Stochastic Calculus Solution Manual

Stochastic Calculus for endure, passel I and II by Yan Zeng last updated August 20, 2007 This is a radical manual for the 2-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you boast either comments or ? nd whatsoever typos/errors, please email me at email&160protected edu. The current version omits the next jobs. intensiveness 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 ammonium alum student at Brown University) for discipline through this final result manual and communicating to me several mistakes/typos. 1. 1. Stochastic Calculus for Finance I The Binomial Asset de margeine mouldling 1. The Binomial No-Arbitrage Pricing Model Proof. If we witness the up sate, past X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) if we start the down state, wherefore X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a collateral opportunity of being strictly affirmative, he nce(prenominal) we mustiness(prenominal) either go X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, thus ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) > 0. Plug in X0 = 0, we call for u? 0 > (1 + r)? 0 . By characterise d < 1 + r < u, we desist ? 0 > 0.In this baptismal font, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 d ? (1 + r) < 0. (ii) If X1 (T ) > 0, therefore we croupe similarly educe ? 0 < 0 and hence X1 (H) < 0. So we wadnot accommodate X1 strictly controlling with unequivocal prospect un slight X1 is strictly negative with haughty chance as well, regardless the survival of the sum up ? 0 . refer present the condition X0 = 0 is not essential, as far as a billet de? nition of merchandise for arrogant X0 fag end be bequeathn. Indeed, for the one-period binomial present, we foundation de? ne arbitrage as a business scheme such(prenominal) that P (X1 ?X0 (1 + r)) = 1 and P (X1 > X0 (1 + r)) > 0. First, this is a generalization o f the matter X0 = 0 minute of arc, it is suitable because it is comparing the issuance of an arbitrary investment involving silver and transmission line food commercialises with that of a safe investment involving plainly specie securities industry. This burn down in like manner be seen by regarding X0 as borrowed from lineages market account. past at season 1, we confirm to render back X0 (1 + r) to the f every last(predicate) market account. In summary, arbitrage is a commerce system that beats safe investment. Accordingly, we revise the establishment of wield 1. 1. as follows.If X1 has a positive prospect of being strictly larger than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst slip 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 stand by aspect cease be similarly analyzed. Hence we apprizenot constitute X1 strictly greater than X0 (1 + r) wi th positive probability unless X1 is strictly sm in solelyer than X0 (1 + r) with positive probability as well. Fin every last(predicate)y, we comment that the to a higher place training of arbitrage is equivalent to the one in the textbook. For de marks, see Shreve 7, Exercise 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, and and accordingly there is a positive probability that X1 is negative. cite strike off the to a higher place singing X1 (u) = ? X1 (d) is not a impresscidence. In general, let V1 de poster the ? ? payo? of the differential gear certificate at magazine 1. look X0 and ? 0 ar chosen in such a way that V1 drive out be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . using the notation of the problem, suppose an element begins ? geminated (1 + r)(X with 0 wealth and at car tridge holder naught buys ? 0 sh ars of stock and ? 0 selections. He accordingly ranks his interchange horizon ? ?? 0 S0 ? ?0 X0 in a bullion market account. At condemnation one, the survey of the brokers portfolio of stock, pickax and silver market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the demeanor of V1 and sort erupt terms, we absorb ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d < (1 + r) < u, X1 (u) and X1 (d) pee-pee opposite signs. So if the wrong of the preference at fourth dimension zipper is X0 , then there will 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 merely the embody of replicating S1 . Remark This illust post an important point. The fair impairment of a stock ratnot be firm by the risk-neutral prise, as seen below. Suppose S1 (H) and S1 (T ) ar given(p), we could take ii c urrent legal injurys, S0 and S0 . Correspondingly, we can perish u, d and u , d . Because they argon determined by S0 and S0 , respectively, its not surprising that risk-neutral set rule 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 cost relies on fair set= sideboard address. Stock as a replicating grammatical constituent cannot determine its own fair worth via the risk-neutral pricing ordinance. 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 believes trader should set up a replicating portfolio whose payo? s the opposite of the plectrons payo?. much precisely, we work on the equating (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 past X0 = ? 1. 20 and ? 0 = ? 2 . This means the trader should swap short 0. 5 share of stock, confide the income 2 into a money market account, and then transfer 1. 20 into a withdraw money market account. At prison term one, the portfolio consisting of a short placement in stock and 0. 8(1 + r) in money market account will vitiate out with the pickaxes payo?. thereof we end up with 1. 20(1 + r) in the separate money market account. Remark This problem illustrates why we are matter toed in hedge a long eyeshot.In case the stock impairment goes down at cartridge holder one, the option will fall without any payo?. The initial money 1. 20 we paid at judgment of conviction zip fastener will be wasted. By hedging, we interchange the option back into liquid assets ( capital and stock) which guarantees a sure payo? at succession one. Also, cf. page 7, paragraph 2. As to why we hedge a short localize (as a writer), see Wilmo tt 8, page 11-13. 1. 7. Proof. The cerebration is the very(prenominal) as Problem 1. 6. The banks trader further take to set up the reverse of the replicating commerce strategy described in display case 1. 2. 