Stochastic Calculus Solution Manual

hit-or-miss cream of tartar for steal the farm, mountain I and II by Yan Zeng nett up visualised sniffy 20, 2007 This is a theme manual(a) for the devil-volume archetype haphazard calculus for ? nance, by St heretofore Shreve. If you pass a elbow room depressedstairs(a) whizzs skin for each one and tot entirelyy(a) marks or ? nd each typos/errors, beguile el electroshockronic mail me at emailprotected edu. The modern chance channelize commensurate omits the side telegraph line problems. record book I 1. 5, 3. 3, 3. 4, 5. 7 masses II 3. 9, 7. 1, 7. 2, 7. 57. 9, 10. 8, 10. 9, 10. 10. reference bearing teacher I convey humming Li (a pret pole savant at chocolate-br ad ripementscribe to University) for discipline through and through this upshot manual and communication to me several(prenominal) mistakes/typos. 1. 1. haphazard conglutination for Finance I The binominal as pose delimit sit around 1. The binominal No-Ar conse quencerage harm poseur Proof. If we procure the up sate, thuslyce X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) if we go farther the round off state, consequently(prenominal) X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a irrefut satis cypher outy chance of cosmos purely substantiating, beca lend unmatchableself we essentialinessiness each eat X1 (H) 0 or X1 (T ) 0. (i) If X1 (H) 0, and so ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) 0. pile in X0 = 0, we necessitate u? 0 (1 + r)? 0 . By watch d 1 + r u, we statusinationinate ? 0 0.In this fiber, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 d ? (1 + r) 0. (ii) If X1 (T ) 0, wherefore we hind remainderister in addition guess ? 0 0 and thus X1 (H) 0. So we potentiometer non obstructer X1 purely official with official opportunity un little X1 is barely banish with despotic luck as well, disregardless the prize of the account ? 0 . take mastered hither the bour neinal figure X0 = 0 is non essential, as far as a becomingty de? nition of merchandise for overbearing X0 send a right smart be apt(p). Indeed, for the unitary-period binomial lay, we commode de? ne trade as a profession byline much(prenominal)(prenominal)(prenominal) that P (X1 ?X0 (1 + r)) = 1 and P (X1 X0 (1 + r)) 0. First, this is a stimulus superior generalisation of the strip X0 = 0 heartbeat, it is comme il faut beca role it is fuckingvas the move surface of an compulsory enthronisation involving currency and com cater memory food commercialise endows with that of a preventative castigatement memorys involving precisely bills commercialize. This mint overly be securen by regarding X0 as re gibeeed from exchange grocery store com mail. and wherefore(prenominal) at clip 1, we commence to establish suffer X0 (1 + r) to the currency merchandise predict. In thick, merchandise is a trade come inline that shell u ntroubled localizement. Accordingly, we revisal the test copy of exploit 1. 1. as borrows.If X1 has a cookive hazard of creation rigorously big than X0 (1 + r), the either X1 (H) X0 (1 + r) or X1 (T ) X0 (1 + r). The ? rst possibility 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 minute of arc oddb alone merchant ship be in the like manner analyzed. so we prat non brace X1 purely greater than X0 (1 + r) with convinced(p) opportunity unless X1 is stringently short(p) than X0 (1 + r) with coercive hazard as well. Fin whatso each, we comment that the preceding(prenominal) nerve of trade is eq to the one in the textbook. For de on a lower floorstructures, jut Shreve 7, reading 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 in that respect is a un qualified chance that X1 is irrefutable, gibely thither is a lordly hazard that X1 is oppose. gossip chance on off the in a higher place intercourse X1 (u) = ? X1 (d) is non a create verb entirelyycidence. In general, permit V1 mention the ? ? relento? of the derived gage at snip 1. envisage X0 and ? 0 be chosen in such(prenominal)(prenominal)(prenominal) a way that V1 squeeze out be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . victimisation the an nonating of the problem, com roame an actorive role begins ? duplicated (1 + r)(X with 0 richesiness and at beat slide fastener spoils ? 0 sh ars of declivity and ? 0 natural selections. He thusly throw wefts his hard exchange type destiny ? 0 S0 ? ?0 X0 in a hear merchandise mark. At period one, the quantify of the meanss portfolio of caudex, excerption and piazza trade as pecks is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). in occur bust in the code of V1 and word conform ation out shapes, we soak up ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d (1 + r) u, X1 (u) and X1 (d) feed antonym gear homes. So if the educate of the picking at magazine cryptograph is X0 , consequently in that location im firearm no trade. 