Stochastic Calculus for Finance, Vol. I and II, Solution

Stochastic Calculus for Finance, Volume I and II by Yan Zeng Last updated: August 20, 2007 This is a solution manual fo...

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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 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 definition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can define 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 involving money and 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 first 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 5.7. 1.2.

1

Proof. X1 (u) = ∆0 × 8 + Γ0 × 3 − 54 (4∆0 + 1.20Γ0 ) = 3∆0 + 1.5Γ0 , and X1 (d) = ∆0 × 2 − 45 (4∆0 + 1.20Γ0 ) = −3∆0 − 1.5Γ0 . 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 ¯ 0 and ∆ ¯ 0 are chosen in such a way that V1 can be payoff of the derivative security at time 1. Suppose X ¯ ¯ ¯ replicated: (1 + r)(X0 − ∆0 S0 ) + ∆0 S1 = V1 . Using the notation of the problem, suppose an agent begins with 0 wealth and at time zero buys ∆0 shares of stock and Γ0 options. He then puts his cash position ¯ 0 in a money market account. At time one, the value of the agent’s portfolio of stock, option −∆0 S0 − Γ0 X and money market assets is ¯ 0 ). X1 = ∆0 S1 + Γ0 V1 − (1 + r)(∆0 S0 + Γ0 X Plug in the expression of V1 and sort out terms, we have ¯ 0 Γ0 )( X1 = S0 (∆0 + ∆

S1 − (1 + r)). S0

¯0, Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X then there will no arbitrage. 1.3. h i h i S0 1 1+r−d u−1−r 1+r−d u−1−r Proof. V0 = 1+r u−d S1 (H) + u−d S1 (T ) = 1+r u−d u + u−d d = S0 . This is not surprising, since 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 S00 . Correspondingly, we can get u, d and u0 , d0 . Because they are determined by S0 and S00 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, S0 =

1+r−d u−d S1 (H)

+

u−1−r u−d S1 (T )

1+r

,

S00

=

1+r−d0 u0 −d0 S1 (H)

+

u0 −1−r u0 −d0 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 p˜Vn+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)) + p˜Vn+1 (H) + q˜Vn+1 (T ) =

= p˜Vn+1 (T ) + q˜Vn+1 (T ) = Vn+1 (T ).

1.6.

2

Proof. The bank’s trader should set up a replicating portfolio whose payoff is the opposite of the option’s payoff. More precisely, we solve the equation (1 + r)(X0 − ∆0 S0 ) + ∆0 S1 = −(S1 − K)+ . Then X0 = −1.20 and ∆0 = − 21 . 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 payoff. 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 payoff. 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 payoff 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 0.6933 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 payoff. After they cancel out, we end up with 1.376(1 + r)3 in the separate money market account. 1.8. (i) Proof. vn (s, y) = 52 (vn+1 (2s, y + 2s) + vn+1 ( 2s , y + 2s )). (ii) Proof. 1.696. (iii) Proof. δn (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 n −dn pen = 1+r en = 1 − pen . Then un −dn and q Vn =

pen Vn+1 (H) + qen Vn+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 ) . (un −dn )Sn

(iii) 3

(T ) (H) = SnS+10 = 1+ S10n and dn = Sn+1 = SnS−10 = 1− S10n . So the risk-neutral probabilities Proof. un = Sn+1 Sn Sn n n n at time n are p˜n = u1−d = 12 and q˜n = 12 . Risk-neutral pricing implies the price of this call at time zero is n −dn 9.375.

2. Probability Theory on Coin Toss Space 2.1. (i) Proof. P (Ac ) + P (A) =

P

ω∈Ac

P (ω) +

P

ω∈A

P (ω) =

P

ω∈Ω

P (ω) = 1.

(ii) Proof. By induction, P Pit suffices to work P on the case N = 2. When A1 and A2 are disjoint, P (A1 ∪ A2 ) = ω∈A1 ∪A2 P (ω) = ω∈A1 P (ω) + ω∈A2 P (ω) = 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) Proof. Pe(S3 = 32) = pe3 = 81 , Pe(S3 = 8) = 3e p2 qe = 38 , Pe(S3 = 2) = 3e pqe2 = 83 , and Pe(S3 = 0.5) = qe3 = 81 . (ii) e 1 ] = 8Pe(S1 = 8) + 2Pe(S1 = 2) = 8e e 2 ] = 16e Proof. E[S p + 2e q = 5, E[S p2 + 4 · 2e pqe + 1 · qe2 = 6.25, and 1 3 3 1 e 3 ] = 32 · + 8 · + 2 · + 0.5 · = 7.8125. So the average rates of growth of the stock price under Pe E[S 8 8 8 8 e2 = 7.8125 are, respectively: re0 = 54 − 1 = 0.25, re1 = 6.25 5 − 1 = 0.25 and r 6.25 − 1 = 0.25. (iii) 8 1 Proof. P (S3 = 32) = ( 23 )3 = 27 , P (S3 = 8) = 3 · ( 23 )2 · 13 = 49 , P (S3 = 2) = 2 · 19 = 29 , and P (S3 = 0.5) = 27 . Accordingly, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13.5. So the average rates of growth of the stock price under P are, respectively: r0 = 46 − 1 = 0.5, r1 = 69 − 1 = 0.5, and r2 = 13.5 9 − 1 = 0.5.

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 Proof. En [ SSn+1 ] = En [eσXn+1 eσ +e −σ ] = n

2 σXn+1 ] eσ +e−σ E[e

= 1.

2.5. (i) Pn−1 Pn−1 Pn−1 Pn−1 Pn−1 Proof. 2In = 2 j=0 Mj (Mj+1 − Mj ) = 2 j=0 Mj Mj+1 − j=1 Mj2 − j=1 Mj2 = 2 j=0 Mj Mj+1 + Pn−1 2 Pn−1 Pn−1 Pn−1 2 − j=0 Mj2 = Mn2 − j=0 (Mj+1 − Mj )2 = Mn2 − j=0 Xj+1 = Mn2 − n. Mn2 − j=0 Mj+1 (ii) Proof. En [f (In+1 )] = En [f (In + Mn (Mn+1 − Mn ))] = En [f (In + Mn Xn+1 )] = 12 [f (In + Mn ) + f (In − Mn )] = √ √ √ g(In ), where g(x) = 12 [f (x + 2x + n) + f (x − 2x + n)], since 2In + n = |Mn |. 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 represented by X = −1. We also suppose P (X = 1) = P (X = −1) = 21 . Define S1 = X1 and Sn+1 = Sn +bn (X1 , · · · , Xn )Xn+1 , where bn (·) is a bounded function on {−1, 1}n , 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 )] = 12 [f (Sn + bn (X1 , · · · , Xn )) + f (Sn − bn (X1 , · · · , Xn ))]. Then 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) 0 Proof. Note Mn = En [MN ] and Mn0 = En [MN ].

(ii) Proof. In the proof of Theorem 1.2.2, we proved by induction that Xn = Vn where Xn is defined 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 e which consists of stock and money market accounts. Since ( (1+r) n )0≤n≤N is a martingale under P (Theorem Vn e 2.4.5), ( n )0≤n≤N is a martingale under P . (1+r)

(iii) Proof.

Vn0 (1+r)n

= En

h

VN (1+r)N

i

, so V00 ,

V10 1+r ,

···,

0 VN −1 , VN (1+r)N −1 (1+r)N

is a martingale under Pe.

(iv) Proof. Combine (ii) and (iii), then use (i). 2.9. (i) S1 (H) = 2, d0 = S1S(H) = 21 , S0 0 and d1 (T ) = SS21(T(TT)) = 1. 1 0 −d0 So pe0 = 1+r e0 = 21 , pe1 (H) u0 −d0 = 2 , q 5 qe1 (T ) = 6 . Therefore Pe(HH) = pe0 pe1 (H) = 14 , 5 qe0 qe1 (T ) = 12 .

Proof. u0 =

u1 (H) = =

S2 (HH) S1 (H)

1+r1 (H)−d1 (H) u1 (H)−d1 (H)

= 1.5, d1 (H) =

S2 (HT ) S1 (H)

= 1, u1 (T ) =

= 12 , qe1 (H) = 21 , pe1 (T ) =

Pe(HT ) = pe0 qe1 (H) =

1 4,

S2 (T H) S1 (T )

1+r1 (T )−d1 (T ) u1 (T )−d1 (T )

Pe(T H) = qe0 pe1 (T ) =

1 12

=4

= 61 , and

and Pe(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 modifications (i.e. Pe would be constructed according to conditional probabilities Pe(ωn+1 = H|ω1 , · · · , ωn ) := pen and Pe(ωn+1 = T |ω1 , · · · , ωn ) := qen . Cf. notes on page 39.). So the time-zero value iof an option that pays off V2 at time two is given by the risk-neutral pricing formula h V2 e V0 = E (1+r0 )(1+r1 ) . (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2.4, V1 (T ) =

p e1 (T )V2 (T H)+e q1 (T )V2 (T T ) 1+r1 (T )

=

1 9,

and V0 =

p e0 V1 (H)+e q0 V1 (T ) 1+r0

5

≈ 1.

p e1 (H)V2 (HH)+e q1 (H)V2 (HT ) 1+r1 (H)

=

(iii) Proof. ∆0 =

V1 (H)−V1 (T ) S1 (H)−S1 (T )

=

2.4− 91 8−2

= 0.4 −

1 54

≈ 0.3815.

(iv) Proof. ∆1 (H) =

V2 (HH)−V2 (HT ) S2 (HH)−S2 (HT )

=

5−1 12−8

= 1.

2.10. (i) (1+r)(Xn −∆n Sn ) ] (1+r)n+1 Xn (1+r)n .

e ∆n Yn+1 Sn en [ Xn+1 Proof. E (1+r)n+1 ] = En [ (1+r)n+1 + de q) +

Xn −∆n Sn (1+r)n

=

∆n Sn +Xn −∆n Sn (1+r)n

=

=

∆n Sn e (1+r)n+1 En [Yn+1 ]

+

Xn −∆n Sn (1+r)n

=

∆n Sn p+ (1+r)n+1 (ue

(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 ).

Xn+1 (H)−Xn+1 (T ) uSn −dSn

en [ Xn+1 ]. To make the portfolio replicate the payoff at time N , we and Xn = E 1+r en [ VNN −n ]. Since (Xn )0≤n≤N is the value process of the en [ XNN −n ] = E must have XN = VN . So Xn = E (1+r) (1+r) unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear en [ VNN −n ]. equations), the no-arbitrage price of VN at time n is Vn = Xn = E (1+r) So ∆n =

(iii) Proof. en [ E

Sn+1 ] (1 + r)n+1

= = < =

1 en [(1 − An+1 )Yn+1 Sn ] E (1 + r)n+1 Sn [e p(1 − An+1 (H))u + qe(1 − An+1 (T ))d] (1 + r)n+1 Sn [e pu + qed] (1 + r)n+1 Sn . (1 + r)n

en [ Sn+1 If An+1 is a constant a, then E (1+r)n+1 ] =

Sn pu+e q d) (1+r)n+1 (1−a)(e

=

Sn (1+r)n (1−a).

Sn (1+r)n (1−a)n .

2.11. (i) Proof. FN + PN = SN − K + (K − SN )+ = (SN − K)+ = CN . (ii) en [ CNN −n ] = E en [ FNN −n ] + E en [ PNN −n ] = Fn + Pn . Proof. Cn = E (1+r) (1+r) (1+r) (iii) e FN N ] = Proof. F0 = E[ (1+r)

1 (1+r)N

e N − K] = S0 − E[S

K (1+r)N

(iv) 6

.

Sn+1 en [ So E (1+r)n+1 (1−a)n+1 ] =

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 −

(1+r)N S0 (1+r)N

= 0, C0 = P0 .

(vi) en [ SN −K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = E ] = Sn − (1+r)N −n So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ≥ 1.

(1+r)N S0 (1+r)N −n

= 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 + + e (SN − K) e (K − SN ) C=E , and P = E . (1 + r)N −m (1 + r)N −m 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 payoff of max(C, P ) at time m. Therefore, its current no-arbitrage ) e max(C,P price should be E[ (1+r)m ]. By the put-call parity, C = Sm − (1+r)KN −m + P . So max(C, P ) = P + (Sm − (1+r)KN −m )+ . Therefore, the time-zero price of a chooser option is # # " " (Sm − (1+r)KN −m )+ (Sm − (1+r)KN −m )+ P (K − SN )+ e e e e E =E . +E +E (1 + r)m (1 + r)m (1 + r)N (1 + r)m The first term stands for the time-zero price of a put, expiring at time N and having strike price K, and the second term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)KN −m . ) e max(C,P If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ m ], (1+r)

due to the economical argument involved (like “the chooser option is equivalent to receiving a payoff 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 payoff at time m is max(C, P ). Fix ω, if C(ω) > P (ω), we can construct a portfolio ∆0m , · · · , ∆0N −1 whose payoff at time N is (SN − K)+ ; if C(ω) < P (ω), we can construct a portfolio ∆00m , · · · , ∆00N −1 whose payoff at time N is (K − SN )+ . By defining (m ≤ k ≤ N − 1) 0 ∆k (ω) if C(ω) > P (ω) ∆k (ω) = ∆00k (ω) if C(ω) < P (ω), we get a portfolio (∆n )0≤n≤N −1 whose payoff 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 ) e Xm m ] = E[ e max(C,P particular, V0 = X0 = E[ (1+r) (1+r)m ]. 2.13. (i) Proof. Note under both actual probability P and risk-neutral probability Pe, coin tosses ωn ’s are i.i.d.. So without loss of generality, we work on P . For any function g, En [g(Sn+1 , Yn+1 )] = En [g( SSn+1 Sn , Yn + n Sn+1 Sn Sn )]

= 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

PN

S

n=0 n Proof. Set vN (s, y) = f ( Ny+1 ). Then vN (SN , YN ) = f ( N +1 ) = VN . Suppose vn+1 is given, then V v (S ,Y ) 1 n+1 n+1 n+1 n+1 en [ e Vn = E ] = 1+r [e pvn+1 (uSn , Yn + uSn ) + qevn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), 1+r ] = En [ 1+r where ven+1 (us, y + us) + ven+1 (ds, y + ds) vn (s, y) = . 1+r

2.14. (i) Proof. For n ≤ M , (Sn , Yn ) = (Sn , 0). Since coin tosses ωn ’s are i.i.d. under Pe, (Sn , Yn )0≤n≤M is Markov en [h(Sn+1 )] = peh(uSn ) + e under Pe. More precisely, for any function h, E h(dSn ), for n = 0, 1, · · · , M − 1. eM [g(SM +1 , YM +1 )] = E eM [g(SM +1 , SM +1 )] = peg(uSM , uSM )+ For any function g of two variables, we have E en [g(Sn+1 , Yn+1 )] = E en [g( Sn+1 Sn , Yn + Sn+1 Sn )] = peg(uSn , Yn +uSn )+ qeg(dSM , dSM ). And for n ≥ M +1, E Sn Sn qeg(dSn , Yn + dSn ), so (Sn , Yn )0≤n≤N is Markov under Pe. (ii) PN

Sk

y +1 Proof. Set vN (s, y) = f ( N −M ). Then vN (SN , YN ) = f ( K=M ) = VN . Suppose vn+1 is already given. N −M en [vn+1 (Sn+1 , Yn+1 )] = pevn+1 (uSn , Yn + uSn ) + qevn+1 (dSn , Yn + dSn ). So vn (s, y) = a) If n > M , then E pevn+1 (us, y + us) + qevn+1 (ds, y + ds). eM [vM +1 (SM +1 , YM +1 )] = pevM +1 (uSM , uSM ) + ven+1 (dSM , dSM ). So vM (s) = b) If n = M , then E pevM +1 (us, us) + qevM +1 (ds, ds). en [vn+1 (Sn+1 )] = pevn+1 (uSn ) + qevn+1 (dSn ). So vn (s) = pevn+1 (us) + qevn+1 (ds). c) If n < M , then E

3. State Prices 3.1. e Proof. Note Z(ω) :=

P (ω) e(ω) P

=

1 Z(ω) .

e we get the Apply Theorem 3.1.1 with P , Pe, Z replaced by Pe, P , Z,

analogous of properties (i)-(iii) of Theorem 3.1.1. 3.2. (i) P P Proof. Pe(Ω) = ω∈Ω Pe(ω) = ω∈Ω Z(ω)P (ω) = E[Z] = 1. (ii) P e ]=P e Proof. E[Y ω∈Ω Y (ω)P (ω) = ω∈Ω Y (ω)Z(ω)P (ω) = E[Y Z]. (iii) P Proof. P˜ (A) = ω∈A Z(ω)P (ω). Since P (A) = 0, P (ω) = 0 for any ω ∈ A. So Pe(A) = 0. (iv) P Proof. IfPPe(A) = ω∈A Z(ω)P (ω) = 0, by P (Z > 0) = 1, we conclude P (ω) = 0 for any ω ∈ A. So P (A) = ω∈A P (ω) = 0. (v) Proof. P (A) = 1 ⇐⇒ P (Ac ) = 0 ⇐⇒ Pe(Ac ) = 0 ⇐⇒ Pe(A) = 1. (vi)

8

Proof. Pick ω0 such that P (ω0 ) > 0, define Z(ω) =

0, 1 P (ω0 ) ,

if ω 6= ω0 Then P (Z ≥ 0) = 1 and E[Z] = if ω = ω0 .

1 P (ω0 )

· P (ω0 ) = 1. P Clearly Pe(Ω \ {ω0 }) = E[Z1Ω\{ω0 } ] = ω6=ω0 Z(ω)P (ω) = 0. But P (Ω \ {ω0 }) = 1 − P (ω0 ) > 0 if P (ω0 ) < 1. Hence in the case 0 < P (ω0 ) < 1, P and Pe are not equivalent. If P (ω0 ) = 1, then E[Z] = 1 if and only if Z(ω0 ) = 1. In this case Pe(ω0 ) = Z(ω0 )P (ω0 ) = 1. And Pe and P have to be equivalent. In summary, if we can find ω0 such that 0 < P (ω0 ) < 1, then Z as constructed above would induce a probability Pe that is not equivalent to P . 3.5. (i) Proof. Z(HH) =

9 16 ,

Z(HT ) = 98 , Z(T H) =

3 8

and Z(T T ) =

15 4 .

(ii) Proof. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (ω2 = H|ω1 = H) + Z2 (HT )P (ω2 = T |ω1 = H) = E1 [Z2 ](T ) = Z2 (T H)P (ω2 = H|ω1 = T ) + Z2 (T T )P (ω2 = T |ω1 = T ) = 23 .

3 4.

Z1 (T ) =

(iii) Proof. V1 (H) =

[Z2 (HH)V2 (HH)P (ω2 = H|ω1 = H) + Z2 (HT )V2 (HT )P (ω2 = T |ω1 = T )] = 2.4, Z1 (H)(1 + r1 (H))

V1 (T ) =

[Z2 (T H)V2 (T H)P (ω2 = H|ω1 = T ) + Z2 (T T )V2 (T T )P (ω2 = T |ω1 = T )] 1 = , Z1 (T )(1 + r1 (T )) 9

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 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 41 )(1 + 12 )

3.6. Proof. U 0 (x) = have XN =

1 x, (1+r)N λZ

so I(x) =

X0 (1 + r)n Z1n En [Z ·

= 1 Z]

1 x.

(1+r)N λZ en [ XNN −n ] E (1+r)

Z (3.3.26) gives E[ (1+r) N

X0 N Z (1 + r) . 0 =X ξn , where

Hence Xn =

] = X0 . So λ = =

n

en [ X0 (1+r) E Z

1 X0 .

By (3.3.25), we

en [ 1 ] = ] = X0 (1 + r)n E Z

the second to last “=” comes from Lemma 3.2.6.

3.7. 1

1

Z λZ p−1 ] = X . Solve it for λ, Proof. U 0 (x) = xp−1 and so I(x) = x p−1 . By (3.3.26), we have E[ (1+r) 0 N ( (1+r)N ) we get  p−1

 λ= 

X0 E

p

Z p−1

  

=

X0p−1 (1 + r)N p p

(E[Z p−1 ])p−1

.

Np

(1+r) p−1 1

λZ p−1 = So by (3.3.25), XN = ( (1+r) N )

1

Np

1

λ p−1 Z p−1 N (1+r) p−1

=

X0 (1+r) p−1 E[Z

3.8. (i)

9

p p−1

]

1

1

Z p−1 N (1+r) p−1

=

(1+r)N X0 Z p−1 E[Z

p p−1

]

.

2

d d (U (x) − yx) = U 0 (x) − y. So x = I(y) is an extreme point of U (x) − yx. Because dx Proof. dx 2 (U (x) − yx) = 00 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.

(ii) Proof. Following the hint of the problem, we have E[U (XN )] − E[XN

λZ λZ λZ λZ ] ≤ E[U (I( ))] − E[ I( )], N N N (1 + r) (1 + r) (1 + r) (1 + r)N

λ ∗ ∗ ∗ ∗ e i.e. E[U (XN )] − λX0 ≤ E[U (XN )] − E[ XN ] = E[U (XN )] − λX0 . So E[U (XN )] ≤ E[U (XN )]. (1+r)N

3.9. (i) en [ XNN −n ]. So if XN ≥ 0, then Xn ≥ 0 for all n. Proof. Xn = E (1+r) (ii) Proof. a) If 0 ≤ x < γ and 0 < y ≤ γ1 , then U (x) − yx = −yx ≤ 0 and U (I(y)) − yI(y) = U (γ) − yγ = 1 − yγ ≥ 0. So U (x) − yx ≤ U (I(y)) − yI(y). b) If 0 ≤ x < γ and y > γ1 , 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). c) If x ≥ γ and 0 < y ≤ γ1 , then U (x) − yx = 1 − yx and U (I(y)) − yI(y) = U (γ) − yγ = 1 − yγ ≥ 1 − yx. So U (x) − yx ≤ U (I(y)) − yI(y). d) If x ≥ γ and y > γ1 , 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) λZ e XN 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 λZ λZ ∗ E[U (XN )] − E[ XN ] ≤ E[U (XN )] − E[ X ∗ ]. (1 + r)N (1 + r)N N ∗ ∗ )]. )] − λX0 . So E[U (XN )] ≤ E[U (XN That is, E[U (XN )] − λX0 ≤ E[U (XN

(iv) Proof. Plug pm and ξm into (3.6.4), we have N

X0 =

2 X

N

pm ξm I(λξm ) =

m=1

So

X0 γ

{m : X0 γ

2 X

pm ξm γ1{λξm ≤ γ1 } .

m=1

P2N

X0 m=1 pm ξm 1{λξm ≤ γ1 } . Suppose there is a solution λ to (3.6.4), note γ > 0, we then can conclude λξm ≤ γ1 } 6= ∅. Let K = max{m : λξm ≤ γ1 }, then λξK ≤ γ1 < λξK+1 . So ξK < ξK+1 and PK N m=1 pm ξm (Note, however, that K could be 2 . In this case, ξK+1 is interpreted as ∞. Also, note

=

= we are looking for positive solution λ > 0). Conversely, suppose there exists some K so that ξK < ξK+1 and PK X0 1 m=1 ξm pm = γ . Then we can find λ > 0, such that ξK < λγ < ξK+1 . For such λ, we have N

2 K X X Z λZ 1 γ = E[ I( )] = p ξ 1 pm ξm γ = X0 . m m {λξ ≤ } m γ (1 + r)N (1 + r)N m=1 m=1

Hence (3.6.4) has a solution. 10

(v) ∗ Proof. XN (ω m ) = I(λξm ) = γ1{λξm ≤ γ1 } =

γ, if m ≤ K . 0, if m ≥ K + 1

4. American Derivative Securities Before proceeding to the exercise problems, we first 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 τ with τ ∈ Sn . The valuation formula for cash flow (Theorem 2.4.8) gives a fair price for the derivative security exercised according to τ : Vn (τ ) =

N X k=n

e En 1{τ =k}

1 1 e Gk = En 1{τ ≤N } Gτ . (1 + r)k−n (1 + r)τ −n

The buyer wants to consider all the possible τ ’s, so that he can find 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 en [1{τ ≤N } Definition 4.4.1: Vn = maxτ ∈Sn E (1+r)τ −n Gτ ]. 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 find (∆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 e measure P , we conclude Cn Vn+1 Vn e En =− ≤ 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 Vn if and only if is a supermartingale under Pe (note this is independent of the requirement n (1+r)

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) Vn is a supermartingale under Pe. (1+r)n 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 specific algorithm for calculating the price, Theorem 4.4.4 establishes the one-to-one correspondence between super-replication and supermartingale property, and finally, 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 eS p e V + V n+1 n+1 VnS = max gS (Sn ), 1+r ) ( C C P P peVn+1 + Ven+1 peVn+1 + Ven+1 ≤ max gP (Sn ) + gC (Sn ), + 1+r 1+r ( ) ( ) P P C C peVn+1 + Ven+1 peVn+1 + Ven+1 ≤ max gP (Sn ), + max gC (Sn ), 1+r 1+r = VnP + VnC . As to when “<” holds, suppose m = max{n : VnS < VnP + VnC }. Then clearly m ≤ N − 1 and it is possible that {n : VnS < VnP + VnC } = ∅. When this set is not empty, m is characterized as m = max{n : gP (Sn ) < P P p eVn+1 +e q Vn+1 1+r

and gC (Sn ) >

C C p eVn+1 +e q Vn+1 1+r

or gP (Sn ) >

P P p eVn+1 +e q Vn+1 1+r

and gC (Sn ) <

C C p eVn+1 +e q Vn+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) =

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 )

and ∆0 =

V1 (H) − V1 (T ) ≈ −0.433. S1 (H) − S1 (T )

The optimal exercise time is τ = inf{n : Vn = Gn }. So τ (HH) = ∞, τ (HT ) = 2, τ (T H) = τ (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 is X1 (T ) = (1 + r)(−1.36 − 0.433S0 ) + 0.433S1 (T ) = (1 + 41 )(−1.36 − 0.433 × 4) + 0.433 × 2 = −3. The agent should exercise the put at time one and get 3 to pay off 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 portfolio 1 shares of is X1 (H) = (1 + r)(−1.36 − 0.433S0 ) + 0.433S1 (H) = −0.4. The agent should borrow to buy 12 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 off 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. For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 32 , G2 (T T ) = 53 , G3 (T HT ) = 1, G3 (T T H) = 1.75, G3 (T T T ) = 2.125. All the other outcomes of G is negative.

