Stochastic Calculus forFinance, Volume I and II by Yan Zeng Last updated: August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you have any comments or ? nd any typos/errors, please email me at[email protected]edu. The current version omits the following problems. Volume I: 1. 5, 3. 3, 3. 4, 5. 7; Volume II: 3. 9, 7. 1, 7. 2, 7. 5–7. 9, 10. 8, 10. 9, 10. 10. Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. 1. 1. Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model Proof. If we get the up sate, then X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ? 0 S0 ); if we get the down state, then X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ? 0 S0 ). If X1 has a positive probability of being strictly positive, then we must either have X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, then ? 0 uS0 + (1 + r)(X0 ? ? 0 S0 ) > 0. Plug in X0 = 0, we get u? 0 > (1 + r)? 0 . By condition d < 1 + r < u, we conclude ? 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, then we can similarly deduce ? 0 < 0 and hence X1 (H) < 0. So we cannot have X1 strictly positive with positive probability unless X1 is strictly negative with positive probability as well, regardless the choice of the number ? 0 . Remark: Here the condition X0 = 0 is not essential, as far as a property de? nition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can de? ne arbitrage as a trading strategy 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 result of an arbitrary investment involvingmoneyand stock markets with that of a safe investment involving only money market. This can also be seen by regarding X0 as borrowed from money market account. Then at time 1, we have to pay back X0 (1 + r) to the money market account. In summary, arbitrage is a trading strategy that beats “ safe” investment. Accordingly, we revise the proof of Exercise 1. 1. as follows.

If X1 has a positive probability of being strictly larger than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) > 0, i. e. ? 0 > 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) < (1 + r)X0 . The second case can be similarly analyzed. Hence we cannot have X1 strictly greater than X0 (1 + r) with positive probability unless X1 is strictly smaller than X0 (1 + r) with positive probability as well. Finally, we comment that the above formulation of arbitrage is equivalent 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 positive probability that X1 is positive, then there is a positive probability that X1 is negative. Remark: Note the above relation X1 (u) = ? X1 (d) is not a coincidence. In general, let V1 denote the ? ? payo? of the derivative security at time 1. Suppose X0 and ? 0 are chosen in such a way that V1 can be ? 0 ? ? 0 S0 ) + ? 0 S1 = V1 . Using the notation of the problem, suppose an agent begins ? replicated: (1 + r)(X with 0 wealth and at time zero buys ? 0 shares of stock and ? 0 options. He then puts his cash position ? ?? 0 S0 ? ? 0 X0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the expression of V1 and sort out terms, we have ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X0 , then there will no arbitrage. 1. 3. S0 1 Proof. V0 = 1+r 1+r? d S1 (H) + u? ? r S1 (T ) = 1+r 1+r? d u + u? 1? r d = S0 . This is not surprising, since u? d u? d u? d u? d this is exactly the cost of replicating S1 . Remark: This illustrates an important point. The “ fair price” of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1 (H) and S1 (T ) are given, we could have two current prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are determined by S0 and S0 , respectively, it’s not surprising that risk-neutral pricing formula always 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 pricing relies on fair price= replication cost. Stock as a replicating component cannot determine its own “ fair” price via the risk-neutral pricing formula. 1. 4. Proof. Xn+1 (T ) = = ? 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 trader should set up a replicating portfolio whose payo? s the opposite of the option’s payo?. More precisely, we solve the equation (1 + r)(X0 ? ? 0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 Then X0 = ? 1. 20 and ? 0 = ? 2 . This means the trader should sell short 0. 5 share of stock, put the income 2 into a money market account, and then transfer 1. 20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0. 8(1 + r) in money market account will cancel out with the option’s payo?. Therefore we end up with 1. 20(1 + r) in the separate money market account. Remark: This problem illustrates why we are interested in hedging a long position.

In case the stock price goes down at time one, the option will expire without any payo?. The initial money 1. 20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payo? at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1. 7. Proof. The idea is the same as Problem 1. 6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1. 2. 4. More precisely, he should short sell 0. 1733 share of stock, invest the income 0. 933 into money market account, and transfer 1. 376 into a separate money market account. The portfolio consisting a short position in stock and 0. 6933-1. 376 in money market account will replicate the opposite of the option’s payo?. After they cancel out, we end up with 1. 376(1 + r)3 in the separate money market account. 1. 8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1. 696. (iii) Proof. ? n (s, y) = vn+1 (us, y + us) ? vn+1 (ds, y + ds) . (u ? d)s 1. 9. (i) Proof. Similar to Theorem 1. 2. 2, but replace r, u and d everywhere with rn , un and dn .

More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . Then 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 pricing implies the price of this call at time zero is ? ? 2 2 n ? d 9. 375. 2. Probability Theory on Coin Toss Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it su? ces to work on the case N = 2.

