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1、Stochastic Calculus for Finance, Volume I and II Solution of Exercise Problems Yan Zeng August 20, 2007 Contents 1Stochastic Calculus for Finance I: The Binomial Asset Pricing Model3 1.1The Binomial No-Arbitrage Pricing Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 1.2Probability
2、Theory on Coin Toss Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 1.3State Prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10 1.4American Derivative Securities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13
3、 1.5Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .16 1.6Interest-Rate-Dependent Assets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19 2Stochastic Calculus for Finance II: Continuous-Time Models23 2.1General Probability T
4、heory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23 2.2Information and Conditioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 2.3Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31 2.4St
5、ochastic Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 2.5Risk-Neutral Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .47 2.6 Connections with Partial Diff erential Equations . . . . . . . . . . . . . . . . . .
6、 . . . . . . . .56 2.7Exotic Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .64 2.8American Derivative Securities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .66 2.9Change of Num eraire . . . . . . . . . . . . . . . . . . . . . .
7、. . . . . . . . . . . . . . . . . . .71 2.10 Term-Structure Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .76 2.11 Introduction to Jump Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83 1 This is a solution manual for the two-volume
8、 textbook Stochastic calculus for fi nance, by Steven Shreve. If you have any comments or fi nd any typos/errors, please email me at yz44cornell.edu. The current version omits the following problems. Volume I: 1.5, 3.3, 3.4, 5.7; Volume II: 3.9, 7.1, 7.2, 7.57.9, 10.8, 10.9, 10.10. Acknowledgment I
9、thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. I also thank Hideki Murakami for pointing out a typo in Exercise 4.3, Volume II. 2 Chapter 1 Stochastic Calculus for Finance I: The Binomial Asset Pricing Mo
10、del 1.1The Binomial No-Arbitrage Pricing Model 1.1. Proof. If we get the up sate, then X1= X1(H) = 0uS0+ (1 + r)(X0 0S0); if we get the down state, then X1= X1(T) = 0dS0+(1+r)(X00S0). If X1has a positive probability of being strictly positive, then we must either have X1(H) 0 or X1(T) 0. (i) If X1(H
11、) 0, then 0uS0+ (1 + r)(X0 0S0) 0. Plug in X0= 0, we get u0 (1 + r)0. By condition d 1 + r 0. In this case, X1(T) = 0dS0+ (1 + r)(X0 0S0) = 0S0d (1 + r) 0, then we can similarly deduce 0 0 and hence X1(H) X0(1+r) 0. First, this is a generalization of the case X0= 0; second, it is “proper” because it
12、 is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X0as borrowed from money market account. Then at time 1, we have to pay back X0(1 + r) to the money market account. In
13、summary, arbitrage is a trading strategy that beats “safe” investment. Accordingly, we revise the proof of Exercise 1.1. as follows. If X1has a positive probability of being strictly larger than X0(1 + r), the either X1(H) X0(1 + r) or X1(T) X0 (1 + r). The fi rst case yields 0S0(u1r) 0, i.e. 0 0. S
14、o X1(T) = (1+r)X0+0S0(d1r) (1+r)X0. The second case can be similarly analyzed. Hence we cannot have X1strictly greater than X0(1 + r) with positive probability unless X1is strictly smaller than X0(1 + r) with positive probability as well. Finally, we comment that the above formulation of arbitrage i