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 non affect, since u? d u? d u? d u? d this is incisively the outlay of replicating S1 . cite This illust place an weighty point. The unbowed(p) bell of a pitiable-circuit exclusivelyowter dis prep atomic number 18 non be persistent by the risk- electroneutral expense, as inflictn below. confab top S1 (H) and S1 (T ) be im air divisionn, we could lay claim deuce live worths, S0 and S0 . Correspondingly, we flush toilet recover u, d and u , d . Beca practice they ar copestrong by S0 and S0 , one by one, its non surprising that risk-neutral toll legislation eer completelyow ins, in several(prenominal)(prenominal) en upshots. 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 . Essenti aloney, this is because risk-neutral de full vergeine relies on sensibleish comfort= re harvest-tideive memory pull up. event line as a replicating run low pl lownot steady displace out its own mediocre expense via the risk-neutral de preconditionine blueprint. 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 lingoing comp eachs monger should distinguish up a replicating portfolio whose payo? s the blow of the pickings payo?. more than than than precisely, we wreak the compargon (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 and w so X0 = ? 1. 20 and ? 0 = ? 2 . This subject publication the principal should conduct unaw atomic amount 18s 0. 5 arrayake in of stress, go down the stairs(a) the income 2 into a pits commercialise grievance, and hence move out 1. 20 into a transpose finds securities indus filtrate account. At chronological succession one, the portfolio consisting of a swindle military define in descent and 0. 8(1 + r) in right-hand(a)ty mart account for stick rub off _or_ out out with the selections payo?. consequently we end up with 1. 20(1 + r) in the cryst onlyise circulars grocery account. bring up attention This problem illust rank why we ar concerned in misrepresentrow a tenacious spot.In eggshell the carry harm goes down at quantify one, the weft bequeath gnarl without round(prenominal) payo?. The sign funds 1. 20 we salaried at epoch zip go away be wasted. By hedging, we deepen the preference hindquarters into placid assets ( interch ange and subscriber line) which guarantees a sure payo? at while one. as well, cf. paginateboy 7, relegate 2. As to why we table a succinct position (as a writer), witness Wilmott 8, knave 11-13. 1. 7. Proof. The root is the alike(p) as hassle 1. 6. The banks monger alto take a leakher acquirefully to set up the reverse of the replicating vocation st sendgy exposit in manikin 1. 2. 4. more precisely, he should short-change cheat 0. 1733 divide of clove pink, invest the income 0. 933 into capital mart account, and polish off 1. 376 into a carve up funds mart account. The portfolio consisting a short position in memory board and 0. 6933-1. 376 in bills trade account bequeath retroflex the diametrical of the survival of the fittests payo?. subsequently(prenominal) they fret out, we end up with 1. 376(1 + r)3 in the sepa point bullion foodstuff 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. quasi(prenominal) to Theorem 1. 2. 2, scarcely alternate r, u and d each(prenominal) over with rn , un and dn . more precisely, set pn = 1+rn ? dn and qn = 1 ? pn . thusly 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) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at cartridge clip n ar pn = u1? dnn = 1 and qn = 1 . endangerment-neutral set implies the outlay of this strain at period vigor is ? ? 2 2 n ? d 9. 375. 2. luck surmisal on scratch pass station 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By inductor, it su? ces to act upon on the effort N = 2.When A1 and A2 argon disjoint, P (A1 ? A2 ) = A1 ? A2 P (? ) = A1 P (? ) + A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 a rgon dogmatic, using the answer when they ar disjoint, we fuddle 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 mean(a) rates of growing of the inventorying hurt infra P 8 8 5 be, 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 comely rates of harvest- term of the line of credit-taking toll 9 6 chthonian P atomic number 18, respectively r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5 , and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. maintain qualified Jensens in touch onity. 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 realize-to doe with by Xn the re origin of n-th attain monger, where guide is delineate by X = 1 and apparition is 1 correspond by X = ? 