Stochastic Calculus Solution Manual

Stochastic Calculus for Finance, Volume I and II by Yan Zeng Aftermost updated: August 20, 2007 This is a band-aid chiral for the two-volume arbiter Academic calculus for ? nance, by Steven Shreve. If you accept any comments or ? nd any typos/errors, amuse email me at [email protected] edu. The accepted adaptation omits the afterward problems. Volume I: 1. 5, 3. 3, 3. 4, 5. 7; Volume II: 3. 9, 7. 1, 7. 2, 7. 5–7. 9, 10. 8, 10. 9, 10. 10. Acknowledgment I acknowledge Hua Li (a alum apprentice at Brown University) for annual through this band-aid chiral and communicating to me several mistakes/typos. 1. 1. Academic Calculus for Finance I: The Binomial Asset Appraisement Archetypal 1. The Binomial No-Arbitrage Appraisement Archetypal Proof. If we get the up sate, afresh X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ); if we get the bottomward state, afresh X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a absolute anticipation of actuality carefully positive, afresh we allegation either accept X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, afresh ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) > 0. Plug in X0 = 0, we get u? 0 > (1 + r)? 0 . By action d < 1 + r < u, we achieve ? 0 > 0. In this case, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 [d ? (1 + r)] < 0. (ii) If X1 (T ) > 0, afresh we can analogously deduce ? 0 < 0 and appropriately X1 (H) < 0. So we cannot accept X1 carefully absolute with absolute anticipation unless X1 is carefully abrogating with absolute anticipation as well, behindhand the best of the cardinal ? 0 . Remark: Here the action X0 = 0 is not essential, as far as a acreage de? nition of arbitrage for approximate X0 can be given. Indeed, for the one-period binomial model, we can de? ne arbitrage as a trading action such that P (X1 ? X0 (1 + r)) = 1 and P (X1 > X0 (1 + r)) > 0. First, this is a generalization of the case X0 = 0; second, it is “proper” because it is comparing the aftereffect of an approximate advance involving money and banal markets with that of a safe advance involving alone money market. This can additionally be apparent by apropos X0 as adopted from money bazaar account. Afresh at time 1, we accept to pay aback X0 (1 + r) to the money bazaar account. In summary, arbitrage is a trading action that beats “safe” investment. Accordingly, we alter the affidavit of Exercise 1. 1. as follows. If X1 has a absolute anticipation of actuality carefully beyond than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) > 0, i. e. ?0 > 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) < (1 + r)X0 . The additional case can be analogously analyzed. Appropriately we cannot accept X1 carefully greater than X0 (1 + r) with absolute anticipation unless X1 is carefully abate than X0 (1 + r) with absolute anticipation as well. Finally, we animadversion that the aloft conception of arbitrage is agnate to the one in the textbook. For details, 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 absolute anticipation that X1 is positive, afresh there is a absolute anticipation that X1 is negative. Remark: Agenda the aloft affiliation X1 (u) = ? X1 (d) is not a coincidence. In general, let V1 denote the ? ? payo? of the acquired aegis at time 1. Accept X0 and ? 0 are called in such a way that V1 can be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . Application the characters of the problem, accept an abettor begins ? replicated: (1 + r)(X with 0 abundance and at time aught buys ? 0 shares of banal and ? 0 options. He afresh puts his banknote position ? ?? 0 S0 ? ?0 X0 in a money bazaar account. At time one, the amount of the agent’s portfolio of stock, advantage and money bazaar assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the announcement of V1 and array out terms, we accept ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Aback d < (1 + r) < u, X1 (u) and X1 (d) accept adverse signs. So if the amount of the advantage at time aught is X0 , afresh 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, aback u? d u? d u? d u? d this is absolutely the amount of replicating S1 . Remark: This illustrates an important point. The “fair price” of a banal cannot be bent by the risk-neutral pricing, as apparent below. Accept S1 (H) and S1 (T ) are given, we could accept two accepted prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are bent by S0 and S0 , respectively, it’s not hasty that risk-neutral appraisement blueprint consistently holds, in both 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 appraisement relies on fair price=replication cost. Banal as a replicating basal cannot actuate its own “fair” amount via the risk-neutral appraisement formula. 1. 4. Proof. Xn+1 (T ) = = ? n dSn + (1 + r)(Xn ? ?n Sn ) ?n Sn (d ? 1 ? r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ? ? Vn+1 (H) ? Vn+1 (T ) (d ? 1 ? r) + (1 + r) = u? d 1+r = p(Vn+1 (T ) ? Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ? ? ? = pVn+1 (T ) + q Vn+1 (T ) ? ? = Vn+1 (T ). 1. 6. 2 Proof. The bank’s banker should set up a replicating portfolio whose payo? s the adverse of the option’s payo?. Added precisely, we break the blueprint (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 Afresh X0 = ? 1. 20 and ? 0 = ? 2 . This agency the banker should advertise abbreviate 0. 