Stochastic Calculus for Finance II: Continuous-Time Models Solution of Exercise Problems Yan Zeng Version 1.0.8, last revised on 2015-03-13. Abstract This is a solution manual for Shreve [14]. If you find any typos/errors or have any comments, please email me at [email protected]. This version skips Exercise 7.1, 7.2, 7.5–7.9. Contents 1 General Probability Theory 2 2 Information and Conditioning 10 3 Brownian Motion 16 4 Stochastic Calculus 26 5 Risk-Neutral Pricing 44 6 Connections with Partial Differential Equations 54 7 Exotic Options 65 8 American Derivative Securities 67 9 Change of Numéraire 72 10 Term-Structure Models 78 11 Introduction to Jump Processes 94 1 1 General Probability Theory F Comments: Example 1.1.4 of the textbook illustrates the paradigm of extending probability measure from a σ-algebra consisting of finitely many elements to a σ-algebra consisting of infinitely many elements. This procedure is typical of constructing probability measures and its validity is justified by Carathéodary’s Extension Theorem (see, for example, Durrett[4, page 402]). I Exercise 1.1. using the properties of Definition 1.1.2 for a probability measure P, show the following. (i) If A 2 F, B 2 F, and A ⊂ B, then P(A) ≤ P(B). Proof. P(B) = P((B − A) [ A) = P(B − A) + P (A) ≥ P(A). 2 F f g1 F P ⊂ (ii) If A and An n=1 is a sequence of sets in with limn!1 (An) = 0 and A An for every n, then P(A) = 0. (This property was used implicitly in Example 1.1.4 when we argued that the sequence of all heads, and indeed any particular sequence, must have probability zero.) Proof. According to (i), P(A) ≤ P(An), which implies P(A) ≤ limn!1 P(An) = 0. So 0 ≤ P(A) ≤ 0. This means P(A) = 0. I Exercise 1.2. The infinite coin-toss space Ω1 of Example 1.1.4 is uncountably infinite. In other words, we cannot list all its elements in a sequence. To see that this is impossible, suppose there were such a sequential list of all elements of Ω1: (1) (1) (1) (1) (1) ··· ! = !1 !2 !3 !4 ; (2) (2) (2) (2) (2) ··· ! = !1 !2 !3 !4 ; (3) (3) (3) (3) (3) ··· ! = !1 !2 !3 !4 ; . (1) An element that does not appear in this list is the sequence whose first component is H is !1 is T and is (1) (2) (2) T if !1 is H, whose second component is H if !2 is T and is T if !2 is H, whose third component is H (3) (3) if !3 is T and is T if !3 is H, etc. Thus, the list does not include every element of Ω1. Now consider the set of sequences of coin tosses in which the outcome on each even-numbered toss matches the outcome of the toss preceding it, i.e., A = f! = !1!2!3!4!5 ··· ; !1 = !2;!3 = !4; · · · g: (i) Show that A is uncountably infinite. Proof. We define a mapping ϕ from A to Ω1 as follows: ϕ(!1!2 ··· ) = !1!3!5 ··· . Then ϕ is one-to-one and onto. So the cardinality of A is the same as that of Ω1, which means in particular that A is uncountably infinite. (ii) Show that, when 0 < p < 1, we have P(A) = 0. Proof. Let An = f! = !1!2 ··· : !1 = !2; ··· ;!2n−1 = !2ng. Then An # A as n ! 1. So 2 2 n P(A) = lim P(An) = lim [P(!1 = !2) ··· P(!2n−1 = !2n)] = lim [p + (1 − p) ] : n!1 n!1 n!1 2 2 2 2 n Since p + (1 − p) ≤ maxfp; 1 − pg[p + (1 − p)] < 1 for 0 < p < 1, we have limn!1(p + (1 − p) ) = 0. This implies P(A) = 0. 