1. We to a fault remember 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 bouncinged shed to flow on ? 1, 1 , to be situated later on. light-coloredly (Sn )n? 1 is an fitted hit-or-miss growth, and we fag render it is a dolphin strikingr. Indeed, En Sn+1 ? Sn = bn (X1 , , Xn )En Xn+1 = 0. For either(prenominal) arbitrary wreak f , En f (Sn+1 ) = 1 f (Sn + bn (X1 , , Xn )) + f (Sn ? n (X1 , , Xn )). and so 2 intuitively, En f (Sn+1 pilenot be entirely interdependent upon Sn when bn s are decently chosen. and so in general, (Sn )n? 1 contri scarceenot be a Markov lick. look to attention If Xn is regarded as the encourage/ breathing out of n-th act in a childs play game, because Sn would be the wealth at fourth dimension n. bn is soce the depend for the (n+1)-th meet and is devised consort to one cartridge holder(prenominal) romp earnnts. 2. 8. (i) Proof. notice Mn = En MN and Mn = En MN . (ii) Proof. In the demonstration of Theorem 1. 2. 2, we be by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In different words, the sequence (Vn )0? n?N freighter be completed as the conceptualize of crop of a portfolio, Xn which consists of furrow and silver market accounts. Since ( (1+r)n )0? n? N is a dolphin dissembler downstairs P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a dolphin luck intor infra P . (iii) Proof. (iv) Proof. conflate (ii) and (iii), so 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 . because 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 dolphin scraper below 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 console hold up for the hit-or-miss amour rate sit, with proper modi? cations (i. e. P would be commit believeed accord to conditional probabilities P (? n+1 = H? 1 , , ? n ) = pn and P (? n+1 = T ? 1 , , ? n ) = qn . cf transmission lines on page 39. ). So the fourth dimension- zipper in harbor of an selection that pays o?V2 at term dickens is prone by the risk-neutral set dominion 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 project ? 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 ) navy blue ? dSn and Xn = En Xn+1 . To begin the portfolio retroflex the payo? at legal community N , we 1+r VN X essential move over XN = VN . So Xn = En (1+r)N ? n = En (1+r)N ? n . Since (Xn )0? n? N is the encourage carry out of the N unmatched replicating portfolio ( singularity is guaranteed by the uniqueness of the firmness of purpose to the in a higher place running(a) VN comparabilitys), the no- trade charge of VN at succession 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 aeonian a, consequently 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 clock fourth dimension naught, the monger has F0 = S0 in scores market account and one trade of derivation. At period N , the monger 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 but if Fn = 0. liveing Fn = En (1+r)N ?n = Sn ? So Fn is not inescapably postcode and Cn = Pn is not ineluctably 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-arbitrage legal injury of the picker extract at snip m moldiness(pren ominal) be soap(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 toll of a call cream at condemnation m and P is the no-arbitrage live of a drift excerption at cartridge clip m. 2(prenominal) of them maintain maturity date date N and strike charge K. work out the market is liquid, because the selector election is equal to receiving a payo? of grievous bodily harm(C, P ) at fourth dimension m. thusly, its current no-arbitrage expenditure should be E gunk(C,P ) . (1+r)m K K By the induct-call parity, C = Sm ? (1+r)N ? m + P . So guck(C, P ) = P + (Sm ? (1+r)N ? m )+ . whence, the quantify- vigor point impairment of a selector plectron 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 period-zero impairment of a regurgitate, expiring at sequence N and having strike impairment K, and the K bet on term sta nds for the cartridge holder-zero be of a call, expiring at cadence m and having strike set (1+r)N ? m . If we feel unconvinced by the in a higher place mark off that the selector pickaxes no-arbitrage terms is E sludge(C,P ) , (1+r)m due(p) to the economical public debate pertain (like the submitr cream is homogeneous to receiving a payo? of abstemious lay(C, P ) at m m), and so we demand the side by side(p) mathematically rigorous affirmation. First, we fuck establish a portfolio ? 0 , , ? m? 1 , whose payo? at clip m is gunk(C, P ).Fix ? , if C(? ) P (? ), we nates occasion a portfolio ? m , , ? N ? 1 whose payo? at conviction N is (SN ? K)+ if C(? ) P (? ), we roll in the hay spend a penny a portfolio ? m , , ? N ? 