12

For the price process, V0 = 0.4, V1 (T ) = 1, V1 (T H) = 32 , V1 (T T ) = 35 , 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 0.4 and the optimal exercise time satisfies ∞ if ω1 = H, τ (ω) = 1 if ω1 = T .

4.4. Proof. 1.36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio sufficient to pay off 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) τ ≡ 0; (2) τ ≡ 1; (3) τ (HT ) = τ (HH) = 1, τ (T H), τ (T T ) ∈ {2, ∞} (4 different ones); (4) τ (HT ), τ (HH) ∈ {2, ∞}, τ (T H) = τ (T T ) = 1 (4 different ones); (5) τ (HT ), τ (HH), τ (T H), τ (T T ) ∈ {2, ∞} (16 different ones). When the option is out of money, the following stopping times do not exercise (i) τ ≡ 0; (ii) τ (HT ) ∈ {2, ∞}, τ (HH) = ∞, τ (T H), τ (T T ) ∈ {2, ∞} (8 different ones); (iii) τ (HT ) ∈ {2, ∞}, τ (HH) = ∞, τ (T H) = τ (T T ) = 1 (2 different ones). e {τ ≤2} ( 4 )τ Gτ ] ≤ E[1 e {τ ∗ ≤2} ( 4 )τ ∗ Gτ ∗ ], where τ ∗ (HT ) = e {τ ≤2} ( 4 )τ Gτ ] = G0 = 1. For (ii), E[1 For (i), E[1 5 5 5 e {τ ∗ ≤2} ( 4 )τ ∗ Gτ ∗ ] = 1 [( 4 )2 · 1 + ( 4 )2 (1 + 4)] = 0.96. For 2, τ ∗ (HH) = ∞, τ ∗ (T H) = τ ∗ (T T ) = 2. So E[1 5 4 5 5 e {τ ≤2} ( 4 )τ Gτ ] has the biggest value when τ satisfies τ (HT ) = 2, τ (HH) = ∞, τ (T H) = τ (T T ) = 1. (iii), E[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 = eN −1 [ VN ]} = max{K − SN −1 , K − SN −1 } = K − SN −1 . The second equality comes from max{K − SN −1 , E 1+r 1+r 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 τ is allowed to be ∞. But intuitively, results in this chapter should still hold for the case τ ≤ 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 off the negative payoff. In effect, 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 K e SN − K ] = V0EC − S0 + . V0AP ≥ V0EP = V0EC − E[ (1 + r)N (1 + r)N

4.7. eN −1 [ VN ]} = max{SN −1 − K, SN −1 − K } = SN −1 − K . Proof. VN = SN − K, VN −1 = max{SN −1 − K, E 1+r 1+r 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[ατ2 ] = E[α(τ2 −τ1 )+τ1 ] = E[α(τ2 −τ1 ) ]E[ατ1 ] = E[ατ1 ]2 . (ii) (m)

(m)

Proof. If we define Mn = Mn+τm − Mτm (m = 1, 2, · · · ), then (M· )m as random functions are i.i.d. with (m) distributions the same as that of M . So τm+1 − τm = inf{n : Mn = 1} are i.i.d. with distributions the same as that of τ1 . Therefore E[ατm ] = E[α(τm −τm−1 )+(τm−1 −τm−2 )+···+τ1 ] = E[ατ1 ]m .

(iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5.2. (i) Proof. f 0 (σ) = peσ − qe−σ , so f 0 (σ) > 0 if and only if σ > f (σ) > f (0) = 1 for all σ > 0.

1 2 (ln q

− ln p). Since

1 2 (ln q

− ln p) < 0,

(ii) 1 1 1 ] = En [eσXn+1 f (σ) Proof. En [ SSn+1 ] = peσ f (σ) + qe−σ f (σ) = 1. n

(iii) 1 )n∧τ1 ≤ eσ·1 , Proof. By optional stopping theorem, E[Sn∧τ1 ] = E[S0 ] = 1. Note Sn∧τ1 = eσMn∧τ1 ( f (σ) by bounded convergence theorem, E[1{τ1 <∞} Sτ1 ] = E[limn→∞ Sn∧τ1 ] = limn→∞ E[Sn∧τ1 ] = 1, that is, 1 1 E[1{τ1 <∞} eσ ( f (σ) )τ1 ] = 1. So e−σ = E[1{τ1 <∞} ( f (σ) )τ1 ]. Let σ ↓ 0, again by bounded convergence theorem, 1 1 = E[1{τ1 <∞} ( f (0) )τ1 ] = P (τ1 < ∞).

(iv) 1 peσ +qe−σ , then as σ varies from 0 to ∞, α can take all the values in (0, 1). √ 1± 1−4pqα2 2 2 (note 4pqα2 < 4( p+q 1). We want to choose σ > in terms of α, we have eσ = 2pα 2 ) ·1 = √ √ 1+ 1−4pqα2 1− 1−4pqα2 should take σ = ln( ). Therefore E[ατ1 ] = √2pα 2 = . 2pα 2qα 1+ 1−4pqα

Proof. Set α =

1 f (σ)

=

14

Write σ 0, so we

(v) Proof.

∂ τ1 ∂α E[α ]

∂ = E[ ∂α ατ1 ] = E[τ1 ατ1 −1 ], and

!0 1 − 4pqα2 2qα i0 h p 1 (1 − 1 − 4pqα2 )α−1 2q p 1 1 1 [− (1 − 4pqα2 )− 2 (−4pq2α)α−1 + (1 − 1 − 4pqα2 )(−1)α2 ]. 2q 2 1−

= = So E[τ1 ] = limα↑1

∂ τ1 ∂α E[α ]

=

p

1 1 2q [− 2 (1

1

− 4pq)− 2 (−8pq) − (1 −

√

1 − 4pq)] =

1 2p−1 .

5.3. (i) √

1−4pq Proof. Solve the equation peσ + qe−σ = 1 and a positive solution is ln 1+ 2p = ln 1−p p = ln q − ln p. Set 0 σ0 = ln q − ln p, then f (σ0 ) = 1 and f (σ) > 0 for σ > σ0 . So f (σ) > 1 for all σ > σ0 .

(ii) 1 1 Proof. As in Exercise 5.2, Sn = eσMn ( f (σ) )n is a martingale, and 1 = E[S0 ] = E[Sn∧τ1 ] = E[eσMn∧τ1 ( f (σ) )τ1 ∧n ]. Suppose σ > σ0 , then by bounded convergence theorem,

1 = E[ lim eσMn∧τ1 ( n→∞

Let σ ↓ σ0 , we get P (τ1 < ∞) = e−σ0 =

p q

1 n∧τ1 1 τ1 ) ] = E[1{τ1 <∞} eσ ( ) ]. f (σ) f (σ)

< 1.

(iii) √ 1± 1−4pqα2 1 1 Proof. From (ii), we can see E[1{τ1 <∞} ( f (σ) )τ1 ] = e−σ , for σ > σ0 . Set α = f (α) , then eσ = . We 2pα √ √ 2 2 1− 1+ 1−4pqα 1−4pqα ), then E[ατ1 1{τ1 <∞} ] = . need to choose the root so that eσ > eσ0 = pq , so σ = ln( 2pα 2qα (iv) Proof. E[τ1 1{τ1 <∞} ] = p 1 q 2q−1 .

∂ τ1 ∂α E[α 1{τ1 <∞} ]|α=1

=

1 √ 4pq 2q [ 1−4pq

− (1 −

√

1 − 4pq)] =

1 4pq 2q [ 2q−1

− 1 + 2q − 1] =

5.4. (i) Proof. E[ατ2 ] =

P∞

k=1

P (τ2 = 2k)α2k =

P∞

α 2k k=1 ( 2 ) P (τ2

= 2k)4k . So P (τ2 = 2k) =

(2k)! . 4k (k+1)!k!

(ii) Proof. P (τ2 = 2) = 14 . For k ≥ 2, P (τ2 = 2k) = P (τ2 ≤ 2k) − P (τ2 ≤ 2k − 2). P (τ2 ≤ 2k)

=

P (M2k = 2) + P (M2k ≥ 4) + P (τ2 ≤ 2k, M2k ≤ 0)

=

P (M2k = 2) + 2P (M2k ≥ 4)

=

P (M2k = 2) + P (M2k ≥ 4) + P (M2k ≤ −4)

=

1 − P (M2k = −2) − P (M2k = 0).

15

Similarly, P (τ2 ≤ 2k − 2) = 1 − P (M2k−2 = −2) − P (M2k−2 = 0). So P (τ2 = 2k)

= = = = =

P (M2k−2 = −2) + P (M2k−2 = 0) − P (M2k = −2) − P (M2k = 0) 1 2k−2 (2k − 2)! (2k − 2)! 1 2k (2k)! (2k)! ( ) + −( ) + 2 k!(k − 2)! (k − 1)!(k − 1)! 2 (k + 1)!(k − 1)! k!k! (2k)! 4 4 2 (k + 1)k(k − 1) + (k + 1)k − k − (k + 1) 4k (k + 1)!k! 2k(2k − 1) 2k(2k − 1) (2k)! 2(k 2 − 1) 2(k 2 + k) 4k 2 − 1 + − 4k (k + 1)!k! 2k − 1 2k − 1 2k − 1 (2k)! . 4k (k + 1)!k!

5.5. (i) Proof. For every path that crosses level m by time n and resides at b at time n, there corresponds a reflected path that resides at time 2m − b. So 1 P (Mn∗ ≥ m, Mn = b) = P (Mn = 2m − b) = ( )n 2 (m +

n! n−b n+b 2 )!( 2

− m)!

.

(ii) Proof. P (Mn∗ ≥ m, Mn = b) =

n! (m +

n−b n+b 2 )!( 2

− m)!

pm+

n−b 2

q

n+b 2 −m

.

5.6. Proof. On the infinite coin-toss space, we define Mn = {stopping times that takes values 0, 1, · · · , n, ∞} and M∞ = {stopping times that takes values 0, 1, 2, · · · }. Then the time-zero value V ∗ of the perpetual + e {τ <∞} (K−Sτ τ) ]. For an American put with American put as in Section 5.4 can be defined as supτ ∈M∞ E[1 (1+r) + e {τ <∞} (K−Sτ τ) ]. the same strike price K that expires at time n, its time-zero value V (n) is maxτ ∈M E[1 n

(1+r)

Clearly (V (n) )n≥0 is nondecreasing and V (n)≤ V ∗ for every n. So limn V (n) exists and limn V (n) ≤ V ∗ . ∞, if τ = ∞ For any given τ ∈ M∞ , we define τ (n) = , then τ (n) is also a stopping time, τ (n) ∈ Mn τ ∧ n, if τ < ∞ and limn→∞ τ (n) = τ . By bounded convergence theorem, + + e 1{τ (n) <∞} (K − Sτ (n) ) e 1{τ <∞} (K − Sτ ) ≤ lim V (n) . = lim E E n→∞ n→∞ (1 + r)τ (1 + r)τ (n) Take sup at the left hand side of the inequality, we get V ∗ ≤ limn→∞ V (n) . Therefore V ∗ = limn V (n) . Remark: In the above proof, rigorously speaking, we should use (K − Sτ ) in the places of (K − Sτ )+ . So this needs some justification. 5.8. (i)

16

Proof. v(Sn ) = Sn ≥ Sn −K = g(Sn ). Under risk-neutral probabilities, by Theorem 2.4.4.

1 (1+r)n v(Sn )

=

Sn (1+r)n

is a martingale

(ii) Proof. If the purchaser chooses to exercises the call at time h i h n, then thei discounted risk-neutral expectation Sn −K K K e of her payoff is E (1+r)n = S0 − (1+r)n . Since limn→∞ S0 − (1+r) = S0 , the value of the call at time n h i K zero is at least supn S0 − (1+r) = S0 . n (iii) o n q v(ds) qv = max{s − K, peu+e Proof. max g(s), pev(us)+e 1+r 1+r s} = max{s − K, s} = s = v(s), so equation (5.4.16) is satisfied. Clearly v(s) = s also satisfies the boundary condition (5.4.18). (iv) h i e Sτ −Kτ 1{τ <∞} ≥ S0 . Then P (τ < ∞) 6= 0 and Proof. Suppose τ is an optimal exercise time, then E (1+r) h i h i h i Sτ −K Sτ Sn K e e e E 1 1 1 > 0. So E < E . Since is a martingale (1+r)τ {τ <∞} (1+r)τ {τ <∞} (1+r)τ {τ <∞} (1+r)n i hn≥0 i h Sτ ∧n Sτ e e ≤ lim inf n→∞ E = under risk-neutral measure, by Fatou’s lemma, E (1+r)τ 1{τ <∞} (1+r)τ ∧n 1{τ <∞} i i h h Sτ ∧n e 0 ] = S0 . Combined, we have S0 ≤ E e Sτ −Kτ 1{τ <∞} < S0 . e = lim inf n→∞ E[S lim inf n→∞ E (1+r)τ ∧n (1+r) Contradiction. So there is no optimal time to exercise the perpetual American call. Simultaneously, we have h i e Sτ −Kτ 1{τ <∞} < S0 for any stopping time τ . Combined with (ii), we conclude S0 is the least shown E (1+r) upper bound for all the prices acceptable to the buyer. 5.9. (i) Proof. Suppose v(s) = sp , then we have sp = 25 2p sp + or p = −1.

2 sp 5 2p .

So 1 =

2p+1 5

+

21−p 5 .

Solve it for p, we get p = 1

(ii) Proof. Since lims→∞ v(s) = lims→∞ (As +

B s)

= 0, we must have A = 0.

(iii) Proof. fB (s) = 0 if and only if B + s2 − 4s = 0. The discriminant ∆ = (−4)2 − 4B = 4(4 − B). So for B ≤ 4, the equation has roots and for B > 4, this equation does not have roots. (iv) Proof. Suppose B ≤ 4, then the equation s2 − 4s + √ B = 0 has solution 2 ± 4 − s and Bs , we should choose B = 4 and sB = 2 + 4 − B = 2.

√

4 − B. By drawing graphs of

(v) Proof. To have continuous derivative, we must have −1 = − sB2 . Plug B = s2B back into s2B − 4sB + B = 0, B we get sB = 2. This gives B = 4. 6. Interest-Rate-Dependent Assets 6.2.

17

Proof. Xk = Sk − Ek [Dm (Sm − K)]Dk−1 − Ek−1 [Dk Xk ]

Sn Bn,m Bk,m

for n ≤ k ≤ m. Then

Sn Bk,m Dk ] Bn,m Sn = Dk−1 Sk−1 − Ek−1 [Dm (Sm − K)] − Ek−1 [Ek [Dm ]] Bn,m Sn −1 = Dk−1 [Sk−1 − Ek−1 [Dm (Sm − K)]Dk−1 − Bk−1,m ] Bn,m = Dk−1 Xk−1 . =

Ek−1 [Dk Sk − Ek [Dm (Sm − K)] −

6.3. Proof. 1 e 1 e en [ Dm − Dm+1 ] = Bn,m − Bn,m+1 . En [Dm+1 Rm ] = En [Dm (1 + Rm )−1 Rm ] = E Dn Dn Dn

6.4.(i) Proof. D1 V1 = E1 [D3 V3 ] = E1 [D2 V2 ] = D2 E1 [V2 ]. So V1 = 4 V1 (H) = 1+R11 (H) V2 (HH)P (w2 = H|w1 = H) = 21 , V1 (T ) = 0.

D2 D1 E1 [V2 ]

=

1 1+R1 E1 [V2 ].

In particular,

(ii) 2 Proof. Let X0 = 21 . Suppose we buy ∆0 shares of the maturity two bond, then at time one, the value of our portfolio is X1 = (1 + R0 )(X0 − ∆B0,2 ) + ∆0 B1,2 . To replicate the value V1 , we must have V1 (H) = (1 + R0 )(X0 − ∆0 B0,2 ) + ∆0 B1,2 (H) V1 (T ) = (1 + R0 )(X0 − ∆0 B0,2 ) + ∆0 B1,2 (T ). V1 (H)−V1 (T ) 4 4 2 20 4 So ∆0 = B1,2 (H)−B1,2 (T ) = 3 . The hedging strategy is therefore to borrow 3 B0,2 − 21 = 21 and buy 3 share of the maturity two bond. The reason why we do not invest in the maturity three bond is that B1,3 (H) = B1,3 (T )(= 47 ) and the portfolio will therefore have the same value at time one regardless the outcome of first coin toss. This makes impossible the replication of V1 , since V1 (H) 6= V1 (T ).

(iii) Proof. Suppose we buy ∆1 share of the maturity three bond at time one, then to replicate V2 at time V2 (HH)−V2 (HT ) 2 two, we must have V2 = (1 + R1 )(X1 − ∆1 B1,3 ) + ∆1 B2,3 . So ∆1 (H) = B2,3 (HH)−B2,3 (HT ) = − 3 , and V2 (T H)−V2 (T T ) ∆1 (T ) = B2,3 (T H)−B2,3 (T T ) = 0. So the hedging strategy is as follows. If the outcome of first coin toss is T , then we do nothing. If the outcome of first coin toss is H, then short 32 shares of the maturity three bond and invest the income into the money market account. We do not invest in the maturity two bond, because at time two, the value of the bond is its face value and our portfolio will therefore have the same value regardless outcomes of coin tosses. This makes impossible the replication of V2 .

6.5. (i)

18

Proof. Suppose 1 ≤ n ≤ m, then e m+1 [Fn,m ] E n−1

en−1 [B −1 (Bn,m − Bn,m+1 )Zn,m+1 Z −1 = E n,m+1 n−1,m+1 ] Bn,m Bn,m+1 Dn en−1 = E −1 Bn,m+1 Bn−1,m+1 Dn−1 Dn en−1 [Dn−1 E en [Dm ] − Dn−1 E en [Dm+1 ]] = E Bn−1,m+1 Dn−1 en−1 [Dm − Dm+1 ] E = Bn−1,m1 Dn−1 Bn−1,m − Bn−1,m+1 = Bn−1,m+1 = Fn−1,m .

6.6. (i) Proof. The agent enters the forward contract at no cost. He is obliged to buy certain asset at time m at n en+1 [Dm (Sm − K)] = . At time n + 1, the contract has the value E the strike price K = F orn,m = BSn,m Sn+1 − KBn+1,m = Sn+1 − flow of Sn+1 −

Sn Bn+1,m . Bn,m

So if the agent sells this contract at time n + 1, he will receive a cash

Sn Bn+1,m Bn,m

(ii) Proof. By (i), the cash flow generated at time n + 1 is Sn Bn+1,m m−n−1 (1 + r) Sn+1 − Bn,m ! Sn =

(1 + r)m−n−1

Sn+1 −

(1+r)m−n−1 1 (1+r)m−n + r)m−n Sn

(1 + r)m−n−1 Sn+1 − (1 en [ Sm ] en [ Sm ] + (1 + r)m E = (1 + r)m E 1 m (1 + r) (1 + r)m = F utn+1,m − F utn,m . =

6.7. Proof. ψn+1 (0)

e n+1 Vn+1 (0)] = E[D Dn e = E[ 1{#H(ω1 ···ωn+1 )=0} ] 1 + rn (0) Dn e = E[ 1{#H(ω1 ···ωn )=0} 1{ωn+1 =T } ] 1 + rn (0) 1e Dn E[ ] = 2 1 + rn (0) ψn (0) = . 2(1 + rn (0)) 19

For k = 1, 2, · · · , n, Dn e ψn+1 (k) = E 1{#H(ω1 ···ωn+1 )=k} 1 + rn (#H(ω1 · · · ωn )) Dn Dn e e = E 1{#H(ω1 ···ωn )=k} 1{ωn+1 =T } + E 1{#H(ω1 ···ωn )=k} 1{ωn+1 =H} 1 + rn (k) 1 + rn (k − 1) e n Vn (k)] 1 E[D e n Vn (k − 1)] 1 E[D + = 2 1 + rn (k) 2 1 + rn (k − 1) ψn (k) ψn (k − 1) = + . 2(1 + rn (k)) 2(1 + rn (k − 1)) Finally, e n+1 Vn+1 (n + 1)] = E e ψn+1 (n + 1) = E[D

Dn ψn (n) 1{#H(ω1 ···ωn )=n} 1{ωn+1 =H} = . 1 + rn (n) 2(1 + rn (n))

Remark: In the above proof, we have used the independence of ωn+1 and (ω1 , · · · , ωn ). This is guaranteed by the assumption that pe = qe = 21 (note ξ ⊥ η if and only if E[ξ|η] = constant). In case the binomial model has stochastic up- and down-factor un and dn , we can use the fact that Pe(ωn+1 = H|ω1 , · · · , ωn ) = pn and u−1−rn n −dn Pe(ωn+1 = T |ω1 , · · · , ωn ) = qn , where pn = 1+r un −dn and qn = un −dn (cf. solution of Exercise 2.9 and e e E[f e (ωn+1 )|Fn ]] = notes on page 39). Then for any X ∈ Fn = σ(ω1 , · · · , ωn ), we have E[Xf (ωn+1 )] = E[X e E[X(p n f (H) + qn f (T ))].

