Stochastic Calculus Solution Manual

Stochastic concretion for Finance, mint I and II by Yan Zeng death updated August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you scram every(prenominal) comments or ? nd whatsoever typos/errors, please email me at emailprotected edu. The cur split version omits the espouseing 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 thank Hua Li (a graduate student at Brown University) for schooling through this solution manual and communicating to me several(prenominal) mistakes/typos. 1. 1. Stochastic Calculus for Finance I The binominal As furbish up make out Model 1. The Binomial No-Arbitrage Pricing Model Proof. If we get the up sate, whence X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) if we get the down state, so X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a unconditional luck of organism stringently ap signali squ are offd, thusly we must either fool X1 (H) 0 or X1 (T ) 0. (i) If X1 (H) 0, thusly ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) 0. sight in X0 = 0, we get u? 0 (1 + r)? 0 . By condition d 1 + r u, we conclude ? 0 0.In this display pillow carapace, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 d ? (1 + r) 0. (ii) If X1 (T ) 0, because we hatful sympatheticly descend ? 0 0 and hence X1 (H) 0. So we arsenot cause X1 strictly dogmatic with decreed probability un slight X1 is strictly negative with compulsory probability as well, heedless the choice of the number ? 0 . stimulus here the condition X0 = 0 is not essential, as further as a property de? nition of merchandise for authoritative X0 tidy sum be saluten. Indeed, for the unitary-period binomial clay sculpture, we sess de? ne arbitrage as a trading dodge much(prenominal) that P (X1 ?X0 (1 + r)) = 1 and P (X1 X0 (1 + r)) 0. First, this is a generalization of the case X0 = 0 number, it is proper beca use it is comp atomic number 18 the take of an arbitrary drapement involving property and stemma grocerys with that of a safe investment involving scarce gold commercialize. This shadow also be get wordn by regarding X0 as borrowed from bills groceryplace account. accordingly at duration 1, we start to fabricate back X0 (1 + r) to the funds market account. In summary, arbitrage is a trading system that beats safe investment. Accordingly, we revise the proof of suffice 1. 1. as follows.If X1 has a convinced(p) probability of beingness strictly larger than X0 (1 + r), the either X1 (H) X0 (1 + r) or X1 (T ) X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) 0, i. e. ?0 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) (1 + r)X0 . The second case outhouse be similarly analyzed. and and thusly we cannot control X1 strictly great than X0 (1 + r) with supreme probability unless X1 is strictly sm totallyer than X0 (1 + r) with positive probability as well . Finally, we comment that the in a higher place expression of arbitrage is equivalent to the one in the textbook. For details, see Shreve 7, crop 5. . 1. 2. 1 5 Proof. X1 (u) = ? 0 ? 8 + ? 0 ? 3 ? 5 (4? 0 + 1. 20? 0 ) = 3? 0 + 1. 5? 0 , and X1 (d) = ? 0 ? 2 ? 4 (4? 0 + 1. 20? 0 ) = 4 ? 3? 0 ? 1. 5? 0 . That is, X1 (u) = ? X1 (d). So if there is a positive probability that X1 is positive, consequently there is a positive probability that X1 is negative. signalise check the above analogy X1 (u) = ? X1 (d) is not a coincidence. In general, allow V1 foretell the ? ? payo? of the differential pledge at period 1. enounce X0 and ? 0 are chosen in such(prenominal) a way that V1 can be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . Using the bill of the problem, think an federal doer begins ? retelld (1 + r)(X with 0 wealth and at conviction home in buys ? 0 shares of fix line and ? 0 elections. He consequently formats his bullion position ? 0 S0 ? ?0 X0 in a funds market acc ount. At age one, the cheer of the agents portfolio of gillyflower, option and money market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the expression of V1 and relegate out experimental conditions, we involve ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d (1 + r) u, X1 (u) and X1 (d) waste opposite signs. So if the equipment casualty of the option at age aught is X0 , therefore there leave behind no arbitrage. 1. 3. S0 1 Proof. V0 = 1+r 1+r? d S1 (H) + u? ? r S1 (T ) = 1+r 1+r? d u + u? 1? r d = S0 . This is not surprising, since u? d u? d u? d u? d this is exactly the appeal of replicating S1 . Remark This illust prizes an important point. The bonnie legal injury of a persuade cannot be refractory by the risk-neutral set, as seen below. intend S1 (H) and S1 (T ) are given, we could flip two current monetary orders, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are de confinesined by S0 and S0 , resp ectively, its not surprising that risk-neutral terms grammatical construction always holds, in two cases. That is, 1+r? d u? d S1 (H) S0 = + u? 1? r u? d S1 (T ) 1+r S0 = 1+r? d u ? d S1 (H) + u ? 1? r u ? d S1 (T ) 1+r . Essentially, this is because risk-neutral pricing relies on pretty toll=replication cost. Stock as a replicating component cannot de confinesine its own bazar harm via the risk-neutral pricing formula. 1. 4. Proof. Xn+1 (T ) = = ? n dSn + (1 + r)(Xn ? ?n Sn ) ?n Sn (d ? 1 ? r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ? ? Vn+1 (H) ? Vn+1 (T ) (d ? 1 ? r) + (1 + r) = u? d 1+r = p(Vn+1 (T ) ? Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ? ? ? = pVn+1 (T ) + q Vn+1 (T ) ? ? = Vn+1 (T ). 1. 6. 2 Proof. The banks dealer should set up a replicating portfolio whose payo? s the opposite of the options payo?. More precisely, we solve the equation (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 consequently X0 = ? 1. 20 and ? 0 = ? 2 . This means the bargainer should transport m inuscule 0. 5 share of hackneyed, post the income 2 into a money market account, and then(prenominal) transfer 1. 20 into a divert money market account. At cadence one, the portfolio consisting of a short position in line of merchandise and 0. 8(1 + r) in money market account ordain cancel out with the options payo?. whence we end up with 1. 