15、s equivalent to the one in the textbook. For details, see Shreve 7, Exercise 5.7. 1.2. Proof. X1(u) = 08+03 5 4(40+1.200) = 30+1.50, and X1(d) = 02 5 4(40+1.200) = 301.50. That is, X1(u) = X1(d). So if there is a positive probability that X1is positive, then there is a positive probability that X1is
16、 negative. Remark: Note the above relation X1(u) = X1(d) is not a coincidence. In general, let V1denote the payoff of the derivative security at time 1. Suppose X0and 0are chosen in such a way that V1can be replicated: (1 + r)(X0 0S0) + 0S1= V1. Using the notation of the problem, suppose an agent be
17、gins 3 with 0 wealth and at time zero buys 0shares of stock and 0options. He then puts his cash position 0S0 0X0in a money market account. At time one, the value of the agents portfolio of stock, option and money market assets is X1= 0S1+ 0V1 (1 + r)(0S0+ 0X0). Plug in the expression of V1and sort o
18、ut terms, we have X1= S0(0+ 00)(S1 S0 (1 + r). Since d (1+r) u, X1(u) and X1(d) have opposite signs. So if the price of the option at time zero is X0, then there will no arbitrage. 1.3. Proof. V0= 1 1+r h 1+rd ud S1(H) + u1r ud S1(T) i = S0 1+r h 1+rd ud u + u1r ud d i = S0. This is not surprising,
19、since this is exactly the cost of replicating S1. Remark: This illustrates an important point. The “fair price” of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1(H) and S1(T) are given, we could have two current prices, S0 and S0 0. Correspondingly, we can get u,
20、 d and u0, d0. Because they are determined by S0and S00, respectively, its not surprising that risk-neutral pricing formula always holds, in both cases. That is, S0= 1+rd ud S1(H) + u1r ud S1(T) 1 + r , S0 0= 1+rd0 u0d0 S1(H) + u01r u0d0 S1(T) 1 + r . Essentially, this is because risk-neutral pricin
21、g relies on fair price=replication cost. Stock as a replicating component cannot determine its own “fair” price via the risk-neutral pricing formula. 1.4. Proof. Xn+1(T)=ndSn+ (1 + r)(Xn nSn) =nSn(d 1 r) + (1 + r)Vn = Vn+1(H) Vn+1(T) u d (d 1 r) + (1 + r) pVn+1(H) + qVn+1(T) 1 + r = p(Vn+1(T) Vn+1(H
22、) + pVn+1(H) + qVn+1(T) = pVn+1(T) + qVn+1(T) =Vn+1(T). 1.6. Proof. The banks trader should set up a replicating portfolio whose payoff is the opposite of the options payoff . More precisely, we solve the equation (1 + r)(X0 0S0) + 0S1= (S1 K)+. Then X0= 1.20 and 0= 1 2. This means the trader should
23、 sell short 0.5 share of stock, put the income 2 into a money market account, and then transfer 1.20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0.8(1 + r) in money market account will cancel out with the options payoff . Therefore we
24、end up with 1.20(1 + r) in the separate money market account. Remark: This problem illustrates why we are interested in hedging a long position. In case the stock price goes down at time one, the option will expire without any payoff . The initial money 1.20 we paid at 4 time zero will be wasted. By
25、 hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payoff at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott 8, page 11-13. 1.7. Proof. The idea is the same as Problem 1.6. The banks trader only nee
26、ds to set up the reverse of the replicating trading strategy described in Example 1.2.4. More precisely, he should short sell 0.1733 share of stock, invest the income 0.6933 into money market account, and transfer 1.376 into a separate money market account. The portfolio consisting a short position