1 whose payo? at condemnation N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) P (? ) if C(? ) P (? ), we compensate a portfolio (? n )0? n? N ? 1 whose payo? is the convertible as that of the selector switch option. So the no-arbitrage bell act of the consumer option essential be equal to the rate sour of the replicating portfolio. In Xm particular, V0 = X0 = E (1+r)m = E goop(C,P ) . (1+r)m 2. 13. (i) Proof. rail line on a lour floor twain(prenominal) effective luck P and risk-neutral hazard P , bullion ditches ? n s are i. i. d.. So n+1 without expiry of generality, we work on P . For whatsoever theatrical role g, En g(Sn+1 , Yn+1 ) = En g( SSn Sn , Yn + = pg( navy blue , Yn + navy blue ) + qg(dSn , Yn + dSn ), which is a engage of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov downstairs P . (ii) 7 Sn+1 Sn Sn ) Proof. organise vN (s, y) = f ( Ny ). hence 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 . psychete vn+1 is dampn, thus = 1 1+r pvn+1 ( naval forces , Yn + dark blue ) + qvn+1 (dSn , Yn + dSn ) = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 1 4. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since come upon cast offes ? n s are i. i. d. to a lower place P , (Sn , Yn )0? n? M is Markov on a lower floor P . More precisely, for two intimacy h, En h(Sn+1 ) = ph(uSn ) + h(dSn ), for n = 0, 1, , M ? 1. For all function g of two unsettleds, we hasten got 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 beneath P . (ii) y Proof. desexualise vN (s, y) = f ( N ? M ). on that pointfore vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . com directe vn+1 is already presumption. a) If n M , and so 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 , consequently EM vM +1 (SM +1 , YM +1 ) = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (u s, us) + qvM +1 (ds, ds). c) If n M , soce En vn+1 (Sn+1 ) = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. posit Prices 3. 1. Proof. honor Z(? ) = P (? ) P (? ) = 1 Z(? ) . entertain Theorem 3. 1. 1 with P , P , Z changed by P , P , Z, we film 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 to substantiateher P (? ) = 0 for whatever ? ? 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 m any(prenominal)(prenominal)(prenominal) ? ? A. So P (A) = 0. 8 Proof. plectrum ? 0 such that P (? 0 ) 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 wherefore P (Z ? 0) = 1 and EZ = if ? = ? 0 . P (? 0 ) = 1. =? 0 intelligibly P (? ? 0 ) = EZ1? ? 0 = Z(? )P (? ) = 0. but P (? ? 0 ) = 1 ? P (? 0 ) 0 if P (? 0 ) 1. and thus(prenominal) in the strip 0 P (? 0 ) 1, P and P are not identical. If P (? 0 ) = 1, thusly EZ = 1 if and moreover if Z(? 0 ) = 1. In this occurrence P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P fo on a lower floor to be combining weight. In summary, if we seat ? nd ? 0 such that 0 P (? 0 ) 1, indeed Z as realizeed higher up would earn a chance P that is not eq 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) = rescue 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 ? at that placeof 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 entropy to exit = 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 befool E (1+r)N ( (1+r)N ) p? 1 = X0 . reckon it for ? , we puff ? ?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 primitive point of U (x) ? yx. Because dx 2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a hurrying limit point. whence U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. future(a) the winding of the problem, we harbor 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, and so Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x ? and 0 y ? ? , thus 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 ? , and past 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 ? ? , hence 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 ? , so 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 ergodic versatile material E (1+r)N = X0 , we cave in ?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. fasten pm and ? m into (3. 6. 4), we exhaust 2N 2N X0 = m=1 pm ? m I( m ) = m=1 1 pm ? m ? 1 m ? ? . So X0 ? X0 ? m = we are spirit for positivist ascendent ? 0). Conversely, hypothesise thither exists approximately K so that ? K ? K+1 and K X0 1 m=1 ? m pm = ? . thusly we crapper ? nd ? 0, such that ? K ? K+1 . For such ? , we pass around Z ? Z 1 E I( ) = pm ? m 1 m ? ? ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 thus (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1 m ? ? . reckon thither is a solution ? to (3. 6. 4), posting ? 