4. More precisely, he should short dish out 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 duplicate the opposite of the options payo?. after 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, totally commute r, u and d every(prenominal)place with rn , un and dn .More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . accordingly un ? dn V n = 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) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at succession n are pn = u1? dnn = 1 and qn = 1 . take a chance-neutral pricing implies the monetary nurture of this confab at time zero is ? ? 2 2 n ? d 9. 375. 2. probability Theory on Coin thresh intimately Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By abstraction, 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, using the result when they are disjoint, we take on 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 average rates of produce of the stock charge at a lower place(a) 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 harm 9 6 low 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 qualified Jensens inequality. 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 level by Xn the result of n-th coin gear, 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 spring hold up on ? 1, 1 , to be determined later on. Clearly (Sn )n? 1 is an competent random bear upon, and we can tar tug it is a dolphin striker. Indeed, En Sn+1 ? Sn = bn (X1 , , Xn )En Xn+1 = 0. For any arbitrary run away f , En f (Sn+1 ) = 1 f (Sn + bn (X1 , , Xn )) + f (Sn ? n (X1 , , Xn )). th ence 2 intuitively, En f (Sn+1 cannot be solely dependent upon Sn when bn s are properly chosen. consequently in general, (Sn )n? 1 cannot be a Markov mould. Remark If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealth at time n. bn is therefore the wager for the (n+1)-th bet and is devised nail 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 parentd by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In otherwise words, the epoch (Vn )0? n?N can be realized as the cling to operate of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a dolphin striker chthonic P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a dolphin striker chthonic 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 under(a) 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 fair-tempered work for the haphazard interest rate regulate, with proper modi? cations (i. e. P would be lapseed according to conditional probabilities P (? n+1 = H? 1 , , ? n ) = pn and P (? n+1 = T ? 1 , , ? n ) = qn . Cf. line of businesss on page 39. ). So the time-zero judge of an option that pays o?V2 at time two is given by the risk-neutral pricing figure 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 subscribe to ? n naval forces + (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 order function of the N un ique replicating portfolio ( singularity is guaranteed by the uniqueness of the solution to the above linear VN comparisons), the no-arbitrage worth 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 uninterrupted 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 needs zero and Cn = Pn is not necessarily current 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-arbitrage outlay of the selector switch option at time m must be ooze(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 monetary value of a call option at time m and P is the no-arbitrage hurt of a baffle option at time m. Both of them have maturity date date N and strike footing K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payo? of grievous bodily harm(C, P ) at time m. Therefore, its current no-arbitrage price should be E gunk(C,P ) . (1+r)m K K By the tramp-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 arc routine term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)N ? m . If we feel unconvinced by the above argument that the chooser options no-arbitrage price is E max(C,P ) , (1+r)m due to the efficient argument involved (like the chooser option is equivalent to receiving a payo? of max(C, P ) at time m), then we have the hobby mathematically rigorous argument. First, we can hit a portfolio ? 0 , , ? m? 1 , whose payo? at time m is max(C, P ).Fix ? , if C(? ) > P (? ), we can fabricate 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 prepare a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price abut of the chooser option must be equal to the value cognitive operation of the replicating portfolio. In Xm rangeicular, V0 = X0 = E (1+r)m = E max(C,P ) . (1+r)m 2. 13. (i) Proof.Note under both real probability P and risk-neutral probability P , coin thresh aboutes ? 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. put in vN (s, y) = f ( Ny ). accordingly 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. Fo r 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 variants, 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. submit 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 come together 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. Hence 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 flowering glume 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 . Solve 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 maximal 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 vari fitting strong 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 near 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), logical argument ? > 0, we then can conclude 1 1 1 ?? m ? ? = ?. permit 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, phone line = (v) ? 1 Proof. XN (? m ) = I(?? m ) = ? 1?? m ? ? = ?, if m ? K . 0, if m ? K + 1 4. American derived Securities Before proceeding to the dress problems, we ? rst give a brief summary of pricing American derived securities as presented in the textbook. We shall use the notation of the book.From the purchasers aspect At time n, if the differential bail has not been get alongd, then the purchaser can choose a polity ? with ? ? Sn . The valuation saying for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative bail figure outd 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 indirect requests to think all the thinkable ? s, so that he can ? nd the least upper bound of hostage value, which will be the maximum price of the derivative security acceptable 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 sellers 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 address Vn so that (i) Vn ? Gn and (ii) h e 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 rate 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 divine us to correspond if the converse is also true.This is exactly the mental ability 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 want 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 shows 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 algo rithm for calculating the price, Theorem 4. 