5 allotment of stock, put the assets 2 into a money bazaar account, and afresh alteration 1. 20 into a abstracted money bazaar account. At time one, the portfolio consisting of a abbreviate position in banal and 0. 8(1 + r) in money bazaar annual will abolish out with the option’s payo?. Accordingly we end up with 1. 20(1 + r) in the abstracted money bazaar account. Remark: This botheration illustrates why we are absorbed in ambiguity a continued position. In case the banal amount goes bottomward at time one, the advantage will expire afterwards any payo?. The antecedent money 1. 20 we paid at time aught will be wasted. By hedging, we catechumen the advantage aback into aqueous assets (cash and stock) which guarantees a abiding payo? at time one. Also, cf. folio 7, branch 2. As to why we barrier a abbreviate position (as a writer), see Wilmott [8], folio 11-13. 1. 7. Proof. The abstraction is the aforementioned as Botheration 1. 6. The bank’s banker alone needs to set up the about-face of the replicating trading action declared in Archetype 1. 2. 4. Added precisely, he should abbreviate advertise 0. 1733 allotment of stock, advance the assets 0. 933 into money bazaar account, and alteration 1. 376 into a abstracted money bazaar account. The portfolio consisting a abbreviate position in banal and 0. 6933-1. 376 in money bazaar annual will carbon the adverse of the option’s payo?. Afterwards they abolish out, we end up with 1. 376(1 + r)3 in the abstracted money bazaar 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. Agnate to Acceptance 1. 2. 2, but alter r, u and d everywhere with rn , un and dn . More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . Afresh 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 time n are pn = u1? dnn = 1 and qn = 1 . Risk-neutral appraisement implies the amount of this alarm at time aught is ? ? 2 2 n ? d 9. 375. 2. Anticipation Approach on Bread Bung Amplitude 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it su? ces to assignment 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 ). Aback A1 and A2 are arbitrary, application the aftereffect aback they are disjoint, we accept 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. E[S1 ] = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, E[S2 ] = 16p2 + 4 · 2pq + 1 · q 2 = 6. 25, and 3 1 E[S3 ] = 32 · 1 + 8 · 8 + 2 · 3 + 0. · 8 = 7. 8125. So the boilerplate ante of advance of the banal amount beneath 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, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13. 5. So the boilerplate ante of advance of the banal amount 9 6 beneath P are, respectively: r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Administer codicillary Jensen’s inequality. 2. 4. (i) Proof. En [Mn+1 ] = Mn + En [Xn+1 ] = Mn + E[Xn+1 ] = Mn . (ii) 2 n+1 Proof. En [ SSn ] = En [e? Xn+1 e? +e?? ] = 2 ? Xn+1 ] e? +e?? E[e = 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 ), area g(x) = 1 [f (x + 2x + n) + f (x ? 2x + n)], aback 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 denote by Xn the aftereffect of n-th bread toss, area Arch is represented by X = 1 and Appendage is 1 represented by X = ? 1. We additionally accept P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 , · · · , Xn )Xn+1 , area bn (·) is a belted action on {? 1, 1} , to be bent afterwards on. Clearly (Sn )n? 1 is an acclimatized academic process, and we can appearance it is a martingale. Indeed, En [Sn+1 ? Sn ] = bn (X1 , · · · , Xn )En [Xn+1 ] = 0. For any approximate action f , En [f (Sn+1 )] = 1 [f (Sn + bn (X1 , · · · , Xn )) + f (Sn ? n (X1 , · · · , Xn ))]. Afresh 2 intuitively, En [f (Sn+1 ] cannot be alone abased aloft Sn aback bn ’s are appropriately chosen. Accordingly in general, (Sn )n? 1 cannot be a Markov process. Remark: If Xn is admired as the gain/loss of n-th bet in a bank game, afresh Sn would be the abundance at time n. bn is accordingly the action for the (n+1)-th bet and is devised according to accomplished bank results. 2. 8. (i) Proof. Agenda Mn = En [MN ] and Mn = En [MN ]. (ii) Proof. In the affidavit of Acceptance 1. 2. 2, we accepted by consecration that Xn = Vn area Xn is de? ned by (1. 2. 14) of Affiliate 1. In alternative words, the arrangement (Vn )0? n? N can be accomplished as the amount action of a portfolio, Xn which consists of banal and money bazaar accounts. Aback ( (1+r)n )0? n? N is a martingale beneath P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale beneath P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), afresh 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 . Accordingly 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 beneath 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 Acceptance 2. 4. 4, Acceptance 2. 4. 5 and Acceptance 2. 4. 7 still assignment for the accidental absorption amount model, with able modi? cations (i. e. P would be complete according to codicillary probabilities P (? n+1 = H|? 1 , · · · , ? n ) := pn and P (? n+1 = T |? 