When A1 and A2 are disjoint, P (A1 ? A2 ) = ?? A1 ? A2 P (? ) = ?? A1 P (? ) + ?? A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, using the result when they are disjoint, we have 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 average rates of growth of the stock price under 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 average rates of growth of the stock price 9 6 under P are, respectively: r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Apply conditional 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 ), 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 denote by Xn the result of n-th coin toss, where Head is represented by X = 1 and Tail is 1 represented by X = ? 1. We also suppose P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 , · · · , Xn )Xn+1 , where bn (·) is a bounded function on {? 1, 1} , to be determined later on. Clearly (Sn )n? 1 is an adapted stochastic process, and we can show it is a martingale. Indeed, En [Sn+1 ? Sn ] = bn (X1 , · · · , Xn )En [Xn+1 ] = 0. For any arbitrary function f , En [f (Sn+1 )] = 1 [f (Sn + bn (X1 , · · · , Xn )) + f (Sn ? n (X1 , · · · , Xn ))]. Then 2 intuitively, En [f (Sn+1 ] cannot be solely dependent upon Sn when bn ’s are properly chosen. Therefore in general, (Sn )n? 1 cannot be a Markov process. Remark: If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealth at time n. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2. 8. (i) Proof. Note Mn = En [MN ] and Mn = En [MN ]. (ii) Proof. In the proof of Theorem 1. 2. 2, we proved by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In other words, the sequence (Vn )0? n?

N can be realized as the value process of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a martingale under P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale under P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), then use (i). 2. 9. (i) (H) S1 (H) 1 = 2, d0 = S1S0 = 2 , S0 (T and d1 (T ) = S21 (TT)) = 1. S 1 1 0 ? d So p0 = 1+r? d0 0 = 2 , q0 = 2 , p1 (H) u0 5 q1 (T ) = 6 . Therefore P (HH) = p0 p1 (H) = 1 , 4 5 q0 q1 (T ) = 12 . Vn (1+r)n = En VN (1+r)N , so V0 , V1 1+r , ···, VN ? 1 , VN (1+r)N ? 1 (1+r)N is a martingale under 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 still work for the random interest rate model, with proper modi? cations (i. e. P would be constructed according to conditional probabilities P (? n+1 = H|? 1 , · · · , ? n ) := pn and P (? n+1 = T |? 1 , · · · , ? n ) := qn . Cf. notes on page 39. ). So the time-zero value of an option that pays o?

V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . )(1+r (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2. 4, V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9, and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ? 1. 5 (iii) Proof. ? 0 = (iv) Proof. ? 1 (H) = 2. 10. (i) Xn+1 Proof. En [ (1+r)n+1 ] = En [ ? n Yn+1 Sn + (1+r)n+1 (1+r)(Xn ?? n Sn ) ] (1+r)n+1 Xn (1+r)n . V2 (HH)? V2 (HT ) S2 (HH)? S2 (HT ) V1 (H)? V1 (T ) S1 (H)? S1 (T ) = 1 2. 4? 9 8? 2 = 0. 4 ? 1 54 ? 0. 3815. = 5? 1 12? 8 = 1. = ? n Sn (1+r)n+1 En [Yn+1 ] + Xn ?? Sn (1+r)n = ? n Sn (1+r)n+1 (up + dq) + Xn ?? n Sn (1+r)n = ? n Sn +Xn ?? n Sn (1+r)n = (ii) Proof. From (2. 8. 2), we have ? n 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 make the portfolio replicate the payo? at time N , we 1+r VN X must have XN = VN . So Xn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ]. Since (Xn )0? n? N is the value process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations), the no-arbitrage price of VN at time n is Vn = Xn = En [ (1+r)N ? ]. (iii) Proof. En [ Sn+1 ] (1 + r)n+1 = = < = 1 En [(1 ? An+1 )Yn+1 Sn ] (1 + r)n+1 Sn [p(1 ? An+1 (H))u + q(1 ? An+1 (T ))d] (1 + r)n+1 Sn [pu + qd] (1 + r)n+1 Sn . (1 + r)n Sn (1+r)n+1 (1? a)(pu+qd) Sn+1 If An+1 is a constant a, then En [ (1+r)n+1 ] = Sn (1+r)n (1? a)n . = Sn (1+r)n (1? a). Sn+1 So En [ (1+r)n+1 (1? a)n+1 ] = 2. 11. (i) Proof. FN + PN = SN ? K + (K ? SN )+ = (SN ? K)+ = CN . (ii) CN FN PN Proof. Cn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ] + En [ (1+r)N ? n ] = Fn + Pn . (iii) FN Proof. F0 = E[ (1+r)N ] = 1 (1+r)N E[SN ? K] = S0 ? K (1+r)N . (iv) 6 Proof.

At time zero, the trader has F0 = S0 in money market account and one share of stock. At time N , the trader has a wealth of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Since F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En [ (1+r)N ? n ] = Sn ? So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ? 1. (1+r)N S0 (1+r)N ? n (1+r)N S0 (1+r)N = 0, C0 = P0 . = Sn ? S0 (1 + r)n . 2. 12. Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P ), where C= E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payo? of max(C, P ) at time m. Therefore, its current no-arbitrage price 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 price of a chooser option is E K (Sm ? (1+r)N ? m )+ P +E (1 + r)m (1 + r)m = E K (Sm ? (1+r)N ? m )+ (K ? SN )+ . +E (1 + r)N (1 + r)m

The ? rst term stands for the time-zero price of a put, expiring at time N and having strike price K, and the K second term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)N ? m . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ max(C, P ) ], (1+r)m due to the economical argument involved (like “ the chooser option is equivalent to receiving a payo? of max(C, P ) at time m”), then we have the following mathematically rigorous argument. First, we can construct a portfolio ? 0 , · · · , ? m? 1 , whose payo? at time m is max(C, P ).

