Calculus Stochastic and Financial Applications Final Take Home Exam (Fall 2006) SOLUTIONS
Instructions. You may consult any books or articles that you find useful. If you use a result that is not from our text, attach a copy of the relevant pages from your source. You may use any software, including the internet, Mathematica, Maple, R, S-Plus, MatLab, etc. Attach any Mathematica (or similar) code that you use. You may NOT consult with any other person about these problems. If you have a question, even one that is just about the meaning of a question, please contact me directly rather than consult with a fellow student. I may post "bug reports" or clarifications on our web page, and you should regularly check for these. You should strive to make your answers as clear and complete as possible. Neatness counts -- especially of thought, but even of handwriting. If I can't read it, I can't grade it. Never, ever, write down anything that you know -- or even vaguely suspect -- to be false. If you understand that your argument is incomplete or only heuristic, this may be fine, but it must be properly labeled as incomplete or heuristic. Don't skip steps. If I can't go from line n to line n + 1 in my head, something is missing. If you use Mathematica or a fact from a table, please say so and document it. Otherwise, I stare and stare at line n wondering how you got to n + 1 in your head while I can't. Use anything from anyplace, but do not steal. If you make use of an argument from some source, give credit to the source. If you find the complete (and correct!) solution to a problem in a book or on the internet, just print out the pages and attach them. You will get full credit. Write on only one side of a page. Use decent, homogeneous, high quality paper. No napkins, hoagie sacks, Indian Chief Tablets, etc. Begin each new solution on a new page. Arrange your solutions in the natural numerical order. If you do not do problem K, then include a self-standing page that says "Problem K was not done." Staple your pages neatly with a high-quality stapler with appropriate length and weight to do a clean and secure job. 1
As discussed in class, you MUST use and complete the cover page given at the website. Self-evaluation is hugely valuable. GENERAL ADVICE 1. Avoid the temptation to just write down things that you think are relevant even though these "things" do not add up to an honest solution. Such near-nonsense lists just waste everyone's time. 2. If you can explain clearly something that you tried that did not work, this sometimes is worth a few points. Please do not abuse this offer. With experience, one learns that many sensible ideas do not work. Almost by definition, this is what separates the trivial from the non-trivial. 3. Try to keep in mind that a good problem requires that one "overcome some objection." What distinguishes a problem from an exercise is that in a good problem a routine plan does not work. The whole point is to go past the place where routine ideas take you. Still, don't shy away from the obvious; many of the "problems" here are exercises. 4. If you do something extra that is valid, you can get "bonus" points. These special rewards cannot be determined in advance. They are usual small, but they can be substantial -- and they do add up. 5. The most common source of bonus points is for saying something particularly well. Clear, well-organize, solutions are gems. They deserve to be acknowledged. THE BIG PICTURE Almost certainly these instructions will seem to be overly detailed to you. It is true that they are detailed, but they evolved case by case. Each rule deals with some previous mess or misunderstanding. When you start teaching (and grading) I encourage you to follow this example. There is no dishonor in a creative eccentricity or two. There is a final -- more important -- motivation for this long list of rules and suggestions. Detailed instructions provide honest coaching for excellence. This is the principal benefit. Nevertheless, I hope that no one minds that these rules will also save many hours of everyone's time. Due Date and Place: The exam with its completed self-evaluation cover sheet is due in my office JMHH 447 at 11am on Tuesday December 19.
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Problems for Everyone
Problem 1: A Construction of Brownian Motion on [0, ) Given that a Brownian motion {B(t) : 0 t 1} has been defined on the unit interval, define an new process on all of 0 t < by setting ~ B(t) (1 + t) B t 1+t - tB(1) 1+t .