20

2

Stochastic Calculus for Finance II: Continuous-Time Models

1. General Probability Theory 1.1. (i) Proof. P (B) = P ((B − A) ∪ A) = P (B − A) + P (A) ≥ P (A). (ii) Proof. P (A) ≤ P (An ) implies P (A) ≤ limn→∞ P (An ) = 0. So 0 ≤ P (A) ≤ 0, which means P (A) = 0. 1.2. (i) Proof. We define a mapping φ from A to Ω as follows: φ(ω1 ω2 · · · ) = ω1 ω3 ω5 · · · . Then φ is one-to-one and onto. So the cardinality of A is the same as that of Ω, which means in particular that A is uncountably infinite. (ii) Proof. Let An = {ω = ω1 ω2 · · · : ω1 = ω2 , · · · , ω2n−1 = ω2n }. Then An ↓ A as n → ∞. So P (A) = lim P (An ) = lim [P (ω1 = ω2 ) · · · P (ω2n−1 = ω2n )] = lim (p2 + (1 − p)2 )n . n→∞

n→∞

n→∞

Since p2 + (1 − p)2 < 1 for 0 < p < 1, we have limn→∞ (p2 + (1 − p)2 )n = 0. This implies P (A) = 0. 1.3. Proof. Clearly P (∅) = 0. For any A and B, if both of them are finite, then A ∪ B is also finite. So P (A ∪ B) = 0 = P (A) + P (B). If at least one of them is infinite, then A ∪ B is also infinite. So P (A ∪ B) = PN ∞ = P (A) + P (B). Similarly, we can prove P (∪N n=1 An ) = n=1 P (An ), even if An ’s are not disjoint. Then A = ∪∞ To see countable additivity property doesn’t hold for P , let An = { n1 }.P n=1 An is an infinite ∞ set and therefore P (A) = ∞. However, P (An ) = 0 for each n. So P (A) 6= n=1 P (An ). 1.4. (i) Proof. By Example 1.2.5, we can construct a random variable X on the coin-toss space, which is uniformly Rx ξ2 distributed on [0, 1]. For the strictly increasing and continuous function N (x) = −∞ √12π e− 2 dξ, we let Z = N −1 (X). Then P (Z ≤ a) = P (X ≤ N (a)) = N (a) for any real number a, i.e. Z is a standard normal random variable on the coin-toss space (Ω∞ , F∞ , P ). (ii) Proof. Define Xn =

Pn

Yi i=1 2i ,

where Yi (ω) =

1, if ωi = H 0, if ωi = T .

Then Xn (ω) → X(ω) for every ω ∈ Ω∞ where X is defined as in (i). So Zn = N −1 (Xn ) → Z = N −1 (X) for every ω. Clearly Zn depends only on the first n coin tosses and {Zn }n≥1 is the desired sequence. 1.5.

21

Proof. First, by the information given by the problem, we have Z ∞Z Z Z ∞ 1[0,X(ω)) (x)dP (ω)dx. 1[0,X(ω)) (x)dxdP (ω) = 0

0

Ω

Ω

The left side of this equation equals to X(ω)

Z Z

Z dxdP (ω) =

Ω

Ω

The right side of the equation equals to Z ∞Z Z 1{x<X(ω)} dP (ω)dx = 0

So E{X} =

R∞ 0

X(ω)dP (ω) = E{X}.

0

Ω

∞

∞

Z

(1 − F (x))dx.

P (x < X)dx =

0

0

(1 − F (x))dx.

1.6. (i) Proof. E{euX }

Z

= = = = =

∞

(x−µ)2 1 eux √ e− 2σ2 dx σ 2π −∞ Z ∞ (x−µ)2 −2σ 2 ux 1 2σ 2 √ e− dx −∞ σ 2π Z ∞ [x−(µ+σ 2 u)]2 −(2σ 2 uµ+σ 4 u2 ) 1 2σ 2 √ e− dx −∞ σ 2π Z ∞ [x−(µ+σ 2 u)]2 σ 2 u2 1 2σ 2 √ e− euµ+ 2 dx −∞ σ 2π

euµ+

σ 2 u2 2

(ii) Proof. E{φ(X)} = E{euX } = euµ+

u2 σ 2 2

≥ euµ = φ(E{X}).

1.7. (i) Proof. Since |fn (x)| ≤

√1 , 2nπ

f (x) = limn→∞ fn (x) = 0.

(ii) Proof. By the change of variable formula,

R∞ −∞

fn (x)dx = Z

−∞

2

x √1 e− 2 2π

dx = 1. So we must have

∞

lim

n→∞

R∞

fn (x)dx = 1. −∞

(iii) Proof. This is not contradictory with the Monotone Convergence Theorem, since {fn }n≥1 doesn’t increase to 0.

22

1.8. (i) tX s X −e n Proof. By (1.9.1), |Yn | = e t−s = |XeθX | = XeθX ≤ Xe2tX . The last inequality is by X ≥ 0 and the n fact that θ is between t and sn , and hence smaller than 2t for n sufficiently large. So by the Dominated Convergence Theorem, ϕ0 (t) = limn→∞ E{Yn } = E{limn→∞ Yn } = E{XetX }. (ii) −

+

+

Proof. Since E[etX 1{X≥0} ] + E[e−tX 1{X<0} ] = E[etX ] < ∞ for every t ∈ R, E[et|X| ] = E[etX 1{X≥0} ] + − E[e−(−t)X 1{X<0} ] < ∞ for every t ∈ R. Similarly, we have E[|X|et|X| ] < ∞ for every t ∈ R. So, similar to (i), we have |Yn | = |XeθX | ≤ |X|e2t|X| for n sufficiently large, So by the Dominated Convergence Theorem, ϕ0 (t) = limn→∞ E{Yn } = E{limn→∞ Yn } = E{XetX }. 1.9. Proof. If g(x) is of the form 1B (x), where B is a Borel subset of R, then the desired equality is just (1.9.3). By the linearity Pn of Lebesgue integral, the desired equality also holds for simple functions, i.e. g of the form g(x) = i=1 1Bi (x), where each Bi is a Borel subset of R. Since any nonnegative, Borel-measurable function g is the limit of an increasing sequence of simple functions, the desired equality can be proved by the Monotone Convergence Theorem. 1.10. (i) Proof. If {Ai }∞ i=1 is a sequence of disjoint Borel subsets of [0, 1], then by the Monotone Convergence Theorem, ∞ e P (∪i=1 Ai ) equals to Z

Z

Z 1∪∞ ZdP = i=1 Ai

lim 1∪ni=1 Ai ZdP = lim

n→∞

n→∞

1∪ni=1 Ai ZdP = lim

n→∞

n Z X i=1

ZdP =

Ai

∞ X

Pe(Ai ).

i=1

Meanwhile, Pe(Ω) = 2P ([ 21 , 1]) = 1. So Pe is a probability measure. (ii) Proof. If P (A) = 0, then Pe(A) =

R A

ZdP = 2

R

A∩[ 21 ,1]

dP = 2P (A ∩ [ 12 , 1]) = 0.

(iii) Proof. Let A = [0, 12 ). 1.11. Proof. 2

2

2

e uY } = E{euY Z} = E{euX+uθ e−θX− θ2 } = euθ− θ2 E{e(u−θ)X } = euθ− θ2 e E{e

(u−θ)2 2

=e

u2 2

.

1.12. R θ2 θ2 θ2 ˆ Pe = Proof. First, Zˆ = eθY − 2 = eθ(X+θ)− 2 = e 2 +θX = Z −1 . Second, for any A ∈ F, Pˆ (A) = A Zd R R ˆ ˆ ˆ (1A Z)ZdP = 1A dP = P (A). So P = P . In particular, X is standard normal under P , since it’s standard normal under P . 1.13. (i)

23

Proof.

1 P (X

∈ B(x, )) =

1

R x+ 2 x− 2

2

u √1 e− 2 2π

2

du is approximately

1 √1 − x2 2π e

X √1 e− 2π

·=

2 (ω) ¯ 2

.

(ii) Proof. Similar to (i). (iii) Proof. {X ∈ B(x, )} = {X ∈ B(y − θ, )} = {X + θ ∈ B(y, )} = {Y ∈ B(y, )}. (iv) Proof. By (i)-(iii),

e(A) P P (A)

is approximately

Y √ e− 2π

√ e 2π

2 (ω) ¯ 2

X 2 (ω) ¯ − 2

= e−

2 (ω) Y 2 (ω)−X ¯ ¯ 2

= e−

2 −X 2 (ω) (X(ω)+θ) ¯ ¯ 2

= e−θX(¯ω)−

θ2 2

.

1.14. (i) Proof. Pe(Ω) =

Z e Z ∞ eZ ∞ λ −(λ−λ)X λ e e e −(λ−λ)x −λx e −λx e dP = e λe dx = λe dx = 1. λ λ 0 0

(ii) Proof. Z Pe(X ≤ a) = {X≤a}

Z ae Z a e λ λ −(λ−λ)x e e e e −(λ−λ)X −λx e −λx e dP = e λe dx = λe dx = 1 − e−λa . λ λ 0 0

1.15. (i) Proof. Clearly Z ≥ 0. Furthermore, we have Z ∞ Z ∞ Z ∞ h(g(X))g 0 (X) h(g(x))g 0 (x) = f (x)dx = E{Z} = E h(g(x))dg(x) = h(u)du = 1. f (X) f (x) −∞ −∞ −∞

(ii) Proof. Z Pe(Y ≤ a) = {g(X)≤a}

h(g(X))g 0 (X) dP = f (X)

Z

g −1 (a)

−∞

h(g(x))g 0 (x) f (x)dx = f (x)

By the change of variable formula, the last equation above equals to Pe.

24

Ra −∞

Z

g −1 (a)

h(g(x))dg(x). −∞

h(u)du. So Y has density h under

2. Information and Conditioning 2.1. Proof. For any real number a, we have {X ≤ a} ∈ F0 = {∅, Ω}. So P (X ≤ a) is either 0 or 1. Since lima→∞ P (X ≤ a) = 1 and lima→∞ P (X ≤ a) = 0, we can find a number x0 such that P (X ≤ x0 ) = 1 and P (X ≤ x) = 0 for any x < x0 . So P (X = x0 ) = lim P (x0 − n→∞

1 1 < X ≤ x0 ) = lim (P (X ≤ x0 ) − P (X ≤ x0 − )) = 1. n→∞ n n

2.2. (i) Proof. σ(X) = {∅, Ω, {HT, T H}, {T T, HH}}. (ii) Proof. σ(S1 ) = {∅, Ω, {HH, HT }, {T H, T T }}. (iii) Proof. Pe({HT, T H} ∩ {HH, HT }) = Pe({HT }) = 41 , Pe({HT, T H}) = Pe({HT }) + Pe({T H}) = and Pe({HH, HT }) = Pe({HH}) + Pe({HT }) = 14 + 14 = 21 . So we have

1 4

+

1 4

= 21 ,

Pe({HT, T H} ∩ {HH, HT }) = Pe({HT, T H})Pe({HH, HT }). Similarly, we can work on other elements of σ(X) and σ(S1 ) and show that Pe(A ∩ B) = Pe(A)Pe(B) for any A ∈ σ(X) and B ∈ σ(S1 ). So σ(X) and σ(S1 ) are independent under Pe. (iv) Proof. P ({HT, T H} ∩ {HH, HT }) = P ({HT }) = 92 , P ({HT, T H}) = 29 + 29 = 49 and P ({HH, HT }) = 4 2 6 9 + 9 = 9 . So P ({HT, T H} ∩ {HH, HT }) 6= P ({HT, T H})P ({HH, HT }). Hence σ(X) and σ(S1 ) are not independent under P . (v) Proof. Because S1 and X are not independent under the probability measure P , knowing the value of X will affect our opinion on the distribution of S1 . 2.3. Proof. We note (V, W ) are jointly Gaussian, so to prove their independence it suffices to show they are uncorrelated. Indeed, E{V W } = E{−X 2 sin θ cos θ +XY cos2 θ −XY sin2 θ +Y 2 sin θ cos θ} = − sin θ cos θ + 0 + 0 + sin θ cos θ = 0. 2.4. (i)

25

Proof. E{euX+vY }

= E{euX+vXZ } = E{euX+vXZ |Z = 1}P (Z = 1) + E{euX+vXZ |Z = −1}P (Z = −1) 1 1 = E{euX+vX } + E{euX−vX } 2 2 (u−v)2 1 (u+v)2 = [e 2 + e 2 ] 2 uv u2 +v 2 e + e−uv = e 2 . 2

(ii) Proof. Let u = 0. (iii) Proof. E{euX } = e be independent.

u2 2

and E{evY } = e

v2 2

. So E{euX+vY } 6= E{euX }E{evY }. Therefore X and Y cannot

2.5. Proof. The density fX (x) of X can be obtained by Z Z Z ξ2 x2 ξ 1 2|x| + y − (2|x|+y)2 2 √ √ e− 2 dξ = √ e− 2 . dy = fX (x) = fX,Y (x, y)dy = e 2π 2π 2π {ξ≥|x|} {y≥−|x|} The density fY (y) of Y can be obtained by Z fY (y) = fXY (x, y)dx Z 2|x| + y − (2|x|+y)2 2 e = 1{|x|≥−y} √ dx 2π Z ∞ Z 0∧y 2x + y − (2x+y)2 −2x + y − (−2x+y)2 2 2 √ √ e dx + e dx = 2π 2π 0∨(−y) −∞ Z ∞ Z 0∨(−y) 2x + y − (2x+y)2 2x + y − (2x+y)2 2 2 √ √ = e dx + e d(−x) 2π 2π 0∨(−y) ∞ Z ∞ ξ2 ξ ξ √ e− 2 d( ) = 2 2 2π |y| 1 − y2 = √ e 2. 2π So both X and Y are standard normal random variables. Since fX,Y (x, y) 6= fX (x)fY (y), X and Y are not

26

independent. However, if we set F (t) = Z E{XY }

∞

Z

R∞ t

2

2 u √u e− 2 2π

∞

xyfX,Y (x, y)dxdy

= −∞ Z ∞

−∞ Z ∞

2|x| + y − (2|x|+y)2 2 xy1{y≥−|x|} √ dxdy e 2π −∞ −∞ Z ∞ Z ∞ 2|x| + y − (2|x|+y)2 2 dy xdx y √ e 2π −∞ −|x| Z ∞ Z ∞ ξ2 ξ (ξ − 2|x|) √ e− 2 dξ xdx 2π |x| −∞ Z ∞ Z ∞ 2 x2 2 ξ e− 2 − ξ2 √ √ xdx( e dξ − 2|x| ) 2π 2π −∞ |x| Z ∞ 2 Z ∞ Z ∞ 2 Z 0 ξ2 ξ2 ξ ξ √ e− 2 dξdx + √ e− 2 dξdx x x 2π 2π x −x 0 −∞ Z ∞ Z 0 xF (x)dx + xF (−x)dx.

= = = = = =

−∞

0

So E{XY } =

R∞ 0

xF (x)dx −

R∞ 0

du, we have

uF (u)du = 0.

2.6. (i) Proof. σ(X) = {∅, Ω, {a, b}, {c, d}}. (ii) Proof. E{Y |X} =

X α∈{a,b,c,d}

E{Y 1{X=α} } 1{X=α} . P (X = α)

(iii) Proof. X

E{Z|X} = X + E{Y |X} = X +

α∈{a,b,c,d}

(iv) Proof. E{Z|X} − E{Y |X} = E{Z − Y |X} = E{X|X} = X. 2.7.

27

E{Y 1{X=α} } 1{X=α} . P (X = α)

Proof. Let µ = E{Y − X} and ξ = E{Y − X − µ|G}. Note ξ is G-measurable, we have V ar(Y − X)

=

E{(Y − X − µ)2 }

=

E{[(Y − E{Y |G}) + (E{Y |G} − X − µ)]2 }

=

V ar(Err) + 2E{(Y − E{Y |G})ξ} + E{ξ 2 }

=

V ar(Err) + 2E{Y ξ − E{Y |G}ξ} + E{ξ 2 }

= V ar(Err) + E{ξ 2 } ≥ V ar(Err).

2.8. Proof. It suffices to prove the more general case. For any σ(X)-measurable random variable ξ, E{Y2 ξ} = E{(Y − E{Y |X})ξ} = E{Y ξ − E{Y |X}ξ} = E{Y ξ} − E{Y ξ} = 0. 2.9. (i) Proof. Consider the dice-toss space similar to the coin-toss space. Then a typical element ω in this space is an infinite sequence ω1 ω2 ω3 · · · , with ωi ∈ {1, 2, · · · , 6} (i ∈ N). We define X(ω) = ω1 and f (x) = 1{odd integers} (x). Then it’s easy to see σ(X) = {∅, Ω, {ω : ω1 = 1}, · · · , {ω : ω1 = 6}} and σ(f (X)) equals to {∅, Ω, {ω : ω1 = 1} ∪ {ω : ω1 = 3} ∪ {ω : ω1 = 5}, {ω : ω1 = 2} ∪ {ω : ω1 = 4} ∪ {ω : ω1 = 6}}. So {∅, Ω} ⊂ σ(f (X)) ⊂ σ(X), and each of these containment is strict. (ii) Proof. No. σ(f (X)) ⊂ σ(X) is always true. 2.10. Proof. Z g(X)dP A

= E{g(X)1B (X)} Z ∞ = g(x)1B (x)fX (x)dx −∞ Z Z yfX,Y (x, y) dy1B (x)fX (x)dx = fX (x) Z Z = y1B (x)fX,Y (x, y)dxdy = E{Y 1B (X)} = E{Y IA } Z = Y dP. A

28

2.11. (i) Proof. We can find a sequence {Wn }n≥1 of σ(X)-measurable simple functions such that Wn ↑ W . Each Wn PKn n ai 1Ani , where Ani ’s belong to σ(X) and are disjoint. So each Ani can be can be written in the form i=1 PKn n PKn n ai 1Bin (X) = gn (X), ai 1{X∈Bin } = i=1 written as {X ∈ Bin } for some Borel subset Bin of R, i.e. Wn = i=1 PKn n n where gn (x) = i=1 ai 1Bi (x). Define g = lim sup gn , then g is a Borel function. By taking upper limits on both sides of Wn = gn (X), we get W = g(X). (ii) Proof. Note E{Y |X} is σ(X)-measurable. By (i), we can find a Borel function g such that E{Y |X} = g(X). 3. Brownian Motion 3.1. Proof. We have Ft ⊂ Fu1 and Wu2 − Wu1 is independent of Fu1 . So in particular, Wu2 − Wu1 is independent of Ft . 3.2. Proof. E[Wt2 − Ws2 |Fs ] = E[(Wt − Ws )2 + 2Wt Ws − 2Ws2 |Fs ] = t − s + 2Ws E[Wt − Ws |Fs ] = t − s. 3.3. 1

2

2

1

2

2

1

2

2

1

Proof. ϕ(3) (u) = 2σ 4 ue 2 σ u +(σ 2 +σ 4 u2 )σ 2 ue 2 σ u = e 2 σ u (3σ 4 u+σ 4 u2 ), and ϕ(4) (u) = σ 2 ue 2 σ 1 2 2 σ 4 u2 ) + e 2 σ u (3σ 4 + 2σ 4 u). So E[(X − µ)4 ] = ϕ(4) (0) = 3σ 4 .

2

u2

(3σ 4 u+

3.4. (i) Pn−1 Proof. Assume there exists A ∈ F, such that P (A) > 0 and for every ω ∈ A, limn j=0 |Wtj+1 − Wtj |(ω) < Pn−1 Pn−1 ∞. Then for every ω ∈ A, j=0 (Wtj+1 −Wtj )2 (ω) ≤ max0≤k≤n−1 |Wtk+1 −Wtk |(ω) j=0 |Wtj+1 −Wtj |(ω) → Pn−1 0, since limn→∞ max0≤k≤n−1 |Wtk+1 − Wtk |(ω) = 0. This is a contradiction with limn→∞ j=0 (Wtj+1 − Wtj )2 = T a.s.. (ii) Pn−1 Pn−1 Proof. Note j=0 (Wtj+1 − Wtj )3 ≤ max0≤k≤n−1 |Wtk+1 − Wtk | j=0 (Wtj+1 − Wtj )2 → 0 as n → ∞, by an argument similar to (i). 3.5.

29

Proof.

=

E[e−rT (ST − K)+ ] Z ∞ −rT e 1 K σ (ln S0

=

e

−rT

= =

(S0 e

−(r− 12 σ 2 )T )

∞

Z

(S0 e σ

=

x2

(r− 12 σ 2 )T +σx

1 √

T

(ln

K S0

e− 2T − K) √ dx 2πT

√ (r− 12 σ 2 )T +σ T y

−(r− 12 σ 2 )T )

y2

e− 2 − K) √ dy 2π

Z ∞ √ y2 y2 1 1 √ e− 2 +σ T y dy − Ke−rT √ e− 2 dy 1 1 2π 2π √ √ (ln SK −(r− 12 σ 2 )T ) (ln SK −(r− 12 σ 2 )T ) σ T σ T 0 0 Z ∞ ξ2 S 1 1 1 √ e− 2 dξ − Ke−rT N √ (ln 0 + (r − σ 2 )T ) S0 √ 1 K 1 2 K 2 2π σ T √ (ln S −(r− 2 σ )T )−σ T σ T 0 1

S0 e− 2 σ

2

T

Z

∞

Ke−rT N (d+ (T, S0 )) − Ke−rT N (d− (T, S0 )).

3.6. (i) Proof. E[f (Xt )|Ft ]

So E[f (Xt )|Fs ] =

R∞ −∞

= E[f (Wt − Ws + a)|Fs ]|a=Ws +µt = E[f (Wt−s + a)]|a=Ws +µt x2 Z ∞ e− 2(t−s) = dx f (x + Ws + µt) p 2π(t − s) −∞ (y−Ws −µs−µ(t−s))2 Z ∞ 2(t−s) e− p = f (y) dy 2π(t − s) −∞ = g(Xs ).

f (y)p(t − s, Xs , y)dy with p(τ, x, y) =

√ 1 e− 2πτ

(y−x−µτ )2 2τ

.

(ii) Proof. E[f (St )|Fs ] = E[f (S0 eσXt )|Fs ] with µ = σv . So by (i), Z ∞ (y−Xs −µ(t−s))2 1 2(t−s) E[f (St )|Fs ] = f (S0 eσy ) p dy e− 2π(t − s) −∞ Z ∞ S ( 1 ln z − 1 ln s −µ(t−s))2 σ S0 σ S0 1 dz S0 eσy =z − 2 e = f (z) p σz 2π(t − s) 0 Z

∞

=

f (z) 0

Z

e

−

(ln z −v(t−s))2 Ss 2σ 2 (t−s)

σz

p

∞

f (z)p(t − s, Ss , z)dz

= 0

=

2π(t − s)

g(Ss ).

3.7. (i)

30

dz

Proof. E

h

Zt Zs |Fs

i

= E[exp{σ(Wt − Ws ) + σµ(t − s) − (σµ +

σ2 2 )(t

− s)}] = 1.

(ii) Proof. By optional stopping theorem, E[Zt∧τm ] = E[Z0 ] = 1, that is, E[exp{σXt∧τm − (σµ + 1.