Fix ? , if C(? ) > P (? ), we can construct a portfolio ? m , · · · , ? N ? 1 whose payo? at time N is (SN ? K)+ ; if C(? ) < P (? ), we can construct a portfolio ? m , · · · , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) > P (? ) if C(? ) < P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In Xm particular, V0 = X0 = E[ (1+r)m ] = E[ max(C, P ) ]. (1+r)m 2. 13. (i) Proof.

Note under both actual probability P and risk-neutral probability P , coin tosses ? n ’s are i. i. d.. So n+1 without loss of generality, we work on P . For any function g, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a function of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov under P . (ii) 7 Sn+1 Sn Sn )] Proof. Set vN (s, y) = f ( Ny ). Then vN (SN , YN ) = f ( +1 Vn = where En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r , Yn+1 ) ] N n= 0 Sn N +1 ) = VN . Suppose vn+1 is given, then = 1 1+r [pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since coin tosses ? n ’s are i. i. d. under P , (Sn , Yn )0? n? M is Markov under P . More precisely, for any function h, En [h(Sn+1 )] = ph(uSn ) + h(dSn ), for n = 0, 1, · · · , M ? 1. For any function g of two variables, we have EM [g(SM +1 , YM +1 )] = EM [g(SM +1 , SM +1 )] = pg(uSM , uSM )+ n+1 n+1 qg(dSM , dSM ). And for n ? M +1, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + SSn Sn )] = pg(uSn , Yn +uSn )+ qg(dSn , Yn + dSn ), so (Sn , Yn )0? n? N is Markov under P . (ii) y Proof. Set vN (s, y) = f ( N ? M ).

Then vN (SN , YN ) = f ( N K= M +1 Sk N ? M ) = VN . Suppose vn+1 is already given. a) If n > M , then En [vn+1 (Sn+1 , Yn+1 )] = pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ). So vn (s, y) = pvn+1 (us, y + us) + qvn+1 (ds, y + ds). b) If n = M , then EM [vM +1 (SM +1 , YM +1 )] = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (us, us) + qvM +1 (ds, ds). c) If n < M , then En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. State Prices 3. 1. Proof. Note Z(? ) := P (? ) P (? ) = 1 Z(? ) . Apply Theorem 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of properties (i)-(iii) of Theorem 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. E[Y ] = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = ?? A Z(? )P (? ) = 0, by P (Z > 0) = 1, we conclude P (? ) = 0 for any ? ? A. So P (A) = ?? A P (? ) = 0. (v) Proof. P (A) = 1 ?? P (Ac ) = 0 ?? P (Ac ) = 0 ?? P (A) = 1. (vi) ?? A ??? ??? P (? ) = ??? Z(? )P (? ) = E[Z] = 1. Y (? )P (? ) = ??? Y (? )Z(? )P (? ) = E[Y Z]. Z(? )P (? ). Since P (A) = 0, P (? ) = 0 for any ? ? A. So P (A) = 0. 8 Proof. Pick ? 0 such that P (? 0 ) > 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 Then P (Z ? 0) = 1 and 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. Hence in the case 0 < P (? 0 ) < 1, P and P are not equivalent. If P (? 0 ) = 1, then E[Z] = 1 if and only if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P have to be equivalent. In summary, if we can ? nd ? 0 such that 0 < P (? 0 ) < 1, then Z as constructed above would induce a probability P that is not equivalent to P . 3. 5. (i) Proof. Z(HH) = (ii) Proof. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )P (? 2 = T |? 1 = H) = 3 E1 [Z2 ](T ) = Z2 (T H)P (? 2 = H|? = T ) + Z2 (T T )P (? 2 = T |? 1 = T ) = 2 . (iii) Proof. V1 (H) = [Z2 (HH)V2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )V2 (HT )P (? 2 = T |? 1 = T )] = 2. 4, Z1 (H)(1 + r1 (H)) [Z2 (T H)V2 (T H)P (? 2 = H|? 1 = T ) + Z2 (T T )V2 (T T )P (? 2 = T |? 1 = T )] 1 = , Z1 (T )(1 + r1 (T )) 9 3 4. 9 16 , Z(HT ) = 9 , Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ? 1. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3. 6. Proof. U (x) = have XN = 1 x, (1+r)N ? Z so I(x) = = 1 Z] 1 x. Z (3. 3. 26) gives E[ (1+r)N 1 X0 (1 + r)n Zn En [Z ·

X0 N Z (1 + r) . 0 = Xn , where ? Hence Xn = (1+r)N ? Z X En [ (1+r)N ? n ] N ] = X0 . So ? = = En [ X0 (1+r) Z n 1 X0 . By (3. 3. 25), we 1 ] = X0 (1 + r)n En [ Z ] = the second to last “=” comes from Lemma 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26), we have E[ (1+r)N ( (1+r)N ) p? 1 ] = X0 . Solve 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 extreme point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a maximum point. Therefore U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Following the hint of the problem, we have 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, then Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x < ? and 0 < y ? ? , then U (x) ? yx = ? yx ? and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 b) If 0 ? x < ? and y > ? , then U (x) ? yx = ? yx ? 0 and U (I(y)) ? yI(y) = U (0) ? y · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 c) If x ? ? and 0 < y ? ? , then U (x) ? yx = 1 ? yx and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 1 ? yx. So U (x) ? yx ? U (I(y)) ? yI(y). 1 d) If x ? ? and y > ? , then U (x) ? yx = 1 ? yx < 0 and U (I(y)) ? yI(y) = U (0) ? y · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). (iii) XN ? Z Proof. Using (ii) and set x = XN , y = (1+r)N , where XN is a random variable satisfying E[ (1+r)N ] = X0 , we have ?