Confirm that this process is indeed a Brownian motion on [0, ). Although this construction of Brownian motion on [0, ) has fewer seams than the one given in the text, it is probably less obvious to most people. Solution 1: A Construction of Brownian Motion on [0, ) ~ ~ It is immediate that B(t) = 0 and that B(t) is a Gaussian process with ~ continuous paths. We also have E(B(t)) = 0, so we just need to check that ~ ~ E(B(s)B(t)) = s whenever 0 s t. One does this simply by substituting the ~ definition of B, multiplying out terms, and using E(B(s )B(t )) = min(s , t ) for s and t in [0, 1]. Problem 2: Intuition about Volatility from Three Perspectives Consider a European call option with the current stock price equal to the current strike price. These are commonly called at-the-money options, though there are more sophisticated definitions of "at-the-money." (a) Now suppose that volatility is zero. If you make the usual Black-Scholes assumptions, explain how one can guess the value of the option by direct reasoning without using either the Black-Scholes formula or the Black-Scholes PDE. (b) Now take the Black-Scholes PDE and let = 0. Does the value that you obtained in Part (a) solve that simplified PDE and its terminal condition? (c) Finally, take the Black-Scholes formula, and calculate what you get when 0. Be sure to handle any indeterminate expressions honestly. Incidentally, Richard Feynman often said that you understand an equation when "without solving the equation you can still say how the solutions behave." Feynman's criterion is worth keeping in mind anytime one meets an equation -- new or old. Solution 2: Intuition about Volatility from Three Perspectives For Part (a) one could reason as follows. "If is zero, then at time T the stock will be worth St e , but if = r then I would have an arbitrage possibility. Hence, letting be zero, forces = r. Hence, at time T my stock will be priced e r St , and an option owner can buy it for K. Hence at time T , the option owner wins precisely w = e r St - K. Right now, that payout has 3
value e- r w = St - Ke- r . Accordingly, this must be the value of the option." To be sure, this is informal, but it is quite sensible. When = 0 the Black-Scholes PDE becomes ft = -rxfx + rf , and it is trivial to check that f (t, x) = x - Ke- r does solve this equation. For Part (c) one just notes that both D+ and D- are asymptotic to r / and thus go to infinity as 0. This gives (D+ ) 1 and (D- ) 1 so the Black-Scholes formula becomes S - e-r K. This gives a second confirmation of the intuitive reasoning in Part (a). Problem 3: Is it Brownian Motion? (a) Consider the process, Xt = B2t - Bt , 0 t < . Is it a Gaussian process? Can you find the mean and variance? Is it Brownian motion? (b) Let Xt and Yt be independent Brownian motions. Let Zt = (Xt +Yt )/ 2. Is it a Gaussian process? Can you find the mean and variance? Is it Brownian motion? Solution 3: Is it Brownian Motion? (a) Yes, Xt is a Gaussian process. We must check that for each (t1 , t2 , ..., tn ) that (Xt1 , Xt2 , ..., Xtn ) is multivariate Gaussian. That is, we need to show that for any 1 , 2 ,.., n the sum S = 1 Xt1 + 2 Xt2 + + n Xtn is univariate Gaussian. Since S can be written as a linear combination of values of Brownian motion at various times and since Brownian motion is a Gaussian process, we see the S is indeed Gaussian. Next, we trivially have E(Xt ) = 0 and Var(Xt ) = t for all t. Finally, since Cov(Xs , Xt ) equals E((B2s - Bs )(B2t - Bt ) = E(B2s B2t - B2s Bt - Bs B2t + Bs Bt ), we see Cov(Xs , Xt ) = 3 min(s, t) - min(2s, t) - min(s, 2t). For s = 2, t = 3 this works out to 1, but the corresponding covariance for Brownian motion is 2. Hence, {Xt } is not Brownian motion. (b) Yes, it's immediate that Zt is a Gaussian process with mean zero and variance t. Moreover, for s t we have E(Zs Zt ) = (1/2)E(Xs Xt + Xs Yt + Ys Xt + Ys Yt ) = (1/2)(s + 0 + 0 + s) = s. Thus, Cov(Zs , Zt ) = min(s, t), so we see {Zt } is a standard Brownian motion. Problem 4: Integrand and Integral Size Suppose that (, t) is a stochastic integrand for which there is a constant C such that (1) E(2 (, t)) Ctp for all t 0. Show that the stochastic integral
t
Xt =
0
(, s) dBs 4
satisfies the bound E(|Xt |) C p+1
1/2
t(p+1)/2 .