σ2 2 )t

∧ τm }] =

(iii) Proof. If µ ≥ 0 and σ > 0, Zt∧τm ≤ eσm . By bounded convergence theorem, E[1{τm <∞} Zτm ] = E[ lim Zt∧τm ] = lim E[Zt∧τm ] = 1, t→∞

1

t→∞

σ2

2

since on the event {τm = ∞}, Zt∧τm ≤ eσm− 2 σ t → 0 as t → ∞. Therefore, E[eσm−(σµ+ 2 2 σ ↓ 0, by bounded convergence theorem, we have P (τm < ∞) = 1. Let σµ + σ2 = α, we get √ 2 E[e−ατm ] = e−σm = emµ−m 2α+µ .

)τm

] = 1. Let

(iv) Proof. We note for α > 0, E[τm e−ατm ] < ∞ since xe−αx is bounded on [0, ∞). So by an argument similar to Exercise 1.8, E[e−ατm ] is differentiable and √ −m ∂ 2 E[e−ατm ] = −E[τm e−ατm ] = emµ−m 2α+µ p . ∂α 2α + µ2 Let α ↓ 0, by monotone increasing theorem, E[τm ] =

m µ

< ∞ for µ > 0.

(v) σ2 2

Proof. By σ > −2µ > 0, we get σµ + e

2 σm−( σ2

+σµ)t

> 0. Then Zt∧τm ≤ eσm and on the event {τm = ∞}, Zt∧τm ≤

→ 0 as t → ∞. Therefore, E[eσm−(σµ+

σ2 2

)τm

1{τm <∞} ] = E[ lim Zt∧τm ] = lim E[Zt∧τm ] = 1. t→∞

t→∞

2

Let σ ↓ −2µ, then we get P (τm < ∞) = e2µm = e−2|µ|m < 1. Set α = σµ + σ2 . So we get √ 2 E[e−ατm ] = E[e−ατm 1{τm <∞} ] = e−σm = emµ−m 2α+µ .

3.8. (i) Proof. ϕn (u)

= =

1

u

u

u

e u √n Mnt,n ] = (E[e e √n X1,n ])nt = (e √n pen + e− √n qen )nt E[e " ! σ !#nt √ − √σn r r n u u + 1 − e − − 1 + e √ √ − n n e n n √σ + e . σ σ σ √ −√ −√ e n −e n e n −e n

(ii) 31

Proof. 2 2 t2 x rx + 1 − e−σx rx + 1 − eσx ux −ux −e ϕ 12 (u) = e . σx −σx σx −σx x e −e e −e So, ln ϕ

1 x2

(u)

= = = =

t x2 t x2 t x2 t x2

(rx2 + 1)(eux − e−ux ) + e(σ−u)x − e−(σ−u)x eσx − e−σx (rx2 + 1) sinh ux + sinh(σ − u)x ln sinh σx 2 (rx + 1) sinh ux + sinh σx cosh ux − cosh σx sinh ux ln sinh σx 2 (rx + 1 − cosh σx) sinh ux ln cosh ux + . sinh σx

ln

(iii) Proof. (rx2 + 1 − cosh σx) sinh ux sinh σx 2 2 (rx2 + 1 − 1 − σ 2x + O(x4 ))(ux + O(x3 )) u2 x2 1+ + O(x4 ) + 2 σx + O(x3 )

cosh ux + =

2

=

(r − σ2 )ux3 + O(x5 ) u2 x2 1+ + + O(x4 ) 2 σx + O(x3 ) 2

(r − σ2 )ux3 (1 + O(x2 )) u2 x2 = 1+ + + O(x4 ) 2 σx(1 + O(x2 )) u2 x2 rux2 1 = 1+ + − σux2 + O(x4 ). 2 σ 2

(iv) Proof. ln ϕ

1 x2

So limx↓0 ln ϕ

=

t u2 x2 ru 2 σux2 t u2 x2 ru 2 σux2 4 ln(1 + + x − + O(x )) = ( + x − + O(x4 )). x2 2 σ 2 x2 2 σ 2 2

1 x2

(u) = t( u2 +

ru σ

−

σu 2 ),

1

e u √n Mnt,n ] = ϕn (u) → 1 tu2 + t( r − σ )u. By the one-to-one and E[e 2 σ 2

correspondence between distribution and moment generating function, ( √1n Mnt,n )n converges to a Gaussian random variable with mean t( σr − σ2 ) and variance t. Hence ( √σn Mnt,n )n converges to a Gaussian random variable with mean t(r −

σ2 2 )

and variance σ 2 t.

4. Stochastic Calculus 4.1.

32

Proof. Fix t and for any s < t, we assume s ∈ [tm , tm+1 ) for some m. Case 1. m = k. Then I(t)−I(s) = ∆tk (Mt −Mtk )−∆tk (Ms −Mtk ) = ∆tk (Mt −Ms ). So E[I(t)−I(s)|Ft ] = ∆tk E[Mt − Ms |Fs ] = 0. Case 2. m < k. Then tm ≤ s < tm+1 ≤ tk ≤ t < tk+1 . So I(t) − I(s)

=

k−1 X

∆tj (Mtj+1 − Mtj ) + ∆tk (Ms − Mtk ) − ∆tm (Ms − Mtm )

j=m

=

k−1 X

∆tj (Mtj+1 − Mtj ) + ∆tk (Mt − Mtk ) + ∆tm (Mtm+1 − Ms ).

j=m+1

Hence E[I(t) − I(s)|Fs ] =

k−1 X

E[∆tj E[Mtj+1 − Mtj |Ftj ]|Fs ] + E[∆tk E[Mt − Mtk |Ftk ]|Fs ] + ∆tm E[Mtm+1 − Ms |Fs ]

j=m+1

=

0.

Combined, we conclude I(t) is a martingale. 4.2. (i) Proof. We follow the simplification in the hint and consider I(tk ) − I(tl ) with tl < tk . Then I(tk ) − I(tl ) = Pk−1 j=l ∆tj (Wtj+1 − Wtj ). Since ∆t is a non-random process and Wtj+1 − Wtj ⊥ Ftj ⊃ Ftl for j ≥ l, we must have I(tk ) − I(tl ) ⊥ Ftl . (ii) Proof. We use the notation in (i) and it is clear I(tk ) − I(tl ) is normal since it is a linear combination of Pk−1 independent normal random variables. Furthermore, E[I(tk ) − I(tl )] = j=l ∆tj E[Wtj+1 − Wtj ] = 0 and Rt Pk−1 Pk−1 V ar(I(tk ) − I(tl )) = j=l ∆2tj V ar(Wtj+1 − Wtj ) = j=l ∆2tj (tj+1 − tj ) = tlk ∆2u du. (iii) Proof. E[I(t) − I(s)|Fs ] = E[I(t) − I(s)] = 0, for s < t. (iv) Proof. For s < t, 2

Z

E[I (t) −

t

∆2u du

2

Z

− (I (s) −

0

s

∆2u du)|Fs ]

0

= E[I 2 (t) − I 2 (s) −

Z

t

∆2u du|Fs ]

s

= E[(I(t) − I(s))2 + 2I(t)I(s) − 2I 2 (s)|Fs ] −

t

Z

∆2u du

s

= E[(I(t) − I(s))2 ] + 2I(s)E[I(t) − I(s)|Fs ] − Z t Z t = ∆2u du + 0 − ∆2u du s

=

s

0.

33

Z s

t

∆2u du

4.3. Proof. I(t) − I(s) = ∆0 (Wt1 − W0 ) + ∆t1 (Wt2 − Wt1 ) − ∆0 (Wt1 − W0 ) = ∆t1 (Wt2 − Wt1 ) = Ws (Wt − Ws ). (i) I(t) − I(s) is not independent of Fs , since Ws ∈ Fs . (ii) E[(I(t) − I(s))4 ] = E[Ws4 ]E[(Wt − Ws )4 ] = 3s · 3(t − s) = 9s(t − s) and 3E[(I(t) − I(s))2 ] = 3E[Ws2 ]E[(Wt − Ws )2 ] = 3s(t − s). Since E[(I(t) − I(s))4 ] 6= 3E[(I(t) − I(s))2 ], I(t) − I(s) is not normally distributed. (iii) E[I(t) − I(s)|Fs ] = Ws E[Wt − Ws |Fs ] = 0. (iv) E[I 2 (t) −

Z

t

∆2u du − (I 2 (s) −

s

Z

∆2u du)|Fs ] Z t = E[(I(t) − I(s))2 + 2I(t)I(s) − 2I 2 (s) − Wu2 du|Fs ] 0

0

=

s 2 2 2 E[Ws (Wt − Ws ) + 2Ws (Wt − Ws ) − Ws (t − s)|Fs ] Ws2 E[(Wt − Ws )2 ] + 2Ws E[Wt − Ws |Fs ] − Ws2 (t − s) Ws2 (t − s) − Ws2 (t − s)

=

0.

= =

4.4. Proof. (Cf. Øksendal [3], Exercise 3.9.) We first note that X B tj +tj+1 (Btj+1 − Btj ) j

=

Xh j

2

i X B tj +tj+1 (Btj+1 − B tj +tj+1 ) + Btj (B tj +tj+1 − Btj ) + (B tj +tj+1 − Btj )2 . 2

2

2

2

j

RT The first term converges in L2 (P ) to 0 Bt dBt . For the second term, we note  2  2 t  X  E  B tj +tj+1 − Btj −   2 2 j  2  2 X t X  j+1 − tj   = E  B tj +tj+1 − Btj −  2 2 j j

tj+1 − tj = E B tj +tj+1 − Btj − 2 2 j, k " # 2 X tj+1 − tj 2 = E B tj+1 −tj − 2 2 j X tj+1 − tj 2 = 2· 2 j X

≤

2

T max |tj+1 − tj | → 0, 2 1≤j≤n

34

B tk +tk+1 − Btk 2

2

tk+1 − tk − 2

since E[(Bt2 − t)2 ] = E[Bt4 − 2tBt2 + t2 ] = 3E[Bt2 ]2 − 2t2 + t2 = 2t2 . So X

T

Z B tj +tj+1 (Btj+1 − Btj ) → 2

j

Bt dBt + 0

T 1 = BT2 in L2 (P ). 2 2

4.5. (i) Proof. d ln St =

dSt 1 dhSit 2St dSt − dhSit 2St (αt St dt + σt St dWt ) − σt2 St2 dt 1 − = = = σt dWt + (αt − σt2 )dt. 2 2 2 St 2 St 2St 2St 2

(ii) Proof. Z ln St = ln S0 +

t

Z σs dWs +

0

0

t

1 (αs − σs2 )ds. 2

Rt Rt So St = S0 exp{ 0 σs dWs + 0 (αs − 12 σs2 )ds}. 4.6. Proof. Without loss of generality, we assume p 6= 1. Since (xp )0 = pxp−1 , (xp )00 = p(p − 1)xp−2 , we have d(Stp )

1 = pStp−1 dSt + p(p − 1)Stp−2 dhSit 2 1 = pStp−1 (αSt dt + σSt dWt ) + p(p − 1)Stp−2 σ 2 St2 dt 2 1 = Stp [pαdt + pσdWt + p(p − 1)σ 2 dt] 2 p−1 2 p = St p[σdWt + (α + σ )dt]. 2

4.7. (i) Proof. dWt4 = 4Wt3 dWt +

1 2

· 4 · 3Wt2 dhW it = 4Wt3 dWt + 6Wt2 dt. So WT4 = 4

RT 0

Wt3 dWt + 6

RT 0

Wt2 dt.

(ii) Proof. E[WT4 ] = 6

RT 0

tdt = 3T 2 .

(iii) Proof. dWt6 = 6Wt5 dWt + 12 · 6 · 5Wt4 dt. So WT6 = 6 15T 3 .

RT 0

Wt5 dWt + 15

4.8.

35

RT 0

Wt4 dt. Hence E[WT6 ] = 15

RT 0

3t2 dt =

Proof. d(eβt Rt ) = βeβt Rt dt + eβt dRt = eβt(αdt+σdWt ) . Hence eβt Rt = R0 +

Z

t

eβs (αds + σdWs ) = R0 +

0 −βt )+σ and Rt = R0 e−βt + α β (1 − e

Rt 0

α βt (e − 1) + σ β

Z

t

eβs dWs ,

0

e−β(t−s) dWs .

4.9. (i) Proof.

Ke

−r(T −t)

0

N (d− )

= = = = =

Ke Ke

−r(T −t) e

−

√

d2 − 2

2π

− −r(T −t) e

(d+ −σ

√

√

T −t)2

2

2π

√ σ 2 (T −t) −r(T −t) σ T −td+ − 2

e e N 0 (d+ ) σ 2 (T −t) σ2 x Ke−r(T −t) e(r+ 2 )(T −t) e− 2 N 0 (d+ ) K xN 0 (d+ ). Ke

(ii) Proof. cx

∂ ∂ d+ (T − t, x) − Ke−r(T −t) N 0 (d− ) d− (T − t, x) ∂x ∂x ∂ 0 ∂ 0 0 = N (d+ ) + xN (d+ ) d+ (T − t, x) − xN (d+ ) d+ (T − t, x) ∂x ∂x = N (d+ ). = N (d+ ) + xN 0 (d+ )

(iii) Proof. ct

∂ ∂ d+ (T − t, x) − rKe−r(T −t) N (d− ) − Ke−r(T −t) N 0 (d− ) d− (T − t, x) ∂x ∂t ∂ ∂ σ = xN 0 (d+ ) d+ (T − t, x) − rKe−r(T −t) N (d− ) − xN 0 (d+ ) d+ (T − t, x) + √ ∂t ∂t 2 T −t σx 0 −r(T −t) = −rKe N (d− ) − √ N (d+ ). 2 T −t

= xN 0 (d+ )

(iv)

36

Proof. 1 ct + rxcx + σ 2 x2 cxx 2 1 σx ∂ N 0 (d+ ) + rxN (d+ ) + σ 2 x2 N 0 (d+ ) d+ (T − t, x) = −rKe−r(T −t) N (d− ) − √ 2 ∂x 2 T −t σx 1 1 0 = rc − √ N (d+ ) + σ 2 x2 N 0 (d+ ) √ 2 2 T −t σ T − tx = rc.

(v) Proof. For x > K, d+ (T − t, x) > 0 and limt↑T d+ (T − t, x)= limτ ↓0 d+ (τ, x) = ∞. limt↑T d− (T − t, x) = √ √ 1 x limτ ↓0 d− (τ, x) = limτ ↓0 σ√ ln K + σ1 (r + 12 σ 2 ) τ − σ τ = ∞. Similarly, limt↑T d± = −∞ for x ∈ τ (0, K). Also it’s clear that limt↑T d± = 0 for x = K. So ( lim c(t, x) = xN (lim d+ ) − KN (lim d− ) = t↑T

t↑T

t↑T

x − K, 0,

if x > K = (x − K)+ . if x ≤ K

(vi) Proof. It is easy to see limx↓0 d± = −∞. So for t ∈ [0, T ], limx↓0 c(t, x) = limx↓0 xN (limx↓ d+ (T − t, x)) − Ke−r(T −t) N (limx↓0 d− (T − t, x)) = 0. (vii) Proof. For t ∈ [0, T ], it is clear limx→∞ d± = ∞. Note ∂ N 0 (d+ ) σ√1T −t d+ N 0 (d+ ) ∂x lim x(N (d+ ) − 1) = lim = lim . x→∞ x→∞ x→∞ −x−2 −x−1 √ By the expression of d+ , we get x = K exp{σ T − td+ − (T − t)(r + 21 σ 2 )}. So we have d2 +

−x e− 2 −Keσ lim x(N (d+ ) − 1) = lim N (d+ ) √ = lim √ x→∞ x→∞ σ T − t d+ →∞ 2π 0

√

T −td+ −(T −t)(r+ 12 σ 2 )

√ σ T −t

Therefore lim [c(t, x) − (x − e−r(T −t) K)]

x→∞

= = = =

lim [xN (d+ ) − Ke−r(T −t)N (d− ) − x + Ke−r(T −t) ]

x→∞

lim [x(N (d+ ) − 1) + Ke−r(T −t) (1 − N (d− ))]

x→∞

lim x(N (d+ ) − 1) + Ke−r(T −t) (1 − N ( lim d− ))

x→∞

x→∞

0.

4.10. (i)

37

= 0.

Proof. We show (4.10.16) + (4.10.9) ⇐⇒ (4.10.16) + (4.10.15), i.e. assuming X has the representation Xt = ∆t St + Γt Mt , “continuous-time self-financing condition” has two equivalent formulations, (4.10.9) or (4.10.15). Indeed, dXt = ∆t dSt +Γt dMt +(St d∆t +dSt d∆t +Mt dΓt +dMt dΓt ). So dXt = ∆t dSt +Γt dMt ⇐⇒ St d∆t + dSt d∆t + Mt dΓt + dMt dΓt = 0, i.e. (4.10.9) ⇐⇒ (4.10.15). (ii) Proof. First, we clarify the problems by stating explicitly the given conditions and the result to be proved. We assume we have a portfolio Xt = ∆t St + Γt Mt . We let c(t, St ) denote the price of call option at time t and set ∆t = cx (t, St ). Finally, we assume the portfolio is self-financing. The problem is to show 1 2 2 rNt dt = ct (t, St ) + σ St cxx (t, St ) dt, 2 where Nt = c(t, St ) − ∆t St . Indeed, by the self-financing property and ∆t = cx (t, St ), we have c(t, St ) = Xt (by the calculations in Subsection 4.5.1-4.5.3). This uniquely determines Γt as Γt =

c(t, St ) − cx (t, St )St Nt Xt − ∆t St = = . Mt Mt Mt

Moreover, dNt

1 = ct (t, St )dt + cx (t, St )dSt + cxx (t, St )dhSt it − d(∆t St ) 2 1 = ct (t, St ) + cxx (t, St )σ 2 St2 dt + [cx (t, St )dSt − d(Xt − Γt Mt )] 2 1 2 2 = ct (t, St ) + cxx (t, St )σ St dt + Mt dΓt + dMt dΓt + [cx (t, St )dSt + Γt dMt − dXt ]. 2

By self-financing property, cx (t, St )dt + Γt dMt = ∆t dSt + Γt dMt = dXt , so 1 2 2 ct (t, St ) + cxx (t, St )σ St dt = dNt − Mt dΓt − dMt dΓt = Γt dMt = Γt rMt dt = rNt dt. 2

4.11. Proof. First, we note c(t, x) solves the Black-Scholes-Merton PDE with volatility σ1 : ∂ ∂ 1 ∂2 + rx + x2 σ12 2 − r c(t, x) = 0. ∂t ∂x 2 ∂x So

1 ct (t, St ) + rSt cx (t, St ) + σ12 St2 cxx (t, St ) − rc(t, St ) = 0, 2

and dc(t, St )

= = =

1 ct (t, St )dt + cx (t, St )(αSt dt + σ2 St dWt ) + cxx (t, St )σ22 St2 dt 2 1 2 2 ct (t, St ) + αcx (t, St )St + σ2 St cxx (t, St ) dt + σ2 St cx (t, St )dWt 2 1 2 2 2 rc(t, St ) + (α − r)cx (t, St )St + St (σ2 − σ1 )cxx (t, St ) dt + σ2 St cx (t, St )dWt . 2 38

Therefore dXt

=

=

1 rc(t, St ) + (α − r)cx (t, St )St + St2 (σ22 − σ12 )σxx (t, St ) + rXt − rc(t, St ) + rSt cx (t, St ) 2 1 2 2 2 − (σ2 − σ1 )St cxx (t, St ) − cx (t, St )αSt dt + [σ2 St cx (t, St ) − cx (t, St )σ2 St ]dWt 2 rXt dt.

This implies Xt = X0 ert . By X0 , we conclude Xt = 0 for all t ∈ [0, T ]. 4.12. (i) Proof. By (4.5.29), c(t, x) − p(t, x) = x − e−r(T −t) K. So px (t, x) = cx (t, x) − 1 = N (d+ (T − t, x)) − 1, pxx (t, x) = cxx (t, x) = σx√1T −t N 0 (d+ (T − t, x)) and pt (t, x)

= ct (t, x) + re−r(T −t) K σx = −rKe−r(T −t) N (d− (T − t, x)) − √ N 0 (d+ (T − t, x)) + rKe−r(T −t) 2 T −t σx N 0 (d+ (T − t, x)). = rKe−r(T −t) N (−d− (T − t, x)) − √ 2 T −t

(ii) Proof. For an agent hedging a short position in the put, since ∆t = px (t, x) < 0, he should short the underlying stock and put p(t, St ) − px (t, St )St (> 0) cash in the money market account. (iii) Proof. By the put-call parity, it suffices to show f (t, x) = x − Ke−r(T −t) satisfies the Black-Scholes-Merton partial differential equation. Indeed, 1 2 2 ∂2 1 ∂ ∂ + rx + σ x − r f (t, x) = −rKe−r(T −t) + σ 2 x2 · 0 + rx · 1 − r(x − Ke−r(T −t) ) = 0. ∂t 2 ∂x2 ∂x 2 Remark: The Black-Scholes-Merton PDE has many solutions. Proper boundary conditions are the key to uniqueness. For more details, see Wilmott [8]. 4.13. Proof. We suppose (W1 , W2 ) is a pair of local martingale defined by SDE ( dW1 (t) = dB1 (t) dW2 (t) = α(t)dB1 (t) + β(t)dB2 (t).

(1)

We want to find α(t) and β(t) such that ( (dW2 (t))2 = [α2 (t) + β 2 (t) + 2ρ(t)α(t)β(t)]dt = dt dW1 (t)dW2 (t) = [α(t) + β(t)ρ(t)]dt = 0. Solve the equation for α(t) and β(t), we have β(t) = √

1 1−ρ2 (t)

and α(t) = − √ ρ(t)2

( W1 (t) = B1 (t) Rt Rt W2 (t) = 0 √−ρ(s) dB1 (s) + 0 √ 2 1−ρ (s)

39

1−ρ (t)

1 dB2 (s) 1−ρ2 (s)

(2) . So

(3)

is a pair of independent BM’s. Equivalently, we have ( B1 (t) = W1 (t) Rt Rtp B2 (t) = 0 ρ(s)dW1 (s) + 0 1 − ρ2 (s)dW2 (s).

(4)

4.14. (i) Proof. Clearly Zj ∈ Ftj+1 . Moreover E[Zj |Ftj ] = f 00 (Wtj )E[(Wtj+1 − Wtj )2 − (tj+1 − tj )|Ftj ] = f 00 (Wtj )(E[Wt2j+1 −tj ] − (tj+1 − tj )) = 0, since Wtj+1 − Wtj is independent of Ftj and Wt ∼ N (0, t). Finally, we have E[Zj2 |Ftj ]

=

[f 00 (Wtj )]2 E[(Wtj+1 − Wtj )4 − 2(tj+1 − tj )(Wtj+1 − Wtj )2 + (tj+1 − tj )2 |Ftj ]

=

[f 00 (Wtj )]2 (E[Wt4j+1 −tj ] − 2(tj+1 − tj )E[Wt2j+1 −tj ] + (tj+1 − tj )2 )

=

[f 00 (Wtj )]2 [3(tj+1 − tj )2 − 2(tj+1 − tj )2 + (tj+1 − tj )2 ]

=

2[f 00 (Wtj )]2 (tj+1 − tj )2 ,

where we used the independence of Browian motion increment and the fact that E[X 4 ] = 3E[X 2 ]2 if X is Gaussian with mean 0. (ii) Pn−1 Pn−1 Proof. E[ j=0 Zj ] = E[ j=0 E[Zj |Ftj ]] = 0 by part (i). (iii) Proof. n−1 X

V ar[

Zj ]

=

n−1 X

E[(

j=0

Zj )2 ]

j=0

=

n−1 X

E[

j=0

=

n−1 X

X

Zj2 + 2

X

E[E[Zj2 |Ftj ]] + 2

j=0

=

Zi Zj ]

0≤i<j≤n−1

E[Zi E[Zj |Ftj ]]

0≤i<j≤n−1

n−1 X

E[2[f 00 (Wtj )]2 (tj+1 − tj )2 ]

j=0

=

n−1 X

2E[(f 00 (Wtj ))2 ](tj+1 − tj )2

j=0

≤

2

max

0≤j≤n−1

|tj+1 − tj | ·

n−1 X

E[(f 00 (Wtj ))2 ](tj+1 − tj )

j=0

→ 0, since

Pn−1 j=0

E[(f 00 (Wtj ))2 ](tj+1 − tj ) →

RT 0

E[(f 00 (Wt ))2 ]dt < ∞.