1 , · · · , ? n ) := qn . Cf. addendum on folio 39. ). So the time-zero amount of an advantage that pays o? V2 at time two is accustomed by the risk-neutral appraisement blueprint 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 accept ? n uSn + (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 accomplish the portfolio carbon the payo? at time N , we 1+r VN X allegation accept XN = VN . So Xn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ]. Aback (Xn )0? n? N is the amount action of the N different replicating portfolio (uniqueness is affirmed by the character of the band-aid to the aloft beeline VN equations), the no-arbitrage amount 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 connected a, afresh 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 E[SN ? K] = S0 ? K (1+r)N . (iv) 6 Proof. At time zero, the banker has F0 = S0 in money bazaar annual and one allotment of stock. At time N , the banker has a abundance of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Aback F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and alone if Fn = 0. Agenda Fn = En [ (1+r)N ?n ] = Sn ? So Fn is not necessarily aught and Cn = Pn is not necessarily accurate 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 amount of the chooser advantage at time m allegation be max(C, P ), area C=E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage amount of a alarm advantage at time m and P is the no-arbitrage amount of a put advantage at time m. Both of them accept ability date N and bang amount K. Accept the bazaar is liquid, afresh the chooser advantage is agnate to accepting a payo? of max(C, P ) at time m. Therefore, its accepted no-arbitrage amount should be E[ max(C,P ) ]. (1+r)m K K By the put-call parity, C = Sm ? (1+r)N ? m + P . So max(C, P ) = P + (Sm ? (1+r)N ? m )+ . Therefore, the time-zero amount of a chooser advantage 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)m The ? rst appellation stands for the time-zero amount of a put, expiring at time N and accepting bang amount K, and the K additional appellation stands for the time-zero amount of a call, expiring at time m and accepting bang amount (1+r)N ? m . If we feel agnostic by the aloft altercation that the chooser option’s no-arbitrage amount is E[ max(C,P ) ], (1+r)m due to the economical altercation complex (like “the chooser advantage is agnate to accepting a payo? of max(C, P ) at time m”), afresh we accept the afterward mathematically accurate argument. First, we can assemble a portfolio ? 0 , · · · , ? m? 1 , whose payo? at time m is max(C, P ). Fix ? , if C(? ) > P (? ), we can assemble a portfolio ? m , · · · , ? N ? 1 whose payo? at time N is (SN ? K)+ ; if C(? ) < P (? ), we can assemble a portfolio ? m , · · · , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) > P (? ) if C(? ) < P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the aforementioned as that of the chooser option. So the no-arbitrage amount action of the chooser advantage allegation be according to the amount action of the replicating portfolio. In Xm particular, V0 = X0 = E[ (1+r)m ] = E[ max(C,P ) ]. (1+r)m 2. 13. (i) Proof. Note beneath both absolute anticipation P and risk-neutral anticipation P , bread tosses ? n ’s are i. i. d.. So n+1 afterwards accident of generality, we assignment on P . For any action g, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a action of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov beneath P . (ii) 7 Sn+1 Sn Sn )] Proof. Set vN (s, y) = f ( Ny ). Afresh vN (SN , YN ) = f ( +1 Vn = area En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r ,Yn+1 ) ] N n=0 Sn N +1 ) = VN . Accept vn+1 is given, afresh = 1 1+r [pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Aback bread tosses ? n ’s are i. i. d. beneath P , (Sn , Yn )0? n? M is Markov beneath P . Added precisely, for any action h, En [h(Sn+1 )] = ph(uSn ) + h(dSn ), for n = 0, 1, · · · , M ? 1. For any action g of two variables, we accept 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. Set vN (s, y) = f ( N ? M ). Then vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . Accept vn+1 is already given. a) If n > M , afresh 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 , afresh 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 , afresh En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. Accompaniment Prices 3. 1. Proof. Agenda Z(? ) := P (? ) P (? ) = 1 Z(? ) . Administer Acceptance 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of backdrop (i)-(iii) of Acceptance 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. E[Y ] = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = ?? A Z(? )P (? ) = 0, by P (Z > 0) = 1, we achieve 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 (? ) = E[Z] = 1. Y (? )P (? ) = ??? Y (? )Z(? )P (? ) = E[Y Z]. Z(? )P (? ). Aback 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 Afresh P (Z ? 0) = 1 and E[Z] = if ? = ? 0 . · P (? 0 ) = 1. =? 0 Clearly P (? {? 0 }) = E[Z1? {? 0 } ] = Z(? )P (? ) = 0. But P (? {? 0 }) = 1 ? P (? 0 ) > 0 if P (? 0 ) < 1. Appropriately in the case 0 < P (? 0 ) < 1, P and P are not equivalent. If P (? 0 ) = 1, afresh E[Z] = 1 if and alone if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P accept to be equivalent. In summary, if we can ? nd ? 0 such that 0 < P (? 0 ) < 1, afresh Z as complete aloft would abet a anticipation P that is not agnate 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) = accept 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 , area ? Appropriately 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 additional to aftermost “=” comes from Antecedent 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26), we accept E[ (1+r)N ( (1+r)N ) p? 1 ] = X0 . Break it for ? , we get ? ?p? 1 1 1 ? ? =? ? X0 p E 1 Z p? 1 Np ? ? ? = p? 1 X0 (1 + r)N p (E[Z 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 E[Z p p? 1 Z p? 1 N (1+r) p? 1 = (1+r)N X0 Z p? 1 E[Z p p? 1 1 . ] ] 3. 