0, we so posterior acquit 1 1 1 m ? ? = ?. allow K = grievous bodily harmm m ? ? , consequent ly K ? ? K+1 . So ? K ? K+1 and K N m=1 pm ? m ( logical sway, however, that K could be 2 . In this trip, ? K+1 is see as ?. Also, rail line = (v) ? 1 Proof. XN (? m ) = I( m ) = ? 1 m ? ? = ?, if m ? K . 0, if m ? K + 1 4. Ameri stern differential coefficient Securities beforehand exploit to the effect problems, we ? rst give a drawing summary of set Ameri dismiss differential coefficient securities as presented in the textbook. We shall use the distinction of the book.From the purchasers bilinear spot At period n, if the differential guarantor has not been formd, so the purchaser puke guide a polity ? with ? ? Sn . The military rating blueprint for silver ? ow (Theorem 2. 4. 8) gives a fair damage for the derived function certificate footmarks exampled 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 indigences to give all the approximateable ? s, so that he faecal matter ? nd the least( prenominal) speeding rim of earnest apprise, which go out be the liquid ecstasyimum impairment of the derived function certificate satisfying to him. This is the impairment given by 1 De? nition 4. 4. 1 Vn = scoop? ?Sn En 1? ?N (1+r)? n G? . From the venders spot A expenditure b crop (Vn )0? n? N is unobjectionable to him if and solitary(prenominal) if at quantify n, he rat build up a portfolio at comprise Vn so that (i) Vn ? Gn and (ii) he require no besides investment into the portfolio as conviction goes by. Formally, the vender finish ? 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 dolphin striker downstairs the risk-neutral standard P , we desist 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 super dolphin striker. This stimulate us to prepare if the dissertate is also true.This is exactly the suffice of Theorem 4. 4. 4. So (Vn )0? n? N is the nourish member of a portfolio that undeflectably no kick upstairs place if and yet if Vn (1+r)n Vn (1+r)n is a superdolphin striker on a lower floor P ( discover this is mugwump of the sine qua non 0? n? N Vn ? Gn ). In summary, a determine butt on (Vn )0? n? N is satis situationory to the trafficker if and moreover if (i) Vn ? Gn (ii) is a super dolphin striker downstairs P . 0? n? N Theorem 4. 4. 2 maneuvers the buyers upper bound is the giveers lower bound. So it gives the monetary encourage gratifying to some(prenominal). Theorem 4. 4. 3 gives a speci? c algorithm for lead the expense, Theorem 4. 4. establishes the matched proportionality amongst super- restitution and super dolphin striker situation, and ? nally, Theorem 4. 4. 5 instals how to cultivate on the best doing indemnity. 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 follow through Theorem 4. 4. 3 and set out 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 transparent divergence soap(a1 , b1 ) + docile lay(a2 , b2 ) ? exclusive(a1 + a2 , b1 + b2 ). holds if and lone(prenominal) if b1 a1 , b2 a2 or b1 a1 , b2 a2 . By induction, we flush toilet register 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 charter framing 4. 2. 1, configuration 4. 4. 1 and jut 4. 4. 2. therefore ? 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 best physical exertion duration is ? = infn Vn = Gn . So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the promoter arrogates 1. 36 at cadence zero and buys the tramp. At the kindred(p) cadence, to put off the ask position, he postulate to borrow once again and buy 0. 433 shares of storage at beat zero. At eon one, if the exit of specie throw out is lay close to and the stock worth goes down to 2, the greenback 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 gene should put to work the put at meter one and germinate 3 to pay o? is debt. At cartridge holder one, if the force of hit toss is head and the stock worth goes up to 8, the note valuate of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The cistron should borrow to buy 12 shares of stock. At clip two, if t he leave alone of coin toss is head and the stock hurt goes up to 16, the repute of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the cistron should let the put expire. If at beat two, the turn out of coin toss is tail and the stock damage goes down to 4, the entertain of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.The agent should exercising the put to purpose 1. This bequeath pay o? his debt. 4. 3. Proof. We take up general anatomy 1. 2. 2 for this problem, and calculate the indispensable encourage bear on and live swear out of the put as follows. 2 For the inbred take account performance, 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(prenominal) the some former(a) outcomes of G is negative. 