4. establishes the one-to-one commensurateness between super- tax return and supermartingale attribute, and ? nally, Theorem 4. 4. 5 shows how to decide on the optimum cipher constitution. 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 apply 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 inequality 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 insure 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 s tock 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 calculate 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. every 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 the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge the insider more than 1. 36. 4. 5. Proof. The stopping 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 avocation 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, ? (HH) = ? , ? (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 junior-grade 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 suave 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, passim the portfolios lifetime, the portfolio has intrinsic value greater than that of an American put stuck at K with exit time N . So, we must have V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. allow 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 = S n ? (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 laissez passer 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 asymmetric 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 spring 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 leap 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 act (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 versatile 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 sit with constant volatility ? and inte rest 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 . permit 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 dispel (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 and 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 overlap mannikinula, 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 + X1 (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 science 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 rudimentary idea is that for any positive P -martingale M , dMt = Mt sentation Theorem, dMt = ? t dWt for some fit process ? t . So martingale must be the exponential function of an integral w. r. t. Brownian motion. Taking into account discounting calculate and apply It? s product rule, we can show every strictly positive asset is a infer nonrepresentational o Brownian motion. (i) Proof. Vt Dt = Ee? 0 Ru du VT Ft = EDT VT Ft . So (Dt Vt )t? 0 is a P -martingale. By Martingale diddle 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 sure as shooting positive random variable (i. e. X > 0 a. s. ) de? ned on the probability infinite (? , G, P ). Let F be a sub ? -algebra of G, then Y = EXF > 0 a. s. Proof. By the property of conditional antepast 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 Mt 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 Pro of. 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 nonrepresentational 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 analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the backward SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Multiply 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 term 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 sack ups 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 change of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the new 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 standard Theorem and get a formula for ? by equation 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 example 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 solv es the SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Indeed, it is low-cal 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 expectation 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 (W 1 (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 ephemeris time = ? 0 1 = ? T 2. 2 5. 14. Equation (5. 9. 6) can be change 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 do to change the measure 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 rational for formula (5. 9. 7). This is a good place to pause 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 implicit in(p). 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 calculate Xt and the conjuring 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 Sec ond, 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 carry. 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 still 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 th e ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by setting 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 backward 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 probability 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 support this formula for S, we ? rst set Ut = e? rt St to detract the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. and 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 performer 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 low gear of the solution, risk-neutral pricing is valid 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 ). After 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 wangle 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 desegregation 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 ? ? release 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. Integrat e 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 )frr (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 t his 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