4.15. (i) 40

Proof. Bi is a local martingale with  (dBi (t))2 = 

d X σij (t)

σi (t)

j=1

2 dWj (t) =

d 2 X σij (t) j=1

σi2 (t)

dt = dt.

So Bi is a Brownian motion. (ii) Proof. dBi (t)dBk (t)

"  # d d X X σ (t) σ (t) kl ij dWj (t) dWl (t) =  σi (t) σk (t) j=1 l=1

X

=

1≤j, l≤d

=

σij (t)σkl (t) dWj (t)dWl (t) σi (t)σk (t)

d X σij (t)σkj (t) j=1

σi (t)σk (t)

dt

= ρik (t)dt.

4.16. Proof. To find the m independent Brownion motion W1 (t), · · · , Wm (t), we need to find A(t) = (aij (t)) so that (dB1 (t), · · · , dBm (t))tr = A(t)(dW1 (t), · · · , dWm (t))tr , or equivalently (dW1 (t), · · · , dWm (t))tr = A(t)−1 (dB1 (t), · · · , dBm (t))tr , and (dW1 (t), · · · , dWm (t))tr (dW1 (t), · · · , dWm (t)) = A(t)−1 (dB1 (t), · · · , dBm (t))tr (dB1 (t), · · · , dBm (t))(A(t)−1 )tr dt = Im×m dt, where Im×m is the m × m unit matrix. By the condition dBi (t)dBk (t) = ρik (t)dt, we get (dB1 (t), · · · , dBm (t))tr (dB1 (t), · · · , dBm (t)) = C(t). So A(t)−1 C(t)(A(t)−1 )tr = Im×m , which gives C(t) = A(t)A(t)tr . This motivates us to define A as the square root of C. Reverse the above analysis, we obtain a formal proof. 4.17. Proof. We will try to solve all the sub-problems in a single, long solution. We start with the general Xi : Z t Z t Xi (t) = Xi (0) + θi (u)du + σi (u)dBi (u), i = 1, 2. 0

0

41

The goal is to show lim p ↓0

C() V1 ()V2 ()

= ρ(t0 ).

First, for i = 1, 2, we have Mi ()

= =

E[Xi (t0 + ) − Xi (t0 )|Ft0 ] Z t0 + Z t0 + Θi (u)du + σi (u)dBi (u)|Ft0 E t0

t0

Z =

t0 +

(Θi (u) − Θi (t0 ))du|Ft0 .

Θi (t0 ) + E t0

By Conditional Jensen’s Inequality, Z t0 + Z E (Θi (u) − Θi (t0 ))du|Ft0 ≤ E

t0 +

|Θi (u) − Θi (t0 )|du|Ft0

t0

t0

R t + R t + Since 1 t00 |Θi (u) − Θi (t0 )|du ≤ 2M and lim→0 1 t00 |Θi (u) − Θi (t0 )|du = 0 by the continuity of Θi , the Dominated Convergence Theorem under Conditional Expectation implies Z t0 + Z 1 1 t0 + |Θi (u) − Θi (t0 )|du|Ft0 = E lim |Θi (u) − Θi (t0 )|du|Ft0 = 0. lim E →0 →0 t t0 0 So Mi () = Θi (t0 ) + o(). This proves (iii). Rt To calculate the variance and covariance, we note Yi (t) = 0 σi (u)dBi (u) is a martingale and by Itˆ o’s Rt formula Yi (t)Yj (t) − 0 σi (u)σj (u)du is a martingale (i = 1, 2). So E[(Xi (t0 + ) − Xi (t0 ))(Xj (t0 + ) − Xj (t0 ))|Ft0 ] Z t0 + Z = E Yi (t0 + ) − Yi (t0 ) + Θi (u)du Yj (t0 + ) − Yj (t0 ) + t0

Θj (u)du |Ft0

t0 t0 +

Z = E [(Yi (t0 + ) − Yi (t0 )) (Yj (t0 + ) − Yj (t0 )) |Ft0 ] + E t0 +

Θj (u)du|Ft0

Z

t0 +

Θi (u)du t0

Z +E (Yi (t0 + ) − Yi (t0 ))

t0 +

Θj (u)du|Ft0

t0

Z + E (Yj (t0 + ) − Yj (t0 ))

t0

t0 +

Θi (u)du|Ft0

t0

= I + II + III + IV. Z

t0 +

I = E[Yi (t0 + )Yj (t0 + ) − Yi (t0 )Yj (t0 )|Ft0 ] = E

σi (u)σj (u)ρij (t)dt|Ft0 .

t0

By an argument similar to that involved in the proof of part (iii), we conclude I = σi (t0 )σj (t0 )ρij (t0 ) + o() and Z t0 + Z t0 + Z t0 + II = E (Θi (u) − Θi (t0 ))du Θj (u)du|Ft0 + Θi (t0 )E Θj (u)du|Ft0 t0

t0

t0

= o() + (Mi () − o())Mj () = Mi ()Mj () + o().

42

By Cauchy’s inequality under conditional expectation (note E[XY |F] defines an inner product on L2 (Ω)), Z t0 + |Θj (u)|du|Ft0 III ≤ E |Yi (t0 + ) − Yi (t0 )| t0 p ≤ M E[(Yi (t0 + ) − Yi (t0 ))2 |Ft0 ] p ≤ M E[Yi (t0 + )2 − Yi (t0 )2 |Ft0 ] s Z t0 + Θi (u)2 du|Ft0 ] ≤ M E[ t0

√

≤ M · M = o() Similarly, IV = o(). In summary, we have E[(Xi (t0 + ) − Xi (t0 ))(Xj (t0 + ) − Xj (t0 ))|Ft0 ] = Mi ()Mj () + σi (t0 )σj (t0 )ρij (t0 ) + o() + o(). This proves part (iv) and (v). Finally, C() ρ(t0 )σ1 (t0 )σ2 (t0 ) + o() lim p = lim p 2 = ρ(t0 ). ↓0 ↓0 V1 ()V2 () (σ1 (t0 ) + o())(σ22 (t0 ) + o()) This proves part (vi). Part (i) and (ii) are consequences of general cases. 4.18. (i) Proof. 1

2

1

2

1

2

d(ert ζt ) = (de−θWt − 2 θ t ) = −e−θWt − 2 θ t θdWt = −θ(ert ζt )dWt , where for the second “=”, we used the fact that e−θWt − 2 θ rert ζt dt + ert dζt , we get dζt = −θζt dWt − rζt dt.

t

solves dXt = −θXt dWt . Since d(ert ζt ) =

(ii) Proof. d(ζt Xt )

=

ζt dXt + Xt dζt + dXt dζt

=

ζt (rXt dt + ∆t (α − r)St dt + ∆t σSt dWt ) + Xt (−θζt dWt − rζt dt) +(rXt dt + ∆t (α − r)St dt + ∆t σSt dWt )(−θζt dWt − rζt dt)

=

ζt (∆t (α − r)St dt + ∆t σSt dWt ) − θXt ζt dWt − θ∆t σSt ζt dt

=

ζt ∆t σSt dWt − θXt ζt dWt .

So ζt Xt is a martingale. (iii) Proof. By part (ii), X0 = ζ0 X0 = E[ζT Xt ] = E[ζT VT ]. (This can be seen as a version of risk-neutral pricing, only that the pricing is carried out under the actual probability measure.) 4.19. (i) Proof. Bt is a local martingale with [B]t = motion.

Rt 0

sign(Ws )2 ds = t. So by L´evy’s theorem, Bt is a Brownian

43

(ii) Proof. d(Bt Wt ) = Bt dWt + sign(Wt )Wt dWt + sign(Wt )dt. Integrate both sides of the resulting equation and the expectation, we get Z t Z t 1 1 E[Bt Wt ] = E[sign(Ws )]ds = E[1{Ws ≥0} − 1{Ws <0} ]ds = t − t = 0. 2 2 0 0

(iii) Proof. By Itˆ o’s formula, dWt2 = 2Wt dWt + dt. (iv) Proof. By Itˆ o’s formula, d(Bt Wt2 )

=

Bt dWt2 + Wt2 dBt + dBt dWt2

=

Bt (2Wt dWt + dt) + Wt2 sign(Wt )dWt + sign(Wt )dWt (2Wt dWt + dt)

=

2Bt Wt dWt + Bt dt + sign(Wt )Wt2 dWt + 2sign(Wt )Wt dt.

So E[Bt Wt2 ]

Z t Z t = E[ Bs ds] + 2E[ sign(Ws )Ws ds] 0 0 Z t Z t = E[Bs ]ds + 2 E[sign(Ws )Ws ]ds 0 0 Z t = 2 (E[Ws 1{Ws ≥0} ] − E[Ws 1{Ws <0} ])ds 0

Z tZ =

4 0 0 Z tr

= 6=

∞

x2

e− 2s dxds x√ 2πs

s ds 2π 0 0 = E[Bt ] · E[Wt2 ]. 4

Since E[Bt Wt2 ] 6= E[Bt ] · E[Wt2 ], Bt and Wt are not independent. 4.20. (i) ( Proof. f (x) =

 (  if x > K 1, x − K, if x ≥ K 0, if x 6= K So f 0 (x) = undefined, if x = K and f 00 (x) =  0, if x < K. undefined, if x = K.  0, if x < K

(ii) Proof. E[f (WT )] =

R∞ K

2 −x

standard normal random variable. If we equal to 0. So (4.10.42) cannot hold.

q

2

T −K 2T 2π e RT suppose 0

e 2T (x − K) √ dx = 2πT

− KΦ(− √KT ) where Φ is the distribution function of f 00 (Wt )dt = 0, the expectation of RHS of (4.10.42) is

(iii)

44

Proof. This is trivial to check. (iv) 1 1 Proof. If x = K, limn→∞ fn (x) = 8n = 0; if x > K, for n large enough, x ≥ K + 2n , so limn→∞ fn (x) = 1 limn→∞ (x − K) = x − K; if x < K, for n large enough, x ≤ K − 2n , so limn→∞ fn (x) = limn→∞ 0 = 0. In summary, limn→∞ fn (x) = (x − K)+ . Similarly, we can show   0, if x < K 0 (5) lim fn (x) = 12 , if x = K n→∞   1, if x > K.

(v) Proof. Fix ω, so that Wt (ω) < K for any t ∈ [0, T ]. Since Wt (ω) can obtain its maximum on [0, T ], there 1 exists n0 , so that for any n ≥ n0 , max0≤t≤T Wt (ω) < K − 2n . So Z LK (T )(ω) = lim n n→∞

0

T 1 (Wt (ω))dt = 0. 1 1(K− 2n ,K+ 2n )

(vi) Proof. Take expectation on both sides of the formula (4.10.45), we have E[LK (T )] = E[(WT − K)+ ] > 0. So we cannot have LK (T ) = 0 a.s.. 4.21. (i) Proof. There are two problems. First, the transaction cost could be big due to active trading; second, the purchases and sales cannot be made at exactly the same price K. For more details, see Hull [2]. (ii) Proof. No. The RHS of (4.10.26) is a martingale, so its expectation is 0. But E[(ST − K)+ ] > 0. So XT 6= (ST − K)+ . 5. Risk-Neutral Pricing 5.1. (i) Proof. df (Xt )

= = = =

1 f 0 (Xt )dt + f 00 (Xt )dhXit 2 1 f (Xt )(dXt + dhXit ) 2

1 1 f (Xt ) σt dWt + (αt − Rt − σt2 )dt + σt2 dt 2 2 f (Xt )(αt − Rt )dt + f (Xt )σt dWt .

This is formula (5.2.20). 45

(ii) Proof. d(Dt St ) = St dDt + Dt dSt + dDt dSt = −St Rt Dt dt + Dt αt St dt + Dt σt St dWt = Dt St (αt − Rt )dt + Dt St σt dWt . This is formula (5.2.20). 5.2. e T VT |Ft ] = E Proof. By Lemma 5.2.2., E[D

h

DT VT ZT Zt

i |Ft . Therefore (5.2.30) is equivalent to Dt Vt Zt =

E[DT VT ZT |Ft ]. 5.3. (i) Proof. cx (0, x)

= = = = = = = =

1 2 d e −rT f E[e (xeσWT +(r− 2 σ )T − K)+ ] dx fT +(r− 1 σ 2 )T −rT d σW e 2 E e h(xe ) dx h i fT +(r− 1 σ 2 )T e e−rT eσW 2 1 σW E f +(r− 1 σ 2 )T T 2 {xe >K} h i fT σW − 21 σ 2 T e E e 1{W e fT > 1 (ln K −(r− 1 σ 2 )T )} σ x 2 √ f WT 1 2 √ T σ −2σ T e T √ √ 1 W e E e f 1 1 2 { √T −σ T > √ (ln K x −(r− 2 σ )T )−σ T } T σ T Z ∞ √ z2 1 2 1 √ e− 2 eσ T z 1{z−σ√T >−d+ (T,x)} dz e− 2 σ T 2π −∞ Z ∞ √ (z−σ T )2 1 2 √ e− 1{z−σ√T >−d+ (T,x)} dz 2π −∞ N (d+ (T, x)).

(ii) 1 2 f bt = E[ e ZbT |Ft ], then Z b is a Pe-martingale, Zbt > 0 and E[ZbT ] = Proof. If we set ZbT = eσWT − 2 σ T and Z 1 2 f σ W − σ T T e 2 E[e ] = 1. So if we define Pb by dPb = ZT dPe on FT , then Pb is a probability measure equivalent to e P , and e ZbT 1{S >K} ] = Pb(ST > K). cx (0, x) = E[ T Rt ct = W ft + (−σ)du = W ft − σt is a Pb-Brownian motion (set Θ = −σ Moreover, by Girsanov’s Theorem, W 0 in Theorem 5.4.1.)

(iii) 1

Proof. ST = xeσWT +(r− 2 σ f

2

)T

1

= xeσWT +(r+ 2 σ c

2

)T

. So

1 2 c Pb(ST > K) = Pb(xeσWT +(r+ 2 σ )T > K) = Pb

46

! cT W √ > −d+ (T, x) = N (d+ (T, x)). T

f (t)” → “σ(t)S(t)dW f (t)”. In the first equation for 5.4. First, a few typos. In the SDE for S, “σ(t)dW e In the second equation for c(0, S(0)), the variable for BSM should be c(0, S(0)), E → E.   s Z T Z T 1 1 r(t)dt, σ 2 (t)dt . BSM T, S(0); K, T 0 T 0 (i) RT RT 1 2 1 1 2 t f f Proof. d ln St = dS St − 2St2 dhSit = rt dt + σt dWt − 2 σt dt. So ST = S0 exp{ 0 (rt − 2 σt )dt + 0 σt dWt }. Let R RT T ft . The first term in the expression of X is a number and the second term X = 0 (rt − 12 σt2 )dt + 0 σt dW RT is a Gaussian random variable N (0, 0 σt2 dt), since both r and σ ar deterministic. Therefore, ST = S0 eX , RT R T with X ∼ N ( 0 (rt − 12 σt2 )dt, 0 σt2 dt),. (ii) Proof. For the standard BSM model with constant volatility Σ and interest rate R, under the risk-neutral fT ∼ N ((R− 1 Σ2 )T, Σ2 T ), and E[(S e 0 eY −K)+ ] = measure, we have ST = S0 eY , where Y = (R− 12 Σ2 )T +ΣW q 2 eRT BSM (T, S0 ; K, R, Σ). Note R = T1 (E[Y ] + 12 V ar(Y )) and Σ = T1 V ar(Y ), we can get 1 T, S0 ; K, T

e 0 eY − K)+ ] = eE[Y ]+ 21 V ar(Y ) BSM E[(S

! r 1 1 V ar(Y ) . E[Y ] + V ar(Y ) , 2 T

So for the model in this problem, c(0, S0 )

= =

e−

RT

e−

RT

0

0

rt dt rt dt

e 0 eX − K)+ ] E[(S e

E[X]+ 21 V

ar(X)

 =

1 BSM T, S0 ; K, T

Z

BSM s

T

rt dt, 0

1 T, S0 ; K, T 1 T

Z

T

! r 1 1 E[X] + V ar(X) , V ar(X) 2 T

 σt2 dt .

0

5.5. (i) Proof. Let f (x) = x1 , then f 0 (x) = − x12 and f 00 (x) = d

1 Zt

2 x3 .

Note dZt = −Zt Θt dWt , so

1 1 1 2 2 2 Θt Θ2 = f 0 (Zt )dZt + f 00 (Zt )dZt dZt = − 2 (−Zt )Θt dWt + Zt Θt dt = dWt + t dt. 3 2 Zt 2 Zt Zt Zt

(ii) fs = E[ eM ft |Fs ] = E Proof. By Lemma 5.2.2., for s, t ≥ 0 with s < t, M fs . So M = Z M f is a P -martingale. Zs M (iii)

47

h

ft Zt M Zs |Fs

i

ft |Fs ] = . That is, E[Zt M

Proof. 1 1 1 1 Γt M t Θt Mt Θ2t Γ t Θt f dMt = d Mt · = dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt Zt

(iv) Proof. In part (iii), we have ft = dM et = Let Γ

Γt M t Θt Mt Θ2t Γt Θt Γt M t Θt dWt + dWt + dt + dt = (dWt + Θt dt) + (dWt + Θt dt). Zt Zt Zt Zt Zt Zt

Γt +Mt Θt , Zt

ft = Γ e t dW ft . This proves Corollary 5.3.2. then dM

5.6. fi (t) is an Ft -martingale under Pe and [W fi , W fj ](t) = tδij Proof. By Theorem 4.6.5, it suffices to show W fi (t) is an Ft -martingale under Pe if and only if W fi (t)Zt is an Ft -martingale (i, j = 1, 2). Indeed, for i = 1, 2, W under P , since " # fi (t)Zt W eW fi (t)|Fs ] = E E[ |Fs . Zs By Itˆ o’s product formula, we have fi (t)Zt ) d(W

fi (t)dZt + Zt dW fi (t) + dZt dW fi (t) = W fi (t)(−Zt )Θ(t) · dWt + Zt (dWi (t) + Θi (t)dt) + (−Zt Θt · dWt )(dWi (t) + Θi (t)dt) = W fi (t)(−Zt ) = W

d X

Θj (t)dWj (t) + Zt (dWi (t) + Θi (t)dt) − Zt Θi (t)dt

j=1

=

fi (t)(−Zt ) W

d X

Θj (t)dWj (t) + Zt dWi (t)

j=1

fi (t)Zt is an Ft -martingale under P . So W fi (t) is an Ft -martingale under Pe. Moreover, This shows W Z · Z · fi , W fj ](t) = Wi + [W Θi (s)ds, Wj + Θj (s)ds (t) = [Wi , Wj ](t) = tδij . 0

0

Combined, this proves the two-dimensional Girsanov’s Theorem. 5.7. (i) Proof. Let a be any strictly positive number. We define X2 (t) = (a + X1 (t))D(t)−1 . Then X2 (0) P X2 (T ) ≥ = P (a + X1 (T ) ≥ a) = P (X1 (T ) ≥ 0) = 1, D(T ) 2 (0) and P X2 (T ) > X = 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 define X1 (t) = X2 (t)D(t) − X2 (0). Then X1 (0) = 0, X2 (0) X2 (0) P (X1 (T ) ≥ 0) = P X2 (T ) ≥ = 1, P (X1 (T ) > 0) = P X2 (T ) > > 0. D(T ) D(T )

5.8. The basic idea is that for any positive Pe-martingale M , dMt = Mt · M1t dMt . By Martingale Repree t dW ft for some adapted process Γ e t . So dMt = Mt ( Γet )dW ft , i.e. any positive sentation Theorem, dMt = Γ Mt martingale must be the exponential of an integral w.r.t. Brownian motion. Taking into account discounting factor and apply Itˆ o’s product rule, we can show every strictly positive asset is a generalized geometric Brownian motion. (i) e T VT |Ft ]. So (Dt Vt )t≥0 is a Pe-martingale. By Martingale Represene − 0T Ru du VT |Ft ] = E[D Proof. Vt Dt = E[e R e t , 0 ≤ t ≤ T , such that Dt Vt = t Γ e s dW fs , or equivalently, tation Theorem, there exists an adapted process Γ R R0 −1 t e f −1 t e f e t dW ft , Vt = Dt Γs dWs . Differentiate both sides of the equation, we get dVt = Rt Dt Γs dWs dt + Dt−1 Γ R

0

i.e. dVt = Rt Vt dt +

0

et Γ Dt dWt .

(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.) defined on the probability space (Ω, G, P ). Let F be a sub σ-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},Pwe shall show P (A) = 0. In∞ 1 deed, note A ∈ F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A∩{X≥1} ] + n=1 E[X1A∩{ n1 >X≥ n+1 }] ≥ P∞ 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, P∞ 1 }) = ∀n ≥ 1. This in turn implies P (A) = P (A ∩ {X > 0}) = P (A ∩ {X ≥ 1}) + n=1 P (A ∩ { n1 > X ≥ n+1 0. e − tT Ru du VT |Ft ] > 0 a.s.. Moreover, By the above lemma, it is clear that for each t ∈ [0, T ], Vt = E[e 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. ω, Vt (ω) > 0 for any t ∈ [0, T ]. R

(iii) Proof. By (ii), V > 0 a.s., so dVt = Vt V1t dVt = Vt V1t Rt Vt dt + ft , where σt = σt Vt dW

et Γ Vt Dt .

et f Γ Dt dWt

ft = Rt Vt dt + = Vt Rt dt + Vt VtΓDt t dW e

This shows V follows a generalized geometric Brownian motion.