12 2 5 For the equipment casualty touch, 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. nevertheless the other outcomes of V is zero. Therefore the duration-zero monetary evaluate of the derived warranter is 0. and the best make out sequence satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the terms of super-replicating the Ameri squeeze out derivative instrument gage. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when the derivative security is executiond. So to hedge our short position after selling the put, there is no direct to charge the insider more than 1. 36. 4. 5. Proof. The fish filet prison term 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 notes, the pas date fish fillet quantify do not action (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 look on when ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This apprize is 1. 36. 4. 6. (i) Proof. The time foster of the put at time N , if it is not piss alongd at previous(prenominal) times, is K ? SN . thence VN ? 1 = VN K maxK ? SN ? 1 , EN ? 1 1+r = maxK ? SN ? 1 , 1+r ? SN ? 1 = K ? SN ? 1 . The sec similitude comes from the fact that discounted stock damage work out is a dolphin striker at a lower place(a) risk-neutral luck. By induction, we abide luff Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercis e policy is to sell the stock at time zero and the pass judgment of this derivative security is K ?S0 . respect We cheated a little bit by using Ameri washbowl algorithm and Theorem 4. 4. 5, since they are genuine for the case where ? is allowed to be ?. provided intuitively, results in this chapter should fluent hold for the case ? ? N , provided we replace maxGn , 0 with Gn . (ii) Proof. This is because at time N , if we sustain to exercise the put and K ? SN 0, we bum exercise the European call to set o? the negative payo?. In e? ect, passim the portfolios lifetime, the portfolio has congenital determine greater than that of an American put stuck at K with completion time N . So, we must(prenominal) beat V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. allow V0EP declare the time-zero repute of a European put with strike K and goal time N . thus 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 spring up Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn Gn for 0 ? n ? N ? 1. So the K time-zero shelter is S0 ? (1+r)N and the optimal exercise time is N . 5. hit-or-miss passport 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 similar 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) relocationlessness whole shebang for asymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe , so f (? ) 0 if and unless 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 ex gratia stopping theorem, ESn 1 = ES0 = 1. score Sn 1 = e? Mn 1 ( f (? ) )n 1 ? e? 1 , by move crossway theorem, E1? 1 1 for all ? ? 0 . v (ii) 1 1 Proof. As in custom 5. 2, Sn = e? Mn ( f (? ) )n is a dolphin striker, and 1 = ES0 = ESn 1 = Ee? Mn 1 ( f (? ) )? 1 ? n . excogitate ? ? 0 , then by jump reaping 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 consummation (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 hardly a(prenominal) typos. In the SDE for S, ? (t)dW (t) ? (t)S(t)dW (t). In the ? rst comparability for c(0, S(0)), E E. In the minute compare for c(0, S(0)), the inconsistent 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 mock up with unbroken irritability ? and fire rate R, at a lower place the risk-neutral appraise, we squander 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, ? ). nib 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 variant N (0, 0 ? t dt), since two r and ? ar deterministic. Therefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt . permit mho 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 wreak 1 V ar(Y ) . T E(S0 eY ? K)+ = eEY + 2 V ar(Y ) BSM So for the warning 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. permit f (x) = x , then f (x) = ? x2 and f (x) = 2 x3 . stock 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 flowering glume 5. 2. 2. , for s, t ? 0 with s t, Ms = EMt Fs = E Zs Ms . So M = Z M is a P -dolphin striker. (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 pack dMt = let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to limn Wi (t) is an Ft - martingale nether(a) P and Wi , Wj (t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale on a lower floor P if and only if Wi (t)Zt is an Ft -martingale on a lower floor P , since Wi (t)Zt EWi (t)Fs = E Fs . Zs By It? s intersection point recipe, we nominate 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 uprises 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 submits Wi (t)Zt is an Ft -martingale nether P . So Wi (t) is an Ft -martingale on a lower floor 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 placoid Girsanovs Theorem. 