Z ? Z ? 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 have 2N 2N X0 = m= 1 pm ? m I(?? m ) = m= 1 1 pm ? m ? 1{?? m ? ? } . So X0 ? X0 ? {m : = we are looking for positive solution ? > 0). Conversely, suppose there exists some K so that ? K < ? K+1 and K X0 1 m= 1 ? m pm = ? . Then we can ? nd ? > 0, such that ? K < ?? < ? K+1 . For such ? , we have Z ? Z 1 E[ I( )] = pm ? m 1{?? m ? ? } ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m= 1 m= 1 Hence (3. 6. 4) has a solution. 0 2N K 2N X0 1 m= 1 pm ? m 1{?? m ? ? } . Suppose there is a solution ? to (3. 6. 4), note ? > 0, we then can conclude 1 1 1 ?? m ? ? } = ?. Let K = max{m : ?? m ? ? }, then ?? K ? ? < ?? K+1 . So ? K < ? K+1 and K N m= 1 pm ? m (Note, however, that K could be 2 . In this case, ? K+1 is interpreted as ?. Also, note = (v) ? 1 Proof. XN (? m ) = I(?? m ) = ? 1{?? m ? ? } = ?, if m ? K . 0, if m ? K + 1 4. American Derivative Securities Before proceeding to the exercise problems, we ? rst give a brief summary of pricing American derivative securities as presented in the textbook. We shall use the notation of the book.

From the buyer’s perspective: At time n, if the derivative security has not been exercised, then the buyer can choose a policy ? with ? ? Sn . The valuation formula for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative security exercised 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 wants to consider all the possible ? ’s, so that he can ? nd the least upper bound of security value, which will be the maximum price of the derivative security acceptable to him. This is the price given by 1 De? nition 4. 4. 1: Vn = max? ? Sn En [1{? ? N } (1+r)? n G? ]. From the seller’s perspective: A price process (Vn )0? n? N is acceptable to him if and only if at time n, he can construct a portfolio at cost Vn so that (i) Vn ? Gn and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can ? nd (? n )0? n? N and (Cn )0? n? N so that Cn ? 0 and Sn Vn+1 = ? n Sn+1 + (1 + r)(Vn ? Cn ? ? n Sn ). Since ( (1+r)n )0? n? N is a martingale under the risk-neutral measure P , we conclude En Cn Vn+1 Vn =? ? 0, ? n+1 n (1 + r) (1 + r) (1 + r)n Vn i. e. ( (1+r)n )0? n? N is a supermartingale. This inspired us to check if the converse is also true.

This is exactly the content of Theorem 4. 4. 4. So (Vn )0? n? N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the requirement 0? n? N Vn ? Gn ). In summary, a price process (Vn )0? n? N is acceptable to the seller if and only if (i) Vn ? Gn ; (ii) is a supermartingale under P . 0? n? N Theorem 4. 4. 2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable to both. Theorem 4. 4. 3 gives a speci? c algorithm for calculating the price, Theorem 4. 4. establishes the one-to-one correspondence between super-replication and supermartingale property, and ? nally, Theorem 4. 4. 5 shows how to decide 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 apply Theorem 4. 4. 3 and have V2S (HH) = 12. 8, V2S (HT ) = V2S (T H) = 2. 4, V2S (T T ) = 3, V1S (H) = 6. 08, V1S (T ) = 2. 16 and V0S = 3. 296. (iv) 11 Proof. First, we note the simple inequality max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). >” holds if and only if b1 > a1 , b2 < a2 or b1 < a1 , b2 > a2 . By induction, we can show S Vn = max gS (Sn ), S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ? max gP (Sn ) + gC (Sn ), ? max gP (Sn ), P C = Vn + Vn . P P pVn+1 + Vn+1 1+r + max gC (Sn ), S P C As to when “ C C pVn+1 +qVn+1 1+r or gP (Sn ) > P P pVn+1 +qVn+1 1+r and gC (Sn ) < C C pVn+1 +qVn+1 }. 1+r 4. 2. Proof. For this problem, we need Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Then ? 1 (H) = and ? 0 = V2 (HH) ? V2 (HT ) 1 V2 (T H) ? V2 (T T ) = ? , ? 1 (T ) = = ? 1, S2 (HH) ? S2 (HT ) 12 S2 (T H) ?

S2 (T T ) V1 (H) ? V1 (T ) ? ? 0. 433. S1 (H) ? S1 (T ) The optimal exercise time is ? = inf{n : Vn = Gn }. So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the agent borrows 1. 36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borrow again and buy 0. 433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio 1 is X1 (T ) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (T ) = (1 + 4 )(? 1. 36 ? 0. 433 ? 4) + 0. 433 ? 2 = ? 3. The agent should exercise the put at time one and get 3 to pay o? is debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The agent should borrow to buy 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.

