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Stochastic Calculus for Finance, 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 finance, by Steven Shreve. If you have any comments or find any typos/errors, please email me at [email protected] 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 Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model 1.1. Proof. If we get the up sate, then X 1 = X 1 (H)=Δ 0 uS 0 + (1 + r)(X 0 - Δ 0 S 0 ); if we get the down state, then X 1 = X 1 (T )=Δ 0 dS 0 + (1 + r)(X 0 - Δ 0 S 0 ). If X 1 has a positive probability of being strictly positive, then we must either have X 1 (H) > 0 or X 1 (T ) > 0. (i) If X 1 (H) > 0, then Δ 0 uS 0 + (1 + r)(X 0 - Δ 0 S 0 ) > 0. Plug in X 0 = 0, we get uΔ 0 > (1 + r0 . By condition d< 1+ r<u, we conclude Δ 0 > 0. In this case, X 1 (T )=Δ 0 dS 0 + (1 + r)(X 0 - Δ 0 S 0 )= Δ 0 S 0 [d - (1 + r)] < 0. (ii) If X 1 (T ) > 0, then we can similarly deduce Δ 0 < 0 and hence X 1 (H) < 0. So we cannot have X 1 strictly positive with positive probability unless X 1 is strictly negative with positive probability as well, regardless the choice of the number Δ 0 . Remark: Here the condition X 0 = 0 is not essential, as far as a property definition of arbitrage for arbitrary X 0 can be given. Indeed, for the one-period binomial model, we can define arbitrage as a trading strategy such that P (X 1 X 0 (1 + r)) = 1 and P (X 1 >X 0 (1 + r)) > 0. First, this is a generalization of the case X 0 = 0; second, it is “proper” because it is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X 0 as borrowed from money market account. Then at time 1, we have to pay back X 0 (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 X 1 has a positive probability of being strictly larger than X 0 (1 + r), the either X 1 (H) >X 0 (1 + r) or X 1 (T ) >X 0 (1 + r). The first case yields Δ 0 S 0 (u - 1 - r) > 0, i.e. Δ 0 > 0. So X 1 (T ) = (1 + r)X 0 0 S 0 (d - 1 - r) < (1 + r)X 0 . The second case can be similarly analyzed. Hence we cannot have X 1 strictly greater than X 0 (1 + r) with positive probability unless X 1 is strictly smaller than X 0 (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 5.7. 1.2. 1

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Stochastic Calculus for Finance, Volume I and IIby Yan ZengLast updated: August 20, 2007This 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] current version omits the following problems. Volume I: 1.5, 3.3, 3.4, 5.7; Volume II: 3.9, 7.1, 7.2,7.57.9, 10.8, 10.9, 10.10.Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through thissolution manual and communicating to me several mistakes/typos.1 Stochastic Calculus for Finance I: The Binomial Asset Pricing Model1. The Binomial No-Arbitrage Pricing Model1.1.Proof. If we get the up sate, then X1 = X1(H) = 0uS0 + (1 + r)(X0 0S0); if we get the down state,then X1 = X1(T) = 0dS0 +(1 +r)(X00S0). 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 0uS0 + (1 + r)(X0 0S0) > 0. Plug in X0 = 0, we get u0 > (1 + r)0.By condition d < 1 + r < u, we conclude 0 > 0. In this case, X1(T) = 0dS0 + (1 + r)(X0 0S0) =0S0[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 positiveprobability as well, regardless the choice of the number 0.Remark: Here the condition X0 = 0 is not essential, as far as a property denition of arbitrage forarbitrary X0 can be given. Indeed, for the one-period binomial model, we can dene arbitrage as a tradingstrategy such that P(X1 X0(1 +r)) = 1 and P(X1 > X0(1 +r)) > 0. First, this is a generalization of thecase X0 = 0; second, it is proper because it is comparing the result of an arbitrary investment involvingmoney and stock markets with that of a safe investment involving only money market. This can also be seenby 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 beingstrictly larger than X0(1 + r), the either X1(H) > X0(1 + r) or X1(T) > X0(1 + r). The rst case yields0S0(u1r) > 0, i.e. 0 > 0. So X1(T) = (1+r)X0+0S0(d1r) < (1+r)X0. The second case canbe similarly analyzed. Hence we cannot have X1 strictly greater than X0(1 + r) with positive probabilityunless 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 X1(u) = 08 +03 54(40+1.200) = 30+1.50, and X1(d) = 02 54(40+1.200) =301.50. That is, X1(u) = X1(d). So if there is a positive probability that X1 is positive, then thereis 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 thepayo of the derivative security at time 1. Suppose X0 and 0 are chosen in such a way that V1 can bereplicated: (1 + r)( X0 0S0) + 0S1 = V1. Using the notation of the problem, suppose an agent beginswith 0 wealth and at time zero buys 0 shares of stock and 0 options. He then puts his cash position0S0 0 X0 in a money market account. At time one, the value of the agents portfolio of stock, optionand money market assets isX1 = 0S1 + 0V1 (1 +r)(0S0 + 0 X0).Plug in the expression of V1 and sort out terms, we haveX1 = S0(0 + 00)(S1S0(1 +r)).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.Proof. V0 = 11+r

1+rdud S1(H) + u1rud S1(T)

= S01+r

1+rdud u + u1rud d

= S0. This is not surprising, sincethis is exactly the cost of replicating S1.Remark: This illustrates an important point. The fair price of a stock cannot be determined by therisk-neutral pricing, as seen below. Suppose S1(H) and S1(T) are given, we could have two current prices, S0and S

0. Correspondingly, we can get u, d and u

, d

. Because they are determined by S0 and S

0, respectively,its not surprising that risk-neutral pricing formula always holds, in both cases. That is,S0 =1+rdud S1(H) + u1rud S1(T)1 +r , S

0 =1+rd



S1(H) + u



S1(T)1 +r .Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicatingcomponent cannot determine its own fair price via the risk-neutral pricing formula.1.4.Proof.Xn+1(T) = ndSn + (1 +r)(Xn nSn)= nSn(d 1 r) + (1 +r)Vn= Vn+1(H) Vn+1(T)u d (d 1 r) + (1 +r) pVn+1(H) + qVn+1(T)1 +r= p(Vn+1(T) Vn+1(H)) + pVn+1(H) + qVn+1(T)= pVn+1(T) + qVn+1(T)= Vn+1(T).1.6.2Proof. The banks trader should set up a replicating portfolio whose payo is the opposite of the optionspayo. More precisely, we solve the equation(1 +r)(X0 0S0) + 0S1 = (S1 K)+.Then X0 = 1.20 and 0 = 12. This means the trader should sell short 0.5 share of stock, put the income2 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 outwith the options 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 stockprice goes down at time one, the option will expire without any payo. The initial money 1.20 we paid attime zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) whichguarantees 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 banks trader only needs to set up the reverse of thereplicating trading strategy described in Example 1.2.4. More precisely, he should short sell 0.1733 share ofstock, invest the income 0.6933 into money market account, and transfer 1.376 into a separate money marketaccount. The portfolio consisting a short position in stock and 0.6933-1.376 in money market account willreplicate the opposite of the options payo. After they cancel out, we end up with 1.376(1 + r)3in theseparate money market account.1.8. (i)Proof. vn(s, y) = 25(vn+1(2s, y + 2s) +vn+1(s2, y + s2)).(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+rndnundnand qn = 1 pn. ThenVn = pnVn+1(H) + qnVn+1(T)1 +rn.(ii)Proof. n = Vn+1(H)Vn+1(T)Sn+1(H)Sn+1(T) = Vn+1(H)Vn+1(T)(undn)Sn.(iii)3Proof. un = Sn+1(H)Sn= Sn+10Sn= 1+ 10Snand dn = Sn+1(T)Sn= Sn10Sn= 1 10Sn. So the risk-neutral probabilitiesat time n are pn = 1dnundn= 12 and qn = 12. Risk-neutral pricing implies the price of this call at time zero is9.375.2. Probability Theory on Coin Toss Space2.1. (i)Proof. P(Ac) +P(A) =Ac P() +AP() =P() = 1.(ii)Proof. By induction, it suces to work on the case N = 2. When A1 and A2 are disjoint, P(A1 A2) =A1A2 P() = A1 P() +A2 P() = P(A1) + P(A2). When A1 and A2 are arbitrary, usingthe 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)Proof. P(S3 = 32) = p3= 18, P(S3 = 8) = 3 p2 q = 38, P(S3 = 2) = 3 p q2= 38, and P(S3 = 0.5) = q3= 18.(ii)Proof. E[S1] = 8P(S1 = 8) + 2P(S1 = 2) = 8 p + 2 q = 5, E[S2] = 16 p2+ 4 2 p q + 1 q2= 6.25, andE[S3] = 32 18 + 8 38 + 2 38 + 0.5 18 = 7.8125. So the average rates of growth of the stock price under Pare, respectively: r0 = 54 1 = 0.25, r1 = 6.255 1 = 0.25 and r2 = 7.81256.25 1 = 0.25.(iii)Proof. P(S3 = 32) = (23)3= 827, P(S3 = 8) = 3 (23)2

13 = 49, P(S3 = 2) = 2 19 = 29, and P(S3 = 0.5) = 127.Accordingly, E[S1] = 6, E[S2] = 9 and E[S3] = 13.5. So the average rates of growth of the stock priceunder P are, respectively: r0 = 64 1 = 0.5, r1 = 96 1 = 0.5, and r2 = 13.59 1 = Apply conditional Jensens inequality.2.4. (i)Proof. En[Mn+1] = Mn +En[Xn+1] = Mn +E[Xn+1] = Mn.(ii)Proof. En[Sn+1Sn] = En[eXn+1 2e+e] = 2e+eE[eXn+1] = 1.2.5. (i)Proof. 2In = 2n1j=0 Mj(Mj+1 Mj) = 2n1j=0 MjMj+1 n1j=1 M2j n1j=1 M2j = 2n1j=0 MjMj+1 +M2n n1j=0 M2j+1 n1j=0 M2j = M2n n1j=0(Mj+1 Mj)2= M2n n1j=0 X2j+1 = M2n n.(ii)Proof. En[f(In+1)] = En[f(In+Mn(Mn+1Mn))] = En[f(In+MnXn+1)] = 12[f(In+Mn)+f(InMn)] =g(In), where g(x) = 12[f(x +2x +n) +f(x 2x +n)], since 2In +n = [Mn[.2.6.4Proof. En[In+1 In] = En[n(Mn+1 Mn)] = nEn[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 isrepresented by X = 1. We also suppose P(X = 1) = P(X = 1) = 12. Dene S1 = X1 and Sn+1 =Sn+bn(X1, , Xn)Xn+1, where bn() is a bounded function on 1, 1n, to be determined later on. Clearly(Sn)n1 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)] = 12[f(Sn+bn(X1, , Xn)) +f(Snbn(X1, , Xn))]. Thenintuitively, En[f(Sn+1] cannot be solely dependent upon Sn when bns are properly chosen. Therefore ingeneral, (Sn)n1 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 wealthat 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 M