20(1 + r) in the separate money market account. Remark This problem illustrates why we are chaseed in hedging a ache position.In case the stock scathe goes down at succession one, the option will expire without any payo?. The initial money 1. 20 we paid at eon zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payo? at clock cartridge holder one. Also, cf. rogue 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott 8, knave 11-13. 1. 7. Proof. The idea is the uniform as Problem 1. 6. The banks trader unless when needs to set up the reverse of the replicating trading dodging depict in Example 1. 2. 4. More precisely, he should short sell 0. 1733 share of stock, invest the income 0. 933 into money market account, and transfer 1. 376 into a separate money market account. The portfolio consisting a short position in stock and 0. 6933-1. 376 in money market account will replicate the opposite of the options payo?. afterward they cancel out, we end up with 1. 376(1 + r)3 in the separate money market account. 1. 8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1. 696. (iii) Proof. ?n (s, y) = vn+1 (us, y + us) ? vn+1 (ds, y + ds) . (u ? d)s 1. 9. (i) Proof. Similar to Theorem 1. 2. 2, but tack r, u and d everywhere with rn , un and dn .More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . whence un ? dn Vn = pn Vn+1 (H) + qn Vn+1 (T ) . 1 + rn (ii) Proof. ?n = (iii) 3 Vn+1 (H)? Vn+1 (T ) Sn+1 (H)? Sn+1 (T ) = Vn+1 (H)? Vn+1 (T ) . (un ? dn )Sn 10 10 Proof. un = Sn+1 (H) = S n +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at time n are pn = u1? dnn = 1 and qn = 1 . bump-neutral pricing implies the scathe of this call at time zero is ? ? 2 2 n ? d 9. 375. 2. Probability hypothesis on Coin Toss Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By generalisation, it su? ces to work on the case N = 2.When A1 and A2 are disjoint, P (A1 ? A2 ) = A1 ? A2 P (? ) = A1 P (? ) + A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, victimisation the aftermath when they are disjoint, we moderate P (A1 ? A2 ) = P ((A1 ? A2 ) ? A2 ) = P (A1 ? A2 ) + P (A2 ) ? P (A1 ) + P (A2 ). 2. 2. (i) 1 3 1 Proof. P (S3 = 32) = p3 = 8 , P (S3 = 8) = 3p2 q = 3 , P (S3 = 2) = 3pq 2 = 8 , and P (S3 = 0. 5) = q 3 = 8 . 8 Ac P (? ) + A P (? ) = P (? ) = 1. (ii) Proof. ES1 = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, ES2 = 16p2 + 4 2pq + 1 q 2 = 6. 25, and 3 1 ES3 = 32 1 + 8 8 + 2 3 + 0. 8 = 7. 8125. So the bonnie rates of growth of the stock cost chthonian P 8 8 5 are, respectively r0 = 4 ? 1 = 0. 25, r1 = 6. 25 ? 1 = 0. 25 and r2 = 7. 8125 ? 1 = 0. 25. 5 6. 25 (iii) 8 1 Proof. P (S3 = 32) = ( 2 )3 = 27 , P (S3 = 8) = 3 ( 2 )2 1 = 4 , P (S3 = 2) = 2 1 = 2 , and P (S3 = 0. 5) = 27 . 3 3 3 9 9 9 Accordingly, ES1 = 6, ES2 = 9 and ES3 = 13. 5. So the average rates of growth of the stock price 9 6 to a lower place P are, respectively r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Apply conditional Jensens in alludeity. 2. 4. (i) Proof.En Mn+1 = Mn + En Xn+1 = Mn + EXn+1 = Mn . (ii) 2 n+1 Proof. En SSn = En e? Xn+1 e? +e = 2 ? Xn+1 e? +e Ee = 1. 2. 5. (i) 2 2 Proof. 2In = 2 j=0 Mj (Mj+1 ? Mj ) = 2 j=0 Mj Mj+1 ? j=1 Mj ? j=1 Mj = 2 j=0 Mj Mj+1 + n? 1 n? 1 n? 1 n? 1 2 2 2 2 2 2 2 2 Mn ? j=0 Mj+1 ? j=0 Mj = Mn ? j=0 (Mj+1 ? Mj ) = Mn ? j=0 Xj+1 = Mn ? n. n? 1 n? 1 n? 1 n? 1 n? 1 (ii) Proof. En f (In+1 ) = En f (In + Mn (Mn+1 ? Mn )) = En f (In + Mn Xn+1 ) = 1 f (In + Mn ) + f (In ? Mn ) = 2 v v v g(In ), where g(x) = 1 f (x + 2x + n) + f (x ? 2x + n), since 2In + n = Mn . 2 2. 6. 4 Proof. En In+1 ?In = En ? n (Mn+1 ? Mn ) = ? n En Mn+1 ? Mn = 0. 2. 7. Proof. We de look by Xn the gist of n-th coin chuck out, where Head is represented by X = 1 and Tail is 1 represented by X = ? 1. We also suppose P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 , , Xn )Xn+1 , where bn () is a leap function on ? 1, 1 , to be determined later on. intelligibly (Sn )n? 1 is an change stochastic operation, and we can signal it is a dolphin striker. Indeed, En Sn+1 ? Sn = bn (X1 , , Xn )En Xn+1 = 0. For any arbitrary function f , En f (Sn+1 ) = 1 f (Sn + bn (X1 , , Xn )) + f (Sn ? n (X1 , , Xn )). Then 2 intuitively, En f (Sn+1 cannot be unaccompanied dependent upon Sn when bn s are right chosen. Therefore in general, (Sn )n? 1 cannot be a Markov subroutine. Remark If Xn is regarded as the gain/lo ss of n-th anticipate in a period of play game, then Sn would be the wealth at time n. bn is and so the wager for the (n+1)-th bet and is devised according to past gambling results. 2. 8. (i) Proof. Note Mn = En MN and Mn = En MN . (ii) Proof. In the proof of Theorem 1. 2. 2, we provoked by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In other words, the sequence (Vn )0? n?N can be realized as the pass judgment lick of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a martingale to a lower place P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale infra P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), then use (i). 2. 9. (i) (H) S1 (H) 1 = 2, d0 = S1S0 = 2 , S0 (T and d1 (T ) = S21 (TT)) = 1. S 1 1 0 ? d So p0 = 1+r? d0 0 = 2 , q0 = 2 , p1 (H) u0 5 q1 (T ) = 6 . Therefore P (HH) = p0 p1 (H) = 1 , 4 5 q0 q1 (T ) = 12 . Vn (1+r)n = En VN (1+r)N , so V0 , V1 1+r , , VN ? 1 , VN (1+r)N ? 1 (1+r)N is a martingale nether P . Proof. u0 = u1 (H) = =S2 (HH) S1 (H) = 1. 5, d1 (H) = S2 (HT ) S1 (H) = 1, u1 (T ) = S2 (T H) S1 (T ) =4 1+r1 (H)? d1 (H) u1 (H)? d1 (H) 1 = 1 , q1 (H) = 2 , p1 (T ) = 2 1 4, 1+r1 (T )? d1 (T ) u1 (T )? d1 (T ) 1 12 1 = 6 , and P (HT ) = p0 q1 (H) = P (T H) = q0 p1 (T ) = and P (T T ) = The proofs of Theorem 2. 4. 4, Theorem 2. 4. 5 and Theorem 2. 4. 7 dumb work for the random interest rate model, with proper modi? cations (i. e. P would be constructed according to conditional probabilities P (? n+1 = H? 1 , , ? n ) = pn and P (? n+1 = T ? 1 , , ? n ) = qn . Cf. visors on page 39. ). So the time-zero entertain of an option that pays o?V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . )(1+r (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2. 4, V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9, and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ? 1. 5 (iii) Proof. ?0 = (iv) Proof. ?1 (H) = 2. 