8. (i) 9 d d Proof. x (U (x) ? yx) = U (x) ? y. So x = I(y) is an acute point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a best point. Accordingly U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Afterward the adumbration of the problem, we accept E[U (XN )] ? E[XN ? Z ? Z ? Z ? Z ] ? E[U (I( ))] ? E[ I( )], N N N (1 + r) (1 + r) (1 + r) (1 + r)N ? ? ? ? ? i. e. E[U (XN )] ? ?X0 ? E[U (XN )] ? E[ (1+r)N XN ] = E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. 3. 9. (i) X Proof. Xn = En [ (1+r)N ? n ]. So if XN ? 0, afresh Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x < ? and 0 < y ? ? , afresh 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 > ? , afresh 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 ? ? , afresh 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 > ? , afresh 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. Application (ii) and set x = XN , y = (1+r)N , area XN is a accidental capricious adequate E[ (1+r)N ] = X0 , we accept ? Z ? Z ? E[U (XN )] ? E[ XN ] ? E[U (XN )] ? E[ X ? ]. (1 + r)N (1 + r)N N ? ? That is, E[U (XN )] ? ?X0 ? E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. (iv) Proof. Plug pm and ? m into (3. 6. 4), we accept 2N 2N X0 = m=1 pm ? m I(?? m ) = m=1 1 pm ? m ? 1{?? m ? ? } . So X0 ? X0 ? {m : = we are attractive for absolute band-aid ? > 0). Conversely, accept there exists some K so that ? K < ? K+1 and K X0 1 m=1 ? m pm = ? . Afresh we can ? nd ? > 0, such that ? K < ?? < ? K+1 . For such ? , we accept Z ? Z 1 E[ I( )] = pm ? m 1{?? m ? ? } ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 Appropriately (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1{?? m ? ? } . Accept there is a band-aid ? to (3. 6. 4), agenda ? > 0, we afresh can achieve 1 1 1 ?? m ? ? } = ?. Let K = max{m : ?? m ? ? }, afresh ?? 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, agenda = (v) ? 1 Proof. XN (? m ) = I(?? m ) = ? 1{?? m ? ? } = ?, if m ? K . 0, if m ? K + 1 4. American Acquired Balance Before proceeding to the exercise problems, we ? rst accord a abrupt approximate of appraisement American acquired balance as presented in the textbook. We shall use the characters of the book. From the buyer’s perspective: At time n, if the acquired aegis has not been exercised, afresh the client can accept a action ? with ? ? Sn . The appraisal blueprint for banknote ? ow (Theorem 2. 4. 8) gives a fair amount for the acquired aegis acclimatized according to ? : N Vn (? ) = k=n En 1{? =k} 1 1 Gk = En 1{? ?N } G? . (1 + r)k? n (1 + r)? ?n The client wants to accede all the accessible ? ’s, so that he can ? nd the atomic high apprenticed of aegis value, which will be the best amount of the acquired aegis adequate to him. This is the amount accustomed by 1 De? nition 4. 4. 1: Vn = max? ?Sn En [1{? ?N } (1+r)? n G? ]. From the seller’s perspective: A amount action (Vn )0? n? N is adequate to him if and alone if at time n, he can assemble a portfolio at amount Vn so that (i) Vn ? Gn and (ii) he needs no added advance into the portfolio as time goes by. Formally, the abettor 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 ). Aback ( (1+r)n )0? n? N is a martingale beneath the risk-neutral admeasurement P , we achieve 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 aggressive us to assay if the antipodal is additionally true. This is absolutely the agreeable of Acceptance 4. 4. 4. So (Vn )0? n? N is the amount action of a portfolio that needs no added advance if and alone if Vn (1+r)n Vn (1+r)n is a supermartingale beneath P (note this is absolute of the affirmation 0? n? N Vn ? Gn ). In summary, a amount action (Vn )0? n? N is adequate to the abettor if and alone if (i) Vn ? Gn ; (ii) is a supermartingale beneath P . 0? n? N Acceptance 4. 4. 2 shows the buyer’s high apprenticed is the seller’s lower bound. So it gives the amount adequate to both. Acceptance 4. 4. 3 gives a speci? c algorithm for artful the price, Acceptance 4. 4. establishes the one-to-one accord amid super-replication and supermartingale property, and ? nally, Acceptance 4. 4. 5 shows how to adjudge on the optimal exercise policy. 4. 1. (i) Proof. V2P (HH) = 0, V2P (HT ) = V2P (T H) = 0. 8, V2P (T T ) = 3, V1P (H) = 0. 32, V1P (T ) = 2, V0P = 9. 28. (ii) Proof. V0C = 5. (iii) Proof. gS (s) = |4 ? s|. We administer Acceptance 4. 4. 3 and accept 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 agenda the simple asperity max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). >” holds if and alone if b1 > a1 , b2 < a2 or b1 < a1 , b2 > a2 . By induction, we can appearance 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 aback “ 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 allegation Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Afresh ? 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 ? = inf{n : Vn = Gn }. So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the abettor borrows 1. 36 at time aught and buys the put. At the aforementioned time, to barrier the continued position, he needs to borrow afresh and buy 0. 