n = En[M

N].(ii)Proof. In the proof of Theorem 1.2.2, we proved by induction that Xn = Vn where Xn is dened by (1.2.14)of Chapter 1. In other words, the sequence (Vn)0nN can be realized as the value process of a portfolio,which consists of stock and money market accounts. Since ( Xn(1+r)n)0nN is a martingale under P (Theorem2.4.5), ( Vn(1+r)n)0nN is a martingale under P.(iii)Proof. V

n(1+r)n = En


, so V

0, V

11+r, , V

N1(1+r)N1, VN(1+r)N is a martingale under P.(iv)Proof. Combine (ii) and (iii), then use (i).2.9. (i)Proof. u0 = S1(H)S0 = 2, d0 = S1(H)S0 = 12, u1(H) = S2(HH)S1(H) = 1.5, d1(H) = S2(HT)S1(H) = 1, u1(T) = S2(TH)S1(T) = 4and d1(T) = S2(TT)S1(T) = 1.So p0 = 1+r0d0u0d0 = 12, q0 = 12, p1(H) = 1+r1(H)d1(H)u1(H)d1(H) = 12, q1(H) = 12, p1(T) = 1+r1(T)d1(T)u1(T)d1(T) = 16, and q1(T) = 56.Therefore P(HH) = p0 p1(H) = 14, P(HT) = p0 q1(H) = 14, P(TH) = q0 p1(T) = 112 and P(TT) = q0 q1(T) = 512.The proofs of Theorem 2.4.4, Theorem 2.4.5 and Theorem 2.4.7 still work for the random interestrate model, with proper modications (i.e. P would be constructed according to conditional probabili-ties P(n+1 = H[1, , n) := pn and P(n+1 = T[1, , n) := qn. Cf. notes on page 39.). Sothe time-zero value of an option that pays o V2 at time two is given by the risk-neutral pricing formulaV0 = E


.(ii)Proof. V2(HH) = 5, V2(HT) = 1, V2(TH) = 1 and V2(TT) = 0. So V1(H) = p1(H)V2(HH)+ q1(H)V2(HT)1+r1(H) =2.4, V1(T) = p1(T)V2(TH)+ q1(T)V2(TT)1+r1(T) = 19, and V0 = p0V1(H)+ q0V1(T)1+r0 1.5(iii)Proof. 0 = V1(H)V1(T)S1(H)S1(T) = 2.41982 = 0.4 154 0.3815.(iv)Proof. 1(H) = V2(HH)V2(HT)S2(HH)S2(HT) = 51128 = 1.2.10. (i)Proof. En[ Xn+1(1+r)n+1] = En[nYn+1Sn(1+r)n+1 + (1+r)(XnnSn)(1+r)n+1 ] = nSn(1+r)n+1 En[Yn+1] + XnnSn(1+r)n = nSn(1+r)n+1(u p +d q) + XnnSn(1+r)n = nSn+XnnSn(1+r)n = Xn(1+r)n.(ii)Proof. From (2.8.2), we have

nuSn + (1 +r)(Xn nSn) = Xn+1(H)ndSn + (1 +r)(Xn nSn) = Xn+1(T).So n = Xn+1(H)Xn+1(T)uSndSnand Xn = En[Xn+11+r ]. To make the portfolio replicate the payo at time N, wemust have XN = VN. So Xn = En[ XN(1+r)Nn] = En[ VN(1+r)Nn]. Since (Xn)0nN is the value process of theunique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linearequations), the no-arbitrage price of VN at time n is Vn = Xn = En[ VN(1+r)Nn].(iii)Proof.En[ Sn+1(1 +r)n+1] = 1(1 +r)n+1En[(1 An+1)Yn+1Sn]= Sn(1 +r)n+1[ p(1 An+1(H))u + q(1 An+1(T))d]< Sn(1 +r)n+1[ pu + qd]= Sn(1 +r)n.If An+1 is a constant a, then En[ Sn+1(1+r)n+1] = Sn(1+r)n+1(1a)( pu+ qd) = Sn(1+r)n(1a). So En[ Sn+1(1+r)n+1(1a)n+1] =Sn(1+r)n(1a)n.2.11. (i)Proof. FN +PN = SN K + (K SN)+= (SN K)+= CN.(ii)Proof. Cn = En[ CN(1+r)Nn] = En[ FN(1+r)Nn] + En[ PN(1+r)Nn] = Fn +Pn.(iii)Proof. F0 = E[ FN(1+r)N ] = 1(1+r)NE[SN K] = S0 K(1+r)N .(iv)6Proof. 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 (1+r)NS0(1+r)N = 0, C0 = P0.(vi)Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En[ SNK(1+r)Nn] = Sn (1+r)NS0(1+r)Nn = SnS0(1 +r)n.So Fn is not necessarily zero and Cn = Pn is not necessarily true for n 1.2.12.Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P), whereC = E(SN K)+(1 +r)Nm

, and P = E(K SN)+(1 +r)Nm

.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 optionat time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then thechooser option is equivalent to receiving a payo of max(C, P) at time m. Therefore, its current no-arbitrageprice should be E[max(C,P)(1+r)m ].By the put-call parity, C = Sm K(1+r)Nm +P. So max(C, P) = P +(Sm K(1+r)Nm)+. Therefore, thetime-zero price of a chooser option isE P(1 +r)m

+ E(Sm K(1+r)Nm)+(1 +r)m= E(K SN)+(1 +r)N

+ E(Sm K(1+r)Nm)+(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 thesecond term stands for the time-zero price of a call, expiring at time m and having strike price K(1+r)Nm.If we feel unconvinced by the above argument that the chooser options no-arbitrage price is E[max(C,P)(1+r)m ],due to the economical argument involved (like the chooser option is equivalent to receiving a payo ofmax(C, P) at time m), then we have the following mathematically rigorous argument. First, we canconstruct a portfolio 0, , m1, whose payo at time m is max(C, P). Fix , if C() > P(), wecan construct a portfolio

m, ,

N1 whose payo at time N is (SN K)+; if C() < P(), we canconstruct a portfolio

m, ,

N1 whose payo at time N is (K SN)+. By dening (m k N 1)k() =

k() if C() > P()

k() if C() < P(),we get a portfolio (n)0nN1 whose payo is the same as that of the chooser option. So the no-arbitrageprice process of the chooser option must be equal to the value process of the replicating portfolio. Inparticular, V0 = X0 = E[ Xm(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 ns are i.i.d.. Sowithout loss of generality, we work on P. For any function g, En[g(Sn+1, Yn+1)] = En[g(Sn+1SnSn, Yn +Sn+1SnSn)] = pg(uSn, Yn + uSn) + qg(dSn, Yn + dSn), which is a function of (Sn, Yn). So (Sn, Yn)0nN isMarkov under P.(ii)7Proof. Set vN(s, y) = f( yN+1). Then vN(SN, YN) = f(

Nn=0 SnN+1 ) = VN. Suppose vn+1 is given, thenVn = En[Vn+11+r ] = En[vn+1(Sn+1,Yn+1)1+r ] = 11+r[ pvn+1(uSn, Yn + uSn) + qvn+1(dSn, Yn + dSn)] = vn(Sn, Yn),wherevn(s, y) = vn+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 ns are i.i.d. under P, (Sn, Yn)0nM is Markovunder 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)+ qg(dSM, dSM). And for n M+1, En[g(Sn+1, Yn+1)] = En[g(Sn+1SnSn, Yn+ Sn+1SnSn)] = pg(uSn, Yn+uSn)+ qg(dSn, Yn +dSn), so (Sn, Yn)0nN is Markov under P.(ii)Proof. Set vN(s, y) = f( yNM). Then vN(SN, YN) = f(

NK=M+1 SkNM ) = 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 Prices3.1.Proof. Note Z() := P()

P() = 1Z(). Apply Theorem 3.1.1 with P, P, Z replaced by P, P, Z, we get theanalogous of properties (i)-(iii) of Theorem (i)Proof. P() =P() =Z()P() = E[Z] = 1.(ii)Proof. E[Y ] =Y ()P() =Y ()Z()P() = E[Y Z].(iii)Proof. P(A) =AZ()P(). Since P(A) = 0, P() = 0 for any A. So P(A) = 0.(iv)Proof. If P(A) = AZ()P() = 0, by P(Z > 0) = 1, we conclude P() = 0 for any A. SoP(A) =AP() = 0.(v)Proof. P(A) = 1 P(Ac) = 0 P(Ac) = 0 P(A) = 1.(vi)8Proof. Pick 0 such that P(0) > 0, dene Z() =

0, if = 01P(0), if = 0. Then P(Z 0) = 1 and E[Z] =1P(0) P(0) = 1.Clearly P( ` 0) = E[Z1\{0}] = =0 Z()P() = 0. But P( ` 0) = 1 P(0) > 0 ifP(0) < 1. Hence in the case 0 < P(0) < 1, P and P are not equivalent. If P(0) = 1, then E[Z] = 1 ifand 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 aprobability P that is not equivalent to P.3.5. (i)Proof. Z(HH) = 916, Z(HT) = 98, Z(TH) = 38 and Z(TT) = 154 .(ii)Proof. Z1(H) = E1[Z2](H) = Z2(HH)P(2 = H[1 = H) + Z2(HT)P(2 = T[1 = H) = 34. Z1(T) =E1[Z2](T) = Z2(TH)P(2 = H[1 = T) +Z2(TT)P(2 = T[1 = T) = 32.(iii)Proof.V1(H) = [Z2(HH)V2(HH)P(2 = H[1 = H) +Z2(HT)V2(HT)P(2 = T[1 = T)]Z1(H)(1 +r1(H)) = 2.4,V1(T) = [Z2(TH)V2(TH)P(2 = H[1 = T) +Z2(TT)V2(TT)P(2 = T[1 = T)]Z1(T)(1 +r1(T)) = 19,andV0 = Z2(HH)V2(HH)(1 + 14)(1 + 14) P(HH) + Z2(HT)V2(HT)(1 + 14)(1 + 14) P(HT) + Z2(TH)V2(TH)(1 + 14)(1 + 12) P(TH) + 0 1.3.6.Proof. U

(x) = 1x, so I(x) = 1x. (3.3.26) gives E[ Z(1+r)N(1+r)NZ ] = X0. So = 1X0. By (3.3.25), wehave XN = (1+r)NZ = X0Z (1 + r)N. Hence Xn = En[ XN(1+r)Nn] = En[X0(1+r)nZ ] = X0(1 + r)nEn[ 1Z] =X0(1 +r)n 1ZnEn[Z 1Z] = X0n, where the second to last = comes from Lemma U

