisticated models. Okay. For all this talk, we still haven't stated the Central Limit Theorem, which is, after all, the title of this lecture. In the -------------------------------------------------------------------------- remainder of this lecture, we'll -------------------------------------------------------------------------- state the 50-50 Central Limit Theorem. Time for the last -------------------------------------------------------------------------- intermission of this lecture. -------------------------------------------------------------------------- Welcome to the Sixth (and final) Act of Lecture 4 of Notes on Financial Mathematics, by Scot Adams and Fernando Reitich. Let's return -------------------------------------------------------------------------- to this slide, which we saw back in Act 4. Some of this -------------------------------------------------------------------------- we won't need. The point of what remains was that, if we have a "reasonable" function -------------------------------------------------------------------------- g, and if we want to compute the -------------------------------------------------------------------------- expected value of g(X), then the answer is closely -------------------------------------------------------------------------- approximated by the -------------------------------------------------------------------------- expression at the bottom of the slide. Remember that -------------------------------------------------------------------------- X was defined as H minus T over root N, where -------------------------------------------------------------------------- H is the number of heads and T is the number of tails, after -------------------------------------------------------------------------- N flips of a (pause...............) -------------------------------------------------------------------------- fair, or 50-50, coin. Eventually, it'll become important to talk about how to adjust this result when the coin is *not* fair, but, in this lecture, we content ourselves with stating the -------------------------------------------------------------------------- 50-50 Central Limit Theorem, which, roughtly speaking, asserts that if -------------------------------------------------------------------------- N is a large number, if -------------------------------------------------------------------------- we flip a 50-50 coin N times, if we -------------------------------------------------------------------------- let H be the number of heads and T the number of tails, and if -------------------------------------------------------------------------- g is any reasonable function -------------------------------------------------------------------------- then the expected value of -------------------------------------------------------------------------- g(X), which we write out as g of H minus T over root N -------------------------------------------------------------------------- is close to -------------------------------------------------------------------------- this value. (pause...............) -------------------------------------------------------------------------- Expected value is usually abbreviated by a -------------------------------------------------------------------------- blackboard bold E and -------------------------------------------------------------------------- "is close to" is usually abbreviated by an -------------------------------------------------------------------------- approximately equal sign. Let's adjust the text in -------------------------------------------------------------------------- this sentence, and fit it on one line, -------------------------------------------------------------------------- like so. As it appears now, while good for conveying the *idea* of the Central Limit Theorem, this is *not* really a mathematical -------------------------------------------------------------------------- theorem because of its imprecision. How large is -------------------------------------------------------------------------- large? What's meant by -------------------------------------------------------------------------- reasonable? How close is -------------------------------------------------------------------------- close? Once we answer those questions, we'll have our -------------------------------------------------------------------------- theorem. Let's tackle the -------------------------------------------------------------------------- second question first. -------------------------------------------------------------------------- Here's the expression g(x) to which we applied the theorem earlier. Remember that -------------------------------------------------------------------------- C and k are known. We leave it as an -------------------------------------------------------------------------- exercise to show, -------------------------------------------------------------------------- for every real number x, that -------------------------------------------------------------------------- this expression of x is -------------------------------------------------------------------------- less than 5,000 C times e to the k absolute value x. Here's a hint about how to do this exercise: Dropping -------------------------------------------------------------------------- this minus 1, -------------------------------------------------------------------------- like so, only increases the LHS. Then we're -------------------------------------------------------------------------- here taking the positive part of a positive number which -------------------------------------------------------------------------- has no effect. Finally, adding in -------------------------------------------------------------------------- these absolute value bars can't decrease the value. So, we can get -------------------------------------------------------------------------- from the LHS to the RHS, by making three changes. -------------------------------------------------------------------------- The first increases the value, and -------------------------------------------------------------------------- the other two don't decrease it. We now make a -------------------------------------------------------------------------- definition. A function g is exponentially bounded, -------------------------------------------------------------------------- abbreviated as "e-x-p dash b-d-d" -------------------------------------------------------------------------- if there are constants A and B such that -------------------------------------------------------------------------- for all real numbers x, -------------------------------------------------------------------------- g(x) is less than A times e to the B absolute value x. Letting A be -------------------------------------------------------------------------- 5000 C and B be -------------------------------------------------------------------------- k, we see then that -------------------------------------------------------------------------- this expression is -------------------------------------------------------------------------- exponentially bounded in x. -------------------------------------------------------------------------- We clear some room, -------------------------------------------------------------------------- rewrite our definition at the top, and -------------------------------------------------------------------------- go back to the 50-50 Central Limit Theorem, *except* that -------------------------------------------------------------------------- "reasonable" has been replaced by "continuous and exponentially bounded", which are precise hypotheses. The meaning of -------------------------------------------------------------------------- large and (pause...............) -------------------------------------------------------------------------- close still need to be clarified. RECERR: I said "verified" not "clarified" Let's replace -------------------------------------------------------------------------- this by -------------------------------------------------------------------------- "for each positive integer N". Then, -------------------------------------------------------------------------- here, we have an expected value that varies as N varies. For example, for -------------------------------------------------------------------------- N equals 1, let's work out a formula for -------------------------------------------------------------------------- this expected value. Note that -------------------------------------------------------------------------- root N is 1, and -------------------------------------------------------------------------- H minus T is either 1 or minus 1, with a -------------------------------------------------------------------------- 50-50 probability of each. Taking -------------------------------------------------------------------------- g of H minus T over root N, we get either -------------------------------------------------------------------------- g(1) or -------------------------------------------------------------------------- g(-1), with probabilities -------------------------------------------------------------------------- point 5 and point 5. To get the expected value, we -------------------------------------------------------------------------- add the two results. Moving on to -------------------------------------------------------------------------- N equals 2, let's work out a new formula. Note that -------------------------------------------------------------------------- root N is now root 2, and -------------------------------------------------------------------------- H minus T is now 2 , 0 or minus 2, with probabilities -------------------------------------------------------------------------- point 25, point 5 and point 25. Taking -------------------------------------------------------------------------- g of H minust T over root N, we get -------------------------------------------------------------------------- g of 2 over root 2, -------------------------------------------------------------------------- g of 0 over root 2, or -------------------------------------------------------------------------- g of minus 2 over root 2, with probabilities -------------------------------------------------------------------------- point 25, point 5 and point 25. To get the expected value, we -------------------------------------------------------------------------- add the three results. As an -------------------------------------------------------------------------- exercise, write out the formula for -------------------------------------------------------------------------- N equals 3. So -------------------------------------------------------------------------- this changes as N changes, giving a *sequence* of expected values, not just one. To state the 50-50 Central Limit Theorem precisely, we -------------------------------------------------------------------------- drop approximations, and say that the -------------------------------------------------------------------------- sequence of expected values -------------------------------------------------------------------------- tends to the expression on the right, -------------------------------------------------------------------------- as N tends to infinity. -------------------------------------------------------------------------- This is a precise mathematical theorem, and, while we won't give a formal mathematical proof of it, we've hinted a great deal at *how* it can be proved, using Fourier transforms and limits of renormalized powers of functions. In the version we're stating here, tail-estimation is another important ingredient. Now, (pause...............) it's not really professional-looking to phrase theorems -------------------------------------------------------------------------- in terms of flipping coins. The meaning is completely precise, and, to make that point, we've even written out exact formulas for the first -------------------------------------------------------------------------- two expected values. Nevertheless, one seeks a more formal mathematical model of coin-flipping. One seeks, it turns out, the concept of a -------------------------------------------------------------------------- sequence of independent random variables, as we'll see in the next lecture. For today, we've gone on long enough. This