10. (i) Xn+1 Proof. En (1+r)n+1 = En ? n Yn+1 Sn + (1+r)n+1 (1+r)(Xn n Sn ) (1+r)n+1 Xn (1+r)n . V2 (HH)? V2 (HT ) S2 (HH)? S2 (HT ) V1 (H)? V1 (T ) S1 (H)? S1 (T ) = 1 2. 4? 9 8? 2 = 0. 4 ? 1 54 ? 0. 3815. = 5? 1 12? 8 = 1. = ?n Sn (1+r)n+1 En Yn+1 + Xn Sn (1+r)n = ?n Sn (1+r)n+1 (up + dq) + Xn n Sn (1+r)n = ?n Sn +Xn n Sn (1+r)n = (ii) Proof. From (2. 8. 2), we have ? n navy blue + (1 + r)(Xn ? ?n Sn ) = Xn+1 (H) ? n dSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (T ). So ? n = Xn+1 (H)? Xn+1 (T ) uSn ? dSn and Xn = En Xn+1 . To make the portfolio replicate the payo? at time N , we 1+r VN X must have XN = VN . So Xn = En (1+r)N ? n = En (1+r)N ? n . Since (Xn )0? n? N is the encourage process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations), the no-arbitrage price of VN at time n is Vn = Xn = En (1+r)N ? . (iii) Proof. En Sn+1 (1 + r)n+1 = = = 1 En ( 1 ? An+1 )Yn+1 Sn (1 + r)n+1 Sn p(1 ? An+1 (H))u + q(1 ? An+1 (T ))d (1 + r)n+1 Sn pu + qd (1 + r)n+1 Sn . (1 + r)n Sn (1+r)n+1 (1? a)(pu+qd) Sn+1 If An+1 is a eternal a, then En (1+r)n+1 = Sn (1+r)n (1? a)n . = Sn (1+r)n (1? a). Sn+1 So En (1+r)n+1 (1? a)n+1 = 2. 11. (i) Proof. FN + PN = SN ? K + (K ? SN )+ = (SN ? K)+ = CN . (ii) CN FN PN Proof. Cn = En (1+r)N ? n = En (1+r)N ? n + En (1+r)N ? n = Fn + Pn . (iii) FN Proof. F0 = E (1+r)N = 1 (1+r)N ESN ? K = S0 ? K (1+r)N . (iv) 6 Proof.At time zero, the trader has F0 = S0 in money market account and one share of stock. At time N , the trader has a wealth of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Since F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En (1+r)N ?n = Sn ? So Fn is not needfully zero and Cn = Pn is not necessarily true for n ? 1. (1+r)N S0 (1+r)N ? n (1+r)N S0 (1+r)N = 0, C0 = P0 . = Sn ? S0 (1 + r)n . 2. 12. Proof. First, the no-arbitr age price of the selector switch option at time m must be liquid ecstasy(C, P ), where C=E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a draw option at time m. two(prenominal) of them have maturity date N and start price K. hypothesize the market is liquid, then the chooser option is equivalent to receiving a payo? of soap(C, P ) at time m. Therefore, its current no-arbitrage price should be E scoop shovel(C,P ) . (1+r)m K K By the put-call parity, C = Sm ? (1+r)N ? m + P . So max(C, P ) = P + (Sm ? (1+r)N ? m )+ . Therefore, the time-zero price of a chooser option is E K (Sm ? (1+r)N ? m )+ P +E (1 + r)m (1 + r)m =E K (Sm ? (1+r)N ? m )+ (K ? SN )+ . +E (1 + r)N (1 + r)mThe ? rst term stands for the time-zero price of a put, expiring at time N and having strike price K, and the K second term stands for the time-zero price of a call, expiring at time m and havi ng strike price (1+r)N ? m . If we feel dubious by the above line of crossings that the chooser options no-arbitrage price is E max(C,P ) , (1+r)m due to the economical argument involved ( equivalent the chooser option is equivalent to receiving a payo? of max(C, P ) at time m), then we have the following mathematically rigorous argument. First, we can construct a portfolio ? 0 , , ? m? 1 , whose payo? at time m is max(C, P ).Fix ? , if C(? ) P (? ), we can construct a portfolio ? m , , ? N ? 1 whose payo? at time N is (SN ? K)+ if C(? ) P (? ), we can construct a portfolio ? m , , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) P (? ) if C(? ) P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In Xm crashicular, V0 = X0 = E (1+r)m = E max(C,P ) . (1+r)m 2. 13. (i) Proof.Note under some(prenominal) actual probability P and risk-neutral probability P , coin tosses ? n s are i. i. d.. So n+1 without loss of generality, we work on P . For any function g, En g(Sn+1 , Yn+1 ) = En g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a function of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov under P . (ii) 7 Sn+1 Sn Sn ) Proof. muckle vN (s, y) = f ( Ny ). Then vN (SN , YN ) = f ( +1 Vn = where En Vn+1 1+r = n+1 En vn+1 (S1+r ,Yn+1 ) N n=0 Sn N +1 ) = VN . Suppose vn+1 is given, then = 1 1+r pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ) = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since coin tosses ? n s are i. i. d. under P , (Sn , Yn )0? n? M is Markov under P . More precisely, for any function h, En h(Sn+1 ) = ph(uSn ) + h(dSn ), for n = 0, 1, , M ? 1. For any function g of two variables, we have EM g(SM +1 , YM +1 ) = EM g(SM +1 , SM +1 ) = pg(uSM , uSM )+ n+1 n+1 qg(dSM , dSM ). And for n ? M +1, En g(Sn+1 , Yn+1 ) = En g( SSn Sn , Yn + SSn Sn ) = pg(uSn , Yn +uSn )+ qg(dSn , Yn + dSn ), so (Sn , Yn )0? n? N is Markov under P . (ii) y Proof. Set vN (s, y) = f ( N ? M ).Then vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . Suppose vn+1 is already given. a) If n M , then En vn+1 (Sn+1 , Yn+1 ) = pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ). So vn (s, y) = pvn+1 (us, y + us) + qvn+1 (ds, y + ds). b) If n = M , then EM vM +1 (SM +1 , YM +1 ) = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (us, us) + qvM +1 (ds, ds). c) If n M , then En vn+1 (Sn+1 ) = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. State Prices 3. 1. Proof. Note Z(? ) = P (? ) P (? ) = 1 Z(? ) . Apply Theorem 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of properties (i)-(iii) of Theorem 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. EY = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = A Z(? )P (? ) = 0, by P (Z 0) = 1, we conclude P (? ) = 0 for any ? ? A. So P (A) = A P (? ) = 0. (v) Proof. P (A) = 1 P (Ac ) = 0 P (Ac ) = 0 P (A) = 1. (vi) A P (? ) = Z(? )P (? ) = EZ = 1. Y (? )P (? ) = Y (? )Z(? )P (? ) = EY Z. Z(? )P (? ). Since P (A) = 0, P (? ) = 0 for any ? ? A. So P (A) = 0. 8 Proof. Pick ? 0 such that P (? 0 ) 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 Then P (Z ? 0) = 1 and EZ = if ? = ? 0 . P (? 0 ) = 1. =? 0 Clearly P (? ? 0 ) = EZ1? ? 0 = Z(? )P (? ) = 0. But P (? ? 0 ) = 1 ? P (? 0 ) 0 if P (? 0 ) 1. because in the case 0 P (? 0 ) 1, P and P are not equivalent. If P (? 0 ) = 1, then EZ = 1 if and only if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P have to be equivalent. In summary, if we can ? nd ? 0 such that 0 P (? 0 ) 1, then Z as constructed above would induce a probability P that is not equivalent to P . 3. 5. (i) Proof. Z(HH) = (ii) Proof. Z1 (H) = E1 Z2 (H) = Z2 (HH)P (? 