The agent should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We need Figure 1. 2. 2 for this problem, and calculate the intrinsic value process and price process of the put as follows. 2 For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the other outcomes of G is negative. 12 2 5 For the price process, V0 = 0. 4, V1 (T ) = 1, V1 (T H) = 3 , V1 (T T ) = 3 , V3 (T HT ) = 1, V3 (T T H) = 1. 75, V3 (T T T ) = 2. 125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge the insider more than 1. 36. 4. 5. Proof. The stopping times in S0 are (1) ? ? 0; (2) ? ? 1; (3) ? (HT ) = ? (HH) = 1, ? (T H), ? (T T ) ? {2, ? } (4 di? erent ones); (4) ? (HT ), ? (HH) ? {2, ? }, ? (T H) = ? (T T ) = 1 (4 di? rent ones); (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? {2, ? } (16 di? erent ones). When the option is out of money, the following stopping times do not exercise (i) ? ? 0; (ii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H), ? (T T ) ? {2, ? } (8 di? erent ones); (iii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H) = ? (T T ) = 1 (2 di? erent ones). ? 4 For (i), E[1{? ? 2} ( 4 )? G? ] = G0 = 1. For (ii), E[1{? ? 2} ( 5 )? G? ] ? E[1{? ? ? 2} ( 4 )? G? ? ], where ? ? (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 biggest value when ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This value is 1. 36. 4. 6. (i) Proof. The value of the put at time N , if it is not exercised at previous times, is K ? SN . Hence VN ? 1 = VN K max{K ? SN ? 1 , EN ? 1 [ 1+r ]} = max{K ? SN ? 1 , 1+r ? SN ? 1 } = K ? SN ? 1 . The secondequalitycomes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K ?

S0 . Remark: We cheated a little bit by using American algorithm and Theorem 4. 4. 5, since they are developed for the case where ? is allowed to be ?. But intuitively, results in this chapter should still hold for the case ? ? N , provided we replace “ max{Gn , 0}” with “ Gn ”. (ii) Proof. This is because at time N , if we have to exercise the put and K ? SN < 0, we can exercise the European call to set o? the negative payo?. In e? ect, throughout the portfolio’s lifetime, the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N . So, we must have V0AP ? V0 + V0EC ? K ?

S0 + V0EC . (iii) 13 Proof. Let V0EP denote the time-zero value of a European put with strike K and expiration time N . Then V0AP ? V0EP = V0EC ? E[ K SN ? K ] = V0EC ? S0 + . (1 + r)N (1 + r)N 4. 7. VN K K Proof. VN = SN ? K, VN ? 1 = 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 value is S0 ? (1+r)N and the optimal exercise time is N . 5. Random Walk 5. 1. (i) Proof. E[?? 2 ] = E[? (? 2 ?? 1 )+? 1 ] = E[? (? 2 ?? 1 ) ]E[?? 1 ] = E[?? 1 ]2 . (ii) Proof. If we de? ne Mn = Mn+? ? M? m (m = 1, 2, · · · ), then (M· )m as random functions are i. i. d. with (m) distributions the same as that of M . So ? m+1 ? ? m = inf{n : Mn = 1} are i. i. d. with distributions the same as that of ? 1 . Therefore E[?? m ] = E[? (? m ?? m? 1 )+(? m? 1 ?? m? 2 )+···+? 1 ] = E[?? 1 ]m . (m) (m) (iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe?? , so f (? ) > 0 if and only if ? > f (? ) > f (0) = 1 for all ? > 0. (ii) 1 1 1 n+1 Proof. En [ SSn ] = En [e? Xn+1 f (? ) ] = pe? f (? ) + qe?? f (? ) = 1. 1 2 (ln q ? ln p). Since 1 2 (ln q ln p) < 0, (iii) 1 Proof. By optional stopping theorem, E[Sn?? 1 ] = E[S0 ] = 1. Note Sn?? 1 = e? Mn?? 1 ( f (? ) )n?? 1 ? e? ·1 , by bounded convergence 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 ]. Suppose ? > ? 0 , then by bounded convergence 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 Theorem 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t (?? )du 0 = Wt ? ? t is a P -Brownian motion (set ? ?? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v > ? d+ (T, x) T = N (d+ (T, x)). P (ST > K) = P (xe? WT +(r+ 2 ? )T > K) = P 46 5. 4. First, a few typos. In the SDE for S, “? (t)dW (t)” > “? (t)S(t)dW (t)”. In the ? rst equation for c(0, S(0)), E > E. In the second equation for c(0, S(0)), the variable for BSM should be ? ? 1 T 2 1 T r(t)dt, ? (t)dt? . BSM ? T, S(0); K, T 0 T 0 (i) Proof. d ln St = X = ? is a Gaussian with X ? N ( (ii) Proof. For the standard BSM model with constant volatility ? and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R? 1 ? 2 )T +? WT ? N ((R? 1 ? )T, ? 2 T ), and E[(S0 eY ? K)+ ] = 2 2 eRT BSM (T, S0 ; K, R, ? ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St ? 2St d S t = rt dt + ? t dWt ? 2 ? t dt. So ST = S0 exp{ 0 (rt ? 2 ? t )dt + 0 T 1 2 2 ? t )dt + 0 ? t dWt . The ? rst term in the expression of X is a number and the T 2 random variable N (0, 0 ? t dt), since both r and ? ar deterministic. Therefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ? t dWt }. Let second term ST = S0 eX , 1 T (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 model 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 , then f (x) = ? x2 and f (x) = 2 x3 . Note dZt = ? Zt ? t dWt , so d 1 Zt 1 1 1 2 2 2 ? t ? 2 t = f (Zt )dZt + f (Zt )dZt dZt = ? 2 (? Zt )? t dWt + 3 Zt ? t dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5. 2. 2. , for s, t ? 0 with s < t, Ms = 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 part (iii), we have dMt = Let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to show Wi (t) is an Ft -martingale under P and [Wi , Wj ](t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt E[Wi (t)| Fs ] = E | Fs . Zs By It? ’s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(? Zt )? (t) · dWt + Zt (dWi (t) + ? i (t)dt) + (? Zt ? t · dWt )(dWi (t) + ? i (t)dt) d t M t ? t M t ? 2 ? t ? t ? t M t ? t t dWt + dWt + dt + dt = (dWt + ? t dt) + (dWt + ? t dt). Zt Zt Zt Zt Zt Zt then dMt = ? t dWt . This proves Corollary 5. 3. 2. ? t +Mt ? t , Zt = Wi (t)(? Zt ) j= 1 d ? j (t)dWj (t) + Zt (dWi (t) + ? i (t)dt) ? Zt ? i (t)dt = Wi (t)(? Zt ) j= 1 ? j (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover, · · [Wi , Wj ](t) = Wi + 0 ? i (s)ds, Wj + 0 ? j (s)ds (t) = [Wi , Wj ](t) = t? ij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5. 7. (i) Proof. Let a be any strictly positive number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . Then P X2 (T ) ? X2 (0) D(T ) = P (a + X1 (T ) ? a) = P (X1 (T ) ? 0) = 1, and P X2 (T ) > X2 (0) = P (X1 (T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Then X1 (0) = 0, P (X1 (T ) ? 0) = P X2 (T ) ? X2 (0) D(T ) = 1, P (X1 (T ) > 0) = P X2 (T ) > X2 (0) D(T ) > 0. 5. 8.