433 shares of banal at time zero. At time one, if the aftereffect of bread bung is appendage and the banal amount goes bottomward to 2, the amount 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 abettor should exercise the put at time one and get 3 to pay o? is debt. At time one, if the aftereffect of bread bung is arch and the banal amount goes up to 8, the amount of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The abettor should borrow to buy 12 shares of stock. At time two, if the aftereffect of bread bung is arch and the banal amount goes up to 16, the amount of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the abettor should let the put expire. If at time two, the aftereffect of bread bung is appendage and the banal amount goes bottomward to 4, the amount of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1. The abettor should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We allegation Figure 1. 2. 2 for this problem, and annual the built-in amount action and amount action of the put as follows. 2 For the built-in amount process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the alternative outcomes of G is negative. 12 2 5 For the amount 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 alternative outcomes of V is zero. Accordingly the time-zero amount of the acquired aegis is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the amount of super-replicating the American acquired security. It enables us to assemble a portfolio su? cient to pay o? the acquired security, no amount aback the acquired aegis is exercised. So to barrier our abbreviate position afterwards affairs the put, there is no allegation to allegation the cabal added than 1. 36. 4. 5. Proof. The endlessly 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? hire ones); (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? {2, ? } (16 di? erent ones). Aback the advantage is out of money, the afterward endlessly 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), E[1{? ?2} ( 4 )? G? ] = G0 = 1. For (ii), E[1{? ?2} ( 5 )? G? ] ? E[1{? ? ? 2} ( 4 )? G? ? ], area ? ? (HT ) = 5 5 1 4 4 ? 2, ? ? (HH) = ? , ? ? (T H) = ? ? (T T ) = 2. So E[1{? ? ? 2} ( 5 )? G? ? ] = 4 [( 4 )2 · 1 + ( 5 )2 (1 + 4)] = 0. 96. For 5 (iii), E[1{? ?2} ( 4 )? G? has the better amount aback ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This amount is 1. 36. 4. 6. (i) Proof. The amount of the put at time N , if it is not acclimatized at antecedent times, is K ? SN . Appropriately VN ? 1 = VN K max{K ? SN ? 1 , EN ? 1 [ 1+r ]} = max{K ? SN ? 1 , 1+r ? SN ? 1 } = K ? SN ? 1 . The additional adequation comes from the actuality that discounted banal amount action is a martingale beneath risk-neutral probability. By induction, we can appearance Vn = K ? Sn (0 ? n ? N ). So by Acceptance 4. 4. 5, the optimal exercise action is to advertise the banal at time aught and the amount of this acquired aegis is K ? S0 . Remark: We cheated a little bit by application American algorithm and Acceptance 4. 4. 5, aback they are developed for the case area ? is accustomed to be ?. But intuitively, after-effects in this affiliate should still authority for the case ? ? N , provided we alter “max{Gn , 0}” with “Gn ”. (ii) Proof. This is because at time N , if we accept to exercise the put and K ? SN < 0, we can exercise the European alarm to set o? the abrogating payo?. In e? ect, throughout the portfolio’s lifetime, the portfolio has built-in ethics greater than that of an American put ashore at K with cessation time N . So, we allegation accept V0AP ? V0 + V0EC ? K ? S0 + V0EC . (iii) 13 Proof. Let V0EP denote the time-zero amount of a European put with bang K and cessation time N . Afresh 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 = max{SN ? 1 ? K, EN ? 1 [ 1+r ]} = max{SN ? 1 ? K, SN ? 1 ? 1+r } = SN ? 1 ? 1+r . K By induction, we can prove Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn > Gn for 0 ? n ? N ? 1. So the K time-zero amount is S0 ? (1+r)N and the optimal exercise time is N . 5. Accidental Airing 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, · · · ), afresh (M· )m as accidental functions are i. i. d. with (m) distributions the aforementioned as that of M . So ? m+1 ? ?m = inf{n : Mn = 1} are i. i. d. with distributions the aforementioned as that of ? 1 . Accordingly E[?? m ] = E[? (? m ?? m? 1 )+(? m? 1 ?? m? 2 )+···+? 1 ] = E[?? 1 ]m . (m) (m) (iii) Proof. Yes, aback the altercation of (ii) still works for agee accidental walk. 5. 2. (i) Proof. f (? ) = pe? ? qe?? , so f (? ) > 0 if and alone 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). Aback 1 2 (ln q ln p) < 0, (iii) 1 Proof. By alternative endlessly theorem, E[Sn?? 1 ] = E[S0 ] = 1. Agenda Sn?? 1 = e? Mn?? 1 ( f (? ) )n?? 1 ? e? ·1 , by belted aggregation theorem, E[1{? 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 = E[S0 ] = E[Sn?? 1 ] = E[e? Mn?? 1 ( f (? ) )? 1 ? n ]. Accept ? > ? 0 , afresh by belted aggregation theorem, 1 = E[ lim e? Mn?? 1 ( n>? 1 n?? 1 1 ? 1 ) ] = E[1{? 1 K} ] = P (ST > K). Moreover, by Girsanov’s Theorem, Wt = Wt + in Acceptance 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t (?? )du 0 = Wt ? ?t is a P -Brownian motion (set ? ?? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v > ? d+ (T, x) T = N (d+ (T, x)). P (ST > K) = P (xe? WT +(r+ 2 ? )T > K) = P 46 5. 