2 = H? 1 = H) + Z2 (HT )P (? 2 = T ? 1 = H) = 3 E1 Z2 (T ) = Z2 (T H)P (? 2 = H? = T ) + Z2 (T T )P (? 2 = T ? 1 = T ) = 2 . (iii) Proof. V1 (H) = Z2 (HH)V2 (HH)P (? 2 = H? 1 = H) + Z2 (HT )V2 (HT )P (? 2 = T ? 1 = T ) = 2. 4, Z1 (H)(1 + r1 (H)) Z2 (T H)V2 (T H)P (? 2 = H? 1 = T ) + Z2 (T T )V2 (T T )P (? 2 = T ? 1 = T ) 1 = , Z1 (T )(1 + r1 (T )) 9 3 4. 9 16 , Z(HT ) = 9 , Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ? 1. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3. 6. Proof. U (x) = have XN = 1 x, (1+r)N ? Z so I(x) = = 1 Z 1 x. Z (3. 3. 26) gives E (1+r)N 1 X0 (1 + r)n Zn En Z X0 N Z (1 + r) . 0 = Xn , where ? Hence Xn = (1+r)N ? Z X En (1+r)N ? n N = X0 . So ? = = En X0 (1+r) Z n 1 X0 . By (3. 3. 25), we 1 = X0 (1 + r)n En Z = the second to last = comes from Lemma 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26 ), we have E (1+r)N ( (1+r)N ) p? 1 = X0 . go it for ? , we get ? ?p? 1 1 1 ? ? =? ? X0 p E 1 Z p? 1 Np ? ? ? = p? 1 X0 (1 + r)N p (EZ p? 1 )p? 1 1 p . (1+r) p? 1 ? Z So by (3. 3. 25), XN = ( (1+r)N ) p? 1 = 1 1 Np ? p? 1 Z p? 1 N (1+r) p? 1 = X0 (1+r) p? 1 EZ p p? 1 Z p? 1 N (1+r) p? 1 = (1+r)N X0 Z p? 1 EZ p p? 1 1 . 3. 8. (i) 9 d d Proof. x (U (x) ? yx) = U (x) ? y. So x = I(y) is an extreme point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a maximum point. Therefore U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Following the hint of the problem, we have EU (XN ) ? EXN ? Z ? Z ? Z ? Z ? EU (I( )) ? E I( ), N N N (1 + r) (1 + r) (1 + r) (1 + r)N ? ? ? ? ? i. e. EU (XN ) ? ?X0 ? EU (XN ) ? E (1+r)N XN = EU (XN ) ? ?X0 . So EU (XN ) ? EU (XN ). 3. 9. (i) X Proof. Xn = En (1+r)N ? n . So if XN ? 0, then Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x ? and 0 y ? ? , then U (x) ? yx = ? yx ? and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 b) If 0 ? x ? and y ? , then U (x) ? yx = ? yx ? 0 and U (I(y)) ? yI(y) = U (0) ? y 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 c) If x ? ? and 0 y ? ? , then U (x) ? yx = 1 ? yx and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 1 ? yx. So U (x) ? yx ? U (I(y)) ? yI(y). 1 d) If x ? ? and y ? , then U (x) ? yx = 1 ? yx 0 and U (I(y)) ? yI(y) = U (0) ? y 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). (iii) XN ? Z Proof. Using (ii) and set x = XN , y = (1+r)N , where XN is a random variable satisfying E (1+r)N = X0 , we have ?Z ? Z ? EU (XN ) ? E XN ? EU (XN ) ? E X ? . (1 + r)N (1 + r)N N ? ? That is, EU (XN ) ? ?X0 ? EU (XN ) ? ?X0 . So EU (XN ) ? EU (XN ). (iv) Proof. Plug pm and ? m into (3. 6. 4), we have 2N 2N X0 = m=1 pm ? m I( m ) = m=1 1 pm ? m ? 1 m ? ? . So X0 ? X0 ? m = we are looking for positive solution ? 0). Conversely, suppose there exists about K so that ? K ? K+1 and K X0 1 m=1 ? m pm = ? . Then we can ? nd ? 0, such that ? K ? K+1 . For such ? , we have Z ? Z 1 E I( ) = pm ? m 1 m ? ? ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 Hence (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1 m ? ? . Suppose there is a solution ? to (3. 6. 4), note ? 0, we then can conclude 1 1 1 m ? ? = ?. allow K = maxm m ? ? , then K ? ? K+1 . So ? K ? K+1 and K N m=1 pm ? m (Note, however, that K could be 2 . In this case, ? K+1 is interpreted as ?. Also, note = (v) ? 1 Proof. XN (? m ) = I( m ) = ? 1 m ? ? = ?, if m ? K . 0, if m ? K + 1 4. American Derivative Securities Before proceeding to the feat problems, we ? rst give a brief summary of pricing American differential securities as presented in the textbook. We shall use the notation of the book.From the emptors stead At time n, if the derivative security has not been exercised, then the vendee can choose a policy ? with ? ? Sn . The valuation formula for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative security exer cised according to ? N Vn (? ) = k=n En 1? =k 1 1 Gk = En 1? ?N G? . (1 + r)k? n (1 + r)? ?n The buyer emergencys to charter all the possible ? s, so that he can ? nd the least upper bound of security value, which will be the maximum price of the derivative security unexceptionable to him. This is the price given by 1 De? nition 4. 4. 1 Vn = max? ?Sn En 1? ?N (1+r)? n G? . From the vendors perspective A price process (Vn )0? n? N is acceptable to him if and only if at time n, he can construct a portfolio at cost Vn so that (i) Vn ? Gn and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can ? nd (? n )0? n? N and (Cn )0? n? N so that Cn ? 0 and Sn Vn+1 = ? n Sn+1 + (1 + r)(Vn ? Cn ? ?n Sn ). Since ( (1+r)n )0? n? N is a martingale under the risk-neutral measure P , we conclude En Cn Vn+1 Vn =? ? 0, ? n+1 n (1 + r) (1 + r) (1 + r)n Vn i. e. ( (1+r)n )0? n? N is a supermartingale. This inspired us to check mark if the converse is also true.This is exactly the content of Theorem 4. 4. 4. So (Vn )0? n? N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the need 0? n? N Vn ? Gn ). In summary, a price process (Vn )0? n? N is acceptable to the seller if and only if (i) Vn ? Gn (ii) is a supermartingale under P . 0? n? N Theorem 4. 4. 2 files the buyers upper bound is the sellers lower bound. So it gives the price acceptable to both. Theorem 4. 4. 3 gives a speci? c algorithm for calculating the price, Theorem 4. 4. establishes the matched correspondence between super-replication and supermartingale property, and ? nally, Theorem 4. 4. 5 shows how to decide on the optimum exercise policy. 4. 1. (i) Proof. V2P (HH) = 0, V2P (HT ) = V2P (T H) = 0. 8, V2P (T T ) = 3, V1P (H) = 0. 32, V1P (T ) = 2, V0P = 9. 28. (ii) Proof. V0C = 5. (iii) Proof. gS (s) = 4 ? s. We generate Theorem 4. 4. 3 and have V2S (HH) = 12. 8, V2S (HT ) = V2S (T H) = 2. 4, V2S (T T ) = 3, V1S (H) = 6. 08, V1S (T ) = 2. 16 and V0S = 3. 296. (iv) 11 Proof. First, we note the simple in comparison max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). holds if and only if b1 a1 , b2 a2 or b1 a1 , b2 a2 . By induction, we can show S Vn = max gS (Sn ), S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ? max gP (Sn ) + gC (Sn ), ? max gP (Sn ), P C = Vn + Vn . P P pVn+1 + Vn+1 1+r + max gC (Sn ), S P C As to when C C pVn+1 +qVn+1 1+r or gP (Sn ) P P pVn+1 +qVn+1 1+r and gC (Sn ) C C pVn+1 +qVn+1 . 1+r 4. 