The basic idea is that for any positive P -martingale M , dMt = Mt · sentation Theorem, dMt = ? t dWt for some adapted process ? t . So martingale must be the exponential of an integral w. r. t. Brownian motion. Taking into account discounting factor and apply It? ’s product rule, we can show every strictly positive asset is a generalized 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 adapted process ? t , 0 ? t ? T , such that Dt Vt = 0 ? s dWs , or equivalently, ? 1 t ? 1 t ? 1 Vt = Dt 0 ? dWs . Di? erentiate both sides of the equation, we get dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i. e. X > 0 a. s. ) de? ned on the probability space (? , G, P ). Let F be a sub ? -algebra of G, then Y = E[X| F] > 0 a. s. Proof. By the property of conditional expectation Yt ? 0 a. s. Let A = {Y = 0}, we shall show P (A) = 0. In? 1 1 deed, note 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 turn implies P (A) = P (A ? {X > 0}) = P (A ? {X ? 1}) + n= 1 P (A ? { n > X ? n+1 }) = 0. ? ? t Dt dWt . T 1 Mt dMt . By Martingale Repre? dMt = Mt ( Mtt )dWt , i. e. any positive By the above lemma, it is clear that for each t ? [0, T ], Vt = E[e? t Ru du VT | Ft ] > 0 a. s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3. 4)), we have the following stronger result: for a. s. ?, Vt (? ) > 0 for any t ? [0, T ]. (iii) 1 1 Proof. By (ii), V > 0 a. s. so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + ? t Dt dWt ? t = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T ? t Vt dWt , where ? t = 5. 9. ? t Vt Dt . This shows V follows a generalized geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d± = then f (y) = ? yf (y), cK (0, T, x, K) = xf (d+ ) 1 v ? T x (ln K + (r ± 1 ? 2 )T ). Let f (y) = 2 y v1 e? 2 2? 2 , ? d+ ? d? ? e? rT N (d? ) ? Ke? rT f (d? ) ? y ? y ? 1 1 = xf (d+ ) v ? e? rT N (d? ) + e? rT f (d? ) v , ? TK ? T 49 and cKK (0, T, x, K) x ? d? e? rT 1 ? d+ d? ? v ? e? rT f (d? ) + v (? d? )f (d? ) xf (d+ ) v f (d+ )(? d+ ) 2 ? y ? y ? y ? TK ? TK ?

T x xd+ ? 1 ? 1 e? rT d? ? 1 v v ? e? rT f (d? ) v ? v f (d? ) v f (d+ ) + v f (d+ ) ? T K2 ? TK K? T K? T ? T K? T x d+ e? rT f (d? ) d? v [1 ? v ] + v f (d+ ) [1 + v ] 2? T K ? T K? T ? T e? rT x f (d? )d+ ? 2 2 f (d+ )d? . K? 2 T K ? T = = = = 5. 10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = 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 pricing 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 term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e? r(T ? t0 ) K. 5. 11. Proof. We ? rst make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the backward SDE dXt = ? t dSt + Rt (Xt ? ? t St )dt ? Ct dt XT = 0. Multiply Dt on both sides of the ? rst equation and apply It? ’s product rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Integrate from 0 to T , we have DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the terminal T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash ? ow, provided we can ? nd a trading strategy ? that solves the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . Take the proper change of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the new measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ? t d(Dt St ) = D0 X0 + 0 ? t (Dt St )? t dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Representation Theorem and get a formula for ? by comparison of the integrands. 50 (Formal proof) Let MT = Xt = ? 1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = E[MT | Ft ]. Then by Martingale Representation Theot 0 rem, we can ? nd an adapted process ? t , so that Mt = M0 + + t 0 ? t dWt . If we set ? t = T 0 ? u d(Du Su ) ? t 0 ? t Dt St ? t , we can check Cu Du du), with X0 = M0 = E[ Ct Dt dt] solves the SDE dXt = ? t dSt + Rt (Xt ? ? t St )dt ? Ct dt XT = 0. Indeed, it is easy to see that X satis? es the ? rst equation.