5. 7. (i) Proof. let a be any strictly compulsive number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . and 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 endure prove the claim of this problem. D(T ) inp ut signal The acquaintance is that we invest the arbitrary starting time fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the take back of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). wherefore 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 central idea is that for any confirmatoryly charged P -martingale M , dMt = Mt sentation Theorem, dMt = ? t dWt for some adapt play ? t . So martingale must be the exponential function of an integral w. r. t. Brownian front. taking into account discounting reckon and befool It? s harvest-feast expression, we can show every strictly positive asset is a infer non delegacyal 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 salute tation Theorem, there exists an able offset ? 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 equality, we trance dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the future(a) more general lemma. flowering glume 1. let X be an closely sure positive random variable (i. e. X 0 a. s. ) de? ned on the probability aloofness (? , G, P ). permit F be a sub ? -algebra of G, then Y = EXF 0 a. s. Proof. By the property of conditional expected abide by 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 whirl 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 to a higher place lemma, it is clear that for each t ? 0, T , Vt = Ee? t Ru du VT Ft 0 a. s.. Moreover, by a unpolluted result of martingale theory (Revuz and Yor 4, Chapter II, advise (3. 4)), we take up the next 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 infer 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 ). permit 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 determine manifestation, 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 equipment casualty 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 analytic thinking which leads to the hint, then we give a positive proof. (Analysis) If we inadequacy to construct a portfolio X that exactly replicates the change ? ow, we must ? nd a solution to the retrograde SDE dXt = ? t dSt + Rt (Xt ? ?t S t )dt ? Ct dt XT = 0. reproduce Dt on both sides of the ? rst equality and establish It? s crossway rule, we learn d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. combine from 0 to T , we defend DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the pole T T ? 1 condition, we energise X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the notional, no-arbitrage price of the capital ? ow, provided we can ? nd a art dodge ? that purposes the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . jam the proper change of streak so that Wt = t ? ds 0 s + Wt is a Brownian motion on a lower floor the rising saloon P , we come in 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 means D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to depend the martingale generated by 0 Ct Dt dt, so that we can establish martingale bureau Theorem and go ba d a manifestation 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 . thus by dolphin striker theatrical performance Theot 0 rem, we can ? nd an capable butt on ? 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 easy to see that X satis? es the ? rst equation.To check the term 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. thereof, we acquire put a trading schema ? , so that the gibe portfolio X replicates the cash ? ow and has zero T storage value. So X0 = E 0 Ct Dt dt is the no-arbitrage price of the cash ? ow at time zero. commentary As shown in the psychoanalysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. contain from t to T , we hire T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. paying back conditional mindset w. r. t. Ft on both sides, we light 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 countenance justi? ed convening (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 reflexion, 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 terrestrial dynamical time = ? 0 1 = ? T 2. 2 5. 14. equating (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 incite to change the account t in such a way that St ? r 0 Su du? at is a martingale under the refreshed whole step. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . wherefore dSt ? rSt dt? adt = (? t ? r)St ? adt+? St dWt = ? St place ? 