2. Proof. For this problem, we need Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Then ? 1 (H) = and ? 0 = V2 (HH) ? V2 (HT ) 1 V2 (T H) ? V2 (T T ) = ? , ? 1 (T ) = = ? 1, S2 (HH) ? S2 (HT ) 12 S2 (T H) ?S2 (T T ) V1 (H) ? V1 (T ) ? ?0. 433. S1 (H) ? S1 (T ) The optimal exercise time is ? = infn Vn = Gn . So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the agent borrows 1. 36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borrow again and buy 0. 433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio 1 is X1 (T ) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (T ) = (1 + 4 )(? 1. 36 ? 0. 433 ? 4) + 0. 433 ? 2 = ? 3. The agent should exercise the put at time one and get 3 to pay o? is debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The agent should borrow to buy 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.The agent should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We need Figure 1. 2. 2 for this problem, and visualize the intrinsic value process and price process of the put as follows. 2 For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the other outcomes of G is negative. 12 2 5 For the price process, V0 = 0. 4, V1 (T ) = 1, V1 (T H) = 3 , V1 (T T ) = 3 , V3 (T HT ) = 1, V3 (T T H) = 1. 75, V3 (T T T ) = 2. 125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when t he derivative security is exercised. So to hedge our short position after merchandising the put, there is no need to charge the insider more than than 1. 36. 4. 5. Proof. The halt times in S0 are (1) ? ? 0 (2) ? ? 1 (3) ? (HT ) = ? (HH) = 1, ? (T H), ? (T T ) ? 2, ? (4 di? erent ones) (4) ? (HT ), ? (HH) ? 2, ? , ? (T H) = ? (T T ) = 1 (4 di? rent ones) (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? 2, ? (16 di? erent ones). When the option is out of money, the following stopping times do not exercise (i) ? ? 0 (ii) ? (HT ) ? 2, ? , ? (HH) = ? , ? (T H), ? (T T ) ? 2, ? (8 di? erent ones) (iii) ? (HT ) ? 2, ? , ? (HH) = ? , ? (T H) = ? (T T ) = 1 (2 di? erent ones). ? 4 For (i), E1? ?2 ( 4 )? G? = G0 = 1. For (ii), E1? ?2 ( 5 )? G? ? E1? ? ? 2 ( 4 )? G? ? , where ? ? (HT ) = 5 5 1 4 4 ? 2, ? ? (HH) = ? , ? ? (T H) = ? ? (T T ) = 2. So E1? ? ? 2 ( 5 )? G? ? = 4 ( 4 )2 1 + ( 5 )2 (1 + 4) = 0. 96. For 5 (iii), E1? ?2 ( 4 )? G? has the biggest value when ? satis? es ? (HT ) = 2, ? (H H) = ? , ? (T H) = ? (T T ) = 1. 5 This value is 1. 36. 4. 6. (i) Proof. The value of the put at time N , if it is not exercised at previous times, is K ? SN . Hence VN ? 1 = VN K maxK ? SN ? 1 , EN ? 1 1+r = maxK ? SN ? 1 , 1+r ? SN ? 1 = K ? SN ? 1 . The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K ?S0 . Remark We cheated a little bit by using American algorithm and Theorem 4. 4. 5, since they are developed for the case where ? is allowed to be ?. But intuitively, results in this chapter should still hold for the case ? ? N , provided we replace maxGn , 0 with Gn . (ii) Proof. This is because at time N , if we have to exercise the put and K ? SN 0, we can exercise the European call to set o? the negative payo?. In e? ect, throughout the portfolios lifetime, the portfolio has intrinsic value greater than that of an American put stuck at K with expiration time N . So, we must have V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. let V0EP denote the time-zero value of a European put with strike K and expiration time N . Then V0AP ? V0EP = V0EC ? E K SN ? K = V0EC ? S0 + . (1 + r)N (1 + r)N 4. 7. VN K K Proof. VN = SN ? K, VN ? 1 = maxSN ? 1 ? K, EN ? 1 1+r = maxSN ? 1 ? K, SN ? 1 ? 1+r = SN ? 1 ? 1+r . K By induction, we can prove Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn Gn for 0 ? n ? N ? 1. So the K time-zero value is S0 ? (1+r)N and the optimal exercise time is N . 5. Random Walk 5. 1. (i) Proof. E 2 = E? (? 2 1 )+? 1 = E? (? 2 1 ) E 1 = E 1 2 . (ii) Proof. If we de? ne Mn = Mn+? ? M? m (m = 1, 2, ), then (M )m as random functions are i. i. d. with (m) distributions the same as that of M . So ? m+1 ? ?m = infn Mn = 1 are i. i. d. with distributions the same as that of ? 1 . Therefore E m = E? (? m m? 1 )+(? m? 1 m? 2 )++? 1 = E 1 m . (m) (m) (iii) Proof. Yes, since the argument of (ii) still works for unsymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe , so f (? ) 0 if and only if ? f (? ) f (0) = 1 for all ? 0. (ii) 1 1 1 n+1 Proof. En SSn = En e? Xn+1 f (? ) = pe? f (? ) + qe f (? ) = 1. 1 2 (ln q ? ln p). Since 1 2 (ln q ln p) 0, (iii) 1 Proof. By optional stopping theorem, ESn 1 = ES0 = 1. Note Sn 1 = e? Mn 1 ( f (? ) )n 1 ? e? 1 , by bounded convergence theorem, E1? 1 1 for all ? ? 0 . v (ii) 1 1 Proof. As in Exercise 5. 2, Sn = e? Mn ( f (? ) )n is a martingale, and 1 = ES0 = ESn 1 = Ee? Mn 1 ( f (? ) )? 1 ? n . Suppose ? ? 0 , then by bounded convergence theorem, 1 = E lim e? Mn 1 ( n? 1 n 1 1 ? 1 ) = E1? 1 K = P (ST K). Moreover, by Girsanovs Theorem, Wt = Wt + in Theorem 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t ( )du 0 = Wt ? ?t is a P -Brownian motion (set ? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v ? d+ (T, x) T = N (d+ (T, x)). P (ST K) = P (xe? WT +(r+ 2 ? )T K) = P 46 5. 4. First, a few typos. In the SDE for S, ? (t)dW (t) ? (t)S(t)dW (t). In the ? rst equation for c(0, S(0)), E E. In the second equation for c(0, S(0)), the variable for BSM should be ? ? 1 T 2 1 T r(t)dt, ? (t)dt? . BSM ? T, S(0) K, T 0 T 0 (i) Proof. d ln St = X = ? is a Gaussian with X ? N ( (ii) Proof. For the standard BSM model with constant volatility ? and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R? 1 ? 2 )T +? WT ? N ((R? 1 ? )T, ? 2 T ), and E(S0 eY ? K)+ = 2 2 eRT BSM (T, S0 K, R, ? ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St ? 2St d S t = rt dt + ? t dWt ? 2 ? t dt. So ST = S0 exp 0 (rt ? 2 ? t )dt + 0 T 1 2 2 ? t )dt + 0 ? t dWt . The ? rst term in the expression of X is a number and the T 2 random variable N (0, 0 ? t dt), since both r and ? ar deterministic. Therefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt . Let second term ST = S0 eX , 1 T (EY + 1 V ar(Y )) and ? = 2 T, S0 K, 1 T 1 T V ar(Y ), we can get 1 V ar(Y ) . T E(S0 eY ? K)+ = eEY + 2 V ar(Y ) BSM So for the model in this problem, c(0, S0 ) = = e? ? T 0 1 EY + V ar(Y ) , 2 rt dt E(S0 eX ? K)+ e BSM T, S0 K, 1 T T 0 T 0 1 rt dt EX+ 2 V ar(X) 1 T ? 