(x) = xp1and so I(x) = x 1p1. By (3.3.26), we have E[ Z(1+r)N ( Z(1+r)N ) 1p1] = X0. Solve it for ,we get =

X0E Zpp1(1+r)Npp1

p1= Xp10 (1 +r)Np(E[Zpp1])p1.So by (3.3.25), XN = ( Z(1+r)N ) 1p1= 1p1Z1p1(1+r)Np1= X0(1+r)Npp1E[Zpp1]Z1p1(1+r)Np1= (1+r)NX0Z1p1E[Zpp1].3.8. (i)9Proof. ddx(U(x) yx) = U

(x) y. So x = I(y) is an extreme point of U(x) yx. Because d2dx2(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 everyx.(ii)Proof. Following the hint of the problem, we haveE[U(XN)] E[XNZ(1 +r)N ] E[U(I( Z(1 +r)N ))] E[ Z(1 +r)N I( Z(1 +r)N )],i.e. E[U(XN)] X0 E[U(XN)] E[ (1+r)NXN] = E[U(XN)] X0. So E[U(XN)] E[U(XN)].3.9. (i)Proof. Xn = En[ XN(1+r)Nn]. So if XN 0, then Xn 0 for all n.(ii)Proof. a) If 0 x < and 0 < y 1, then U(x) yx = yx 0 and U(I(y)) yI(y) = U() y =1 y 0. So U(x) yx U(I(y)) yI(y).b) If 0 x < and y > 1, then U(x) yx = yx 0 and U(I(y)) yI(y) = U(0) y 0 = 0. SoU(x) yx U(I(y)) yI(y).c) If x and 0 < y 1, 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).d) If x and y > 1, then U(x) yx = 1 yx < 0 and U(I(y)) yI(y) = U(0) y 0 = 0. SoU(x) yx U(I(y)) yI(y).(iii)Proof. Using (ii) and set x = XN, y = Z(1+r)N , where XN is a random variable satisfying E[ XN(1+r)N ] = X0,we haveE[U(XN)] E[ Z(1 +r)N XN] E[U(XN)] E[ Z(1 +r)N XN].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 haveX0 =2Nm=1pmmI(m) =2Nm=1pmm1{m1}.So X0 =2Nm=1pmm1{m1}. Suppose there is a solution to (3.6.4), note X0 > 0, we then can concludem : m 1 = . Let K = maxm : m 1, then K 1 < K+1. So K < K+1 andX0 =Km=1pmm (Note, however, that K could be 2N. In this case, K+1 is interpreted as . Also, notewe are looking for positive solution > 0). Conversely, suppose there exists some K so that K < K+1 andKm=1mpm = X0 . Then we can nd > 0, such that K < 1 < K+1. For such , we haveE[ Z(1 +r)N I( Z(1 +r)N )] =2Nm=1pmm1{m1} =Km=1pmm = X0.Hence (3.6.4) has a solution.10(v)Proof. XN(m) = I(m) = 1{m1} =

, if m K0, if m K + 1.4. American Derivative SecuritiesBefore proceeding to the exercise problems, we rst give a brief summary of pricing American derivativesecurities as presented in the textbook. We shall use the notation of the book.From the buyers perspective: At time n, if the derivative security has not been exercised, then the buyercan choose a policy with on. The valuation formula for cash ow (Theorem 2.4.8) gives a fair pricefor the derivative security exercised according to :Vn() =Nk=nEn1{=k}1(1 +r)knGk

= En1{N}1(1 +r)nG

.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 byDenition 4.4.1: Vn = maxSnEn[1{N}1(1+r)nG].From the sellers perspective: A price process (Vn)0nN 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 theportfolio as time goes by. Formally, the seller can nd (n)0nN and (Cn)0nN so that Cn 0 andVn+1 = nSn+1 + (1 + r)(Vn Cn nSn). Since ( Sn(1+r)n)0nN is a martingale under the risk-neutralmeasure P, we concludeEn Vn+1(1 +r)n+1

Vn(1 +r)n = Cn(1 +r)n 0,i.e. ( Vn(1+r)n)0nN is a supermartingale. This inspired us to check if the converse is also true. This is exactlythe content of Theorem 4.4.4. So (Vn)0nN is the value process of a portfolio that needs no further investingif and only if


0nNis a supermartingale under P (note this is independent of the requirementVn Gn). In summary, a price process (Vn)0nN is acceptable to the seller if and only if (i) Vn Gn; (ii)


0nNis a supermartingale under P.Theorem 4.4.2 shows the buyers upper bound is the sellers lower bound. So it gives the price acceptableto both. Theorem 4.4.3 gives a specic algorithm for calculating the price, Theorem 4.4.4 establishes theone-to-one correspondence between super-replication and supermartingale property, and nally, Theorem4.4.5 shows how to decide on the optimal exercise policy.4.1. (i)Proof. VP2 (HH) = 0, VP2 (HT) = VP2 (TH) = 0.8, VP2 (TT) = 3, VP1 (H) = 0.32, VP1 (T) = 2, VP0 = 9.28.(ii)Proof. VC0 = 5.(iii)Proof. gS(s) = [4 s[. We apply Theorem 4.4.3 and have VS2 (HH) = 12.8, VS2 (HT) = VS2 (TH) = 2.4,VS2 (TT) = 3, VS1 (H) = 6.08, VS1 (T) = 2.16 and VS0 = 3.296.(iv)11Proof. First, we note the simple inequalitymax(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 showVSn = max

gS(Sn), pVSn+1 + VSn+11 +r max

gP(Sn) +gC(Sn), pVPn+1 + VPn+11 +r + pVCn+1 + VCn+11 +r max

gP(Sn), pVPn+1 + VPn+11 +r+ max

gC(Sn), pVCn+1 + VCn+11 +r= VPn +VCn .As to when pVCn+1+ qVCn+11+r or gP(Sn) > pVPn+1+ qVPn+11+r and gC(Sn) < pVCn+1+ qVCn+11+r .4.2.Proof. For this problem, we need Figure 4.2.1, Figure 4.4.1 and Figure 4.4.2. Then1(H) = V2(HH) V2(HT)S2(HH) S2(HT) = 112, 1(T) = V2(TH) V2(TT)S2(TH) S2(TT) = 1,and0 = V1(H) V1(T)S1(H) S1(T) 0.433.The optimal exercise time is = infn : Vn = Gn. So(HH) = , (HT) = 2, (TH) = (TT) = 1.Therefore, the agent borrows 1.36 at time zero and buys the put. At the same time, to hedge the longposition, 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 portfoliois X1(T) = (1 +r)(1.36 0.433S0) +0.433S1(T) = (1 + 14)(1.36 0.433 4) +0.433 2 = 3. The agentshould exercise the put at time one and get 3 to pay o his debt.At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfoliois X1(H) = (1 + r)(1.36 0.433S0) + 0.433S1(H) = 0.4. The agent should borrow to buy 112 shares ofstock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of theportfolio is X2(HH) = (1+r)(X1(H) 112S1(H)) + 112S2(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 isX2(HT) = (1 +r)(X1(H) 112S1(H)) + 112S2(HT) = 1. The agent should exercise the put to get 1. Thiswill 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 ofthe put as follows.For the intrinsic value process, G0 = 0, G1(T) = 1, G2(TH) = 23, G2(TT) = 53, G3(THT) = 1,G3(TTH) = 1.75, G3(TTT) = 2.125. All the other outcomes of G is negative.12For the price process, V0 = 0.4, V1(T) = 1, V1(TH) = 23, V1(TT) = 53, V3(THT) = 1, V3(TTH) = 1.75,V3(TTT) = 2.125. All the other outcomes of V is zero.Therefore the time-zero price of the derivative security is 0.4 and the optimal exercise time satises() =

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 aportfolio sucient to pay o the derivative security, no matter when the derivative security is exercised. Soto hedge our short position after selling the put, there is no need to charge the insider more than The stopping times in o0 are(1) 0;(2) 1;(3) (HT) = (HH) = 1, (TH), (TT) 2, (4 dierent ones);(4) (HT), (HH) 2, , (TH) = (TT) = 1 (4 dierent ones);(5) (HT), (HH), (TH), (TT) 2, (16 dierent ones).When the option is out of money, the following stopping times do not exercise(i) 0;(ii) (HT) 2, , (HH) = , (TH), (TT) 2, (8 dierent ones);(iii) (HT) 2, , (HH) = , (TH) = (TT) = 1 (2 dierent ones).For (i), E[1{2}(45)G] = G0 = 1. For (ii), E[1{2}(45)G] E[1{2}(45)G], where (HT) =2, (HH) = , (TH) = (TT) = 2. So E[1{2}(45)G] = 14[(45)2 1 + (45)2(1 + 4)] = 0.96. For(iii), E[1{2}(45)G] has the biggest value when satises (HT) = 2, (HH) = , (TH) = (TT) = 1.This value is (i)Proof. The value of the put at time N, if it is not exercised at previous times, is K SN. Hence VN1 =maxKSN1, EN1[ VN1+r] = maxKSN1, K1+r SN1 = KSN1. The second equality comes fromthe fact that discounted stock price process is a martingale under risk-neutral probability. By induction, wecan show Vn = K Sn (0 n N). So by Theorem 4.4.5, the optimal exercise policy is to sell the stockat 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 developedfor the case where is allowed to be . But intuitively, results in this chapter should still hold for the case N, provided we replace maxGn, 0 with Gn.(ii)Proof. This is because at time N, if we have to exercise the put and K SN < 0, we can exercise theEuropean call to set o the negative payo. In eect, throughout the portfolios lifetime, the portfolio hasintrinsic values greater than that of an American put stuck at K with expiration time N. So, we must haveVAP0 V0 +VEC0 K S0 +VEC0 .(iii)13Proof. Let VEP0 denote the time-zero value of a European put with strike K and expiration time N. ThenVAP0 VEP0 = VEC0 E[ SN K(1 +r)N ] = VEC0 S0 + K(1 +r)N .4.7.Proof. VN = SN K, VN1 = maxSN1K, EN1[ VN1+r] = maxSN1K, SN1 K1+r = SN1 K1+r.By induction, we can prove Vn = Sn K(1+r)Nn (0 n N) and Vn > Gn for 0 n N 1. So thetime-zero value is S0 K(1+r)N and the optimal exercise time is N.5. Random Walk5.1. (i)Proof. E[2] = E[(21)+1] = E[(21)]E[1] = E[1]2.(ii)Proof. If we dene M(m)n = Mn+mMm (m = 1, 2, ), then (M(m) )m as random functions are i.i.d. withdistributions the same as that of M. So m+1 m = infn : M(m)n = 1 are i.i.d. with distributions thesame as that of 1. ThereforeE[m] = E[(mm1)+(m1m2)++1] = E[1]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 > 12(ln q ln p). Since 12(ln q ln p) < 0,f() > f(0) = 1 for all > 0.(ii)Proof. En[Sn+1Sn] = En[eXn+1 1f()] = pe 1f() +qe 1f() = 1.(iii)Proof. By optional stopping theorem, E[Sn1] = E[S0] = 1. Note Sn1 = eMn1( 1f())n1 e1,by bounded convergence theorem, E[1{1 1 for all > 0.(ii)Proof. As in Exercise 5.2, Sn = eMn( 1f())nis a martingale, and 1 = E[S0] = E[Sn1] = E[eMn1( 1f())1n].Suppose > 0, then by bounded convergence theorem,1 = E[ limneMn1( 1f())n1] = E[1{1 e0= qp, so = ln(1+14pq22p ), then E[11{1 X2(0)D(T)