5.9. Proof. c(0, T, x, K) = xN (d+ ) − Ke−rT N (d− ) with d± = then f 0 (y) = −yf (y), cK (0, T, x, K)

1 √ σ T

x (ln K + (r ± 12 σ 2 )T ). Let f (y) =

∂d+ ∂d− − e−rT N (d− ) − Ke−rT f (d− ) ∂y ∂y −1 1 = xf (d+ ) √ − e−rT N (d− ) + e−rT f (d− ) √ , σ TK σ T = xf (d+ )

49

2

y √1 e− 2 2π

,

and

= = = =

cKK (0, T, x, K) x ∂d− e−rT 1 ∂d+ d− − √ − e−rT f (d− ) + √ (−d− )f (d− ) xf (d+ ) √ f (d+ )(−d+ ) 2 ∂y ∂y ∂y σ TK σ TK σ T −rT x xd+ −1 −1 e d −1 √ √ − e−rT f (d− ) √ − √ − f (d− ) √ f (d+ ) + √ f (d+ ) σ T K2 σ TK Kσ T Kσ T σ T Kσ T x d e−rT f (d− ) d− √ [1 − √+ ] + √ f (d+ ) [1 + √ ] 2 K σ T σ T Kσ T σ T e−rT x f (d− )d+ − 2 2 f (d+ )d− . Kσ 2 T K σ 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) e −rt0 V (t0 )] = E[e e −rt0 C(t0 )+(e−rT K −e−rt0 S(t0 )+ ] = Proof. By the risk-neutral pricing formula, V (0) = E[e −rt0 −r(T −t0 ) + e C(0) + E[e (e K − S(t0 )) ]. The first 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 first 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 flow, we must find 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 first equation and apply Itˆo’s product rule, we get d(Dt Xt ) = ∆t d(Dt St ) − RT RT Ct Dt dt. Integrate from 0 to T , we have DT XT − D0 X0 = 0 ∆t d(Dt St ) − 0 Ct Dt dt. By the terminal RT RT condition, we get X0 = D0−1 ( 0 Ct Dt dt − 0 ∆t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash flow, provided we can find a trading strategy ∆ that solves the BSDE. Note the SDE for S t . Take the proper change of measure so that gives d(Dt St ) = (Dt St )σt (θt dt + dWt ), where θt = αtσ−R t Rt f Wt = θs ds + Wt is a Brownian motion under the new measure Pe, we get 0

Z

T

Z Ct Dt dt = D0 X0 +

0

T

Z ∆t d(Dt St ) = D0 X0 +

0

T

ft . ∆t (Dt St )σt dW 0

RT RT ft . This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ∆t Dt St σt dW RT 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

RT

e T |Ft ]. Then by Martingale Representation TheoCt Dt dt, and Mt = E[M R e t , so that Mt = M0 + t Γ e dW ft . If we set ∆t = Γet , we can check rem, we can find an adapted process Γ Dt St σt 0 t Rt Rt R e T Ct Dt dt] solves the SDE Xt = Dt−1 (D0 X0 + 0 ∆u d(Du Su ) − 0 Cu Du du), with X0 = M0 = E[ 0 ( dXt = ∆t dSt + Rt (Xt − ∆t St )dt − Ct dt XT = 0. (Formal proof) Let MT =

0

Indeed, it is easy to see that X satisfies the first equation. To check the terminal condition, we note RT R R ft − T Ct Dt dt = M0 + T Γ e t dW ft − MT = 0. So XT = 0. Thus, we have XT DT = D0 X0 + 0 ∆t Dt St σt dW 0 0 found a trading strategy ∆, so that the corresponding portfolio X replicates the cash flow and has zero R e T Ct Dt dt] is the no-arbitrage price of the cash flow at time zero. terminal value. So X0 = E[ 0 Remark: As shown in the analysis, d(Dt Xt ) = ∆t d(Dt St ) − Ct Dt dt. Integrate from t to T , we get RT RT 0 − Dt Xt = t ∆u d(Du Su ) − t Cu Du du. Take conditional expectation w.r.t. Ft on both sides, we get R R e T Cu Du du|Ft ]. So Xt = Dt−1 E[ e T Cu Du du|Ft ]. This is the no-arbitrage price of the cash −Dt Xt = −E[ t t flow at time t, and we have justified formula (5.6.10) in the textbook. 5.12. (i) fj (t). So Bi is a ei (t) = dBi (t) + γi (t)dt = Pd σij (t) dWj (t) + Pd σij (t) Θj (t)dt = Pd σij (t) dW Proof. dB j=1 σi (t) j=1 σi (t) j=1 σi (t) 2 P ei (t)dB ei (t) = d σij (t)2 dt = dt, by L´evy’s Theorem, B ei is a Brownian motion under martingale. Since dB j=1 σi (t) Pe. (ii) Proof. dSi (t)

=

ei (t) + (αi (t) − R(t))Si (t)dt − σi (t)Si (t)γi (t)dt R(t)Si (t)dt + σi (t)Si (t)dB

=

ei (t) + R(t)Si (t)dt + σi (t)Si (t)dB

d X

σij (t)Θj (t)Si (t)dt − Si (t)

j=1

=

d X

σij (t)Θj (t)dt

j=1

ei (t). R(t)Si (t)dt + σi (t)Si (t)dB

(iii) ei (t)dB ek (t) = (dBi (t) + γi (t)dt)(dBj (t) + γj (t)dt) = dBi (t)dBj (t) = ρik (t)dt. Proof. dB (iv) Proof. By Itˆ o’s product rule and martingale property, Z t Z t Z t E[Bi (t)Bk (t)] = E[ Bi (s)dBk (s)] + E[ Bk (s)dBi (s)] + E[ dBi (s)dBk (s)] 0 0 0 Z t Z t = E[ ρik (s)ds] = ρik (s)ds. 0

0

eB ei (t)B ek (t)] = Similarly, by part (iii), we can show E[

Rt 0

(v)

51

ρik (s)ds.

Proof. By Itˆ o’s product formula, Z t Z t [P (W1 (u) ≥ 0) − P (W1 (u) < 0)]du = 0. sign(W1 (u))du] = E[B1 (t)B2 (t)] = E[ 0

0

Meanwhile, eB e1 (t)B e2 (t)] E[

Z t e E[ sign(W1 (u))du 0 Z t [Pe(W1 (u) ≥ 0) − Pe(W1 (u) < 0)]du 0 Z t f1 (u) ≥ u) − Pe(W f1 (u) < u)]du [Pe(W 0 Z t 1 f1 (u) < u) du 2 − Pe(W 2 0 0,

= = = = <

eB e1 (t)B e2 (t)] for all t > 0. for any t > 0. So E[B1 (t)B2 (t)] = E[ 5.13. (i) e 1 (t)] = E[ eW f1 (t)] = 0 and E[W e 2 (t)] = E[ eW f2 (t) − Proof. E[W

Rt 0

f1 (u)du] = 0, for all t ∈ [0, T ]. W

(ii) Proof. e Cov[W 1 (T ), W2 (T )]

e 1 (T )W2 (T )] = E[W "Z Z T e = E W1 (t)dW2 (t) + 0

"Z

T

# W2 (t)dW1 (t)

0 T

e = E

# "Z f f f e W1 (t)(dW2 (t) − W1 (t)dt) + E

T

# f W2 (t)dW1 (t)

0

0

"Z

#

T

e = −E

f1 (t)2 dt W

0

Z = −

T

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−rt Xt a martingale, we are motivated to change the measure Rt in such a way that St −r 0 Su du−at is a martingale under the new measure. To do this,hwe note the SDE for iS

t −a is dSt = αt St dt+σSt dWt . Hence dSt −rSt dt−adt = [(αt −r)St −a]dt+σSt dWt = σSt (αt −r)S dt + dWt . σSt R t t −a ft = θs ds + Wt , we can find an equivalent probability measure Pe, under which Set θt = (αt −r)S and W σSt 0 ft + adt and W ft is a BM. This is the rational for formula (5.9.7). S satisfies the SDE dSt = rSt dt + σSt dW This is a good place to pause and think about the meaning of “martingale measure.” What is to be a martingale? The new measure Pe 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 Pe because we suppose that X is able to replicate the derivative payoff at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ∈ [0, T ]. The difficulty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: e T XT |Ft ] = Dt−1 E[D e T VT |Ft ]. You can think of martingale measure as a calculational Vt = Xt = Dt−1 E[D convenience. That is all about martingale measure! Risk neutral is a just perception, referring to the actual effect of constructing a hedging portfolio! Second, we note when the portfolio is self-financing, the discounted price process of the underlying is a martingale under Pe, 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 Pe is more or less equivalent to Dt St being a martingale under Pe. However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ∆t d(Dt St ) no longer holds, as shown in formula (5.9.6). The portfolio is no longer self-financing, but self-financing 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. e −rT VT |Ft ], by Martingale Representation Let VT be a payoff at time T , then for the martingale Mt = E[e R e t , so that Mt = M0 + t Γ e dW fs . If we let ∆t = Γet ert , then the Theorem, we can find an adapted process Γ σSt 0 s e t dW ft . So by setting X0 = M0 = E[e e −rT VT ], value of the corresponding portfolio X satisfies d(e−rt Xt ) = Γ −rt we must have e Xt = Mt , for all t ∈ [0, T ]. In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justifies the risk-neutral pricing of European-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is different 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 find a probability measure Pe, under which e−rt Xt is Ra martingale, e −rT VT |Ft ] := Mt . Martingale Representation Theorem gives Mt = M0 + t Γ e dW fu for then e−rt Xt = E[e 0 u e This would give us a theoretical representation of ∆ by comparison of integrands, some adapted process Γ. hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5.9.7) under Pe, then d(e−rt Xt ) = ∆t [d(e−rt St ) − ft . So (e−rt Xt )t≥0 , where X is given by (5.9.6), is a Pe-martingale. ae−rt dt] = ∆t e−rt σSt dW (ii) ft + (r − 1 σ 2 )dt] + 1 Yt σ 2 dt = Yt (σdW ft + rdt). So d(e−rt Yt ) = Proof. By Itˆ o’s formula, dYt = Yt [σdW 2 2 R t a ft and e−rt Yt is a Pe-martingale. Moreover, if St = S0 Yt + Yt σe−rt Yt dW ds, then 0 Ys Z dSt = S0 dYt + 0

t

a dsdYt + adt = Ys

Z S0 + 0

t

a ft + rdt) + adt = St (σdW ft + rdt) + adt. ds Yt (σdW Ys

This shows S satisfies (5.9.7). Remark: To obtain this formula for S, we first set Ut = e−rt St to remove the rSt dt term. The SDE for ft + ae−rt dt. Just like solving linear ODE, to remove U in the dW ft term, we consider U is dUt = σUt dW ft −σ W Vt = Ut e . Itˆ o’s product formula yields 1 2 1 2 ft ft ft −σ W −σ W −σ W f f dVt = e dUt + Ut e (−σ)dWt + σ dt + dUt · e (−σ)dWt + σ dt 2 2 1 f = e−σWt ae−rt dt − σ 2 Vt dt. 2 53

1

2

Note V appears only in the dt term, so multiply the integration factor e 2 σ t on both sides of the equation, we get 1 2 f 1 2 d(e 2 σ t Vt ) = ae−rt−σWt + 2 σ t dt. Rt 1 2 f Set Yt = eσWt +(r− 2 σ )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + 0 ads Ys ). (iii) Proof. e T |Ft ] E[S

" Z e e = S0 E[YT |Ft ] + E YT = = = =

# a ds|Ft Ys t 0 Z t Z T a e T |Ft ] + e T |Ft ] + a e YT |Ft ds S0 E[Y dsE[Y E Ys 0 Ys t Z T Z t a e T −t ] + a e T −s ]ds e T −t ] + dsYt E[Y S0 Yt E[Y E[Y Y s t 0 Z T Z t a r(T −t) ds Yt e +a er(T −s) ds S0 + t 0 Ys Z t a ads Yt er(T −t) − (1 − er(T −t) ). S0 + r 0 Ys t

a ds + YT Ys

Z

T

e T ] = S0 erT − a (1 − erT ). In particular, E[S r (iv) Proof. e T |Ft ] dE[S

Z t a ads (er(T −t) dYt − rYt er(T −t) dt) + er(T −t) (−r)dt = aer(T −t) dt + S0 + r 0 Ys Z t ads r(T −t) ft . = S0 + e σYt dW 0 Ys

e T |Ft ] is a Pe-martingale. As we have argued at the beginning of the solution, risk-neutral pricing is So E[S e T |Ft ] valid even in the presence of cost of carry. So by an argument similar to that of §5.6.2, the process E[S is the futures price process for the commodity. (v) e −r(T −t) (ST − K)|Ft ] = 0 for K, and get K = E[S e T |Ft ]. So F orS (t, T ) = Proof. We solve the equation E[e 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−rt Xt ) = d(e−rt St ) − ae−rt dt. Integrate from 0 to t on both sides, we get Xt = St − S0 ert + ar (1 − ert ) = St − S0 ert − ar (ert XT = ST − S0 erT − ar (erT − 1). − 1).R In particular, e T |Ft ] = S0 + t ads Yt er(T −t) − a (1−er(T −t) ). So F orS (0, T ) = Meanwhile, F orS (t, T ) = F uts (t, T ) = E[S 0 Ys

r

S0 erT − ar (1 − erT ) and hence XT = ST − F orS (0, T ). After the agent delivers the commodity, whose value is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54

6. Connections with Partial Differential 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 Itˆ o’s formula, it is easy to obtain dZu = bu Zu du + σu 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 − σu γu γu Yu (bu Zu du + σu Zu dWu ) + Zu du + dWu Zu Zu [Yu bu Zu + (au − σu γu ) + σu γu ]du + (σu Zu Yu + γu )dWu

=

(bu Xu + au )du + (σu Xu + γu )dWu .

=

+ σu Z u

γu du Zu

Remark: To see how to find the above solution, we manipulate the equation (6.2.4) Ras follows. First, to u remove the term bu Xu du, we multiply on both sides of (6.2.4) the integrating factor e− t bv dv . Then d(Xu e− ¯ u = e− Let X

Ru t

bv dv

Xu , a ¯ u = e−

Ru

Ru t

t

bv dv

bv dv

) = e−

Ru t

bv dv

(au du + (γu + σu Xu )dWu ).

au and γ¯u = e−

Ru t

bv dv

¯ satisfies the SDE γu , then X

¯u = a ¯ u )dWu = (¯ ¯ u dWu . dX ¯u du + (¯ γu + σu X au du + γ¯u dWu ) + σu X ¯ u dWu , we consider X ˆu = X ¯ u e− To deal with the term σu X ˆu dX

=

e−

Ru t

σv dWv

Ru t

σv dWv

¯ u dWu ] + X ¯ u e− [(¯ au du + γ¯u dWu ) + σu X

¯ u )(−σu )e− +(¯ γu + σu X

Ru t

σv dWv

Ru t

. Then σv dWv

1 Ru (−σu )dWu + e− t σv dWv σu2 du 2

du

ˆ u dWu − σu X ˆ u dWu + 1 X ˆ u σu2 du − σu (ˆ ˆ u )du = a ˆu du + γˆu dWu + σu X γu + σu X 2 1 ˆ 2 = (ˆ au − σu γˆu − X ˆu dWu , u σu )du + γ 2 where a ˆu = a ¯ u e−

Ru t

σv dWv

and γˆu = γ¯u e−

Ru t

σv dWv

. Finally, use the integrating factor e

Ru t

1 2 2 σv dv

, we have

R R R ˆ u e 12 tu σv2 dv = e 12 tu σv2 dv (dX ˆu + X ˆ u · 1 σ 2 du) = e 21 tu σv2 dv [(ˆ d X au − σu γˆu )du + γˆu dWu ]. u 2 Write everything back into the original X, a and γ, we get Ru 2 Ru 2 Ru Ru Ru Ru 1 1 d Xu e− t bv dv− t σv dWv + 2 t σv dv = e 2 t σv dv− t σv dWv − t bv dv [(au − σu γu )du + γu dWu ], i.e.

d

Xu Zu

=

1 [(au − σu γu )du + γu dWu ] = dYu . Zu

This inspired us to try Xu = Yu Zu . 6.2. (i)

55

Proof. The portfolio is self-financing, 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 )γ 2 (t, Rt )dt 2 1 +∆2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )γ 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 ) + α(t, Rt )fr (t, Rt , T1 ) + γ 2 (t, Rt )frr (t, Rt , T1 )]dt 2 1 +∆2 (t)Dt [−Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + α(t, Rt )fr (t, Rt , T2 ) + γ 2 (t, Rt )frr (t, Rt , T2 )]dt 2 +Dt γ(t, Rt )[Dt γ(t, Rt )[∆1 (t)fr (t, Rt , T1 ) + ∆2 (t)fr (t, Rt , T2 )]]dWt =

∆1 (t)Dt [α(t, Rt ) − β(t, Rt , T1 )]fr (t, Rt , T1 )dt + ∆2 (t)Dt [α(t, Rt ) − β(t, Rt , T2 )]fr (t, Rt , T2 )dt +Dt γ(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 [β(t, Rt , T2 ) − β(t, Rt , T1 )]fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt |[β(t, Rt , T1 ) − β(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 Z T DT XT − D0 X0 = Dt |[β(t, Rt , T1 ) − β(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. 0

If β(t, Rt , T1 ) 6= β(t, Rt , T2 ) for some t ∈ [0, T ], under the assumption that fr (t, r, T ) 6= 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. ω, β(t, Rt , T1 ) = β(t, Rt , T2 ), ∀t ∈ [0, T ]. This implies β(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 get 1 d(Dt Xt ) = ∆(t)Dt −Rt f (t, Rt , T ) + ft (t, Rt , T ) + α(t, Rt )fr (t, Rt , T ) + γ 2 (t, Rt )frr (t, Rt , T ) dt 2 +Dt γ(t, Rt )∆(t)fr (t, Rt , T )dWt . This is formula (6.9.5). If fr (t, r, T ) = 0, then d(Dt Xt ) = ∆(t)Dt −Rt f (t, Rt , T ) + ft (t, Rt , T ) + 21 γ 2 (t, Rt )frr (t, Rt , T ) dt. We choose ∆(t) = sign −Rt f (t, Rt , T ) + ft (t, Rt , T ) + 12 γ 2 (t, Rt )frr (t, Rt , T ) . To avoid arbitrage in this case, we must have ft (t, Rt , T ) + 12 γ 2 (t, Rt )frr (t, Rt , T ) = Rt f (t, Rt , T ), or equivalently, for any r in the range of Rt , ft (t, r, T ) + 12 γ 2 (t, r)frr (t, r, T ) = rf (t, r, T ). 56

6.3. Proof. We note i Rs Rs d h − R s bv dv e 0 C(s, T ) = e− 0 bv dv [C(s, T )(−bs ) + bs C(s, T ) − 1] = −e− 0 bv dv . ds So integrate on both sides of the equation from t to T, we obtain e

−

RT 0

bv dv

C(T, T ) − e

−

Rt 0

bv dv

Z

T

e−

C(t, T ) = −

Rs 0

bv dv

ds.

t Rt

Since C(T, T ) = 0, we have C(t, T ) = e −a(s)C(s, T ) + 21 σ 2 (s)C 2 (s, T ), we get

0

Z A(T, T ) − A(t, T ) = −

bv dv

RT t

e−

Rs 0

bv dv

T

a(s)C(s, T )ds + t

RT

Since A(T, T ) = 0, we have A(t, T ) =

t

ds =

(a(s)C(s, T ) −

1 2

Z

RT t

e

Rt s

bv dv

ds. Finally, by A0 (s, T ) =

T

σ 2 (s)C 2 (s, T )ds.

t

1 2 2 2 σ (s)C (s, T ))ds.

6.4. (i) Proof. By the definition of ϕ, we have 1

ϕ0 (t) = e 2 σ

2

RT t

C(u,T )du 1

2

1 σ 2 (−1)C(t, T ) = − ϕ(t)σ 2 C(t, T ). 2

0

2ϕ (t) 1 0 2 So C(t, T ) = − φ(t)σ 2 . Differentiate both sides of the equation ϕ (t) = − 2 ϕ(t)σ C(t, T ), we get

1 − σ 2 [ϕ0 (t)C(t, T ) + ϕ(t)C 0 (t, T )] 2 1 1 = − σ 2 [− ϕ(t)σ 2 C 2 (t, T ) + ϕ(t)C 0 (t, T )] 2 2 1 4 1 = σ ϕ(t)C 2 (t, T ) − σ 2 ϕ(t)C 0 (t, T ). 4 2 00 (t) So C 0 (t, T ) = 41 σ 4 ϕ(t)C 2 (t, T ) − ϕ00 (t) / 12 ϕ(t)σ 2 = 21 σ 2 C 2 (t, T ) − 2ϕ σ 2 ϕ(t) . ϕ00 (t)

=

(ii) Proof. Plug formulas (6.9.8) and (6.9.9) into (6.5.14), we get −

2ϕ00 (t) 1 2 2 2ϕ0 (t) 1 + σ C (t, T ) = b(−1) 2 + σ 2 C 2 (t, T ) − 1, 2 σ ϕ(t) 2 σ ϕ(t) 2

i.e. ϕ00 (t) − bϕ0 (t) − 21 σ 2 ϕ(t) = 0. (iii) Proof. The characteristic equation of ϕ00 (t) − bϕ0 (t) − 21 σ 2 ϕ(t) = 0 is λ2 − bλ − 21 σ 2 = 0, which gives two √ √ roots 12 (b ± b2 + 2σ 2 ) = 12 b ± γ with γ = 21 b2 + 2σ 2 . Therefore by standard theory of ordinary differential 1 equations, a general solution of ϕ is ϕ(t) = e 2 bt (a1 eγt + a2 e−γt ) for some constants a1 and a2 . It is then easy to see that we can choose appropriate constants c1 and c2 so that ϕ(t) =

1 1 c1 c2 e−( 2 b+γ)(T −t) − 1 e−( 2 b−γ)(T −t) . +γ b − γ 2

1 2b

57

(iv) 1

1

Proof. From part (iii), it is easy to see ϕ0 (t) = c1 e−( 2 b+γ)(T −t) − c2 e−( 2 b−γ)(T −t) . In particular, 0 = C(T, T ) = −

2ϕ0 (T ) 2(c1 − c2 ) =− 2 . σ 2 ϕ(T ) σ ϕ(T )

So c1 = c2 . (v) Proof. We first recall the definitions and properties of sinh and cosh: sinh z =

ez − e−z ez + e−z , cosh z = , (sinh z)0 = cosh z, and (cosh z)0 = sinh z. 2 2

Therefore ϕ(t)

1

= c1 e− 2 b(T −t) = c1 e = =

− 12 b(T −t)

eγ(T −t) e−γ(T −t) − 1 1 2b + γ 2b − γ

1 2b 1 2 4b

1 − γ −γ(T −t) 2b + γ e − eγ(T −t) 1 2 2 2 −γ b − γ 4

1 2c1 − 1 b(T −t) 1 −γ(T −t) γ(T −t) 2 e b − γ)e + ( b + γ)e −( σ2 2 2 2c1 − 1 b(T −t) e 2 [b sinh(γ(T − t)) + 2γ cosh(γ(T − t))]. σ2

and ϕ0 (t)

1 2c1 − 1 b(T −t) [b sinh(γ(T − t)) + 2γ cosh(γ(T − t))] b· 2 e 2 2 σ 2c1 1 + 2 e− 2 b(T −t) [−γb cosh(γ(T − t)) − 2γ 2 sinh(γ(T − t))] σ 2 1 bγ bγ 2γ 2 b = 2c1 e− 2 b(T −t) sinh(γ(T − t)) + cosh(γ(T − t)) − cosh(γ(T − t)) − sinh(γ(T − t)) 2σ 2 σ2 σ2 σ2 1 b2 − 4γ 2 = 2c1 e− 2 b(T −t) sinh(γ(T − t)) 2σ 2 1 = −2c1 e− 2 b(T −t) sinh(γ(T − t)). =

This implies C(t, T ) = −

2ϕ0 (t) sinh(γ(T − t)) . = σ 2 ϕ(t) γ cosh(γ(T − t)) + 12 b sinh(γ(T − t))

(vi) Proof. By (6.5.15) and (6.9.8), A0 (t, T ) =

2aϕ0 (t) σ 2 ϕ(t) .