1 EX + V ar(X) , 2 1 V ar(X) T ? = 1 BSM ? T, S0 K, T 0 T rt dt, 2 ? t dt? . 5. 5. (i) 1 1 Proof. Let f (x) = x , then f (x) = ? x2 and f (x) = 2 x3 . Note dZt = ? Zt ? t dWt , so d 1 Zt 1 1 1 2 2 2 ? t ? 2 t = f (Zt )dZt + f (Zt )dZt dZt = ? 2 (? Zt )? t dWt + 3 Zt ? t dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5. 2. 2. , for s, t ? 0 with s t, Ms = EMt Fs = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs Fs . That is, EZt Mt Fs = 47 Proof. dMt = d Mt 1 Zt = 1 1 1 ? M t ? t M t ? 2 ? t ? t t dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt (iv) Proof. In part (iii), we have dMt = Let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to show Wi (t) is an Ft -martingale under P an d Wi , Wj (t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt EWi (t)Fs = E Fs . Zs By It? s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(? Zt )? (t) dWt + Zt (dWi (t) + ? i (t)dt) + (? Zt ? t dWt )(dWi (t) + ? i (t)dt) d t M t ? t M t ? 2 ? t ? t ? t M t ? t t dWt + dWt + dt + dt = (dWt + ? t dt) + (dWt + ? t dt). Zt Zt Zt Zt Zt Zt then dMt = ? t dWt . This proves Corollary 5. 3. 2. ?t +Mt ? t , Zt = Wi (t)(? Zt ) j=1 d ?j (t)dWj (t) + Zt (dWi (t) + ? i (t)dt) ? Zt ? i (t)dt = Wi (t)(? Zt ) j=1 ?j (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover, Wi , Wj (t) = Wi + 0 ?i (s)ds, Wj + 0 ?j (s)ds (t) = Wi , Wj (t) = t? ij . Combined, this proves the two-dimensional Girsanovs Theorem. 5. 7. (i) Proof. Let a be any strictly positive number. We de? e X2 (t) = (a + X 1 (t))D(t)? 1 . Then P X2 (T ) ? X2 (0) D(T ) = P (a + X1 (T ) ? a) = P (X1 (T ) ? 0) = 1, and P X2 (T ) X2 (0) = P (X1 (T ) 0) 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Then X1 (0) = 0, P (X1 (T ) ? 0) = P X2 (T ) ? X2 (0) D(T ) = 1, P (X1 (T ) 0) = P X2 (T ) X2 (0) D(T ) 0. 5. 8.The basic idea is that for any positive P -martingale M , dMt = Mt sentation Theorem, dMt = ? t dWt for some adapted process ? t . So martingale must be the exponential of an entire w. r. t. Brownian motion. Taking into account discounting cistron and apply It? s product rule, we can show every strictly positive asset is a generalized nonrepresentational o Brownian motion. (i) Proof. Vt Dt = Ee? 0 Ru du V T Ft = EDT VT Ft . So (Dt Vt )t? 0 is a P -martingale. By dolphin striker Represent tation Theorem, there exists an adapted process ? t , 0 ? t ? T , such that Dt Vt = 0 ? s dWs , or equivalently, ? 1 t ? 1 t ? 1 Vt = Dt 0 ? dWs . Di? erentiate both sides of the equation, we get dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i. e. X 0 a. s. ) de? ned on the probability space (? , G, P ). Let F be a sub ? -algebra of G, then Y = EXF 0 a. s. Proof. By the property of conditional view Yt ? 0 a. s. Let A = Y = 0, we shall show P (A) = 0. In? 1 1 deed, note A ? F, 0 = EY IA = EEXFIA = EXIA = EX1A? X? 1 + n=1 EX1A? n X? n+1 ? 1 1 1 1 1 P (A? X ? 1)+ n=1 n+1 P (A? n X ? n+1 ). So P (A? X ? 1) = 0 and P (A? n X ? n+1 ) = 0, ? 1 1 ? n ? 1. This in turn implies P (A) = P (A ? X 0) = P (A ? X ? 1) + n=1 P (A ? n X ? n+1 ) = 0. ? ? t Dt dWt . T 1 M t dMt . By martingale Repre? dMt = Mt ( Mtt )dWt , i. e. any positive By the above lemma, it is clear that for each t ? 0, T , Vt = Ee? t Ru du VT Ft 0 a. s.. Moreover, by a classical result of martingale theory (Revuz and Yor 4, Chapter II, Proposition (3. 4)), we have the following stronger result for a. s. ?, Vt (? ) 0 for any t ? 0, T . (iii) 1 1 Proof. By (ii), V 0 a. s. so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + ? t Dt dWt ? t = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T ?t Vt dWt , where ? t = 5. 9. ?t Vt Dt . This shows V follows a generalized geometrical Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d = then f (y) = ? yf (y), cK (0, T, x, K) = xf (d+ ) 1 v ? T x (ln K + (r 1 ? 2 )T ). Let f (y) = 2 y v1 e? 2 2? 2 , ?d+ ? d? ? e? rT N (d? ) ? Ke? rT f (d? ) ? y ? y ? 1 1 = xf (d+ ) v ? e? rT N (d? ) + e? rT f (d? ) v , ? TK ? T 49 and cKK (0, T, x, K) x ? d? e? rT 1 ? d+ d? ? v ? e? rT f (d? ) + v (? d? )f (d? ) xf (d+ ) v f (d+ )(? d+ ) 2 ? y ? y ? y ? TK ? TK ?T x xd+ ? 1 ? 1 e? rT d? ?1 v v ? e? rT f (d? ) v ? v f (d? ) v f (d+ ) + v f (d+ ) ? T K2 ? TK K? T K? T ? T K? T x d+ e? rT f (d? ) d? v 1 ? v + v f (d+ ) 1 + v 2? T K ? T K? T ? T e? rT x f (d? )d+ ? 2 2 f (d+ )d? . K? 2 T K ? T = = = = 5. 10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = maxC(t0 ), P (t0 ) = maxC(t0 ), C(t0 ) ? F (t0 ) = C(t0 ) + max0, ? F (t0 ) = C(t0 ) + (e? r(T ? t0 ) K ? S(t0 ))+ . (ii) Proof. By the risk-neutral pricing formula, V (0) = Ee? rt0 V (t0 ) = Ee? rt0 C(t0 )+(e? rT K ? e? rt0 S(t0 )+ = C(0) + Ee? rt0 (e? r(T ? t0 ) K ? S(t0 ))+ . The ? st term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e? r(T ? t0 ) K. 5. 11. Proof. We ? rst make an digest which leads to the hint, then we give a formal proof. (Analysis) If we fate to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the s elf-referent SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. work out Dt on both sides of the ? rst equation and apply It? s product rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Integrate from 0 to T , we have DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the perch T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash ? ow, provided we can ? nd a trading strategy ? that solves the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . Take the proper wobble of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the impudently measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ?t d(Dt St ) = D0 X0 + 0 ?t (Dt St )? t dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Represe ntation Theorem and get a formula for ? by comparison of the integrands. 50 (Formal proof) Let MT = Xt = ?1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = EMT Ft . Then by Martingale Representation Theot 0 rem, we can ? nd an adapted process ? t , so that Mt = M0 + + t 0 ?t dWt . If we set ? t = T 0 ?u d(Du Su ) ? t 0 ?t Dt St ? t , we can check Cu Du du), with X0 = M0 = E Ct Dt dt solves the SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Indeed, it is diffused to see that X satis? es the ? rst equation.To check the terminal condition, we note T T T XT DT = D0 X0 + 0 ? t Dt St ? t dWt ? 