To check the terminal condition, we note T T T XT DT = D0 X0 + 0 ? t Dt St ? t dWt ? 0 Ct Dt dt = M0 + 0 ? t dWt ? MT = 0. So XT = 0. Thus, we have found a trading strategy ? , so that the corresponding portfolio X replicates the cash ? ow and has zero T terminal value. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage price of the cash ? ow at time zero. Remark: As shown in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Integrate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take conditional expectation w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E[ t Cu Du du| Ft ]. So Xt = Dt E[ t Cu Du du| Ft ].

This is the no-arbitrage price of the cash ? ow at time t, and we have justi? ed formula (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Since dBi (t)dBi (t) = P. (ii) Proof. dSi (t) = = = R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + (? i (t) ? R(t))Si (t)dt ? ? i (t)Si (t)? i (t)dt d d ? ij (t) ? ij (t) d d j= 1 ? i (t) ? j (t)dt = j= 1 ? i (t) dWj (t) + ? ij (t)2 d e j= 1 ? i (t)2 dt = dt, by L? vy’s 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 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 part (iii), we can show E[Bi (t)Bk (t)] = (v) ? ik (s)ds. 51 Proof. By It? ’s product 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. Equation (5. 9. 6) can be transformed into d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt [dSt ? rSt dt ? adt]. So, to make the discounted portfolio value e? t Xt a martingale, we are motivated to change the measure t in such a way that St ? r 0 Su du? at is a martingale under the new measure. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . Hence dSt ? rSt dt? adt = [(? t ? r)St ? 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 equivalent probability measure P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the rational for formula (5. 9. 7). This is a good place to pause and think about the meaning of “ martingale measure. ” What is to be a martingale?

The new measure P should be such that the discounted value process of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payo? at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ? [0, T ]. The di? culty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: ? 1 ? 1 Vt = Xt = Dt E[DT XT | Ft ] = Dt E[DT VT | Ft ]. You can think of martingale measure as a calculational convenience.

That is all about martingale measure! Risk neutral is a just perception, referring to the actual e? ect of constructing a hedging portfolio! Second, we note when the portfolio is self-? nancing, the discounted price process of the underlying is a martingale under P , as in the classical Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case the relation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no longer holds, as shown in formula (5. 9. 6). The portfolio is no longer self-? nancing, but self-? nancing with consumption. What we still want to retain is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the above paragraph. Let VT be a payo? at time T , then for the martingale Mt = E[e? rT VT | Ft ], by Martingale Representation rt t Theorem, we can ? nd an adapted process ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , then the ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by setting X0 = M0 = E[e? T VT ], we must have e? rt Xt = Mt , for all t ? [0, T ]. In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justi? es the risk-neutral pricing of European-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is di? erent from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the backward SDE dXt = ? t dSt ? a? t dt + r(Xt ? ? t St )dt XT = VT for two unknowns, X and ?. To do so, we ? nd a probability measure P , under which e? rt Xt is a martingale, t then e? rt Xt = E[e? T VT | Ft ] := Mt . Martingale Representation Theorem gives Mt = M0 + 0 ? u dWu for some adapted process ?. This would give us a theoretical representation of ? by comparison of integrands, hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5. 9. 7) under P , then d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , where X is given by (5. 9. 6), is a P -martingale. (ii) 1 1 Proof. By It? ’s formula, dYt = Yt [? dWt + (r ? 2 ? 2 )dt] + 2 Yt ? 2 dt = Yt (? dWt + rdt). So d(e? rt Yt ) = o t a ? e? rt Yt dWt and e? rt Yt is a P -martingale.

Moreover, if St = S0 Yt + Yt 0 Ys ds, then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (? dWt + rdt) + adt = St (? dWt + rdt) + adt. Ys This shows S satis? es (5. 9. 7). Remark: To obtain this formula for S, we ? rst set Ut = e? rt St to remove the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Just like solving linear ODE, to remove U in the dWt term, we consider Vt = Ut e?? Wt . It? ’s product formula yields o dVt = = e?? Wt dUt + Ut e?? Wt 1 (?? )dWt + ? 2 dt + dUt · e?? Wt 2 1 (?? )dWt + ? 2 dt 2 1 e?? Wt ae? rt dt ? ? 2 Vt dt. 2 53 Note V appears only in the dt term, so multiply the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt?? Wt + 2 ? t dt. Set Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys 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 have argued at the beginning of the solution, risk-neutral pricing is valid even in the presence of cost of carry. So by an argument similar to that of §5. 6. 2, the process E[ST | Ft ] is the futures price process for the commodity. (v) Proof. We solve the equation 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 follow the hint. First, we solve the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our analysis in part (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Integrate from 0 to t on both sides, we get Xt = St ? S0 ert + a (1 ? ert ) = St ? S0 ert ? a (ert ? 1). In particular, XT = ST ? S0 erT ? a (erT ? 1). r r r Meanwhile, F orS (t, T ) = F uts (t, T ) = 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 hence XT = ST ? F orS (0, T ). After the agent delivers the commodity, whose value r is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious.