Solution 4: Integrand and Integral Size We combine the Cauchy-Schwarz inequality with the It^ isometry. Specifio cally, we have
2 E(|Xt |) E(Xt )1/2 = t 1/2
E
0
2 (, s) ds
.
Now, change the order of integration and expectation and replace the expectation E(2 (, s) by the bound Csp . Problem 5: A Greek Bound on the Call Price Let f (t, St ) denotes the time t arbitrage-free price of a call option under the usual BlackScholes model, and explain why one has the bound f (t, St ) St t + 2 2 S t , 2r t
where, as usual, we have t = fx (t, St ) and t = fxx t, St ). Solution 5: A Greek Bound on the Call Price This is almost free. The BlackScholes PDE can be written as 1 2 rf (t, St ) = ft (t, St ) + 2 St fxx (t, St ) + rSt fx (t, St ). 2 The key observation is that ft (t, St ) 0, which is financially obvious since, if the only thing that changes is time, then the option must become steadily less valuable. After guessing this relation, one gives a formal derivation by differentiating the Black-Scholes formula an collecting terms. Problem 6: A Text Book Blooper A recently published text asserts that for each fixed > 0 the collection of random variables C = {Xt = exp(Bt - 2 t/2) : 0 t < } is uniformly integrable. Prove that this statement is false. Curiously enough, C is almost a "poster child" for a collection that fails to be uniformly integrable. Dozens of papers offer conditions on a stopping time that suffice to give E[exp(B - /2)] = 1, and all of these would be irrelevant if C were a uniformly integrable collection. 5
Solution 6: A Text Book Blooper Take any > 0. By the law of large numbers for Brownian motion, the exponent Bt - 2 t/2 = t(Bt /t - /2) diverges to minus infinity as t . Hence for Xt = eBt -
t
2
t/2
we see that
lim Xt = 0 with probability one, yet E(Xt ) = 1 for all t.
On the other hand, for any uniformly integrable family {Yt } such that Yt converges to Y in probability (or with probability one) we have E(Yt ) E(Y ). Indeed, this property largely accounts for the usefulness of the notion of uniform integrability. Problem 7: A Hitting Time Identity Let = min{t : Bt = 1}, and show that we have the nice identity E[ exp(- /2) ] = 1 . e
We know the density of , but direct integration would not be the most pleasing way to obtain this expectation. What's wanted here is a simple martingale argument. Incidentally, this formula adds something to our intuition about . We know that E( ) = , but this new formula says that "properly discounted" is "around" 2. Solution 7: A Hitting Time Identity We know that Xt = exp(Bt - t/2) is a martingale, so by Doob's stopping time theorem, so is Xt . This gives us E(Xt ) = 1 for all t 0. Since Xt is nonnegative and bounded by e, the fact that P ( < ) = 1 and the DCT then gives us E(X ) = 1. Since X = exp(1 - /2), taking the expectation gives us our identity. Problem 8: A Martingale and an Integral Show that if g is a continuously differentiable function, then the process
t
Mt = g(t)Bt -
0
g (s)Bs ds
is a martingale. Next, show that for = min{t : Bt = A or Bt = -B} one has
E[ sin( )B ] = E
0
cos(s)Bs ds .
(2)
Solution 8: A Martingale and an Integral 6
For f (t, x) g(t)x we have f C 1,2 (R+ R), so It^'s formula can be applied. o Since fx = g, fxx = 0, and ft = xg we then find
t t
g(t)Bt =
0
g (s)Bs ds +
0
g(s) dBs .
Now, the second integral equals Mt , and, since g is bounded on any interval [0, T ], this integral is a martingale. Finally, to prove the identity (2), we first note by Doob's stopping time theorem that Mt is also a martingale, so we have E(Mt ) = 0. In other words, we have
t
E[B t sin( t)] = E
0
cos(s)Bs ds .