0 Ct Dt dt = M0 + 0 ? t dWt ? MT = 0. So XT = 0. Thus, we have found a trading strategy ? , so that the corresponding portfolio X replicates the cash ? ow and has zero T terminal value. So X0 = E 0 Ct Dt dt is the no-arbitrage price of the cash ? ow at time zero. Remark As shown in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Integrate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take conditional prediction w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E t Cu Du duFt . So Xt = Dt E t Cu Du duFt .This is the no-arbitrage price of the cash ? ow at time t, and we have justi? ed formula (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Since dBi (t)dBi (t) = P. (ii) Proof. dSi (t) = = = R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + (? i (t) ? R(t))Si (t)dt ? ?i (t)Si (t)? i (t)dt d d ? ij (t) ? ij (t) d d j=1 ? i (t) ? j (t)dt = j=1 ? i (t) dWj (t) + ? ij (t)2 d e j=1 ? i (t)2 dt = dt, by L? vys Theorem, Bi ? ij (t) d j=1 ? i (t) dWj (t). So Bi is a is a Brownian motion under R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + j=1 ?ij (t)? j (t)Si (t)dt ? Si (t) j=1 ?ij (t)? j (t)dt R(t)Si (t)dt + ? (t)Si (t)dBi (t). (iii) Proof. dBi (t)dBk (t) = (dBi (t) + ? i (t)dt)(dBj (t) + ? j (t)dt) = dBi (t)dBj (t) = ? ik (t)dt. (iv) Proof. By It? s product rule and martingale property, o t t t EBi (t)Bk (t) = E 0 t Bi (s)dBk (s) + E 0 t Bk (s)dBi (s) + E 0 dBi (s)dBk (s) = E 0 ?ik (s)ds = 0 ?ik (s)ds. t 0 Similarly, by part (iii), we can show EBi (t)Bk (t) = (v) ?ik (s)ds. 51 Proof. By It? s product formula, o t t EB1 (t)B2 (t) = E 0 sign(W1 (u))du = 0 P (W1 (u) ? 0) ? P (W1 (u) 0)du = 0. Meanwhile, t EB1 (t)B2 (t) = E 0 t sign(W1 (u))du P (W1 (u) ? 0) ? P (W1 (u) 0)du = 0 t = 0 t P (W1 (u) ? ) ? P (W1 (u) u)du 2 0 = 0, 1 ? P (W1 (u) u) du 2 for any t 0. So EB1 (t)B2 (t) = EB1 (t)B2 (t) for all t 0. 5. 13. (i) Proof. EW1 (t) = EW1 (t) = 0 and EW2 (t) = EW2 (t) ? (ii) Proof. CovW1 (T ), W2 (T ) = EW1 (T )W2 (T ) T T t 0 W1 (u)du = 0, for all t ? 0, T . = E 0 T W1 (t)dW2 (t) + 0 W2 (t)dW1 (t) T = E 0 W1 (t)(dW2 (t) ? W1 (t)dt) + E 0 T W2 (t)dW1 (t) = ? E 0 T W1 (t)2 dt tdt = ? 0 1 = ? T 2. 2 5. 14. Equation (5. 9. 6) can be transformed into d(e? rt Xt ) = ? t d(e? rt St ) ? ae? rt dt = ? t e? rt dSt ? rSt dt ? adt. So, to make the discounted portfolio value e? t Xt a martingale, we are motivated to change the mea sure t in such a way that St ? r 0 Su du? at is a martingale under the new measure. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . Hence dSt ? rSt dt? adt = (? t ? r)St ? adt+? St dWt = ? St Set ? t = (? t ? r)St ? a ? St (? t ? r)St ? a dt ? St + dWt . and Wt = t ? ds 0 s + Wt , we can ? nd an equivalent probability measure P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the acute for formula (5. 9. 7). This is a good place to suspension system and think about the meaning of martingale measure. What is to be a martingale?The new measure P should be such that the discounted value process of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payo? at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ? 0, T . The di? culty is how to ca lculate Xt and the magic is brought by the martingale measure in the following line of reasoning ? 1 ? 1 Vt = Xt = Dt EDT XT Ft = Dt EDT VT Ft . You can think of martingale measure as a calculational convenience.That is all about martingale measure Risk neutral is a just perception, referring to the actual e? ect of constructing a hedging portfolio Second, we note when the portfolio is self-? nancing, the discounted price process of the underlying is a martingale under P , as in the classical Black-Scholes-Merton model without dividends or cost of support. This is not a coincidence. Indeed, we have in this case the relation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no longer holds, as shown in formula (5. 9. 6). The portfolio is no longer self-? nancing, but self-? nancing with consumption. What we sti ll want to retain is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the above paragraph. Let VT be a payo? at time T , then for the martingale Mt = Ee? rT VT Ft , by Martingale Representation rt t Theorem, we can ? nd an adapted process ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , then the ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by scope X0 = M0 = Ee? T VT , we must have e? rt Xt = Mt , for all t ? 0, T . In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justi? es the risk-neutral pricing of European-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is di? erent from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the indisposed SDE dXt = ? t dSt ? a? t dt + r(Xt ? ?t St )dt XT = VT for two unknowns, X and ?. To do so, we ? nd a probabil ity measure P , under which e? rt Xt is a martingale, t then e? rt Xt = Ee? T VT Ft = Mt . Martingale Representation Theorem gives Mt = M0 + 0 ? u dWu for some adapted process ?. This would give us a theoretical representation of ? by comparison of integrands, hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5. 9. 7) under P , then d(e? rt Xt ) = ? t d(e? rt St ) ? ae? rt dt = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , where X is given by (5. 9. 6), is a P -martingale. (ii) 1 1 Proof. By It? s formula, dYt = Yt ? dWt + (r ? 2 ? 2 )dt + 2 Yt ? 2 dt = Yt (? dWt + rdt). So d(e? rt Yt ) = o t a ? e? rt Yt dWt and e? rt Yt is a P -martingale.Moreover, if St = S0 Yt + Yt 0 Ys ds, then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (? dWt + rdt) + adt = St (? dWt + rdt) + adt. Ys This shows S satis? es (5. 9. 7). Remark To obtain this formula for S, we ? rst set Ut = e? rt St to remove the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Just like solving linear ODE, to remove U in the dWt term, we consider Vt = Ut e Wt . It? s product formula yields o dVt = = e Wt dUt + Ut e Wt 1 ( )dWt + ? 2 dt + dUt e Wt 2 1 ( )dWt + ? 2 dt 2 1 e Wt ae? rt dt ? ? 