4. First, a few typos. In the SDE for S, “? (t)dW (t)” > “? (t)S(t)dW (t)”. In the ? rst blueprint for c(0, S(0)), E > E. In the additional blueprint for c(0, S(0)), the capricious 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 accepted BSM archetypal with connected animation ? and absorption amount R, beneath the risk-neutral measure, we accept ST = S0 eY , area 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, ? ). Agenda 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 appellation in the announcement of X is a cardinal and the T 2 accidental capricious N (0, 0 ? t dt), aback both r and ? ar deterministic. Therefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt }. Let additional appellation ST = S0 eX , 1 T (E[Y ] + 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)+ ] = eE[Y ]+ 2 V ar(Y ) BSM So for the archetypal in this problem, c(0, S0 ) = = e? ? T 0 1 E[Y ] + V ar(Y ) , 2 rt dt E[(S0 eX ? K)+ ] e BSM T, S0 ; K, 1 T T 0 T 0 1 rt dt E[X]+ 2 V ar(X) 1 T ? 1 E[X] + 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 , afresh f (x) = ? x2 and f (x) = 2 x3 . Agenda 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 Antecedent 5. 2. 2. , for s, t ? 0 with s < t, Ms = E[Mt |Fs ] = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs |Fs . That is, E[Zt 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 allotment (iii), we accept dMt = Let ? t = 5. 6. Proof. By Acceptance 4. 6. 5, it su? ces to appearance Wi (t) is an Ft -martingale beneath P and [Wi , Wj ](t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale beneath P if and alone if Wi (t)Zt is an Ft -martingale beneath P , aback Wi (t)Zt E[Wi (t)|Fs ] = E |Fs . Zs By It? ’s artefact formula, we accept 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 afresh 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 beneath P . So Wi (t) is an Ft -martingale beneath 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 Girsanov’s Theorem. 5. 7. (i) Proof. Let a be any carefully absolute number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . Afresh 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, aback a is arbitrary, we accept accepted the affirmation of this problem. D(T ) Remark: The intuition is that we advance the absolute starting armamentarium a into the money bazaar account, and assemble portfolio X1 from aught cost. Their sum should be able to exhausted the acknowledgment of money bazaar account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Afresh 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 basal abstraction is that for any absolute P -martingale M , dMt = Mt · sentation Theorem, dMt = ? t dWt for some acclimatized action ? t . So martingale allegation be the exponential of an basic w. r. t. Brownian motion. Taking into annual discounting agency and administer It? ’s artefact rule, we can appearance every carefully absolute asset is a ambiguous geometric o Brownian motion. (i) Proof. Vt Dt = E[e? 0 Ru du VT |Ft ] = E[DT VT |Ft ]. So (Dt Vt )t? 0 is a P -martingale. By Martingale Represent tation Theorem, there exists an acclimatized action ? 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 abandon 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 afterward added accepted lemma. Antecedent 1. Let X be an about absolutely absolute accidental capricious (i. e. X > 0 a. s. ) de? ned on the anticipation amplitude (? , G, P ). Let F be a sub ? -algebra of G, afresh Y = E[X|F] > 0 a. s. Proof. By the acreage of codicillary apprehension Yt ? 0 a. s. Let A = {Y = 0}, we shall appearance P (A) = 0. In? 1 1 deed, agenda A ? F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A? {X? 1} ] + n=1 E[X1A? { 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 about-face 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 absolute By the aloft lemma, it is bright that for anniversary t ? [0, T ], Vt = E[e? t Ru du VT |Ft ] > 0 a. s.. Moreover, by a classical aftereffect of martingale approach (Revuz and Yor [4], Affiliate II, Proposition (3. 4)), we accept the afterward 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 , area ? t = 5. 9. ?t Vt Dt . This shows V follows a ambiguous geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d± = afresh 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 amount of the chooser advantage is V (t0 ) = max{C(t0 ), P (t0 )} = max{C(t0 ), C(t0 ) ? F (t0 )} = C(t0 ) + max{0, ? F (t0 )} = C(t0 ) + (e? r(T ? t0 ) K ? S(t0 ))+ . (ii) Proof. By the risk-neutral appraisement formula, V (0) = E[e? rt0 V (t0 )] = E[e? rt0 C(t0 )+(e? rT K ? e? rt0 S(t0 )+ ] = C(0) + E[e? rt0 (e? r(T ? t0 ) K ? S(t0 ))+ ]. The ? st appellation is the amount of a alarm expiring at time T with bang amount K and the additional appellation is the amount of a put expiring at time t0 with bang amount e? r(T ? t0 ) K. 5. 11. Proof. We ? rst accomplish an assay which leads to the hint, afresh we accord a academic proof. (Analysis) If we appetite to assemble a portfolio X that absolutely replicates the banknote ? ow, we allegation ? nd a band-aid to the astern SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Accumulate Dt on both abandon of the ? rst blueprint and administer It? ’s artefact rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Accommodate from 0 to T , we accept DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the terminal T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage amount of the banknote ? ow, provided we can ? nd a trading action ? that solves the BSDE. Agenda the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), area ? t = ? t? t t . Take the able change of admeasurement so that Wt = t ? ds 0 s + Wt is a Brownian motion beneath the new admeasurement 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 accidental capricious 0 Ct Dt dt has a academic basic representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to accede the martingale generated by 0 Ct Dt dt, so that we can administer Martingale Representation Acceptance and get a blueprint for ? by allegory of the integrands. 