= P(X1(T) > 0) > 0, since a is arbitrary, we have proved the claim of this problem.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 marketaccount.(ii)48Proof. We dene 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 1MtdMt. By Martingale Repre-sentation Theorem, dMt = td

Wt for some adapted process t. So dMt = Mt( tMt)d

Wt, i.e. any positivemartingale must be the exponential of an integral w.r.t. Brownian motion. Taking into account discountingfactor and apply Itos product rule, we can show every strictly positive asset is a generalized geometricBrownian motion.(i)Proof. VtDt = E[e

T0 RuduVT[Tt] = E[DTVT[Tt]. So (DtVt)t0 is a P-martingale. By Martingale Represen-tation Theorem, there exists an adapted process t, 0 t T, such that DtVt =


Ws, or equivalently,Vt = D1t


Ws. Dierentiate both sides of the equation, we get dVt = RtD1t


Wsdt +D1t td

Wt,i.e. dVt = RtVtdt + tDtdWt.(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.) dened on the probabilityspace (, (, P). Let T be a sub -algebra of (, then Y = E[X[T] > 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-deed, note A T, 0 = E[Y IA] = E[E[X[T]IA] = E[XIA] = E[X1A{X1}] +n=1E[X1A{1n>X 1n+1}] P(AX 1)+n=11n+1P(A1n > X 1n+1). So P(AX 1) = 0 and P(A1n > X 1n+1) = 0,n 1. This in turn implies P(A) = P(AX > 0) = P(AX 1) +n=1P(A1n > X 1n+1) =0.By the above lemma, it is clear that for each t [0, T], Vt = E[e

Tt RuduVT[Tt] > 0 a.s.. Moreover,by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3.4)), we have thefollowing stronger result: for a.s. , Vt() > 0 for any t [0, T].(iii)Proof. By (ii), V > 0 a.s., so dVt = Vt1VtdVt = Vt1Vt

RtVtdt + tDtd


= VtRtdt + Vt


Wt = RtVtdt +tVtd

Wt, where t = tVtDt. This shows V follows a generalized geometric Brownian motion.5.9.Proof. c(0, T, x, K) = xN(d+) KerTN(d) with d = 1T (ln xK + (r 122)T). Let f(y) = 12ey22 ,then f

(y) = yf(y),cK(0, T, x, K) = xf(d+)d+y erTN(d) KerTf(d)dy= xf(d+) 1TKerTN(d) +erTf(d) 1T,49andcKK(0, T, x, K)= xf(d+) 1TK2 xTKf(d+)(d+)d+y erTf(d)dy + erTT(d)f(d)dy= xTK2f(d+) + xd+TKf(d+) 1KTerTf(d) 1KT erTdTf(d) 1KT= f(d+) xK2T[1 d+T] + erTf(d)KT[1 + dT]= erTK2T f(d)d+ xK22T f(d+)d.5.10. (i)Proof. At time t0, the value of the chooser option is V (t0) = maxC(t0), P(t0) = maxC(t0), C(t0) F(t0) = C(t0) + max0, F(t0) = C(t0) + (er(Tt0)K S(t0))+.(ii)Proof. By the risk-neutral pricing formula, V (0) = E[ert0V (t0)] = E[ert0C(t0)+(erTKert0S(t0)+] =C(0) + E[ert0(er(Tt0)K S(t0))+]. The rst term is the value of a call expiring at time T with strikeprice K and the second term is the value of a put expiring at time t0 with strike price er(Tt0)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 asolution to the backward SDE

dXt = tdSt +Rt(Xt tSt)dt CtdtXT = 0.Multiply Dt on both sides of the rst equation and apply Itos product rule, we get d(DtXt) = td(DtSt) CtDtdt. Integrate from 0 to T, we have DTXT D0X0 = T0 td(DtSt)

T0 CtDtdt. By the terminalcondition, we get X0 = D10 (

T0 CtDtdt

T0 td(DtSt)). X0 is the theoretical, no-arbitrage price ofthe cash ow, provided we can nd a trading strategy that solves the BSDE. Note the SDE for Sgives d(DtSt) = (DtSt)t(tdt + dWt), where t = tRtt. Take the proper change of measure so that

Wt =

t0 sds +Wt is a Brownian motion under the new measure P, we get

T0CtDtdt = D0X0 +

T0td(DtSt) = D0X0 +


Wt.This says the random variable

T0 CtDtdt has a stochastic integral representation D0X0+

T0 tDtSttd

Wt.This inspires us to consider the martingale generated by T0 CtDtdt, so that we can apply Martingale Rep-resentation Theorem and get a formula for by comparison of the integrands.50(Formal proof) Let MT = T0 CtDtdt, and Mt = E[MT[Tt]. Then by Martingale Representation Theo-rem, we can nd an adapted process t, so that Mt = M0 +


Wt. If we set t = tDtStt, we can checkXt = D1t (D0X0 +

t0 ud(DuSu)

t0 CuDudu), with X0 = M0 = E[

T0 CtDtdt] solves the SDE

dXt = tdSt +Rt(Xt tSt)dt CtdtXT = 0.Indeed, it is easy to see that X satises the rst equation. To check the terminal condition, we noteXTDT = D0X0 +

T0 tDtSttd


T0 CtDtdt = M0 +


Wt MT = 0. So XT = 0. Thus, we havefound a trading strategy , so that the corresponding portfolio X replicates the cash ow and has zeroterminal value. So X0 = E[

T0 CtDtdt] is the no-arbitrage price of the cash ow at time zero.Remark: As shown in the analysis, d(DtXt) = td(DtSt) CtDtdt. Integrate from t to T, we get0 DtXt = Tt ud(DuSu)

Tt CuDudu. Take conditional expectation w.r.t. Tt on both sides, we getDtXt = E[

Tt CuDudu[Tt]. So Xt = D1t E[

Tt CuDudu[Tt]. This is the no-arbitrage price of the cashow at time t, and we have justied formula (5.6.10) in the textbook.5.12. (i)Proof. dBi(t) = dBi(t) +i(t)dt =dj=1ij(t)i(t) dWj(t) +dj=1ij(t)i(t) j(t)dt =dj=1ij(t)i(t) d

Wj(t). So Bi is amartingale. Since dBi(t)dBi(t) = dj=1ij(t)2i(t)2 dt = dt, by Levys Theorem, Bi is a Brownian motion underP.(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= R(t)Si(t)dt +i(t)Si(t)dBi(t) +dj=1ij(t)j(t)Si(t)dt Si(t)dj=1ij(t)j(t)dt= R(t)Si(t)dt +i(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 os product rule and martingale property,E[Bi(t)Bk(t)] = E[

t0Bi(s)dBk(s)] +E[

t0Bk(s)dBi(s)] +E[

t0dBi(s)dBk(s)]= E[

t0ik(s)ds] =

t0ik(s)ds.Similarly, by part (iii), we can show E[Bi(t)Bk(t)] =

t0 ik(s)ds.(v)51Proof. By It os product formula,E[B1(t)B2(t)] = E[

t0sign(W1(u))du] =

t0[P(W1(u) 0) P(W1(u) < 0)]du = 0.Meanwhile,E[B1(t)B2(t)] = E[


t0[P(W1(u) 0) P(W1(u) < 0)]du=


W1(u) u) P(

W1(u) < u)]du=


12 P(

W1(u) < u)

du< 0,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[



W1(u)du] = 0, for all t [0, T].(ii)Proof.Cov[W1(T), W2(T)] = E[W1(T)W2(T)]= E

T0W1(t)dW2(t) +

T0W2(t)dW1(t)= E




W1(t)dt)+ E


W1(t)= E



T0tdt= 12T2.5.14. Equation (5.9.6) can be transformed into d(ertXt) = t[d(ertSt) aertdt] = tert[dStrStdt adt]. So, to make the discounted portfolio value ertXt a martingale, we are motivated to change the measurein such a way that Str

t0 Suduat is a martingale under the new measure. To do this, we note the SDE for Sis dSt = tStdt+StdWt. Hence dStrStdtadt = [(tr)Sta]dt+StdWt = St

(tr)StaStdt +dWt

.Set t = (tr)StaStand Wt =

t0 sds +Wt, we can nd an equivalent probability measure P, under whichS satises the SDE dSt = rStdt +Std

Wt +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 bea martingale? The new measure P should be such that the discounted value process of the replicating52portfolio is a martingale, not the discounted price process of the underlying. First, we want DtXt to be amartingale 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 diculty is howto calculate Xt and the magic is brought by the martingale measure in the following line of reasoning:Vt = Xt = D1t E[DTXT[Tt] = D1t E[DTVT[Tt]. You can think of martingale measure as a calculationalconvenience. That is all about martingale measure! Risk neutral is a just perception, referring to theactual eect of constructing a hedging portfolio! Second, we note when the portfolio is self-nancing, thediscounted price process of the underlying is a martingale under P, as in the classical Black-Scholes-Mertonmodel without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case therelation d(DtXt) = td(DtSt). So DtXt being a martingale under P is more or less equivalent to DtStbeing a martingale under P. However, when the underlying pays dividends, or there is cost of carry,d(DtXt) = td(DtSt) 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 DtXt, notthat of DtSt. 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[erTVT[Tt], by Martingale RepresentationTheorem, we can nd an adapted process t, so that Mt = M0 +


Ws. If we let t = tertSt, then thevalue of the corresponding portfolio X satises d(ertXt) = td

Wt. So by setting X0 = M0 = E[erTVT],we must have ertXt = Mt, for all t [0, T]. In particular, XT = VT. Thus the portfolio perfectly hedgesVT. This justies the risk-neutral pricing of European-type contingent claims in the model where cost ofcarry exists. Also note the risk-neutral measure is dierent 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 = tdSt atdt +r(Xt tSt)dtXT = VTfor two unknowns, X and . To do so, we nd a probability measure P, under which ertXt is a martingale,then ertXt = E[erTVT[Tt] := Mt. Martingale Representation Theorem gives Mt = M0 +