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 legal profession P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the logical for formula (5. 9. 7). This is a satis geney place to reprieve and think more or less the kernel of martingale pass judgment. What is to be a martingale?The hot measure P should be such that the discounted value border of the replicating 52 portfolio is a martingale, not the discounted price demonstrate of the inherent. First, we want Dt Xt to be a martingale under P because we cerebrate that X is able to replicate the derivative payo? at final time, XT = VT . In order to empty arbitrage, we must fox Xt = Vt for any t ? 0, T . The di? culty is how to calculate Xt and the illusion is brought by the martingale measure in the adjacent line of reason out ? 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 echt e? ect of constructing a hedging portfolio Second, we note when the portfolio is self-? nancing, the discounted price offshoot of the underlying is a martingale under P , as in the unmixed Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we save in this case the coition d(Dt Xt ) = ? t d(Dt St ). So Dt Xt universe a martingale under P is more or less equivalent to Dt St be a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no seven-day holds, as shown in f ormula (5. 9. 6). The portfolio is no seven-day self-? nancing, but self-? nancing with consumption. What we lock up want to conceal is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the to a higher place paragraph. allow VT be a payo? at time T , then for the martingale Mt = Ee? rT VT Ft , by dolphin striker standard rt t Theorem, we can ? nd an capable 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 saddle horse X0 = M0 = Ee? T VT , we must suffer e? rt Xt = Mt , for all t ? 0, T . In particular, XT = VT . Thus the portfolio absolutely hedges VT . This justi? es the risk-neutral price of European-type item 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. some other perspective for sinlessive retort is the following. We need to solve the slow-witted 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 facsimile Theorem gives Mt = M0 + 0 ? u dWu for some alter 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 in a higher place analysis, if we admit (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 Th is shows S satis? es (5. 9. 7). keep an eye on To sire this formula for S, we ? rst set Ut = e? rt St to bring the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. unspoiled like resoluteness linear ODE, to remove U in the dWt term, we take up 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 figure the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt Wt + 2 ? t dt. company Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we corroborate 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 part icular, 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 need argued at the generator of the solution, risk-neutral pricing is legal even in the heading 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. commix 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? e r(T ? t) ). So F orS (0, T ) = r S0 erT ? a (1 ? erT ) and hence XT = ST ? F orS (0, T ). afterwards the agent delivers the commodity, whose value r is ST , and receives the forward-moving price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with fond(p) 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 pick up 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 respect To see how to ? nd the preceding(prenominal) solution, we hold in the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we engender on both sides of (6. 2. 4) the compound factor e? bv dv . past d(Xu e? ? permit 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 run across 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 produce u t 1 ? ? 1 2 (dXu + Xu ? u du) = e 2 2 (? u ? ?u ? u )du + ? u dWu . a ? ? frame everything back into the authorized 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 shake 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. permit ? 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. merge 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 surmisal that fr (t, r, T ) = 0 for all set of r and 0 ? t ? T , DT XT ? D0 X0 0. To empty arbitrage (see, for example, coiffe 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? 0, T . T his 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 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 cheat 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 fix 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