2 I Exercise 1.3. Consider the set function P defined for every subset of [0; 1] by the formula that P(A) = 0 if A is a finite set and P(A) = 1 if A is an infinite set. Show that P satisfies (1.1.3)-(1.1.5), but P does not have the countable additivity property (1.1.2). We see then that the finite additivity property (1.1.5) does not imply the countable additivity property (1.1.2). Proof. Clearly P(;) = 0. For any A and B, if both of them are finite, then A [ B is also finite. So P [ P P [ P [ (A B) = 0 = (A) + (B). If at least one of them is infinite,P then A B is also infinite. So (A B) = 1 P P P [N N P = (A) + (B). Similarly, we can prove ( n=1An) = n=1 (An), even if An’s are not disjoint. P f 1 g [1 To see countable additivity property doesn’t hold for , let An = n P. Then A = n=1An is an infinite P 1 P P 6 1 P set and therefore (A) = . However, (An) = 0 for each n. So (A) = n=1 (An). I Exercise 1.4. (i) Construct a standard normal random variable Z on the probability space (Ω1; F1; P) of Example 1.1.4 1 under the assumption that the probability for head is p = 2 . (Hint: Consider Examples 1.2.5 and 1.2.6) Solution. By Example 1.2.5, we can construct a random variable X on the coin-toss space, which is uniformly R 2 x 1 − ξ distributed on [0; 1]. For the strictly increasing and continuous function N(x) = p e 2 dξ, we let −∞ 2π Z = N −1(X). Then P(Z ≤ a) = P(X ≤ N(a)) = N(a) for any real number a, i.e. Z is a standard normal random variable on the coin-toss space (Ω1; F1; P). f g1 (ii) Define a sequence of random variables Zn n=1 on Ω1 such that lim Zn(!) = Z(!) for every ! 2 Ω1 n!1 and, for each n, Zn depends only on the first n coin tosses. (This gives us a procedure for approximating a standard normal random variable by random variables generated by a finite number of coin tosses, a useful algorithm for Monte Carlo simulation). Solution. Define Xn 1 X = 1f g: n 2i !i=H i=1 −1 Then Xn(!) ! X(!) for every ! 2 Ω1 where X is defined as in Example 1.2.5. So Zn = N (Xn) ! −1 Z = N (X) for every !. Clearly Zn depends only on the first n coin tosses and fZngn≥1 is the desired sequence. I Exercise 1.5. When dealing with double Lebesgue integrals, just as with double Riemann integrals, the order of integration can be reversed. The only assumption required is that the function being integrated be either nonnegative or integrable. Here is an application of this fact. Let X be a nonnegative random variable with cumulative distribution function F (x) = PfX ≤ xg. Show that Z 1 EX = (1 − F (x))dx 0 by showing that Z Z 1 1[0;X(!))(x)dxdP(!) R Ω 0 E 1 − is equal to both X and 0 (1 F (x))dx. Proof. First, by the information given by the problem, we have Z Z 1 Z 1 Z 1[0;X(!))(x)dxdP(!) = 1[0;X(!))(x)dP(!)dx: Ω 0 0 Ω 3 The left side of this equation equals to Z Z Z X(!) dxdP (!) = X(!)dP (!) = P[X]: Ω 0 Ω The right side of the equation equals to Z 1 Z Z 1 Z 1 1fx<X(!)gdP(!)dx = P(x < X)dx = (1 − F (x))dx: 0 Ω 0 0 R E 1 − So [X] = 0 (1 F (x))dx. I Exercise 1.6. Let u be a fixed number in R, and define the convex function '(x) = eux for all x 2 R. 2 1 Let X be a normal random variable with mean µ = EX and standard deviation σ = [E(X − µ) ] 2 , i.e., with density 2 1 − (x−µ) f(x) = p e 2σ2 : σ 2π (i) Verify that uX uµ+ 1 u2σ2 Ee = e 2 : Proof. Z 1 [ ] − 2 uX ux 1 − (x µ) E e = e p e 2σ2 dx Z−∞ σ 2π 1 2 2 1 − (x−µ) −2σ ux = p e 2σ2 dx Z−∞ σ 2π 1 2 2 2 4 2 1 − [x−(µ+σ u)] −(2σ uµ+σ u ) = p e 2σ2 dx −∞ σ 2π Z 1 2 2 − 2 2 uµ+ σ u 1 − [x (µ+σ u)] = e 2 p e 2σ2 dx −∞ σ 2π 2 2 uµ+ σ u = e 2 (ii) Verify that Jensen’s inequality holds (as it must): E'(X) ≥ '(EX): [ ] 2 2 uX uµ+ u σ uµ Proof. E['(X)] = E e = e 2 ≥ e = '(E[X]). I Exercise 1.7. For each positive integer n, define fn to be the normal density with mean zero and variance n, i.e., 1 − x2 fn(x) = p e 2n : 2nπ (i) What is the function f(x) = limn!1 fn(x)? 1 Solution. Since jf (x)j ≤ p , f(x) = lim !1 f (x) = 0. n 2nπ n n R 1 (ii) What is limn!1 −∞ fn(x)dx? 4 R R 2 1 1 1 − x Solution. By the change of variable formula, f (x)dx = p e 2 dx = 1. So we must have −∞ n −∞ 2π Z 1 lim fn(x)dx = 1: n!1 −∞ (iii) Note that Z 1 Z 1 lim fn(x)dx =6 f(x)dx: n!1 −∞ −∞ Explain why this does not violate the Monotone Convergence Theorem, Theorem 1.4.5.