Hence Z

A(T, T ) − A(t, T ) = t

and

T

2aϕ0 (s) 2a ϕ(T ) ds = 2 ln , 2 σ ϕ(s) σ ϕ(t)

# " 1 2a ϕ(T ) 2a γe 2 b(T −t) A(t, T ) = − 2 ln = − 2 ln . σ ϕ(t) σ γ cosh(γ(T − t)) + 12 b sinh(γ(T − t))

58

6.5. (i) Proof. Since g(t, X1 (t), X2 (t)) = E[h(X1 (T ), X2 (T ))|Ft ] and e−rt f (t, X1 (t), X2 (t)) = E[e−rT h(X1 (T ), X2 (T ))|Ft ], iterated conditioning argument shows g(t, X1 (t), X2 (t)) and e−rt f (t, X1 (t), X2 (t)) ar both martingales. (ii) and (iii) Proof. We note dg(t, X1 (t), X2 (t)) 1 1 = gt dt + gx1 dX1 (t) + gx2 dX2 (t) + gx1 x2 dX1 (t)dX1 (t) + gx2 x2 dX2 (t)dX2 (t) + gx1 x2 dX1 (t)dX2 (t) 2 2 1 2 2 = gt + gx1 β1 + gx2 β2 + gx1 x1 (γ11 + γ12 + 2ργ11 γ12 ) + gx1 x2 (γ11 γ21 + ργ11 γ22 + ργ12 γ21 + γ12 γ22 ) 2 1 2 2 + gx2 x2 (γ21 + γ22 + 2ργ21 γ22 ) dt + martingale part. 2 So we must have 1 2 2 gt + gx1 β1 + gx2 β2 + gx1 x1 (γ11 + γ12 + 2ργ11 γ12 ) + gx1 x2 (γ11 γ21 + ργ11 γ22 + ργ12 γ21 + γ12 γ22 ) 2 1 2 2 + γ22 + 2ργ21 γ22 ) = 0. + gx2 x2 (γ21 2 Taking ρ = 0 will give part (ii) as a special case. The PDE for f can be similarly obtained. 6.6. (i) 1

Proof. Multiply e 2 bt on both sides of (6.9.15), we get 1 1 1 b 1 1 1 d(e 2 bt Xj (t)) = e 2 bt Xj (t) bdt + (− Xj (t)dt + σdWj (t) = e 2 bt σdWj (t). 2 2 2 2 R R 1 1 t 1 t 1 So e 2 bt Xj (t) − Xj (0) = 12 σ 0 e 2 bu dWj (u) and Xj (t) = e− 2 bt Xj (0) + 12 σ 0 e 2 bu dWj (u) . By Theorem Rt −bt 2 1 4.4.9, Xj (t) is normally distributed with mean Xj (0)e− 2 bt and variance e 4 σ 2 0 ebu du = σ4b (1 − e−bt ). (ii) Proof. Suppose R(t) =

Pd

j=1

dR(t)

Xj2 (t), then =

d X

(2Xj (t)dXj (t) + dXj (t)dXj (t))

j=1

=

d X j=1

=

d X j=1

=

1 2 2Xj (t)dXj (t) + σ dt 4 −bXj2 (t)dt

1 2 + σXj (t)dWj (t) + σ dt 4

d X p d 2 X (t) pj σ − bR(t) dt + σ R(t) dWj (t). 4 R(t) j=1

Pd Xj2 (t) then B is a local martingale with dB(t)dB(t) = j=1 R(t) dt = dt. So p by L´evy’s Theorem, B is a Brownian motion. Therefore dR(t) = (a − bR(t))dt + σ R(t)dB(t) (a := d4 σ 2 ) and R is a CIR interest rate process. Let B(t) =

R t Xj (s) √ dWj (s), j=1 0 R(s)

Pd

59

(iii) 1

Proof. By (6.9.16), Xj (t) is dependent on Wj only and is normally distributed with mean e− 2 bt Xj (0) and 2 variance σ4b [1 − e−bt ]. So X1 (t), · · · , Xd (t) are i.i.d. normal with the same mean µ(t) and variance v(t). (iv) Proof. h

uXj2 (t)

E e

i

Z

(x−µ(t))2

∞

=

e

ux2

−∞

Z

∞

=

e−

e− 2v(t) dx p 2πv(t)

(1−2uv(t))x2 −2µ(t)x+µ2 (t) 2v(t)

p

−∞

Z

∞

= −∞

Z

∞

= −∞

dx

2πv(t)

µ2 (t) µ2 (t) − 1−2uv(t) (1−2uv(t))2 2v(t)/(1−2uv(t)) 2

1

µ(t) (x− 1−2uv(t) )

+

p e− dx 2πv(t) p 2 µ2 (t)(1−2uv(t))−µ2 (t) µ(t) (x− 1−2uv(t) ) 1 − 2uv(t) − 2v(t)/(1−2uv(t)) e− 2v(t)(1−2uv(t)) p p e dx · 2πv(t) 1 − 2uv(t) uµ2 (t)

e− 1−2uv(t) p . 1 − 2uv(t)

=

(v) Proof. By R(t) =

Pd

j=1

Xj2 (t) and the fact X1 (t), · · · , Xd (t) are i.i.d., 2

d

udµ2 (t)

2a

E[euR(t) ] = (E[euX1 (t) ])d = (1 − 2uv(t))− 2 e 1−2uv(t) = (1 − 2uv(t))− σ2 e−

e−bt uR(0) 1−2uv(t)

.

6.7. (i) e −rT (ST − K)+ |Ft ] is a martingale by iterated conditioning argument. Since Proof. e−rt c(t, St , Vt ) = E[e d(e−rt c(t, St , Vt )) 1 −rt c(t, St , Vt )(−r) + ct (t, St , Vt ) + cs (t, St , Vt )rSt + cv (t, St , Vt )(a − bVt ) + css (t, St , Vt )Vt St2 + = e 2 1 2 cvv (t, St , Vt )σ Vt + csv (t, St , Vt )σVt St ρ dt + martingale part, 2 we conclude rc = ct + rscs + cv (a − bv) + 12 css vs2 + 21 cvv σ 2 v + csv σsvρ. This is equation (6.9.26). (ii)

60

Proof. Suppose c(t, s, v) = sf (t, log s, v) − e−r(T −t) Kg(t, log s, v), then ct = sft (t, log s, v) − re−r(T −t) Kg(t, log s, v) − e−r(T −t) Kgt (t, log s, v), 1 1 cs = f (t, log s, v) + sfs (t, log s, v) − e−r(T −t) Kgs (t, log s, v) , s s cv = sfv (t, log s, v) − e−r(T −t) Kgv (t, log s, v), 1 1 1 1 css = fs (t, log s, v) + fss (t, log s, v) − e−r(T −t) Kgss (t, log s, v) 2 + e−r(T −t) Kgs (t, log s, v) 2 , s s s s K csv = fv (t, log s, v) + fsv (t, log s, v) − e−r(T −t) gsv (t, log s, v), s cvv = sfvv (t, log s, v) − e−r(T −t) Kgvv (t, log s, v). So 1 1 ct + rscs + (a − bv)cv + s2 vcss + ρσsvcsv + σ 2 vcvv 2 2 = sft − re−r(T −t) Kg − e−r(T −t) Kgt + rsf + rsfs − rKe−r(T −t) gs + (a − bv)(sfv − e−r(T −t) Kgv ) 1 1 gs K K 1 + s2 v − fs + fss − e−r(T −t) 2 gss + e−r(T −t) K 2 + ρσsv fv + fsv − e−r(T −t) gsv 2 s s s s s 1 2 + σ v(sfvv − e−r(T −t) Kgvv ) 2 1 1 1 1 = s ft + (r + v)fs + (a − bv + ρσv)fv + vfss + ρσvfsv + σ 2 vfvv − Ke−r(T −t) gt + (r − v)gs 2 2 2 2 1 1 2 +(a − bv)gv + vgss + ρσvgsv + σ vgvv + rsf − re−r(T −t) Kg 2 2 = rc. That is, c satisfies the PDE (6.9.26). (iii) Proof. First, by Markov property, f (t, Xt , Vt ) = E[1{XT ≥log K} |Ft ]. So f (T, Xt , Vt ) = 1{XT ≥log K} , which implies f (T, x, v) = 1{x≥log K} for all x ∈ R, v ≥ 0. Second, f (t, Xt , Vt ) is a martingale, so by differentiating f and setting the dt term as zero, we have the PDE (6.9.32) for f . Indeed, 1 1 df (t, Xt , Vt ) = ft (t, Xt , Vt ) + fx (t, Xt , Vt )(r + Vt ) + fv (t, Xt , Vt )(a − bvt + ρσVt ) + fxx (t, Xt , Vt )Vt 2 2 1 + fvv (t, Xt , Vt )σ 2 Vt + fxv (t, Xt , Vt )σVt ρ dt + martingale part. 2 So we must have ft + (r + 21 v)fx + (a − bv + ρσv)fv + 12 fxx v + 12 fvv σ 2 v + σvρfxv = 0. This is (6.9.32). (iv) Proof. Similar to (iii). (v) Proof. c(T, s, v) = sf (T, log s, v) − e−r(T −t) Kg(T, log s, v) = s1{log s≥log K} − K1{log s≥log K} = 1{s≥K} (s − K) = (s − K)+ . 6.8. 61

Proof. We follow the hint. Suppose h is smooth and compactly supported, then it is legitimate to exchange integration and differentiation: Z ∞ Z ∞ ∂ h(y)p(t, T, x, y)dy = h(y)pt (t, T, x, y)dy, gt (t, x) = ∂t 0 0 Z ∞ gx (t, x) = h(y)px (t, T, x, y)dy, 0 Z ∞ gxx (t, x) = h(y)pxx (t, T, x, y)dy. 0

R∞

So (6.9.45) implies 0 h(y) pt (t, T, x, y) + β(t, x)px (t, T, x, y) + 21 γ 2 (t, x)pxx (t, T, x, y) dy = 0. By the arbitrariness of h and assuming β, pt , px , v, pxx are all continuous, we have 1 pt (t, T, x, y) + β(t, x)px (t, T, x, y) + γ 2 (t, x)pxx (t, T, x, y) = 0. 2 This is (6.9.43). 6.9. Proof. We first note dhb (Xu ) = h0b (Xu )dXu + 21 h00b (Xu )dXu dXu = h0b (Xu )β(u, Xu ) + 21 γ 2 (u, Xu )h00b (Xu ) du+ h0b (Xu )γ(u, Xu )dWu . Integrate on both sides of the equation, we have Z

T

hb (XT ) − hb (Xt ) = t

1 2 0 00 hb (Xu )β(u, Xu ) + γ (u, Xu )hb (Xu ) du + martingale part. 2

Take expectation on both sides, we get Z

E t,x [hb (XT ) − hb (Xt )]

∞

hb (y)p(t, T, x, y)dy − h(x)

= −∞ Z T

= t

Z = t

T

1 E t,x [h0b (Xu )β(u, Xu ) + γ 2 (u, Xu )h00b (Xu )]du 2 Z ∞ 1 2 00 0 hb (y)β(u, y) + γ (u, y)hb (y) p(t, u, x, y)dydu. 2 −∞

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 Z b Z b ∂ 0 b β(u, y)p(t, u, x, y)hb (y)dy = hb (y)β(u, y)p(t, u, x, y)|0 − hb (y) (β(u, y)p(t, u, x, y))dy ∂y 0 0 Z b ∂ = − hb (y) (β(u, y)p(t, u, x, y))dy, ∂y 0 and Z b Z γ 2 (u, y)p(t, u, x, y)h00b (y)dy = − 0

0

b

∂ 2 (γ (u, y)p(t, u, x, y))h0b (y)dy = ∂y

b

Z 0

∂2 2 (γ (u, y)p(t, u, x, y))hb (y)dy. ∂y

Plug these formulas into (6.9.48), we get (6.9.49). Differentiate w.r.t. T on both sides of (6.9.49), we have Z

b

hb (y) 0

∂ p(t, T, x, y)dy = − ∂T

Z 0

b

∂ 1 [β(T, y)p(t, T, x, y)]hb (y)dy + ∂y 2 62

Z 0

b

∂2 2 [γ (T, y)p(t, T, x, y)]hb (y)dy, ∂y 2

that is, Z 0

b

∂ ∂ 1 ∂2 2 (γ (T, y)p(t, T, x, y)) dy = 0. hb (y) p(t, T, x, y) + (β(T, y)p(t, T, x, y)) − ∂T ∂y 2 ∂y 2

This is (6.9.50). By (6.9.50) and the arbitrariness of hb , we conclude for any y ∈ (0, ∞), ∂ ∂ 1 ∂2 2 (γ (T, y)p(t, T, x, y)) = 0. p(t, T, x, y) + (β(T, y)p(t, T, x, y)) − ∂T ∂y 2 ∂y 2

6.10. Proof. Under the assumption that limy→∞ (y − K)rye p(0, T, x, y) = 0, we have Z ∞ Z ∞ Z ∞ ∂ − (y−K) (rye p(0, T, x, y))dy = −(y−K)rye p(0, T, x, y)|∞ + rye p (0, T, x, y)dy = rye p(0, T, x, y)dy. K ∂y K K K If we further assume (6.9.57) and (6.9.58), then use integration-by-parts formula twice, we have Z 1 ∞ ∂2 (y − K) 2 (σ 2 (T, y)y 2 pe(0, T, x, y))dy 2 K ∂y Z ∞ ∂ 2 1 ∂ 2 2 ∞ 2 (y − K) (σ (T, y)y pe(0, T, x, y))|K − (σ (T, y)y pe(0, T, x, y))dy = 2 ∂y K ∂y 1 = − (σ 2 (T, y)y 2 pe(0, T, x, y)|∞ K) 2 1 2 = σ (T, K)K 2 pe(0, T, x, K). 2 Therefore, cT (0, T, x, K)

Z ∞ −rc(0, T, x, K) + e−rT (y − K)e pT (0, T, x, y)dy K Z ∞ Z ∞ = −re−rT (y − K)e p(0, T, x, y)dy + e−rT (y − K)e pT (0, T, x, y)dy ZK∞ ZK∞ ∂ p(t, T, x, y))dy = −re−rT (y − K)e p(0, T, x, y)dy − e−rT (y − K) (rye ∂y K K Z ∞ 1 ∂2 2 +e−rT (y − K) (σ (T, y)y 2 pe(t, T, x, y))dy 2 ∂y 2 K Z ∞ Z ∞ −rT −rT = −re (y − K)e p(0, T, x, y)dy + e rye p(0, T, x, y)dy =

K

K

1 2 +e σ (T, K)K 2 pe(0, T, x, K) 2Z ∞ 1 = re−rT K pe(0, T, x, y)dy + e−rT σ 2 (T, K)K 2 pe(0, T, x, K) 2 K 1 2 = −rKcK (0, T, x, K) + σ (T, K)K 2 cKK (0, T, x, K). 2 −rT

7. Exotic Options 7.1. (i) 63

Proof. Since δ± (τ, s) =

1 √ [log s σ τ

+ (r ± 12 σ 2 )τ ] =

∂ δ± (τ, s) ∂t

log s − 12 σ τ

+

r± 21 σ 2 √ τ, σ

r ± 12 σ 2 1 − 1 ∂τ log s 1 − 3 ∂τ (− )τ 2 + τ 2 σ 2 ∂t σ 2 ∂t r ± 21 σ 2 √ 1 log s 1 √ (−1) − = − τ (−1) 2τ σ σ τ 1 1 1 = − · √ − log ss + (r ± σ 2 )τ ) 2τ σ τ 2 1 1 = − δ± (τ, ). 2τ s =

(ii) Proof. ∂ x ∂ 1 x 1 1 √ log + (r ± σ 2 )τ √ , δ± (τ, ) = = ∂x c ∂x σ τ c 2 xσ τ c ∂ 1 1 2 1 c ∂ √ log + (r ± σ )τ δ± (τ, ) = =− √ . ∂x x ∂x σ τ x 2 xσ τ

(iii) Proof. (log s+rτ )2 ±σ 2 τ (log s+rτ )+ 1 σ 4 τ 2 δ± (τ,s) 4 1 1 2σ 2 τ . N 0 (δ± (τ, s)) = √ e− 2 = √ e− 2π 2π

Therefore

2σ 2 τ (log s+rτ ) N 0 (δ+ (τ, s)) e−rτ − 2σ 2 τ = = e N 0 (δ− (τ, s)) s

and e−rτ N 0 (δ− (τ, s)) = sN 0 (δ+ (τ, s)). (iv) Proof. [(log s+rτ ) N 0 (δ± (τ, s)) = e− 0 −1 N (δ± (τ, s ))

]

2 −(log 1 +rτ )2 ±σ 2 τ (log s−log 1 ) s s 2σ 2 τ

= e−

4rτ log s±2σ 2 τ log s 2σ 2 τ

2r

2r

= e−( σ2 ±1) log s = s−( σ2 ±1) .

2r

So N 0 (δ± (τ, s−1 )) = s( σ2 ±1) N 0 (δ± (τ, s)). (v) Proof. δ+ (τ, s) − δ− (τ, s) =

1 √ σ τ

log s + (r + 21 σ 2 )τ −

1 √ σ τ

log s + (r − 12 σ 2 )τ =

1 √ σ2 τ σ τ

√ = σ τ.

(vi) Proof. δ± (τ, s) − δ± (τ, s−1 ) =

1 √ σ τ

log s + (r ± 21 σ 2 )τ −

1 √ σ τ

log s−1 + (r ± 12 σ 2 )τ =

(vii) Proof. N 0 (y) =

2

y √1 e− 2 2π

, so N 00 (y) =

2

y √1 e− 2 2π

2

(− y2 )0 = −yN 0 (y).

64

2 log √ s. σ τ

To be continued ... 7.3. c c c cT − W ct = (W fT − W ft ) + α(T − t) is independent of Ft , Proof. We note ST = S0 eσWT = St eσ(WT −Wt ) , W cu − W ct ) is independent of Ft , and supt≤u≤T (W

YT

=

S0 eσMT

=

S0 eσ supt≤u≤T Wu 1{M ct ≤sup

c

t≤u≤T

=

St eσ supt≤u≤T (Wu −Wt ) 1 c

c

Y

+ S0 eσMt 1{M ct >sup c

c

{ St ≤e

ct } W

t≤u≤T

cu −W c ) σ supt≤u≤T (W t

t

}

+ Yt 1

Y

{ St ≤e

cu } W

cu −W c ) σ supt≤u≤T (W t

t

}

.

So E[f (ST , YT )|Ft ] = E[f (x STS0−t , x YTS−t 1{ y ≤ YT −t } + y1{ y ≤ YT −t } )], where x = St , y = Yt . Therefore 0 x

S0

x

S0

E[f (ST , YT )|Ft ] is a Borel function of (St , Yt ). 7.4. Proof. By Cauchy’s inequality and the monotonicity of Y , we have |

m X (Ytj − Ytj−1 )(Stj − Stj−1 )|

≤

j=1

m X

|Ytj − Ytj−1 ||Stj − Stj−1 |

j=1

v v uX uX um um 2 t (Ytj − Ytj−1 ) t (Stj − Stj−1 )2 ≤ j=1

≤

r

j=1

v uX um max |Ytj − Ytj−1 |(YT − Y0 )t (Stj − Stj−1 )2 .

1≤j≤m

j=1

If we increase the number of partition points qP to infinity and letpthe length of the longest subinterval m 2 max1≤j≤m |tj − tj−1 | approach zero, then [S]T − [S]0 < ∞ and max1≤j≤m |Ytj − j=1 (Stj − Stj−1 ) → Pm Ytj−1 | → 0 a.s. by the continuity of Y . This implies j=1 (Ytj − Ytj−1 )(Stj − Stj−1 ) → 0. 8. American Derivative Securities 8.1. x − σ2r2 −1 1 0 Proof. vL (L+) = (K − L)(− σ2r2 )( L ) L − σ2r 2 L (K − L) = −1. Solve for L, we get L =

0 0 = − σ2r So vL (L+) = vL (L−) if and only if 2 L (K − L).

x=L 2rK 2r+σ 2 .

8.2. Proof. By the calculation in Section 8.3.3, we can see v2 (x) ≥ (K2 − x)+ ≥ (K1 − x)+ , rv2 (x) − rxv20 (x) − 1 2 2 00 2 σ x v2 (x) ≥ 0 for all x ≥ 0, and for 0 ≤ x < L1∗ < L2∗ , 1 rv2 (x) − rxv20 (x) − σ 2 x2 v200 (x) = rK2 > rK1 > 0. 2 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) 65

Proof. Suppose x takes its values in a domain bounded away from 0. By the general theory of linear differential equations, if we can find two linearly independent solutions v1 (x), v2 (x) of (8.8.4), then any solution of (8.8.4) can be represented in the form of C1 v1 +C2 v2 where C1 and C2 are constants. So it suffices to find two linearly independent special solutions of (8.8.4). Assume v(x) = xp for some constant p to be determined, (8.8.4) yields xp (r−pr− 21 σ 2 p(p−1)) = 0. Solve the quadratic equation 0 = r−pr− 21 σ 2 p(p−1) = 2r (− 21 σ 2 p − r)(p − 1), we get p = 1 or − σ2r2 . So a general solution of (8.8.4) has the form C1 x + C2 x− σ2 . (ii) Proof. Assume there is an interval [x1 , x2 ] where 0 < x1 < x2 < ∞, such that v(x) 6≡ 0 satisfies (8.3.19) with equality on [x1 , x2 ] and satisfies (8.3.18) with equality for x at and immediately to the left of x1 and 2r for x at and immediately to the right of x2 , then we can find some C1 and C2 , so that v(x) = C1 x + C2 x− σ2 on [x1 , x2 ]. If for some x0 ∈ [x1 , x2 ], v(x0 ) = v 0 (x0 ) = 0, by the uniqueness of the solution of (8.8.4), we would 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 0 (x2 )=the right derivative of (K − x)+ at x2 , which is 0). 1 Thus we have four equations for C1 and C2 :  − 2r2  C1 x1 + C2 x1 σ = K − x1    C x + C x− σ2r2 = K − x 1 2

  C1 −     C1 −

2 2 − σ2r2 −1 2r σ 2 C2 x1 − σ2r2 −1 2r σ 2 C2 x2

2

= −1 = −1.

Since x1 6= x2 , the last two equations imply C2 = 0. Plug C2 = 0 into the first two equations, we have 1 2 C1 = K−x = K−x x1 x2 ; plug C2 = 0 into the last two equations, we have C1 = −1. Combined, we would have x1 = x2 . Contradiction. Therefore our initial assumption is incorrect, and the only solution v that satisfies the specified conditions in the problem is the zero solution. (iii) Proof. If in a right neighborhood of 0, v satisfies (8.3.19) with equality, then part (i) implies v(x) = C1 x + 2r C2 x− σ2 for some constants C1 and C2 . Then v(0) = limx↓0 v(x) = 0 < (K − 0)+ , i.e. (8.3.18) will be violated. So we must have rv − rxv 0 − 21 σ 2 x2 v 00 > 0 in a right neighborhood of 0. According to (8.3.20), v(x) = (K − x)+ 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 first 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 (K − x)+ with equality for all x ≥ 0, then v cannot have a continuous derivative as stated in the problem. This is a contradiction. (vi) 1 Note we have interpreted the condition “v(x) satisfies (8.3.18) with equality for x at and immediately to the right of x ” 2 as “v(x2 ) = (K − x2 )+ and v 0 (x2 ) =the right derivative of (K − x)+ at x2 .” This is weaker than “v(x) = (K − x) in a right neighborhood of x2 .”