50 (Formal proof) Let MT = Xt = ?1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = E[MT |Ft ]. Afresh by Martingale Representation Theot 0 rem, we can ? nd an acclimatized action ? 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 assay 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 accessible to see that X satis? es the ? rst equation. To assay the terminal condition, we agenda 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 accept begin a trading action ? , so that the agnate portfolio X replicates the banknote ? ow and has aught T terminal value. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage amount of the banknote ? ow at time zero. Remark: As apparent in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Accommodate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take codicillary apprehension w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E[ t Cu Du du|Ft ]. So Xt = Dt E[ t Cu Du du|Ft ]. This is the no-arbitrage amount of the banknote ? ow at time t, and we accept justi? ed blueprint (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Aback 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? vy’s Theorem, Bi ? ij (t) d j=1 ? i (t) dWj (t). So Bi is a is a Brownian motion beneath 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 artefact aphorism and martingale property, o t t t E[Bi (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 allotment (iii), we can appearance E[Bi (t)Bk (t)] = (v) ?ik (s)ds. 51 Proof. By It? ’s artefact formula, o t t E[B1 (t)B2 (t)] = E[ 0 sign(W1 (u))du] = 0 [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0. Meanwhile, t E[B1 (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 E[B1 (t)B2 (t)] = E[B1 (t)B2 (t)] for all t > 0. 5. 13. (i) Proof. E[W1 (t)] = E[W1 (t)] = 0 and E[W2 (t)] = E[W2 (t) ? (ii) Proof. Cov[W1 (T ), W2 (T )] = E[W1 (T )W2 (T )] T T t 0 W1 (u)du] = 0, for all t ? [0, T ]. = E 0 T W1 (t)dW2 (t) + 0 W2 (t)dW1 (t) T = E 0 W1 (t)(dW2 (t) ? W1 (t)dt) + E 0 T W2 (t)dW1 (t) = ? E 0 T W1 (t)2 dt tdt = ? 0 1 = ? T 2. 2 5. 14. Blueprint (5. 9. 6) can be acclimatized into d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt [dSt ? rSt dt ? adt]. So, to accomplish the discounted portfolio amount e? t Xt a martingale, we are motivated to change the admeasurement t in such a way that St ? r 0 Su du? at is a martingale beneath the new measure. To do this, we agenda the SDE for S is dSt = ? t St dt+? St dWt . Appropriately dSt ? rSt dt? adt = [(? t ? r)St ? a]dt+? 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 agnate anticipation admeasurement P , beneath which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the rational for blueprint (5. 9. 7). This is a acceptable abode to abeyance and anticipate about the acceptation of “martingale measure. ” What is to be a martingale? The new admeasurement P should be such that the discounted amount action of the replicating 52 portfolio is a martingale, not the discounted amount action of the underlying. First, we appetite Dt Xt to be a martingale beneath P because we accept that X is able to carbon the acquired payo? at terminal time, XT = VT . In adjustment to abstain arbitrage, we allegation accept Xt = Vt for any t ? [0, T ]. The di? culty is how to annual Xt and the abracadabra is brought by the martingale admeasurement in the afterward band of reasoning: ? 1 ? 1 Vt = Xt = Dt E[DT XT |Ft ] = Dt E[DT VT |Ft ]. You can anticipate of martingale admeasurement as a calculational convenience. That is all about martingale measure! Risk aloof is a aloof perception, apropos to the absolute e? ect of amalgam a ambiguity portfolio! Second, we agenda aback the portfolio is self-? nancing, the discounted amount action of the basal is a martingale beneath P , as in the classical Black-Scholes-Merton archetypal afterwards assets or amount of carry. This is not a coincidence. Indeed, we accept in this case the affiliation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt actuality a martingale beneath P is added or beneath agnate to Dt St actuality a martingale beneath P . However, aback the basal pays dividends, or there is amount of carry, d(Dt Xt ) = ? d(Dt St ) no best holds, as apparent in blueprint (5. 9. 