27、in stock and 0.6933-1.376 in money market account will replicate the opposite of the options payoff . After they cancel out, we end up with 1.376(1 + r)3in the separate money market account. 1.8. (i) Proof. vn(s,y) = 2 5(vn+1(2s,y + 2s) + vn+1( s 2,y + s 2). (ii) Proof. 1.696. (iii) Proof. n(s,y) =
28、vn+1(us,y + us) vn+1(ds,y + ds) (u d)s . 1.9. (i) Proof. Similar to Theorem 1.2.2, but replace r, u and d everywhere with rn, unand dn. More precisely, set e pn= 1+rndn undn and e qn= 1 e pn. Then Vn= e pnVn+1(H) + e qnVn+1(T) 1 + rn . (ii) Proof. n= Vn+1(H)Vn+1(T) Sn+1(H)Sn+1(T) = Vn+1(H)Vn+1(T) (u
29、ndn)Sn . (iii) Proof. un= Sn+1(H) Sn = Sn+10 Sn = 1+ 10 Sn and dn= Sn+1(T) Sn = Sn10 Sn = 1 10 Sn. So the risk-neutral probabilities at time n are pn= 1dn undn = 1 2 and qn= 1 2. Risk-neutral pricing implies the price of this call at time zero is 9.375. 5 1.2Probability Theory on Coin Toss Space 2.1
30、. (i) Proof. P(Ac) + P(A) = P AcP() + P AP() = P P() = 1. (ii) Proof. By induction, it suffi ces to work on the case N = 2. When A1and A2are disjoint, P(A1 A2) = P A1A2P() = P A1P() + P A2P() = P(A1) + P(A2). When A1 and A2are arbitrary, using the result when they are disjoint, we have P(A1 A2) = P(
31、A1 A2) A2) = P(A1 A2) + P(A2) P(A1) + P(A2). 2.2. (i) Proof. e P(S3= 32) = e p3= 1 8, e P(S3= 8) = 3e p2e q = 3 8, e P(S3= 2) = 3e pe q2= 3 8, and e P(S3= 0.5) = e q3= 1 8. (ii) Proof. e ES1 = 8eP(S1= 8) + 2eP(S1= 2) = 8e p + 2e q = 5, e ES2 = 16e p2+ 4 2e pe q + 1 e q2= 6.25, and e ES3 = 32 1 8 + 8
32、 3 8 + 2 3 8 + 0.5 1 8 = 7.8125. So the average rates of growth of the stock price under e P are, respectively: e r0= 5 4 1 = 0.25, e r1= 6.25 5 1 = 0.25 and e r2= 7.8125 6.25 1 = 0.25. (iii) Proof. P(S3= 32) = (2 3) 3 = 8 27, P(S3 = 8) = 3 (2 3) 2 1 3 = 4 9, P(S3 = 2) = 2 1 9 = 2 9, and P(S3 = 0.5)
33、 = 1 27. Accordingly, ES1 = 6, ES2 = 9 and ES3 = 13.5. So the average rates of growth of the stock price under P are, respectively: r0= 6 4 1 = 0.5, r1= 9 6 1 = 0.5, and r2= 13.5 9 1 = 0.5. 2.3. Proof. Apply conditional Jensens inequality. 2.4. (i) Proof. EnMn+1 = Mn+ EnXn+1 = Mn+ EXn+1 = Mn. (ii) P
34、roof. EnSn+1 Sn = EneXn+1 2 e+e = 2 e+eEe Xn+1 = 1. 2.5. (i) Proof. 2In= 2 Pn1 j=0 Mj(Mj+1 Mj) = 2 Pn1 j=0 MjMj+1 Pn1 j=1 M2 j Pn1 j=1 M2 j = 2 Pn1 j=0 MjMj+1+ M2 n Pn1 j=0 M2 j+1 Pn1 j=0 M2 j = M2 n Pn1 j=0(Mj+1 Mj) 2 = M2 n Pn1 j=0 X2 j+1= M 2 n n. (ii) Proof. Enf(In+1) = Enf(In+Mn(Mn+1Mn) = Enf(I
35、n+MnXn+1) = 1 2f(In+Mn)+f(InMn) = g(In), where g(x) = 1 2f(x + 2x + n) + f(x 2x + n), since2I n+ n = |Mn|. 2.6. Proof. EnIn+1 In = Enn(Mn+1 Mn) = nEnMn+1 Mn = 0. 2.7. 6 Proof. We denote by Xnthe result of n-th coin toss, where Head is represented by X = 1 and Tail is represented by X = 1. We also su
36、ppose P(X = 1) = P(X = 1) = 1 2 . Defi ne S1= X1and Sn+1= Sn+bn(X1, ,Xn)Xn+1, where bn() is a bounded function on 1,1n, to be determined later on. Clearly (Sn)n1is an adapted stochastic process, and we can show it is a martingale. Indeed, EnSn+1 Sn = bn(X1, ,Xn)EnXn+1 = 0. For any arbitrary function
37、 f, Enf(Sn+1) = 1 2f(Sn+bn(X1, ,Xn)+f(Snbn(X1, ,Xn). Then intuitively, Enf(Sn+1 cannot be solely dependent upon Snwhen bns are properly chosen. Therefore in general, (Sn)n1cannot be a Markov process. Remark 1. If Xnis regarded as the gain/loss of n-th bet in a gambling game, then Snwould be the weal
38、th at time n. bnis therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2.8. (i) Proof. Note Mn= EnMN and M0 n= EnM 0 N. (ii) Proof. In the proof of Theorem 1.2.2, we proved by induction that Xn= Vnwhere Xn is defi ned by (1.2.14) of Chapter 1. In other words,