2 Vt dt. 2 53 Note V appears only in the dt term, so multiply the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt Wt + 2 ? t dt. Set Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys EST Ft = S0 EYT Ft + E YT 0 t a ds + YT Ys T t T a dsFt Ys E YT Ft ds Ys EYT ? s ds t = S0 EYT Ft + 0 a dsEYT Ft + a Ys t t T = S0 Yt EYT ? t + 0 t a dsYt EYT ? t + a Ys T t = = S0 + 0 t a ds Yt er(T ? t) + a Ys ads Ys er(T ? s) ds S0 + 0 a Yt er(T ? t) ? (1 ? er(T ? t) ). r In particular, EST = S0 erT ? a (1 ? erT ). r (iv) Proof. t dEST Ft = aer(T ? t) dt + S0 + 0 t ads Ys a (er(T ? ) dYt ? rYt er(T ? t) dt) + er(T ? t) (? r)dt r = S0 + 0 ads Ys er(T ? t) ? Yt dWt . So EST Ft is a P -martingale. As we have argued at the beginning of the solution, risk-neutral pricing is binding even in the presence of cost of carry. So by an argument similar to that of 5. 6. 2, the process EST Ft is the futures price process for the commodity. (v) Proof. We solve the equation Ee? r(T ? t) (ST ? K)Ft = 0 for K, and get K = EST Ft . So F orS (t, T ) = F utS (t, T ). (vi) Proof. We follow the hint. First, we solve the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our analysis in part (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Integrate from 0 to t on both sides, we get Xt = St ? S0 ert + a (1 ? ert ) = St ? S0 ert ? a (ert ? 1). In particular, XT = ST ? S0 erT ? a (erT ? 1). r r r Meanwhile, F orS (t, T ) = F uts (t, T ) = EST Ft = S0 + t ads 0 Ys Yt er(T ? t) ? a (1? er(T ? t) ). So F orS (0, T ) = r S0 erT ? a (1 ? erT ) and hence XT = ST ? F orS (0, T ). later on the agent delivers the commodity, whose value r is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious.Note the form of Z is similar to that of a geometric Brownian motion. So by It? s o formula, it is easy to obtain dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), then Xt = Yt Zt = x 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au ? ?u ? u ? u du + dWu Zu Zu Yu bu Zu + (au ? ?u ? u ) + ? u ? u du + (? u Zu Yu + ? u )dWu Yu (bu Zu du + ? u Zu dWu ) + Zu (bu Xu + au )du + (? u Xu + ? u )dWu . + ? u Z u ? u du Zu Remark To see how to ? nd the above solution, we manipulate the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we multiply on both sides of (6. 2. 4) the integrating factor e? bv dv . Then d(Xu e? ? Let Xu = e? u t u t bv dv ) = e? u t bv dv (au du + (? u + ? u Xu )dWu ). u t bv dv Xu , au = e? ? u t bv dv au and ? u = e? ? bv dv ? ? u , then X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To deal with the term ? u Xu dWu , we consider Xu = Xu e? ? dXu = e? u t u t ?v dWv . Then ?v dWv ?v dWv ? ? (? u du + ? u dWu ) + ? u Xu dWu + Xu e? a ? u t u t 1 ( u )dWu + e? 2 u t ?v dWv 2 ? u du ? +(? u + ? u Xu )( u )e? ? ?v dWv du 1 ? 2 ? ? ? = au du + ? u dWu + ? u Xu dWu ? ?u Xu dWu + Xu ? u du ? ?u (? u + ? u Xu )du ? ? ? 1 ? 2 = (? u ? ?u ? u ? Xu ? u )du + ? u dWu , a ? ? 2 where au = au e? ? ? ? 1 d Xu e 2 u t ?v dWv 2 ? v dv and ? u = ? u e? ? ? = e2 1 u t 2 ? v dv u t ?v dWv . Finally, use the integrating factor e u t 2 ? v dv u 1 2 ? dv t 2 v , we have u t 1 ? ? 1 2 (dXu + Xu ? u du) = e 2 2 (? u ? ?u ? u )du + ? u dWu . a ? ? spell everything back into the original X, a and ? , we get d Xu e? i. e. d u t bv dv? u t 1 ? v dWv + 2 u t 2 ? v dv = e2 1 u t 2 ? v dv? u t ?v dWv ? u t bv dv (au ? ?u ? u )du + ? u dWu , Xu Zu = 1 (au ? ?u ? u )du + ? u dWu = dYu . Zu This inspired us to try Xu = Yu Zu . 6. 2. (i) 55 Proof.The portfolio is self-? nancing, so for any t ? T1 , we have dXt = ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) + Rt (Xt ? ?1 (t)f (t, Rt , T1 ) ? ?2 (t)f (t, Rt , T2 ))dt, and d(Dt Xt ) = ? Rt Dt Xt dt + Dt dXt = Dt ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt 1 = Dt ? 1 (t) ft (t, Rt , T1 )dt + fr (t, Rt , T1 )dRt + frr (t, Rt , T1 )? 2 (t, Rt )dt 2 1 +? 2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )? 2 (t, Rt )dt 2 ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt 1 = ? 1 (t)Dt ? Rt f (t, Rt , T1 ) + ft (t, Rt , T1 ) + ? t, Rt )fr (t, Rt , T1 ) + ? 2 (t, Rt )frr (t, Rt , T1 )dt 2 1 +? 2 (t)Dt ? Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + ? (t, Rt )fr (t, Rt , T2 ) + ? 2 (t, Rt )frr (t, Rt , T2 )dt 2 +Dt ? (t, Rt )Dt ? (t, Rt )? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )dWt = ? 1 (t)Dt ? (t, Rt ) ? ?(t, Rt , T1 )fr (t, Rt , T1 )dt + ? 2 (t)Dt ? (t, Rt ) ? ?(t, Rt , T2 )fr (t , Rt , T2 )dt +Dt ? (t, Rt )? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )dWt . (ii) Proof. Let ? 1 (t) = St fr (t, Rt , T2 ) and ? 2 (t) = ? St fr (t, Rt , T1 ), then d(Dt Xt ) = Dt St ? (t, Rt , T2 ) ? ?(t, Rt , T1 )fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt ? t, Rt , T1 ) ? ?(t, Rt , T2 )fr (t, Rt , T1 )fr (t, Rt , T2 )dt. Integrate from 0 to T on both sides of the above equation, we get T DT XT ? D0 X0 = 0 Dt ? (t, Rt , T1 ) ? ?(t, Rt , T2 )fr (t, Rt , T1 )fr (t, Rt , T2 )dt. If ? (t, Rt , T1 ) = ? (t, Rt , T2 ) for some t ? 0, T , under the assumption that fr (t, r, T ) = 0 for all values of r and 0 ? t ? T , DT XT ? D0 X0 0. To avoid arbitrage (see, for example, Exercise 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? 0, T . This implies ? (t, r, T ) does not depend on T . (iii) Proof. In (6. 9. 4), let ? 1 (t) = ? (t), T1 = T and ? (t) = 0, we get d(Dt Xt ) = 1 ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + ? (t, Rt )fr (t, Rt , T ) + ? 2 (t, Rt )f rr (t, Rt , T ) dt 2 +Dt ? (t, Rt )? (t)fr (t, Rt , T )dWt . This is formula (6. 9. 5). 1 If fr (t, r, T ) = 0, then d(Dt Xt ) = ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? 2 (t, Rt )frr (t, Rt , T ) dt. We 1 2 choose ? (t) = sign ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? (t, Rt )frr (t, Rt , T ) . To avoid arbitrage in this case, we must have ft (t, Rt , T ) + 1 ? 2 (t, Rt )frr (t, Rt , T ) = Rt f (t, Rt , T ), or equivalently, for any r in the 2 range of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We note d ? e ds s 0 bv dv C(s, T ) = e? s 0 bv dv C(s, T )(? bs ) + bs C(s, T ) ? 1 = ? e? s 0 bv dv . So integrate on both sides of the equation from t to T, we obtain e? T 0 bv dv C(T, T ) ? e? t 0 t 0 T bv dv C(t, T ) = ? t s 0 e? T t s 0 bv dv ds. Since C(T, T ) = 0, we have C(t, T ) = e 1 ? a(s)C(s, T ) + 2 ? 2 (s)C 2 (s, T ), we get A(T, T ) ? A(t, T ) = ? bv dv T t e? bv dv ds = T e t s bv dv ds. Finally, by A (s, T ) = T a(s)C(s, T )ds + t 1 2 ? 2 (s)C 2 (s, T )ds. t