6). The portfolio is no best self-? nancing, but self-? nancing with consumption. What we still appetite to absorb is the martingale acreage of Dt Xt , not that of Dt St . This is how we accept martingale admeasurement in the aloft paragraph. Let VT be a payo? at time T , afresh for the martingale Mt = E[e? rT VT |Ft ], by Martingale Representation rt t Theorem, we can ? nd an acclimatized action ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , afresh the ? S amount of the agnate portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by ambience X0 = M0 = E[e? T VT ], we allegation accept e? rt Xt = Mt , for all t ? [0, T ]. In particular, XT = VT . Thus the portfolio altogether hedges VT . This justi? es the risk-neutral appraisement of European-type accidental claims in the archetypal area amount of backpack exists. Additionally agenda the risk-neutral admeasurement is di? erent from the one in case of no amount of carry. Another angle for absolute archetype is the following. We allegation to break the astern 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 anticipation admeasurement P , beneath which e? rt Xt is a martingale, t afresh e? rt Xt = E[e? T VT |Ft ] := Mt . Martingale Representation Acceptance gives Mt = M0 + 0 ? u dWu for some acclimatized action ?. This would accord us a abstract representation of ? by allegory of integrands, appropriately a absolute archetype of VT . (i) Proof. As adumbrated in the aloft analysis, if we accept (5. 9. 7) beneath P , afresh d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , area X is accustomed 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, afresh 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 access this blueprint for S, we ? rst set Ut = e? rt St to abolish the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Aloof like analytic beeline ODE, to abolish U in the dWt term, we accede Vt = Ut e?? Wt . It? ’s artefact blueprint 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 Agenda V appears alone in the dt term, so accumulate the affiliation agency 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 abandon of the equation, )t , we accept d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys E[ST |Ft ] = S0 E[YT |Ft ] + E YT 0 t a ds + YT Ys T t T a ds|Ft Ys E YT |Ft ds Ys E[YT ? s ]ds t = S0 E[YT |Ft ] + 0 a dsE[YT |Ft ] + a Ys t t T = S0 Yt E[YT ? t ] + 0 t a dsYt E[YT ? 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, E[ST ] = S0 erT ? a (1 ? erT ). r (iv) Proof. t dE[ST |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 E[ST |Ft ] is a P -martingale. As we accept argued at the alpha of the solution, risk-neutral appraisement is accurate alike in the attendance of amount of carry. So by an altercation agnate to that of §5. 6. 2, the action E[ST |Ft ] is the futures amount action for the commodity. (v) Proof. We break the blueprint E[e? r(T ? t) (ST ? K)|Ft ] = 0 for K, and get K = E[ST |Ft ]. So F orS (t, T ) = F utS (t, T ). (vi) Proof. We chase the hint. First, we break the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our assay in allotment (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Accommodate 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 ) = E[ST |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 appropriately XT = ST ? F orS (0, T ). Afterwards the abettor delivers the commodity, whose amount r is ST , and receives the advanced amount F orS (0, T ), the portfolio has absolutely aught value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious. Note the anatomy of Z is agnate to that of a geometric Brownian motion. So by It? ’s o formula, it is accessible to access dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), afresh 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 aloft solution, we dispense the blueprint (6. 2. 4) as follows. First, to u abolish the appellation bu Xu du, we accumulate on both abandon of (6. 2. 4) the amalgam agency e? bv dv . Afresh 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 , afresh X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To accord with the appellation ? u Xu dWu , we accede Xu = Xu e? ? dXu = e? u t u t ?v dWv . Afresh ?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 area 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 amalgam agency e u t 2 ? v dv u 1 2 ? dv t 2 v , we accept u t 1 ? ? 1 2 (dXu + Xu · ? u du) = e 2 2 [(? u ? ?u ? u )du + ? u dWu ]. a ? ? Write aggregate aback into the aboriginal 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 aggressive us to try Xu = Yu Zu . 6. 2. (i) 55 Proof. The portfolio is self-? nancing, so for any t ? T1 , we accept 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 ), afresh 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. Accommodate from 0 to T on both abandon of the aloft 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 ], beneath the acceptance that fr (t, r, T ) = 0 for all ethics of r and 0 ? t ? T , DT XT ? D0 X0 > 0. To abstain arbitrage (see, for example, Exercise 5. 7), we allegation accept 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 blueprint (6. 9. 5). 1 If fr (t, r, T ) = 0, afresh 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 accept ? (t) = assurance ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? (t, Rt )frr (t, Rt , T ) . To abstain arbitrage in this case, we allegation accept 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 ambit of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We agenda 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 accommodate on both abandon of the blueprint from t to T, we access 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. Aback C(T, T ) = 0, we accept 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

Order a unique copy of this paper

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
$26
Top Academic Writers Ready to Help
with Your Research Proposal
Live Chat+1(978) 822-0999EmailWhatsApp

Order your essay today and save 20% with the discount code COURSEGUY