Wu forsome 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(ertXt) = t[d(ertSt) aertdt] = tertStd

Wt. So (ertXt)t0, where X is given by (5.9.6), is a P-martingale.(ii)Proof. By Itos formula, dYt = Yt[d

Wt + (r 122)dt] + 12Yt2dt = Yt(d

Wt + rdt). So d(ertYt) =ertYtd

Wt and ertYt is a P-martingale. Moreover, if St = S0Yt +Yt

t0aYsds, thendSt = S0dYt +

t0aYsdsdYt +adt =

S0 +



Wt +rdt) +adt = St(d

Wt +rdt) +adt.This shows S satises (5.9.7).Remark: To obtain this formula for S, we rst set Ut = ertSt to remove the rStdt term. The SDE forU is dUt = Utd

Wt + aertdt. Just like solving linear ODE, to remove U in the d

Wt term, we considerVt = Ute

Wt. Itos product formula yieldsdVt = e

WtdUt +Ute



Wt + 122dt

+dUt e



Wt + 122dt

= e

Wtaertdt 122Vtdt.53Note V appears only in the dt term, so multiply the integration factor e122ton both sides of the equation,we getd(e122tVt) = aert

Wt+122tdt.Set Yt = e

Wt+(r122)t, we have d(St/Yt) = adt/Yt. So St = Yt(S0 +

t0adsYs).(iii)Proof.E[ST[Tt] = S0E[YT[Tt] + EYT

t0aYsds +YT

TtaYsds[Tt= S0E[YT[Tt] +

t0aYsdsE[YT[Tt] +a


ds= S0YtE[YTt] +

t0aYsdsYtE[YTt] +a


S0 +




S0 +


Yter(Tt) ar(1 er(Tt)).In particular, E[ST] = S0erT ar(1 erT).(iv)Proof.dE[ST[Tt] = aer(Tt)dt +

S0 +


(er(Tt)dYt rYter(Tt)dt) + arer(Tt)(r)dt=

S0 +



Wt.So E[ST[Tt] is a P-martingale. As we have argued at the beginning of the solution, risk-neutral pricing isvalid even in the presence of cost of carry. So by an argument similar to that of '5.6.2, the process E[ST[Tt]is the futures price process for the commodity.(v)Proof. We solve the equation E[er(Tt)(ST K)[Tt] = 0 for K, and get K = E[ST[Tt]. So ForS(t, T) =FutS(t, T).(vi)Proof. We follow the hint. First, we solve the SDE

dXt = dSt adt +r(Xt St)dtX0 = 0.By our analysis in part (i), d(ertXt) = d(ertSt) aertdt. Integrate from 0 to t on both sides, we getXt = St S0ert+ ar(1 ert) = St S0ert ar(ert 1). In particular, XT = ST S0erT ar(erT 1).Meanwhile, ForS(t, T) = Futs(t, T) = E[ST[Tt] =

S0 +


Yter(Tt)ar(1er(Tt)). So ForS(0, T) =S0erT ar(1 erT) and hence XT = ST ForS(0, T). After the agent delivers the commodity, whose valueis ST, and receives the forward price ForS(0, T), the portfolio has exactly zero value.546. Connections with Partial Dierential Equations6.1. (i)Proof. Zt = 1 is obvious. Note the form of Z is similar to that of a geometric Brownian motion. So by Itosformula, it is easy to obtain dZu = buZudu +uZudWu, u t.(ii)Proof. If Xu = YuZu (u t), then Xt = YtZt = x 1 = x anddXu = YudZu +ZudYu +dYuZu= Yu(buZudu +uZudWu) +Zu

au uuZudu + uZudWu

+uZuuZudu= [YubuZu + (au uu) +uu]du + (uZuYu +u)dWu= (buXu +au)du + (uXu +u)dWu.Remark: To see how to nd the above solution, we manipulate the equation (6.2.4) as follows. First, toremove the term buXudu, we multiply on both sides of (6.2.4) the integrating factor e

ut bvdv. Thend(Xue

ut bvdv) = e

ut bvdv(audu + (u +uXu)dWu).Let Xu = e

ut bvdvXu, au = e

ut bvdvau and u = e

ut bvdvu, then X satises the SDEd Xu = audu + ( u +u Xu)dWu = ( audu + udWu) +u XudWu.To deal with the term u XudWu, we consider Xu = Xue

ut vdWv. Thend Xu = e

ut vdWv[( audu + udWu) +u XudWu] + Xu


ut vdWv(u)dWu + 12e

ut vdWv2udu

+( u +u Xu)(u)e

ut vdWvdu= audu + udWu +u XudWu u XudWu + 12 Xu2udu u( u +u Xu)du= ( au u u 12 Xu2u)du + udWu,where au = aue

ut vdWvand u = ue

ut vdWv. Finally, use the integrating factor e

ut122vdv, we haved


ut 2vdv

= e12

ut 2vdv(d Xu + Xu 122udu) = e12

ut 2vdv[( au u u)du + udWu].Write everything back into the original X, a and , we getd


ut bvdv

ut vdWv+12

ut 2vdv

= e12

ut 2vdv

ut vdWv

ut bvdv[(au uu)du +udWu],i.e.d


= 1Zu[(au uu)du +udWu] = dYu.This inspired us to try Xu = YuZu.6.2. (i)55Proof. The portfolio is self-nancing, so for any t T1, we havedXt = 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,andd(DtXt)= RtDtXtdt +DtdXt= 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]= Dt[1(t)

ft(t, Rt, T1)dt +fr(t, Rt, T1)dRt + 12frr(t, Rt, T1)2(t, Rt)dt


ft(t, Rt, T2)dt +fr(t, Rt, T2)dRt + 12frr(t, Rt, T2)2(t, Rt)dt

Rt(1(t)f(t, Rt, T1) + 2(t)f(t, Rt, T2))dt]= 1(t)Dt[Rtf(t, Rt, T1) +ft(t, Rt, T1) +(t, Rt)fr(t, Rt, T1) + 122(t, Rt)frr(t, Rt, T1)]dt+2(t)Dt[Rtf(t, Rt, T2) +ft(t, Rt, T2) +(t, Rt)fr(t, Rt, T2) + 122(t, Rt)frr(t, Rt, T2)]dt+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) = Stfr(t, Rt, T2) and 2(t) = Stfr(t, Rt, T1), thend(DtXt) = DtSt[(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 getDTXT D0X0 =

T0Dt[[(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 rand 0 t T, DTXT D0X0 > 0. To avoid arbitrage (see, for example, Exercise 5.7), we must have fora.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 2(t) = 0, we getd(DtXt) = (t)DtRtf(t, Rt, T) +ft(t, Rt, T) +(t, Rt)fr(t, Rt, T) + 122(t, Rt)frr(t, Rt, T)

dt+Dt(t, Rt)(t)fr(t, Rt, T)dWt.This is formula (6.9.5).If fr(t, r, T) = 0, then d(DtXt) = (t)Dt

Rtf(t, Rt, T) +ft(t, Rt, T) + 122(t, Rt)frr(t, Rt, T)

dt. Wechoose (t) = signRtf(t, Rt, T) +ft(t, Rt, T) + 122(t, Rt)frr(t, Rt, T). To avoid arbitrage in thiscase, we must have ft(t, Rt, T) + 122(t, Rt)frr(t, Rt, T) = Rtf(t, Rt, T), or equivalently, for any r in therange of Rt, ft(t, r, T) + 122(t, r)frr(t, r, T) = rf(t, r, T).566.3.Proof. We notedds


s0 bvdvC(s, T)

= e

s0 bvdv[C(s, T)(bs) +bsC(s, T) 1] = e

s0 bvdv.So integrate on both sides of the equation from t to T, we obtaine

T0 bvdvC(T, T) e

t0 bvdvC(t, T) =


s0 bvdvds.Since C(T, T) = 0, we have C(t, T) = e

t0 bvdv

Tt e

s0 bvdvds = Tt e

ts bvdvds. Finally, by A

(s, T) =a(s)C(s, T) + 122(s)C2(s, T), we getA(T, T) A(t, T) =

Tta(s)C(s, T)ds + 12

Tt2(s)C2(s, T)ds.Since A(T, T) = 0, we have A(t, T) =

Tt (a(s)C(s, T) 122(s)C2(s, T))ds.6.4. (i)Proof. By the denition of , we have

(t) = e122 Tt C(u,T)du122(1)C(t, T) = 12(t)2C(t, T).So C(t, T) = 2

(t)(t)2. Dierentiate both sides of the equation

(t) = 12(t)2C(t, T), we get

(t) = 122[

(t)C(t, T) +(t)C

(t, T)]= 122[12(t)2C2(t, T) +(t)C

(t, T)]= 144(t)C2(t, T) 122(t)C

(t, T).So C

(t, T) =

144(t)C2(t, T)


/12(t)2= 122C2(t, T) 2

(t)2(t).(ii)Proof. Plug formulas (6.9.8) and (6.9.9) into (6.5.14), we get2

(t)2(t) + 122C2(t, T) = b(1) 2

(t)2(t) + 122C2(t, T) 1,i.e.

(t) b

(t) 122(t) = 0.(iii)Proof. The characteristic equation of

(t) b

(t) 122(t) = 0 is 2 b 122= 0, which gives tworoots 12(b b2+ 22) = 12b with = 12b2+ 22. Therefore by standard theory of ordinary dierentialequations, a general solution of is (t) = e12bt(a1et+ a2et) for some constants a1 and a2. It is theneasy to see that we can choose appropriate constants c1 and c2 so that(t) = c112b +e(12b+)(Tt) c212b e(12b)(Tt).57(iv)Proof. From part (iii), it is easy to see

(t) = c1e(12b+)(Tt)c2e(12b)(Tt). In particular,0 = C(T, T) = 2

(T)2(T) = 2(c1 c2)2(T) .So c1 = c2.(v)Proof. We rst recall the denitions and properties of sinh and cosh:sinh z = ezez2 , cosh z = ez+ez2 , (sinh z)

= cosh z, and (cosh z)

= sinh z.Therefore(t) = c1e12b(Tt)e(Tt)12b + e(Tt)12b

= c1e12b(Tt) 12b 14b22e(Tt)12b +14b22e(Tt)

= 2c12 e12b(Tt)(12b )e(Tt)+ (12b +)e(Tt)

= 2c12 e12b(Tt)[b sinh((T t)) + 2 cosh((T t))].and

(t) = 12b 2c12 e12b(Tt)[b sinh((T t)) + 2 cosh((T t))]+2c12 e12b(Tt)[b cosh((T t)) 22sinh((T t))]= 2c1e12b(Tt) b222 sinh((T t)) + b2 cosh((T t)) b2 cosh((T t)) 222 sinh((T t))