Note the form of Z is similar to that of a geometric Brownian motion. So by It? ’s o formula, it is easy to obtain dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), then Xt = Yt Zt = x · 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au ? ? u ? u ? u du + dWu Zu Zu [Yu bu Zu + (au ? ? u ? u ) + ? u ? u ]du + (? u Zu Yu + ? u )dWu Yu (bu Zu du + ? u Zu dWu ) + Zu (bu Xu + au )du + (? u Xu + ? u )dWu . + ? u Z u ? u du Zu Remark: To see how to ? nd the above solution, we manipulate the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we multiply on both sides of (6. 2. 4) the integrating factor e? bv dv . Then d(Xu e? ? Let Xu = e? u t u t bv dv ) = e? u t bv dv (au du + (? u + ? u Xu )dWu ). u t bv dv Xu , au = e? ? u t bv dv au and ? u = e? ? bv dv ? ? u , then X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To deal with the term ? u Xu dWu , we consider Xu = Xu e? ? dXu = e? u t u t ? v dWv . Then ? v dWv ? v dWv ? ? [(? u du + ? u dWu ) + ? u Xu dWu ] + Xu e? a ? u t u t 1 (?? u )dWu + e? 2 u t ? v dWv 2 ? u du ? +(? u + ? u Xu )(?? u )e? ? ? v dWv du 1 ? 2 ? ? ? = au du + ? u dWu + ? u Xu dWu ? ? u Xu dWu + Xu ? u du ? ? u (? u + ? u Xu )du ? ? ? 1 ? 2 = (? u ? ? u ? u ? Xu ? u )du + ? u dWu , a ? ? 2 where au = au e? ? ? ? 1 d Xu e 2 u t ? v dWv 2 ? v dv and ? u = ? u e? ? ? = e2 1 u t 2 ? v dv u t ? v dWv . Finally, use the integrating factor e u t 2 ? v dv u 1 2 ? dv t 2 v , we have u t 1 ? ? 1 2 (dXu + Xu · ? u du) = e 2 2 [(? u ? ? u ? u )du + ? u dWu ]. a ? ? Write everything back into the original X, a and ? , we get d Xu e? i. e. d u t bv dv? u t 1 ? v dWv + 2 u t 2 ? v dv = e2 1 u t 2 ? v dv? u t ? v dWv ? u t bv dv [(au ? ? u ? u )du + ? u dWu ], Xu Zu = 1 [(au ? ? u ? u )du + ? u dWu ] = dYu . Zu This inspired us to try Xu = Yu Zu . 6. 2. (i) 55 Proof.

The portfolio is self-? nancing, so for any t ? T1 , we have dXt = ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) + Rt (Xt ? ? 1 (t)f (t, Rt , T1 ) ? ? 2 (t)f (t, Rt , T2 ))dt, and d(Dt Xt ) = ? Rt Dt Xt dt + Dt dXt = Dt [? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = Dt [? 1 (t) ft (t, Rt , T1 )dt + fr (t, Rt , T1 )dRt + frr (t, Rt , T1 )? 2 (t, Rt )dt 2 1 +? 2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )? 2 (t, Rt )dt 2 ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = ? 1 (t)Dt [? Rt f (t, Rt , T1 ) + ft (t, Rt , T1 ) + ? t, Rt )fr (t, Rt , T1 ) + ? 2 (t, Rt )frr (t, Rt , T1 )]dt 2 1 +? 2 (t)Dt [? Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + ? (t, Rt )fr (t, Rt , T2 ) + ? 2 (t, Rt )frr (t, Rt , T2 )]dt 2 +Dt ? (t, Rt )[Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]]dWt = ? 1 (t)Dt [? (t, Rt ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )dt + ? 2 (t)Dt [? (t, Rt ) ? ?(t, Rt , T2 )]fr (t, Rt , T2 )dt +Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]dWt . (ii) Proof. Let ? 1 (t) = St fr (t, Rt , T2 ) and ? 2 (t) = ? St fr (t, Rt , T1 ), then d(Dt Xt ) = Dt St [? (t, Rt , T2 ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt |[? t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )| dt. Integrate from 0 to T on both sides of the above equation, we get T DT XT ? D0 X0 = 0 Dt |[? (t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )| dt. If ? (t, Rt , T1 ) = ? (t, Rt , T2 ) for some t ? [0, T ], under the assumption that fr (t, r, T ) = 0 for all values of r and 0 ? t ? T , DT XT ? D0 X0 > 0. To avoid arbitrage (see, for example, Exercise 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? [0, T ]. This implies ? (t, r, T ) does not depend on T . (iii) Proof. In (6. 9. 4), let ? 1 (t) = ? (t), T1 = T and ? (t) = 0, we get d(Dt Xt ) = 1 ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + ? (t, Rt )fr (t, Rt , T ) + ? 2 (t, Rt )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 range of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We note d ? e ds s 0 bv dv C(s, T ) = e? s 0 bv dv [C(s, T )(? bs ) + bs C(s, T ) ? 1] = ? e? s 0 bv dv . So integrate on both sides of the equation from t to T, we obtain e? T 0 bv dv C(T, T ) ? e? t 0 t 0 T bv dv C(t, T ) = ? t s 0 e? T t s 0 bv dv ds. Since C(T, T ) = 0, we have C(t, T ) = e 1 ? a(s)C(s, T ) + 2 ? 2 (s)C 2 (s, T ), we get A(T, T ) ? A(t, T ) = ? bv dv T t e? bv dv ds = T e t s bv dv ds. Finally, by A (s, T ) = T a(s)C(s, T )ds + t 1 2 ? 2 (s)C 2 (s, T )ds. t