39、the sequence (Vn)0nNcan be realized as the value process of a portfolio, which consists of stock and money market accounts. Since ( Xn (1+r)n)0nN is a martingale under e P (Theorem 2.4.5), ( Vn (1+r)n)0nN is a martingale under e P. (iii) Proof. V 0 n (1+r)n = En h VN (1+r)N i , so V 0 0, V 0 1 1+r,
40、, V 0 N1 (1+r)N1, VN (1+r)N is a martingale under e P. (iv) Proof. Combine (ii) and (iii), then use (i). 2.9. (i) Proof. u0= S1(H) S0 = 2, d0= S1(H) S0 = 1 2, u1(H) = S2(HH) S1(H) = 1.5, d1(H) = S2(HT) S1(H) = 1, u1(T) = S2(TH) S1(T) = 4 and d1(T) = S2(TT) S1(T) = 1. So e p0= 1+r0d0 u0d0 = 1 2, e q0
41、 = 1 2, e p1(H) = 1+r1(H)d1(H) u1(H)d1(H) = 1 2, e q1(H) = 1 2, e p1(T) = 1+r1(T)d1(T) u1(T)d1(T) = 1 6, and e q1(T) = 5 6. Therefore e P(HH) = e p0e p1(H) = 1 4, e P(HT) = e p0e q1(H) = 1 4, e P(TH) = e q0e p1(T) = 1 12 and e P(TT) = e q0e q1(T) = 5 12. The proofs of Theorem 2.4.4, Theorem 2.4.5 an
42、d Theorem 2.4.7 still work for the random interest rate model, with proper modifi cations (i.e. e P would be constructed according to conditional probabili- ties e P(n+1= H|1, ,n) := e pnand e P(n+1= T|1, ,n) := e qn.Cf.notes on page 39.).So the time-zero value of an option that pays off V2at time t
43、wo is given by the risk-neutral pricing formula V0= e E h V2 (1+r0)(1+r1) i . (ii) Proof. V2(HH) = 5, V2(HT) = 1, V2(TH) = 1 and V2(TT) = 0. So V1(H) = e p1(H)V2(HH)+e q1(H)V2(HT) 1+r1(H) = 2.4, V1(T) = e p1(T)V2(TH)+e q1(T)V2(TT) 1+r1(T) = 1 9, and V0 = e p0V1(H)+e q0V1(T) 1+r0 1. (iii) Proof. 0= V
44、1(H)V1(T) S1(H)S1(T) = 2.41 9 82 = 0.4 1 54 0.3815. 7 (iv) Proof. 1(H) = V2(HH)V2(HT) S2(HH)S2(HT) = 51 128 = 1. 2.10. (i) Proof. e En Xn+1 (1+r)n+1 = e EnnYn+1Sn (1+r)n+1 + (1+r)(XnnSn) (1+r)n+1 = nSn (1+r)n+1 e EnYn+1 + XnnSn (1+r)n = nSn (1+r)n+1(ue p + de q) + XnnSn (1+r)n = nSn+XnnSn (1+r)n = X
45、n (1+r)n. (ii) Proof. From (2.8.2), we have ? nuSn+ (1 + r)(Xn nSn) = Xn+1(H) ndSn+ (1 + r)(Xn nSn) = Xn+1(T). So n= Xn+1(H)Xn+1(T) uSndSn and Xn= e EnXn+1 1+r . To make the portfolio replicate the payoff at time N, we must have XN= VN. So Xn= e En XN (1+r)Nn = e En VN (1+r)Nn. Since (Xn)0nN is the
46、value process of the unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear equations), the no-arbitrage price of VNat time n is Vn= Xn= e En VN (1+r)Nn. (iii) Proof. e En Sn+1 (1 + r)n+1 = 1 (1 + r)n+1 e En(1 An+1)Yn+1Sn = Sn (1 + r)n+1 e p(1 An
47、+1(H)u + e q(1 An+1(T)d P(), we can construct a portfolio 0m, , 0 N1 whose payoff at time N is (SN K)+; if C() P() 00 k() if C() M, then e Envn+1(Sn+1,Yn+1) = e pvn+1(uSn,Yn+ uSn) + e qvn+1(dSn,Yn+ dSn). So vn(s,y) = e pvn+1(us,y + us) + e qvn+1(ds,y + ds). b) If n = M, then e EMvM+1(SM+1,YM+1) = e
48、pvM+1(uSM,uSM) + e vn+1(dSM,dSM).So vM(s) = e pvM+1(us,us) + e qvM+1(ds,ds). c) If n 0) = 1, we conclude P() = 0 for any A. So P(A) = P AP() = 0. (v) Proof. P(A) = 1 P(Ac) = 0 e P(Ac) = 0 e P(A) = 1. (vi) 10 Proof. Pick 0such that P(0 ) 0, defi ne Z() = ? 0,if 6= 0 1 P(0), if = 0. Then P(Z 0) = 1 an
49、d EZ = 1 P(0) P(0) = 1. Clearly e P( 0) = EZ10 = P 6=0Z()P() = 0. But P( 0) = 1 P(0) 0 if P(0) 1. Hence in the case 0 P(0) 1, P and e P are not equivalent. If P(0) = 1, then EZ = 1 if and only if Z(0) = 1. In this case e P(0) = Z(0)P(0) = 1. And e P and P have to be equivalent. In summary, if we can fi nd 0such that 0 P(0) 1, then Z as constructed above would induce a probability e P that is not equivalent to P. 3.5. (i) Proof. Z(HH) = 9 16, Z(HT) = 9 8, Z(TH) = 3 8 and Z(TT) = 15 4