Since |Bt | max(A, B) for t , the DCT permits us to take limits inside both of these expectations. Doing so completes the proof. Problem 9: All that Glitters is not Brownian Motion Suppose that Xt and Yt are two continuous processes such that for each pair of constants and with 2 + 2 = 1 the process Zt = Xt + Yt is a standard Brownian motion. First, show that for each t the random variables Xt and Yt are independent. Second, show by example that the processes {Xt } and {Yt } need not be independent. Big Hint: OK, for part two, I'll show you the example, but you still have to check that it works. Consider independent Brownian motions B1 (t) and B2 (t), and then set Xt = B1 (2t/3) - B2 (t/3) and Yt = B1 (t/3) + B2 (2t/3). Solution 9: All that Glitters is not Brownian Motion By taking = 1 and = 0 we see that Xt is a standard Brownian motion and taking = 0 and = 1 shows that Yt is a standard Brownian motion. The issue is to prove independence. By our hypothesis and the definition of the multivariate Gaussian distribution, we see that for each s and t the pair (Xs , Yt ) is bivariate Gaussian; thus to show independence, we need to show just E(Xs Yt ) = 0. Take = = 1/ 2. We have Var[(Xs + Yt )/ 2] = t from our hypothesis, so we have t= 1 1 Var(Xt ) + Var(Yt ) + Cov(Xt , Yt ). 2 2
Since Var(Xt ) = Var(Yt ) = t this gives Cov(Xt , Yt ) = 0. Since (Xs , Yt ) is bivariate Gaussian this implies that Xt and Yt are independent. 2 2 For the example, it is immediate that E(Xt ) = t, E(Xt ) = t, and one easily checks E(Xt Yt ) = 0. Since (Xt , Yt ) bivariate Gaussian, we see that Zt is Gaussian, mean zero, and Var(Zt ) = t. The definition also shows that Zt has independent increments, so Zt is a Brownian motion. To see that the processes
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{Xt } and {Yt } are dependent, just compute E(Xs Yt ) for some 0 < s < t. Taking s = 1 and t = 2 gives 1/3, so the processes {Xt } and {Yt } are not independent. D. [Clyde Hansen, Jr. (1985)"A Spurious Brownian Motion," Proceedings of the American Mathematical Society, 93 (2), 350.]
Problems with a Challenge, but Still Solidly in Range
Problem 10: The BlackScholes "Lower Case" Transformation Take the BlackScholes formula, replace by , and simplify. I mean really simplify!. What do you get? I promise that you will remember the stunning answer for the rest of your life. It's also financially informative. Show your work! Solution 10: The BlackScholes "Lower Case" Transformation The answer is zero. You can give a proof that parallels our proof that = (D+ ), but, with afterthought, there is a more elegant arrangement. First we note (D+ ) 1 2 2 = exp - (D+ - D- ) (D- ) 2 log(S/K) + r 1 = exp - (D+ + D- )(D+ - D- ) , 2 = . 2 S Ke-r
and, as before, we have the two relations D+ + D- = 2 From these we see 1 (D+ + D- )(D+ - D- ) = log(S/K) + r = log 2 (D+ ) Ke-r = . (D- ) S , and D+ - D- =
and hence one finds the lovely -- and conceptually informative -- identity (3)
When the identity (3) is rewritten without fractions, it tells us that changing to lower case in the Black-Scholes formula gives us zero. Moreover, the formula (3) offers what is perhaps the best way to make intuitive sense of the ubiquitous factors D- and D+ . Among other things it suggests that "moneyness" can be viewed as a kind of likelihood ratio. Problem 11: Texas Chain Rule Massacre Suppose that f (t, x) solves the Black-Scholes PDE with the usual terminal condition f (T, x) = (x-K)+ . Consider the new variable y and the new function g(t, y) that are defined by the relations: y = er(T -t) x and g(t, y) = er(T -t) f (t, x). 8