completes our lecture on the Central Limit Theorem, and we now outline a plan of seven -------------------------------------------------------------------------- future lectures. First, we need to have a better idea of what -------------------------------------------------------------------------- random variables are, in a mathematical context. Here, we'll also define and discuss -------------------------------------------------------------------------- independence and sequences of independent random variables. We'll also talk about -------------------------------------------------------------------------- expectation -------------------------------------------------------------------------- variance and -------------------------------------------------------------------------- standard deviation. Our discussion of standard deviation will explain why the odd-looking factors appeared in the 50-50 volatility equation. Second, we want to -------------------------------------------------------------------------- return to the Central Limit Theorem. We'll state it more formally, and not just in the 50-50 case. Next'll come an important theorem called -------------------------------------------------------------------------- Girsanov's Theorem, part of which asserts that, in -------------------------------------------------------------------------- continuous hedging, the -------------------------------------------------------------------------- volatility in the risk-neutral world is -------------------------------------------------------------------------- the same as in the real world. This is important, because, it solves the problem of computing -------------------------------------------------------------------------- risk-neutral volatility from observable real-world data. In fact, anything that helps us to see into that imaginary risk-neutral world helps us to price. You may wonder if, under continuous hedging, -------------------------------------------------------------------------- risk-neutral drift is equal to -------------------------------------------------------------------------- real-world drift. Well, in fact, it's typically not, but -------------------------------------------------------------------------- risk-neutral drift is nevertheless easy to calculate from real-world data. You see, by *definition* of the risk-neutral world, all portfolios have the same drift, so the drift of any asset is just the drift of the bank. So to find the -------------------------------------------------------------------------- risk-neutral drift, we simply ask our banker about the interest rate. Then -------------------------------------------------------------------------- risk-neutral drift is determined by the bank, and -------------------------------------------------------------------------- risk-neutral volatility is, by Girsanov, real-world volatility. Neither depends on -------------------------------------------------------------------------- real-world drift. That's why the real-world -------------------------------------------------------------------------- drift parameter, mu, is unused in continuous hedging, and it -------------------------------------------------------------------------- doesn't appear in the Black-Scholes formula. With Girsanov's Theorem and the Central Limit Theorem as our allies, we'll be in a position to -------------------------------------------------------------------------- derive the Black-Scholes Option Pricing Formula, a major milestone. We'll then devote a lecture to the formalism of -------------------------------------------------------------------------- stochastic processes, which are simply random variables that evolve over time. Whenever something is evolving, understanding its rate of change is important, and there's a whole -------------------------------------------------------------------------- calculus specifically designed for stochastic processes. Ordinary Differential Equations also have a stochastic analogue, called -------------------------------------------------------------------------- Stochastic Differential Equations, and we'll develop the basics of that subject. We'll separate out, into another lecture, the most important -------------------------------------------------------------------------- Stochastic Calculus result, which is a stochastic version of the -------------------------------------------------------------------------- Chain Rule. Remember that the ordinary Chain Rule is used when we see an expression inside a function, and we want to compute the derivative, or rate of change. There's something similar for stochastic processes: If you plug a stochastic process into a function, and you want to know the rate of change of the resulting stochastic process, there's a generalization of the Chain Rule that comes into play, and it's called -------------------------------------------------------------------------- Ito's lemma, which can -------------------------------------------------------------------------- also be called the Stochastic Chain Rule. With the -------------------------------------------------------------------------- Stochastic Calculus and Stochastic Differential Equations as our allies, we'll be ready to -------------------------------------------------------------------------- rederive the Black-Scholes Option Pricing formula in a simpler way, but using these sophisticated tools. One big advantage of this new derivation is that it points the way for how quantitative analysts can study *new* models that are more complicated than Black-Scholes. However, they must first learn how to handle the -------------------------------------------------------------------------- Stochastic Calculus and Stochastic Differential Equations. So, to do quantitative finance at the highest levels, one has to learn a lot of -------------------------------------------------------------------------- *rules* of advanced mathematics, and the bottom line is that you can do almost anything, if you know -------------------------------------------------------------------------- math rules. See you next time. --------------------------------------------------------------------------