= 2c1e12b(Tt)b24222 sinh((T t))= 2c1e12b(Tt)sinh((T t)).This impliesC(t, T) = 2

(t)2(t) = sinh((T t)) cosh((T t)) + 12b sinh((T t)).(vi)Proof. By (6.5.15) and (6.9.8), A

(t, T) = 2a

(t)2(t) . HenceA(T, T) A(t, T) =


(s)2(s) ds = 2a2 ln (T)(t) ,andA(t, T) = 2a2 ln (T)(t) = 2a2 ln e12b(Tt) cosh((T t)) + 12b sinh((T t)).586.5. (i)Proof. Since g(t, X1(t), X2(t)) = E[h(X1(T), X2(T))[Tt] and ertf(t, X1(t), X2(t)) = E[erTh(X1(T), X2(T))[Tt],iterated conditioning argument shows g(t, X1(t), X2(t)) and ertf(t, X1(t), X2(t)) ar both martingales.(ii) and (iii)Proof. We notedg(t, X1(t), X2(t))= gtdt +gx1dX1(t) +gx2dX2(t) + 12gx1x2dX1(t)dX1(t) + 12gx2x2dX2(t)dX2(t) +gx1x2dX1(t)dX2(t)=gt +gx11 +gx22 + 12gx1x1(211 +212 + 21112) +gx1x2(1121 +1122 +1221 +1222)+12gx2x2(221 +222 + 22122)

dt + martingale part.So we must havegt +gx11 +gx22 + 12gx1x1(211 +212 + 21112) +gx1x2(1121 +1122 +1221 +1222)+12gx2x2(221 +222 + 22122) = 0.Taking = 0 will give part (ii) as a special case. The PDE for f can be similarly obtained.6.6. (i)Proof. Multiply e12bton both sides of (6.9.15), we getd(e12btXj(t)) = e12bt

Xj(t)12bdt + (b2Xj(t)dt + 12dWj(t)

= e12bt12dWj(t).So e12btXj(t) Xj(0) = 12

t0 e12budWj(u) and Xj(t) = e12bt

Xj(0) + 12

t0 e12budWj(u)

. By Theorem4.4.9, Xj(t) is normally distributed with mean Xj(0)e12btand variance ebt4 2

t0 ebudu = 24b(1 ebt).(ii)Proof. Suppose R(t) =dj=1X2j(t), thendR(t) =dj=1(2Xj(t)dXj(t) +dXj(t)dXj(t))=dj=1

2Xj(t)dXj(t) + 142dt


bX2j(t)dt +Xj(t)dWj(t) + 142dt



dt +


R(t)dWj(t).Let B(t) = dj=1

t0Xj(s)R(s)dWj(s), then B is a local martingale with dB(t)dB(t) = dj=1X2j(t)R(t) dt = dt. Soby Levys Theorem, B is a Brownian motion. Therefore dR(t) = (a bR(t))dt +

R(t)dB(t) (a := d42)and R is a CIR interest rate process.59(iii)Proof. By (6.9.16), Xj(t) is dependent on Wj only and is normally distributed with mean e12btXj(0) andvariance 24b[1 ebt]. So X1(t), , Xd(t) are i.i.d. normal with the same mean (t) and variance v(t).(iv)Proof.E



eux2 e(x(t))22v(t)dx



2v(t) dx=


2v(t)e(x (t)12uv(t))2+ 2(t)12uv(t) 2(t)(12uv(t))22v(t)/(12uv(t))dx=

1 2uv(t)

2v(t) e(x (t)12uv(t))22v(t)/(12uv(t))dx e2(t)(12uv(t))2(t)2v(t)(12uv(t))

1 2uv(t)= e u2(t)12uv(t)

1 2uv(t).(v)Proof. By R(t) =dj=1X2j(t) and the fact X1(t), , Xd(t) are i.i.d.,E[euR(t)] = (E[euX21(t)])d= (1 2uv(t))d2eud2(t)12uv(t)= (1 2uv(t))2a2eebtuR(0)12uv(t).6.7. (i)Proof. ertc(t, St, Vt) = E[erT(ST K)+[Tt] is a martingale by iterated conditioning argument. Sinced(ertc(t, St, Vt))= ertc(t, St, Vt)(r) +ct(t, St, Vt) +cs(t, St, Vt)rSt +cv(t, St, Vt)(a bVt) + 12css(t, St, Vt)VtS2t+12cvv(t, St, Vt)2Vt +csv(t, St, Vt)VtSt

dt + martingale part,we conclude rc = ct +rscs +cv(a bv) + 12cssvs2+ 12cvv2v +csvsv. This is equation (6.9.26).(ii)60Proof. Suppose c(t, s, v) = sf(t, log s, v) er(Tt)Kg(t, log s, v), thenct = sft(t, log s, v) rer(Tt)Kg(t, log s, v) er(Tt)Kgt(t, log s, v),cs = f(t, log s, v) +sfs(t, log s, v)1s er(Tt)Kgs(t, log s, v)1s,cv = sfv(t, log s, v) er(Tt)Kgv(t, log s, v),css = fs(t, log s, v)1s +fss(t, log s, v)1s er(Tt)Kgss(t, log s, v) 1s2 +er(Tt)Kgs(t, log s, v) 1s2,csv = fv(t, log s, v) +fsv(t, log s, v) er(Tt)Ks gsv(t, log s, v),cvv = sfvv(t, log s, v) er(Tt)Kgvv(t, log s, v).Soct +rscs + (a bv)cv + 12s2vcss +svcsv + 122vcvv= sft rer(Tt)Kg er(Tt)Kgt +rsf +rsfs rKer(Tt)gs + (a bv)(sfv er(Tt)Kgv)+12s2v1sfs + 1sfss er(Tt)Ks2gss +er(Tt)Kgss2


fv +fsv er(Tt)Ks gsv

+122v(sfvv er(Tt)Kgvv)= sft + (r + 12v)fs + (a bv +v)fv + 12vfss +vfsv + 122vfvv

Ker(Tt)gt + (r 12v)gs+(a bv)gv + 12vgss +vgsv + 122vgvv

+rsf rer(Tt)Kg= rc.That is, c satises the PDE (6.9.26).(iii)Proof. First, by Markov property, f(t, Xt, Vt) = E[1{XTlog K}[Tt]. So f(T, Xt, Vt) = 1{XTlog K}, whichimplies f(T, x, v) = 1{xlog K} for all x R, v 0. Second, f(t, Xt, Vt) is a martingale, so by dierentiatingf and setting the dt term as zero, we have the PDE (6.9.32) for f. Indeed,df(t, Xt, Vt) =ft(t, Xt, Vt) +fx(t, Xt, Vt)(r + 12Vt) +fv(t, Xt, Vt)(a bvt +Vt) + 12fxx(t, Xt, Vt)Vt+12fvv(t, Xt, Vt)2Vt +fxv(t, Xt, Vt)Vt

dt + martingale part.So we must have ft + (r + 12v)fx + (a bv +v)fv + 12fxxv + 12fvv2v +vfxv = 0. This is (6.9.32).(iv)Proof. Similar to (iii).(v)Proof. c(T, s, v) = sf(T, log s, v) er(Tt)Kg(T, log s, v) = s1{log slog K} K1{log slog K} = 1{sK}(s K) = (s K)+.6.8.61Proof. We follow the hint. Suppose h is smooth and compactly supported, then it is legitimate to exchangeintegration and dierentiation:gt(t, x) = t

0h(y)p(t, T, x, y)dy =

0h(y)pt(t, T, x, y)dy,gx(t, x) =

0h(y)px(t, T, x, y)dy,gxx(t, x) =

0h(y)pxx(t, T, x, y)dy.So (6.9.45) implies 0 h(y)

pt(t, T, x, y) +(t, x)px(t, T, x, y) + 122(t, x)pxx(t, T, x, y)

dy = 0. By the ar-bitrariness of h and assuming , pt, px, v, pxx are all continuous, we havept(t, T, x, y) +(t, x)px(t, T, x, y) + 122(t, x)pxx(t, T, x, y) = 0.This is (6.9.43).6.9.Proof. We rst note dhb(Xu) = h


b(Xu)dXudXu =


b(Xu)(u, Xu) + 122(u, Xu)h



b(Xu)(u, Xu)dWu. Integrate on both sides of the equation, we havehb(XT) hb(Xt) =


b(Xu)(u, Xu) + 122(u, Xu)h


du + martingale part.Take expectation on both sides, we getEt,x[hb(XT) hb(Xt)] =

hb(y)p(t, T, x, y)dy h(x)=


b(Xu)(u, Xu) + 122(u, Xu)h




b(y)(u, y) + 122(u, y)h


p(t, u, x, y)dydu.Since hb vanishes outside (0, b), the integration range can be changed from (, ) to (0, b), which gives(6.9.48).By integration-by-parts formula, we have

b0(u, y)p(t, u, x, y)h

b(y)dy = hb(y)(u, y)p(t, u, x, y)[b0

b0hb(y) y((u, y)p(t, u, x, y))dy=

b0hb(y) y((u, y)p(t, u, x, y))dy,and

b02(u, y)p(t, u, x, y)h

b(y)dy =

b0y(2(u, y)p(t, u, x, y))h

b(y)dy =

b02y(2(u, y)p(t, u, x, y))hb(y)dy.Plug these formulas into (6.9.48), we get (6.9.49).Dierentiate w.r.t. T on both sides of (6.9.49), we have

b0hb(y) T p(t, T, x, y)dy =

b0y[(T, y)p(t, T, x, y)]hb(y)dy + 12

b02y2[2(T, y)p(t, T, x, y)]hb(y)dy,62that is,

b0hb(y) T p(t, T, x, y) + y((T, y)p(t, T, x, y)) 122y2(2(T, y)p(t, T, x, y))

dy = 0.This is (6.9.50).By (6.9.50) and the arbitrariness of hb, we conclude for any y (0, ),T p(t, T, x, y) + y((T, y)p(t, T, x, y)) 122y2(2(T, y)p(t, T, x, y)) = 0.6.10.Proof. Under the assumption that limy(y K)ry p(0, T, x, y) = 0, we have

K(yK) y(ry p(0, T, x, y))dy = (yK)ry p(0, T, x, y)[K+

Kry p(0, T, x, y)dy =

Kry p(0, T, x, y)dy.If we further assume (6.9.57) and (6.9.58), then use integration-by-parts formula twice, we have12

K(y K) 2y2(2(T, y)y2 p(0, T, x, y))dy= 12(y K) y(2(T, y)y2 p(0, T, x, y))[K

Ky(2(T, y)y2 p(0, T, x, y))dy

= 12(2(T, y)y2 p(0, T, x, y)[K)= 122(T, K)K2 p(0, T, x, K).Therefore,cT(0, T, x, K) = rc(0, T, x, K) +erT