Show that g satisfies the more pleasant terminal value PDE g 1 2 2 2 g + y =0 t 2 y 2 and g(T, y) = (y - K)+ . (4)
This certainly a nice simplification of the Black-Scholes equation, and it suggests that the new variables are "better" variables. With time on our hands, we could see what other simplifications might be achieved with using y and g in place of x and f . Hint: This is only a chain rule exercise, but it is easily messed up. If you make a mistake in your calculations and still arrive at the target formula, you may have engaged in a willful fiction -- which is not polite. To succeed here is easy, if you start out right. The right way to start is to first write the "old stuff" as a function of the "new stuff." You can then do straightforward chain rule calculations of the derivatives associated with the "old stuff," plug these into the "old equation," and get the "new equation." Less systematic approaches are often fought both with error and irrelevant algebra. Solution 11: Texas Chain Rule Massacre We have x = e-r(T -t) y and f (t, x) = e-r(T -t) g(t, y). We can't forget that y/x = er(T -t) and y/t = -e-r(T -t) xr = -ry. Now we just work out the various derivatives f using the chain rule: ft (t, x) = re-r(T -t) g(t, y) + e-r(T -t) gt (t, y) - e-r(T -t) (-ry)gy (t, y) fx (t, x) = e-r(T -t) gy (t, y)(y/x) = gy (t, y) fxx (t, x) = gyy (t, y)(y/x) = er(T -t) gyy (t, y) Substitution into the BlackScholes PDE ft = -(1/2)x2 2 fxx - rxfx + rf now gives re-r(T -t) g(t, y) + e-r(T -t) gt (t, y) - e-r(T -t) (-ry)gy (t, y) = - (1/2)(e-r(T -t) y)2 2 er(T -t) gyy (t, y) - re-r(T -t) ygy (t, y) + re-r(T -t) g(t, y). Direct cancelations now give use the simple PDE (4). Incidentally, y and g are sometimes called the "forward" stock price and option price, although this terminology can collide with other uses of the word "forward." One mathematical motivation for using a change of variables of the form y = A(t)x and g(t, y) = A(t)f (t, x) is that the relation fx (t, x) = gy (t, y) is baked into the cake. Also, with A(t) = er(T -t) we have A(T ) = 1 so the boundary condition is unchanged. Problem 12: One of Two Lines 9
Consider the standard Brownian motion Zt = (Xt , Yt ) in R2 , and let denote the first time that either Yt = aXt + b or Yt = aXt - b. That is, we let denote the first time that Zt hits the boundary of the strip S defined by the parallel lines y = ax + b and y = ax - b. Show that one has E( ) = b2 /(1 + a2 ), then pose -- and prove -- a three dimensional generalization. Solution: One of Two Lines Solution This is almost a freebie. The key observation is that the process Bt = (Yt - aXt )/ 1 + a2 is a standard Brownian motion, and is equal to the first time that Bt hits A = b/ 1 + a2 or hits -B = -b/ 1 + a2 . We've known forever that E( ) = AB and here AB = b2 /(1 + a2 ). This method easily generalizes to give the expected time for a standard Brownian motion (Xt , Yt , Zt ) in R3 to hit the boundary of the slab S determined by z = ax + by + c and z = ax + by - d where c > 0 and d > 0. This time we have the standard 1-dimensional Brownian motion Bt = (Zt - aXt - bYt )/ 1 + a2 + b2 so upon setting A = c/ 1 + a2 + b2 and B = d/ 1 + a2 + b2 we find E( ) = AB = cd/(1 + a2 + b2 ).