K(y K) pT(0, T, x, y)dy= rerT

K(y K) p(0, T, x, y)dy +erT

K(y K) pT(0, T, x, y)dy= rerT

K(y K) p(0, T, x, y)dy erT

K(y K) y(ry p(t, T, x, y))dy+erT

K(y K)122y2(2(T, y)y2 p(t, T, x, y))dy= rerT

K(y K) p(0, T, x, y)dy +erT

Kry p(0, T, x, y)dy+erT 122(T, K)K2 p(0, T, x, K)= rerTK

K p(0, T, x, y)dy + 12erT2(T, K)K2 p(0, T, x, K)= rKcK(0, T, x, K) + 122(T, K)K2cKK(0, T, x, K).7. Exotic Options7.1. (i)63Proof. Since (, s) = 1[log s + (r 122)] = log s 12 + r122,t(, s) = log s (12)32t + r 1221212t= 12log s1 (1) r 122(1)

= 12 1log ss + (r 122))

= 12 (, 1s).(ii)Proof.x(, xc) = x

1log xc + (r 122)

= 1x ,x(, cx) = x

1log cx + (r 122)

= 1x .(iii)Proof.N

((, s)) = 12e(,s)2= 12e(log s+r)22(log s+r)+144222 .ThereforeN

(+(, s))N

((, s)) = e22(log s+r)22 = ersand erN

((, s)) = sN

(+(, s)).(iv)Proof.N

((, s))N

((, s1)) = e[(log s+r)2(log 1s+r)2]2(log slog 1s)22 = e4r log s22 log s22 = e( 2r21) log s= s( 2r21).So N

((, s1)) = s( 2r21)N

((, s)).(v)Proof. +(, s) (, s) = 1

log s + (r + 122)


log s + (r 122)

= 12 = .(vi)Proof. (, s) (, s1) = 1

log s + (r 122)


log s1+ (r 122)

= 2 log s .(vii)Proof. N

(y) = 12ey22 , so N

(y) = 12ey22 (y22 )

= yN

(y).64To be continued ...7.3.Proof. We note ST = S0e

WT= Ste(


Wt), WT Wt = (

WT Wt) + (T t) is independent of Tt,suptuT(Wu Wt) is independent of Tt, andYT = S0e

MT= S0e suptuT Wu1{

MtsuptuT Wt} +S0e


Mt>suptuT Wu}= Ste suptuT(


Wt)1{YtSte suptuT(


Wt)} +Yt1{YtSte suptuT(


Wt)}.So E[f(ST, YT)[Tt] = E[f(xSTtS0 , xYTtS0 1{yxYTtS0 } + y1{yxYTtS0 })], where x = St, y = Yt. ThereforeE[f(ST, YT)[Tt] is a Borel function of (St, Yt).7.4.Proof. By Cauchys inequality and the monotonicity of Y , we have[mj=1(Ytj Ytj1)(Stj Stj1)[ mj=1[Ytj Ytj1[[Stj Stj1[

mj=1(Ytj Ytj1)2

mj=1(Stj Stj1)2

max1jm[Ytj Ytj1[(YT Y0)

mj=1(Stj Stj1)2.If we increase the number of partition points to innity and let the length of the longest subintervalmax1jm[tjtj1[ approach zero, then

mj=1(Stj Stj1)2

[S]T [S]0 < and max1jm[Ytj Ytj1[ 0 a.s. by the continuity of Y . This implies mj=1(Ytj Ytj1)(Stj Stj1) 0.8. American Derivative Securities8.1.Proof. v

L(L+) = (K L)(2r2)(xL)2r21 1L

x=L= 2r2L(K L). So v

L(L+) = v

L(L) if and only if 2r2L(K L) = 1. Solve for L, we get L = 2rK2r+2.8.2.Proof. By the calculation in Section 8.3.3, we can see v2(x) (K2 x)+ (K1 x)+, rv2(x) rxv

2(x) 122x2v

2(x) 0 for all x 0, and for 0 x < L1 < L2,rv2(x) rxv

2(x) 122x2v

2(x) = rK2 > rK1 > 0.So the linear complementarity conditions for v2 imply v2(x) = (K2 x)+= K2 x > K1 x = (K1 x)+on [0, L1]. Hence v2(x) does not satisfy the third linear complementarity condition for v1: for each x 0,equality holds in either (8.8.1) or (8.8.2) or both.8.3. (i)65Proof. Suppose x takes its values in a domain bounded away from 0. By the general theory of lineardierential equations, if we can nd two linearly independent solutions v1(x), v2(x) of (8.8.4), then anysolution of (8.8.4) can be represented in the form of C1v1+C2v2 where C1 and C2 are constants. So it sucesto nd two linearly independent special solutions of (8.8.4). Assume v(x) = xpfor some constant p to bedetermined, (8.8.4) yields xp(rpr122p(p1)) = 0. Solve the quadratic equation 0 = rpr122p(p1) =(122p r)(p 1), we get p = 1 or 2r2. So a general solution of (8.8.4) has the form C1x +C2x2r2.(ii)Proof. Assume there is an interval [x1, x2] where 0 < x1 < x2 < , such that v(x) 0 satises (8.3.19)with equality on [x1, x2] and satises (8.3.18) with equality for x at and immediately to the left of x1 andfor x at and immediately to the right of x2, then we can nd some C1 and C2, so that v(x) = C1x+C2x2r2on [x1, x2]. If for some x0 [x1, x2], v(x0) = v

(x0) = 0, by the uniqueness of the solution of (8.8.4), wewould conclude v 0. This is a contradiction. So such an x0 cannot exist. This implies 0 < x1 < x2 < K(if K x2, v(x2) = (K x2)+= 0 and v

(x2)=the right derivative of (K x)+at x2, which is 0). 1Thuswe have four equations for C1 and C2:

C1x1 +C2x2r21 = K x1C1x2 +C2x2r22 = K x2C1 2r2C2x2r211 = 1C1 2r2C2x2r212 = 1.Since x1 = x2, the last two equations imply C2 = 0. Plug C2 = 0 into the rst two equations, we haveC1 = Kx1x1 = Kx2x2 ; plug C2 = 0 into the last two equations, we have C1 = 1. Combined, we would havex1 = x2. Contradiction. Therefore our initial assumption is incorrect, and the only solution v that satisesthe specied conditions in the problem is the zero solution.(iii)Proof. If in a right neighborhood of 0, v satises (8.3.19) with equality, then part (i) implies v(x) = C1x +C2x2r2for some constants C1 and C2. Then v(0) = limx0v(x) = 0 < (K 0)+, i.e. (8.3.18) will beviolated. So we must have rv rxv


> 0 in a right neighborhood of 0. According to (8.3.20),v(x) = (Kx)+near o. So v(0) = K. We have thus concluded simultaneously that v cannot satisfy (8.3.19)with equality near 0 and v(0) = K, starting from rst principles (8.3.18)-(8.3.20).(iv)Proof. This is already shown in our solution of part (iii): near 0, v cannot satisfy (8.3.19) with equality.(v)Proof. If v satisfy (Kx)+with equality for all x 0, then v cannot have a continuous derivative as statedin the problem. This is a contradiction.(vi)1Note we have interpreted the condition v(x) satises (8.3.18) with equality for x at and immediately to the right of x2as v(x2) = (K x2)+and v

(x2) =the right derivative of (K x)+at x2. This is weaker than v(x) = (K x) in a rightneighborhood of x2.66Proof. By the result of part (i), we can start with v(x) = (K x)+on [0, x1] and v(x) = C1x +C2x2r2on[x1, ). By the assumption of the problem, both v and v

are continuous. Since (Kx)+is not dierentiableat K, we must have x1 K.This gives us the equations

K x1 = (K x1)+= C1x1 +C2x2r211 = C1 2r2C2x2r211 .Because v is assumed to be bounded, we must have C1 = 0 and the above equations only have two unknowns:C2 and x1. Solve them for C2 and x1, we are done.8.4. (i)Proof. This is already shown in part (i) of Exercise 8.3.(ii)Proof. We solve for A, B the equations

AL2r2+BL = K L2r2AL2r21+B = 1,and we obtain A = 2KL2r22+2r , B = 2rKL(2+2r) 1.(iii)Proof. By (8.8.5), B > 0. So for x K, f(x) BK > 0 = (K x)+. If L x < K,f(x) (K x)+= 2KL2r22+ 2r x2r2+ 2rKxL(2+ 2r) K = x2r2KL2r2

2+ 2r(xL)2r2 +1(2+ 2r)(xL)2r2

(2+ 2r)L .Let g() = 2+ 2r2r2 +1 (2+ 2r)2r2with 1. Then g(1) = 0 and g

() = 2r(2r2 + 1)2r2 (2+2r)2r22r21= 2r2(2+ 2r)2r21( 1) 0. So g() 0 for any 1. This shows f(x) (K x)+forL x < K. Combined, we get f(x) (K x)+for all x L.(iv)Proof. Since limxv(x) = limxf(x) = and limxvL(x) = limx(K L)( xL)2r2= 0, v(x)and vL(x) are dierent. By part (iii), v(x) (K x)+. So v satises (8.3.18). For x L, rv rxv


= rf rxf 122x2f

= 0. For 0 x L, rv rxv


= r(Kx) +rx = rK. Combined,rv rxv


0 for x 0. So v satises (8.3.19). Along the way, we also showed v satises (8.3.20).In summary, v satises the linear complementarity condition (8.3.18)-(8.3.20), but v is not the function vLgiven by (8.3.13).(v)Proof. By part (ii), B = 0 if and only if 2rKL(2+2r) 1 = 0, i.e. L = 2rK2r+2. In this case, v(x) = Ax2r2=2K2+2r(xL)2r2= (K L)(xL)2r2= vL(x), on the interval [L, ).8.5. The diculty of the dividend-paying case is that from Lemma 8.3.4, we can only obtain E[e(ra)L],not E[erL]. So we have to start from Theorem 8.3.2.(i)67Proof. By (8.8.9), St = S0e

Wt+(ra122)t. Assume S0 = x, then St = L if and only if

Wt 1(r a 122)t = 1 log xL. By Theorem 8.3.2,E[erL] = e1 log xL


12 (ra122)2+2r

.If we set = 12(r a 122) + 1

12(r a 12)2+ 2r, we can write E[erL] as e log xL = (xL). Sothe risk-neutral expected discounted pay o of this strategy isvL(x) =

K x, 0 x L(K L)(xL), x > L.(ii)Proof. LvL(x) = (xL)(1 (KL)L ). Set LvL(x) = 0 and solve for L, we have L = K+1.(iii)Proof. By It os formula, we haved


= ertrvL(St) +v

L(St)(r a)St + 12v


dt +ertv


Wt.If x > L,rvL(x) +v

L(x)(r a)x + 12v

L(x)2x2= r(K L)