Problem 13: BM without Assuming Independence Let {Xt } denote a mean zero Gaussian process and let (s, t) = Xt - Xs denote the increment in the process from time s to time t. Assume that for each choice of four times 0 t1 t2 t3 t4 the variance of the sum S = (t1 , t2 ) + (t2 , t3 ) + (t3 , t4 ) is equal to t4 - t1 . Show that {Xt } is Brownian motion. The point here is that our assumption on S is enough to show that {Xt } has independent increments. Solution 13: BM without Assuming Independence First observe that by taking t1 = s and t = t2 = t3 = t4 we see that E(2 (s, t) = t - s for all s t. Next, since we have ((t1 , t2 ) + (t2 , t3 ))2 = 2 (t1 , t2 ), we can expand and take expectations to get t2 - t1 + 2E((t1 , t2 )(t2 , t3 )) + t3 - t2 = t3 - t1 . Cancelation shows E((t1 , t2 )(t2 , t3 )) = 0, so we see that adjacent increments of {Xt } are independent. Now we use this fact together with our full assumption 10
about S. Specifically, squaring S gives S 2 = 2 (t1 , t2 ) + 2 (t2 , t3 ) + 2 (t3 , t4 ) + 2(t1 , t2 )(t2 , t3 ) + 2(t1 , t2 )(t3 , t4 ) + 2(t2 , t3 )(t3 , t4 ). Hence, when we take expectations, we get t4 - t1 = t4 - t1 + 2E((t1 , t2 )(t3 , t4 )), and cancelation shows (t1 , t2 ) and (t3 , t4 ) are uncorrelated. Since these are two arbitrary time-disjoint increments, we see that the Gaussian process {Xt } has independent increments. Source: Rnyi, A. (1967). Remarks on the Poisson Process. Studia Sciene tiarum Mathematicarum Hungarica, 2, 119123. Problem 14: A Stopping Time Inequality Prove that for any stopping time and any t 0 one has the inequality 1 E exp( Bt ) 2 {E(exp( /2))}
1/2
.
(5)
Hint: You'll want to get one of your favorite martingales into the game, and, when a square root is in sight, it never hurts to consider Cauchy-Schwarz. Solution 14: A Stopping Time Inequality Since exp(Bt - t/2) is a martingale, Cauchy-Schwarz and Doob's stopping time theorem give us 1 E exp( Bt ) 2 1 1 1 = E exp( Bt - t) exp( t) 2 4 4 = 1 E exp(Bt - t) 2 1 E(exp( t) 2
1/2 1/2
1 E(exp( t) 2
1/2
1/2
1 E(exp( ) 2
.
Remark: This shows that Kazamaki's condition is weaker than Novikov's condition. I learned this from "On Criteria for the Uniform Integrability of Brownian Stochastic Exponentials" by A.S. Cherny and A.N. Shiryaev (unpublished manuscript), but the result probably goes back to Kazamaki. Problem 15: BM and a Stochastic Integral Consider the process
t
Xt =
0
sign(Bs ) dBs 11
where sign(x) = 1 for x 0 and sign(x) = -1 for x < 0. Use Levy's theorem to check that Xt is again a Brownian motion. Next, confirm that the two processes Xt and Bt are uncorrelated. Finally show that 2 E(Xt Bt ) = 25/2 t3/2 /3 , (6) and explain why this implies that the processes Xt and Bt are not independent -- despite being uncorrelated and Gaussian processes. Solution 15: BM and a Stochastic Integral Since the integrand sign(Bs ) is bounded, the stochastic integral that defines Xt is a martingale, and its quadratic variation is given by
t
X
t
=
0
sign2 (Bs ) ds =
0
t
1 ds = t.
Hence, Xt is a continuous martingale with quadratic variation t, so by Lvy's e characterization we see that Xt is a Brownian motion. Next, since both Xt and Bt have mean zero and both are in H2 , their covariance is given by the polarized It^ Isometry o
t t t
E(Bt Xt ) = E
0
dBs
0
sign(Bs ) dBs
=E
0
1 sign(Bs ) ds .
By symmetry, E(sign(Bs )) = 0 for each s = 0, so, when we take the expectation 2 inside the integral, we get E(Bt Xt ) = 0. Finally, to compute E(Xt Bt ) we still 2 want to use the It^ Isometry, so we replace Bt by its It^ integral representation. o o We then compute with the polarized It...