+ (r a)x(K L)()x1L+ 122x2()( 1)(K L)x2L= (K L)


r (r a) + 122( + 1)

.By the denition of , if we dene u = r a 122, we haver + (r a) 122( + 1)= r 1222+(r a 122)= r 122

u2 + 1

u22 + 2r


u2 + 1

u22 + 2r

u= r 122

u24 + 2u3

u22 + 2r + 12

u22 + 2r

+ u22 + u

u22 + 2r= r u222 u

u22 + 2r 12

u22 + 2r

+ u22 + u

u22 + 2r= 0.If x < L, rvL(x) +v

L(x)(r a)x + 12v

L(x)2x2= r(K x) +(1)(r a)x = rK +ax. Combined,we getd


= ert1{St r2r2 0 > r 2.Since 2< 0, we have obtained a contradiction. So our initial assumption is incorrect, and rK aL 0must be true.(iv)Proof. The proof is similar to that of Corollary 8.3.6. Note the only properties used in the proof of Corollary8.3.6 are that ertvL(St) is a supermartingale, ertLvL(StL) is a martingale, and vL(x) (Kx)+.Part (iii) already proved the supermartingale-martingale property, so it suces to show vL(x) (K x)+in our problem. Indeed, by 0, L = K+1 < K. For x K > L, vL(x) > 0 = (Kx)+; for 0 x < L,vL(x) = K x = (K x)+; nally, for L x K,ddx(vL(x) (K x)) = (K L)x1L+ 1 (K L)L1L+ 1 = (K K + 1) 1K+1+ 1 = 0.and (vL(x) (K x))[x=L = 0. So for L x K, vL(x) (K x)+ 0. Combined, we havevL(x) (K x)+ 0 for all x 0.8.6.Proof. By Lemma 8.5.1, Xt = ert(St K)+is a submartingale. For any 0,T, Theorem 8.8.1 impliesE[erT(ST K)+] E[erT(ST K)+] E[er(S K)+1{ 0. Take logarithm on both sides and plug in the expression of A(t, T), we getlog B(0, T) = Y1(0)C1(0, T) Y2(0)C2(0, T) +

T012(C21(s, T) +C22(s, T)) 0(s)

ds.Taking derivative w.r.t. T, we haveT log B(0, T) = Y1(0) T C1(0, T) Y2(0) T C2(0, T) + 12C21(T, T) + 12C22(T, T) 0(T).78So0(T) = Y1(0) T C1(0, T) Y2(0) T C2(0, T) T log B(0, T)=


1e1T 2122 e2T

Y2(0)2e2T T log B(0, T) if 1 = 2Y1(0)


Y2(0)2e2T T log B(0, T) if 1 = 2.10.4. (i)Proof.dXt = dXt +KeKt

t0eKu(u)dudt (t)dt= KXtdt + dBt +KeKt

t0eKu(u)dudt= K Xtdt + dBt.(ii)Proof.

Wt = CBt =

11 0 1121212

1 00 2

Bt =

1 0 12112


W is a martingale with '

W1`t = 'B1`t = t, '

W2`t = ' 12B1+ 112B2`t = 2t12 + t122 12t =2+12212 t = t, and '


W2`t = 'B1, 12B1+ 112B2`t = t12 + t12 = 0. Therefore W is atwo-dimensional BM. Moreover, dYt = CdXt = CK Xtdt+CdBt = CKC1Ytdt+d

Wt = Ytdt+d

Wt,where = CKC1=

11 0 1121212

1 01 2


1212 011211


11 0 1112 12122212

1 02 2

1 2


1 02(21)1212 2

.(iii)Proof.Xt = Xt +eKt

t0eKu(u)du = C1Yt +eKt


1 02 2

1 2




1Y1(t)2Y1(t) +2

1 2Y2(t)


t0eKu(u)du.79So Rt = X2(t) = 2Y1(t)+2

1 2Y2(t)+0(t), where 0(t) is the second coordinate of eKt

t0 eKu(u)duand can be derived explicitly by Lemma 10.2.3. Then 1 = 2 and 2 = 2

1 2.10.5.Proof. We note C(t, T) and A(t, T) are dependent only on T t. So C(t, t + ) and A(t, t + ) aare constantswhen is xed. SoddtLt = B(t, t + )[C(t, t + )R

(t) A(t, t + )] B(t, t + )= 1 [C(t, t + )R

(t) +A(t, t + )]= 1 [C(0, )R

(t) +A(0, )].Hence L(t2)L(t1) = 1 C(0, )[R(t2)R(t1)]+1 A(0, )(t2t1). Since L(t2)L(t1) is a linear transformation,it is easy to verify that their correlation is 1.10.6. (i)Proof. If 2 = 0, then dRt = 1dY1(t) = 1(1Y1(t)dt +d

W1(t)) = 1

(01 Rt1 )1dt +d


= (01 1Rt)dt +1d

W1(t). So a = 01 and b = 1.(ii)Proof.dRt = 1dY1(t) +2dY2(t)= 11Y1(t)dt +1d

W1(t) 221Y1(t)dt 22Y2(t)dt +2d

W2(t)= Y1(t)(11 +221)dt 22Y2(t)dt +1d

W1(t) +2d

W2(t)= Y1(t)21dt 22Y2(t)dt +1d

W1(t) +2d

W2(t)= 2(Y1(t)1 +Y2(t)2)dt +1d

W1(t) +2d

W2(t)= 2(Rt 0)dt +

21 +22 1

21 +22d

W1(t) + 2

21 +22d

W2(t).So a = 20, b = 2, =

21 +22 and Bt = 121+22

W1(t) + 221+22

W2(t).10.7. (i)Proof. We use the canonical form of the model as in formulas (10.2.4)-(10.2.6). By (10.2.20),dB(t, T) = df(t, Y1(t), Y2(t))= deY1(t)C1(Tt)Y2(t)C2(Tt)A(Tt)= dt term +B(t, T)[C1(T t)d

W1(t) C2(T t)d

W2(t)]= dt term +B(t, T)(C1(T t), C2(T t))




.So the volatility vector of B(t, T) under P is (C1(T t), C2(T t)). By (9.2.5), WTj (t) = t0 Cj(T u)du +

Wj(t) (j = 1, 2) form a two-dimensional PTBM.(ii)80Proof. Under the T-forward measure, the numeraire is B(t, T). By risk-neutral pricing, at time zero therisk-neutral price V0 of the option satisesV0B(0, T) = ET 1B(T, T)(eC1(TT)Y1(T)C2(TT)Y2(T)A(TT)K)+

.Note B(T, T) = 1, we get (10.7.19).(iii)Proof. We can rewrite (10.2.4) and (10.2.5) as

dY1(t) = 1Y1(t)dt +d

WT1 (t) C1(T t)dtdY2(t) = 21Y1(t)dt 2Y2(t)dt +d

WT2 (t) C2(T t)dt.Then

Y1(t) = Y1(0)e1t+

t0 e1(st)d

WT1 (s)

t0 C1(T s)e1(st)dsY2(t) = Y0e2t21

t0 Y1(s)e2(st)ds +

t0 e2(st)d


t0 C2(T s)e2(st)ds.So (Y1, Y2) is jointly Gaussian and X is therefore Gaussian.(iv)Proof. First, we recall the Black-Scholes formula for call options: if dSt = Stdt +Std

Wt, thenE[eT(S0eWT+(122)TK)+] = S0N(d+) KeTN(d)with d = 1T (log S0K +( 122)T). Let T = 1, S0 = 1 and = W1+( 122), then d= N( 122, 2)andE[(eK)+] = eN(d+) KN(d),where d = 1(log K + ( 122)) (dierent from the problem. Check!). Since under PT, X d=N( 122, 2), we haveB(0, T)ET[(eXK)+] = B(0, T)(eN(d+) KN(d)).10.11.Proof. On each payment date Tj, the payo of this swap contract is (K L(Tj1, Tj1)). Its no-arbitrageprice at time 0 is (KB(0, Tj) B(0, Tj)L(0, Tj1)) by Theorem 10.4. So the value of the swap isn+1j=1[KB(0, Tj) B(0, Tj)L(0, Tj1)] = Kn+1j=1B(0, Tj) n+1j=1B(0, Tj)L(0, Tj1).10.12.81Proof. Since L(T, T) = 1B(T,T+)B(T,T+) TT, we haveE[D(T +)L(T, T)] = E[E[D(T +)L(T, T)[TT]]= E1 B(T, T +)B(T, T +) E[D(T +)[TT]

= E1 B(T, T +)B(T, T +) D(T)B(T, T +)

= ED(T) D(T)B(T, T +)

= B(0, T) B(0, T +)= B(0, T +)L(0, T).11. Introduction to Jump Processes11.1. (i)Proof. First, M2t = N2t 2tNt + 2t2. So E[M2t ] < . f(x) = x2is a convex function. So by conditionalJensens inequality,E[f(Mt)[Ts] f(E[Mt[Ts]) = f(Ms), s t.So M2t is a submartingale.(ii)Proof. We note Mt has independent and stationary increment. So s t, E[M2t M2s[Ts] = E[(Mt Ms)2[Ts] + E[(Mt Ms) 2Ms[Ts] = E[M2ts] + 2MsE[Mts] = V ar(Nts) + 0 = (t s). That is,E[M2t t[Ts] = M2s s.11.2.Proof. P(Ns+t = k[Ns = k) = P(Ns+t Ns = 0[Ns = k) = P(Nt = 0) = et= 1 t + O(t2). Similarly,we have P(Ns+t = k + 1[Ns = k) = P(Nt = 1) = (t)11! et= t(1 t + O(t2)) = t + O(t2), andP(Ns+t k + 2[N2 = k) = P(Nt 2) =k=2(t)kk! et= O(t2).11.3.Proof. For any t u, we haveESuSt


= E[( + 1)NtNue(tu)[Tt]= e(tu)E[( + 1)Ntu]= e(tu)E[eNtu log(+1)]= e(tu)e(tu)(elog(+1)1)(by (11.3.4))= e(tu)e(tu)= 1.So St = E[Su[Tt] and S is a martingale.11.4.82Proof. The problem is ambiguous in that the relation between N1 and N2 is not clearly stated. Accordingto page 524, paragraph 2, we would guess the condition should be that N1 and N2 are independent.Suppose N1 and N2 are independent. Dene M1(t) = N1(t) 1t and M2(t) = N2(t) 2t. Then byindependence E[M1(t)M2(t)] = E[M1(t)]E[M2(t)] = 0. Meanwhile, by Itos product formula, M1(t)M2(t) =

t0 M1(s)dM2(s) +

t0 M2(s)dM1(s) +[M1, M2]t. Both

t0 M1(s)dM2(s) and

t0 M2(s)dM1(s) are mar-tingales. So taking expectation on both sides, we get 0 = 0 + E[M1, M2]t = E[0