66

2r

Proof. By the result of part (i), we can start with v(x) = (K − x)+ on [0, x1 ] and v(x) = C1 x + C2 x− σ2 on [x1 , ∞). By the assumption of the problem, both v and v 0 are continuous. Since (K −x)+ is not differentiable at K, we must have x1 ≤ K.This gives us the equations  K − x = (K − x )+ = C x + C x− σ2r2 1 1 1 1 2 1 2r −1 = C − 2r C x− σ2 −1 . 1

σ2

2 1

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 ( 2r AL− σ2 + BL = K − L 2r − σ2r2 AL− σ2 −1 + B = −1, 2r

and we obtain A =

σ 2 KL σ2 σ 2 +2r

,B=

2rK L(σ 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, h i 2r x σ2r2 x σ2r2 +1 2 2 2r σ2 ) − (σ + 2r)( ) KL σ + 2r( 2 2 2r L L 2rKx σ KL σ − 2r2 x σ + − K = x− σ2 . f (x) − (K − x)+ = 2 σ + 2r L(σ 2 + 2r) (σ 2 + 2r)L 2r

2r

2r

Let g(θ) = σ 2 + 2rθ σ2 +1 − (σ 2 + 2r)θ σ2 with θ ≥ 1. Then g(1) = 0 and g 0 (θ) = 2r( σ2r2 + 1)θ σ2 − (σ 2 + 2r 2r 2r) σ2r2 θ σ2 −1 = σ2r2 (σ 2 + 2r)θ σ2 −1 (θ − 1) ≥ 0. So g(θ) ≥ 0 for any θ ≥ 1. This shows f (x) ≥ (K − x)+ for L ≤ x < K. Combined, we get f (x) ≥ (K − x)+ for all x ≥ L. (iv) 2r

Proof. Since limx→∞ v(x) = limx→∞ f (x) = ∞ and limx→∞ vL∗ (x) = limx→∞ (K − L∗ )( Lx∗ )− σ2 = 0, v(x) and vL∗ (x) are different. By part (iii), v(x) ≥ (K − x)+ . So v satisfies (8.3.18). For x ≥ L, rv − rxv 0 − 1 2 2 00 1 2 2 00 1 2 2 00 0 2 σ x v = rf − rxf − 2 σ x f = 0. For 0 ≤ x ≤ L, rv − rxv − 2 σ x v = r(K − x) + rx = rK. Combined, 1 2 2 00 0 rv − rxv − 2 σ x v ≥ 0 for x ≥ 0. So v satisfies (8.3.19). Along the way, we also showed v satisfies (8.3.20). In summary, v satisfies the linear complementarity condition (8.3.18)-(8.3.20), but v is not the function vL∗ given by (8.3.13). (v) Proof. By part (ii), B = 0 if and only if σ2 K x − σ2r2 σ 2 +2r ( L )

2rK L(σ 2 +2r)

− 1 = 0, i.e. L =

2rK 2r+σ 2 .

2r

In this case, v(x) = Ax− σ2 =

2r

x − σ2 = (K − L)( L ) = vL∗ (x), on the interval [L, ∞).

e −(r−a)τL ], 8.5. The difficulty of the dividend-paying case is that from Lemma 8.3.4, we can only obtain E[e e −rτL ]. So we have to start from Theorem 8.3.2. not E[e (i)

67

1

Proof. By (8.8.9), St = S0 eσWt +(r−a− 2 σ 1 x 1 2 2 σ )t = σ log L . By Theorem 8.3.2, f

2

)t

ft − 1 (r − a − . Assume S0 = x, then St = L if and only if −W σ

1

x

h

1

1

2

q

1

1

2 2

i

e −rτL ] = e− σ log L σ (r−a− 2 σ )+ σ2 (r−a− 2 σ ) +2r . E[e q e −rτL ] as e−γ log Lx = ( x )−γ . So If we set γ = σ12 (r − a − 12 σ 2 ) + σ1 σ12 (r − a − σ12 )2 + 2r, we can write E[e L the risk-neutral expected discounted pay off of this strategy is ( K − x, 0≤x≤L vL (x) = x −γ (K − L)( L ) , x > L.

(ii) ∂ ∂L vL (x)

Proof.

x −γ = −( L ) (1 −

γ(K−L) ). L

Set

∂ ∂L vL (x)

= 0 and solve for L∗ , we have L∗ =

γK γ+1 .

(iii) Proof. By Itˆ o’s formula, we have −rt 1 00 −rt 0 2 2 0 ft . d e vL∗ (St ) = e −rvL∗ (St ) + vL∗ (St )(r − a)St + vL∗ (St )σ St dt + e−rt vL (St )σSt dW ∗ 2 If x > L∗ , 1 00 0 −rvL∗ (x) + vL (x)(r − a)x + vL (x)σ 2 x2 ∗ 2 ∗ −γ x 1 x−γ−2 x−γ−1 = −r(K − L∗ ) + (r − a)x(K − L∗ )(−γ) −γ + σ 2 x2 (−γ)(−γ − 1)(K − L∗ ) −γ L∗ 2 L∗ L∗ −γ x 1 = (K − L∗ ) −r − (r − a)γ + σ 2 γ(γ + 1) . L∗ 2 By the definition of γ, if we define u = r − a − 21 σ 2 , we have

= =

= = =

1 r + (r − a)γ − σ 2 γ(γ + 1) 2 1 2 2 1 r − σ γ + γ(r − a − σ 2 ) 2 2 !2 ! r r 1 2 u 1 u2 u 1 u2 r− σ + + 2r + + + 2r u 2 σ2 σ σ2 σ2 σ σ2 r r ! 2u u2 1 u2 u2 u u2 1 2 u2 r− σ + 3 + 2r + 2 + 2r + 2+ + 2r 2 σ4 σ σ2 σ σ2 σ σ σ2 r r u2 u u2 1 u2 u2 u u2 r− 2 − + 2r − + 2r + 2 + + 2r 2σ σ σ2 2 σ2 σ σ σ2 0.

0 00 If x < L∗ , −rvL∗ (x) + vL (x)(r − a)x + 21 vL (x)σ 2 x2 = −r(K − x) + (−1)(r − a)x = −rK + ax. Combined, ∗ ∗ we get 0 ft . d e−rt vL∗ (St ) = −e−rt 1{St
68

Following the reasoning in the proof of Theorem 8.3.5, we only need to show 1{x
⇐⇒

⇐⇒

(r − a)γ + r < 0 " # r 1 1 1 1 2 (r − a) (θ − σ) + (θ − σ) + 2r + r < 0 σ 2 σ 2 r 1 1 θ(θ − σ) + θ (θ − σ)2 + 2r + r < 0 2 2 r 1 2 1 θ (θ − σ) + 2r < −r − θ(θ − σ)(< 0) 2 2 1 2 1 1 2 2 2 θ [(θ − σ) + 2r] > r + θ (θ − σ)2 + 2θr(θ − σ 2 ) 2 2 2 0 > r2 − θrσ 2

⇐⇒

0 > r − θσ 2 .

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

Since θσ 2 < 0, we have obtained a contradiction. So our initial assumption is incorrect, and rK − aL∗ ≥ 0 must be true. (iv) Proof. The proof is similar to that of Corollary 8.3.6. Note the only properties used in the proof of Corollary 8.3.6 are that e−rt vL∗ (St ) is a supermartingale, e−rt∧τL∗ vL∗ (St ∧τL∗ ) is a martingale, and vL∗ (x) ≥ (K −x)+ . Part (iii) already proved the supermartingale-martingale property, so it suffices to show vL∗ (x) ≥ (K − x)+ γK < K. For x ≥ K > L∗ , vL∗ (x) > 0 = (K − x)+ ; for 0 ≤ x < L∗ , in our problem. Indeed, by γ ≥ 0, L∗ = γ+1 vL∗ (x) = K − x = (K − x)+ ; finally, for L∗ ≤ x ≤ K, x−γ−1 1 d L∗−γ−1 γK (vL∗ (x) − (K − x)) = −γ(K − L∗ ) −γ + 1 ≥ −γ(K − L∗ ) −γ + 1 = −γ(K − ) γK + 1 = 0. dx γ + 1 γ+1 L∗ L∗ and (vL∗ (x) − (K − x))|x=L∗ = 0. So for L∗ ≤ x ≤ K, vL∗ (x) − (K − x)+ ≥ 0. Combined, we have vL∗ (x) ≥ (K − x)+ ≥ 0 for all x ≥ 0. 8.6. Proof. By Lemma 8.5.1, Xt = e−rt (St − K)+ is a submartingale. For any τ ∈ Γ0,T , Theorem 8.8.1 implies e −rT (ST − K)+ ] ≥ E[e e −rτ ∧T (Sτ ∧T − K)+ ] ≥ E[e−rτ (Sτ − K)+ 1{τ <∞} ] = E[e−rτ (Sτ − K)+ ], E[e e −rT (ST − where we take the convention that e−rτ (Sτ −K)+ = 0 when τ = ∞. Since τ is arbitrarily chosen, E[e + −rτ + e K) ] ≥ maxτ ∈Γ0,T E[e (Sτ − K) ]. The other direction “≤” is trivial since T ∈ Γ0,T . 8.7.

69

Proof. Suppose λ ∈ [0, 1] and 0 ≤ x1 ≤ x2 , we have f ((1 − λ)x1 + λx2 ) ≤ (1 − λ)f (x1 ) + λf (x2 ) ≤ (1 − λ)h(x1 ) + λh(x2 ). Similarly, g((1 − λ)x1 + λx2 ) ≤ (1 − λ)h(x1 ) + λh(x2 ). So h((1 − λ)x1 + λx2 ) = max{f ((1 − λ)x1 + λx2 ), g((1 − λ)x1 + λx2 )} ≤ (1 − λ)h(x1 ) + λh(x2 ). That is, h is also convex. 9. Change of Num´ eraire To provide an intuition for change of num´eraire, we give a summary of results for change of num´ eraire in discrete case. This summary is based on Shiryaev [5]. e B, ¯ S) as in [1] Definition 2.1.1 or [5] page 383. Here B e and Consider a model of financial market (B, ¯ are both one-dimensional while S could be a vector price process. Suppose B e and B ¯ are both strictly B positive, then both of them can be chosen as num´eaire. Several results hold under this model. First, no-arbitrage and completeness properties of market are independent of the choice of num´eraire (see, for example, Shiryaev [5] page 413 Remark and page 481). e (resp. B), ¯ there is an equivalent probability Second, if the market is arbitrage-free, then corresponding to B e ¯ S B S B , (resp. ¯ , ¯ ) is a martingale under Pe (resp. P¯ ). Third, if the market is Pe (resp. P¯ ), such that e B

B

e B

B

both arbitrage-free and complete, we have the relation dP¯ =

¯T B 1 h i dPe. e BT E Be¯0 B 0

Finally, if fT is a European contingent claim with maturity N and the market is both arbitrage-free and complete, then ¯t E ¯ fT |Ft = B et E e fT |Ft . B ¯T eT B B That is, the price of fT is independent of the choice of num´eraire. The above theoretical results can be applied to market involving foreign money market account. We consider the following market: a domestic money market account M (M0 = 1), a foreign money market account M f (M0f = 1), a (vector) asset price process S called stock. Suppose the domestic vs. foreign currency exchange rate is Q. Note Q is not a traded asset. Denominated by domestic currency, the traded assets are (M, M f Q, S), where M f Q can as the price process of one unit foreign currency. Domestic risk fbe seen M Q S e is a Pe-martingale. Denominated by foreign currency, the traded neutral measure P is such that M , M S M ef is such that assets are M f , M , S is a Pef -martingale. Q , Q . Foreign risk-neutral measure P QM f QM f This is a change of num´eraire in the market denominated by domestic currency, from M to M f Q. If we assume the market is arbitrage-free and complete, the foreign risk-neutral measure is dPef =

QT MTf QT DT MTf e e h i d P = dP Q0 M0f Q0 MT E M 0

on FT . Under the above set-up, for a European contingent claim fT , denominated h f i in domestic currency, its f DT fT e payoff in foreign currency is fT /QT . Therefore its foreign price is E Df Q |Ft . Convert this price into T t h f i f f DT fT e e domestic currency, we have Qt E Df Q |Ft . Use the relation between P and Pe on FT and the Bayes T t formula, we get " # f D f DT fT T f T e e Qt E |Ft = E |Ft . Dt Dtf QT The RHS is exactly the price of fT in domestic market if we apply risk-neutral pricing. 9.1. (i) 70

Proof. For any 0 ≤ t ≤ T , by Lemma 5.5.2, M2 (T ) M1 (T ) E[M1 (T )|Ft ] M1 (t) (M2 ) M1 (T ) E Ft = E Ft = = . M2 (T ) M2 (t) M2 (T ) M2 (t) M2 (t) So

M1 (t) M2 (t)

is a martingale under P M2 .

(ii) Proof. Let M1 (t) = Dt St and M2 (t) = Dt Nt /N0 . Then Pe(N ) as defined in (9.2.6) is P (M2 ) as defined in St 1 (t) e(N ) , which implies St(N ) = St is a martingale Remark 9.2.5. Hence M M2 (t) = Nt N0 is a martingale under P Nt (N ) e under P . 9.2. (i) 1

Proof. Since Nt−1 = N0−1 e−ν Wt −(r− 2 ν f

1

2

d(Nt−1 ) = N0−1 e−ν Wt −(r− 2 ν f

)t

2

, we have

)t

ct − rdt). ft − (r − 1 ν 2 )dt + 1 ν 2 dt] = Nt−1 (−νdW [−νdW 2 2

(ii) Proof. ct = Mt d dM

1 Nt

1 + dMt + d Nt

1 Nt

ct (−νdW ct − rdt) + rM ct dt = −ν M ct dW ct . dMt = M

Remark: This can also be obtained directly from Theorem 9.2.2. (iii) Proof.

bt dX

1 1 1 + dXt + d dXt Nt Nt Nt 1 1 1 = (∆t St + Γt Mt )d + (∆t dSt + Γt dMt ) + d (∆t dSt + Γt dMt ) Nt Nt Nt 1 1 1 1 1 1 + dSt + d dSt + Γt Mt d + dMt + d dMt = ∆t St d Nt Nt Nt Nt Nt Nt b c = ∆t dSt + Γt dMt . = d

Xt Nt

= Xt d

9.3. To avoid singular cases, we need to assume −1 < ρ < 1. (i) 1

Proof. Nt = N0 eν W3 (t)+(r− 2 ν f

dNt−1

2

)t

. So 1

= d(N0−1 e−ν W3 (t)−(r− 2 ν

2

)t ) 2 1 f f3 (t) − (r − 1 ν 2 )dt + 1 ν 2 dt = N0−1 e−ν W3 (t)−(r− 2 ν )t −νdW 2 2 −1 2 f = Nt [−νdW3 (t) − (r − ν )dt], f

71

and (N )

dSt

= Nt−1 dSt + St dNt−1 + dSt dNt−1 f1 (t)) + St Nt−1 [−νdW f3 (t) − (r − ν 2 )dt] = Nt−1 (rSt dt + σSt dW (N )

f1 (t)) + St(N ) [−νdW f3 (t) − (r − ν 2 )dt] − σSt(N ) ρdt (rdt + σdW (N ) (N ) f1 (t) − νdW f3 (t)). = St (ν 2 − σρ)dt + St (σdW

= St

Define γ = variation

p

f4 (t) = σ 2 − 2ρσν + ν 2 and W f4 ]t = [W

σf γ W1 (t)

−

νf γ W3 (t),

f4 is a martingale with quadratic then W

σν ν2 σ2 t − 2 ρt + t = t. γ2 γ2 r2

f4 is a BM and therefore, St(N ) has volatility γ = By L´evy’s Theorem, W

p σ 2 − 2ρσν + ν 2 .

(ii) f2 (t) = √−ρ W f1 (t) + √ 1 Proof. This problem is the same as Exercise 4.13, we define W 2

1−ρ2

1−ρ

f3 (t), then W f2 W

is a martingale, with f2 (t))2 = (dW

f1 (t) + p 1 f3 (t) dW dW −p 2 1−ρ 1 − ρ2 ρ

!2

=

1 2ρ2 ρ2 + − 1 − ρ2 1 − ρ2 1 − ρ2

dt = dt,

f2 (t)dW f1 (t) = − √ ρ and dW

f2 is a BM independent of W f1 , and dNt = rNt dt + dt + √ ρ 2 dt = 0. So W 1−ρ p f3 (t) = rNt dt + νNt [ρdW f1 (t) + 1 − ρ2 dW f2 (t)]. νNt dW 1−ρ2

(iii) f1 , W f2 ) is a two-dimensional BM, and Proof. Under Pe, (W    f   dSt = rSt dt + σSt dW1 (t) = rSt dt + St (σ, 0) ·

! f1 (t) dW f2 (t) dW p   f3 (t) = rNt dt + Nt (νρ, ν 1 − ρ2 ) ·  dN = rN dt + νN d W  t t t 

! f1 (t) dW . f2 (t) dW

p So under Pe, the volatility vector for S is (σ, 0), and the volatility vector for N is (νρ, ν p1 − ρ2 ). By Theorem 9.2.2, under the measure Pe(N ) , the volatility vector for S (N ) is (v1 , v2 ) = (σ − νρ, −ν 1 − ρ2 . In particular, the volatility of S (N ) is q q p p v12 + v22 = (σ − νρ)2 + (−ν 1 − ρ2 )2 = σ 2 − 2νρσ + ν 2 , consistent with the result of part (i). 9.4. Proof. From (9.3.15), we have Mtf Qt = M0f Q0 e

Rt 0

R f3 (s)+ t (Rs − 1 σ 2 (s))ds σ2 (s)dW 2 2 0

. So

R R Dtf f3 (s)− t (Rs − 1 σ 2 (s))ds − 0t σ2 (s)dW 2 2 0 = D0f Q−1 0 e Qt

72

and d

Dtf Qt

! =

f Dtf f3 (t) − (Rt − 1 σ22 (t))dt + 1 σ22 (t)dt] = Dt [−σ2 (t)dW f3 (t) − (Rt − σ22 (t))dt]. [−σ2 (t)dW Qt 2 2 Qt

To get (9.3.22), we note d

Mt Dtf Qt

To get (9.3.23), we note ! Dtf St = d Qt =

!

Dtf Qt

!

Df + t dMt + dMt d Qt

Dtf Qt

!

=

Mt d

=

f Mt Dtf f3 (t) − (Rt − σ22 (t))dt] + Rt Mt Dt dt [−σ2 (t)dW Qt Qt

=

−

Mt Dtf f3 (t) − σ 2 (t)dt) (σ2 (t)dW 2 Qt

=

−

Mt Dtf f f (t). σ2 (t)dW 3 Qt

Dtf dSt + St d Qt

Dtf Qt

! + dSt d

Dtf Qt

!

f Dtf f1 (t)) + St Dt [−σ2 (t)dW f3 (t) − (Rt − σ22 (t))dt] St (Rt dt + σ1 (t)dW Qt Qt f

f1 (t) Dt (−σ2 (t))dW f3 (t) +St σ1 (t)dW Qt =

Dtf St f1 (t) − σ2 (t)dW f3 (t) + σ22 (t)dt − σ1 (t)σ2 (t)ρt dt] [σ1 (t)dW Qt

=

Dtf St f f (t) − σ2 dW f f (t)]. [σ1 (t)dW 1 3 Qt

9.5. Proof. We combine the solutions of all the sub-problems into a single solution as follows. The payoff of a ST quanto call is ( Q − K)+ units of domestic currency at time T . By risk-neutral pricing formula, its price at T −r(T −t) e time t is E[e ( ST − K)+ |Ft ]. So we need to find the SDE for St under risk-neutral measure Pe. By QT

Qt

1

2

formula (9.3.14) and (9.3.16), we have St = S0 eσ1 W1 (t)+(r− 2 σ1 )t and √ 2 f f 1 2 1 2 f f f Qt = Q0 eσ2 W3 (t)+(r−r − 2 σ2 )t = Q0 eσ2 ρW1 (t)+σ2 1−ρ W2 (t)+(r−r − 2 σ2 )t . √ f1 (t)−σ2 1−ρ2 W f2 (t)+(r f + 1 σ 2 − 1 σ 2 )t St S0 (σ1 −σ2 ρ)W 2 2 2 1 . Define So Q = e Q0 t f

σ4 =

q

(σ1 − σ2 ρ)2 + σ22 (1 − ρ2 ) =

q

f4 is a martingale with [W f4 ]t = Then W f

if we set a = r − r + ρσ1 σ2 −

σ22 ,

p 2 f4 (t) = σ1 − σ2 ρ W f1 (t) − σ2 1 − ρ W f2 (t). σ12 − 2ρσ1 σ2 + σ22 and W σ4 σ4 2 (σ1 −σ2 ρ)2 ) t + σ2 (1−ρ t + t. σ42 σ42

f4 is a Brownian motion under Pe. So So W

we have

St S0 σ4 W f4 (t)+(r−a− 1 σ 2 )t 2 4 = e and d Qt Q0

73

St Qt

=

St f4 (t) + (r − a)dt]. [σ4 dW Qt

St behaves like dividend-paying stock and the price of the quanto call option is like the Therefore, under Pe, Q t price of a call option on a dividend-paying stock. Thus formula (5.5.12) gives us the desired price formula for quanto call option.

9.6. (i) Proof. d+ (t) − d− (t) =

√1 σ 2 (T σ T −t

√ √ − t) = σ T − t. So d− (t) = d+ (t) − σ T − t.

(ii) Proof. d+ (t)+d− (t) =

√2 σ T −t

log ForSK(t,T ) . So d2+ (t)−d2− (t) = (d+ (t)+d− (t))(d+ (t)−d− (t)) = 2 log ForSK(t,T ) .

(iii) Proof. 2

2

ForS (t, T )e−d+ (t)/2 − Ke−d− (t)

2

2

2

=

e−d+ (t)/2 [ForS (t, T ) − Ked+ (t)/2−d− (t)/2 ] ForS (t,T ) 2 K ] e−d+ (t)/2 [ForS (t, T ) − Kelog

=

0.

=

(iv) Proof.

= = =

dd+ (t) 1 1 1√ p ForS (t, T ) 1 2 dForS (t, T ) (dForS (t, T ))2 3 + σ (T − t)]dt + √ − − σdt 1σ (T − t) [log 2 K 2 2ForS (t, T )2 2 σ T − t ForS (t, T ) 1 ForS (t, T ) σ 1 1 1 f T (t) − σ 2 dt − σ 2 dt) p log dt + √ dt + √ (σdW K 2 2 4 T −t σ T −t 2σ (T − t)3 f T (t) dW 1 ForS (t, T ) 3σ √ √ dt + . log dt − K 2σ(T − t)3/2 4 T −t T −t

(v) √ Proof. dd− (t) = dd+ (t) − d(σ T − t) = dd+ (t) +

σdt √ . 2 T −t

(vi) Proof. By (iv) and (v), (dd− (t))2 = (dd+ (t))2 =

dt T −t .

(vii) Proof. dN (d+ (t)) = N 0 (d+ (t))dd+ (t)+ 21 N 00 (d+ (t))(dd+ (t))2 = (viii)

74

√1 e− 2π

d2 + (t) 2

dd+ (t)+ 12 √12π e−

d2 + (t) 2

(−d+ (t)) Tdt −t .

Proof. dN (d− (t))

1 = N 0 (d− (t))dd− (t) + N 00 (d− (t))(dd− (t))2 2 d2 − (t) dt σdt 1 e− 2 1 − d2− (t) √ (−d− (t)) dd+ (t) + √ + = √ e 2 2 T −t 2 T −t 2π 2π 2

d2 − (t)(σ

√ T −t−d+ (t))

=

2 2 1 e− σe−d− (t)/2 √ e−d− (t)/2 dd+ (t) + p √ dt + 2π 2(T − t) 2π 2 2π(T − t)

=

2 1 σe−d− (t)/2 d+ (t)e− 2 √ dt. √ e−d− (t)/2 dd+ (t) + p dt − 2π 2(T − t) 2π 2π(T − t)

2

dt

d2 − (t)

(ix) Proof. −d2+ (t)/2

f T (t) e √ dForS (t, T )dN (d+ (t)) = σForS (t, T )dW

2π

2

−d (t)/2 1 S (t, T )e + c T (t) = σForp √ dW dt. T −t 2π(T − t)

(x) Proof. ForS (t, T )dN (d+ (t)) + dForS (t, T )dN (d+ (t)) − KdN (d− (t)) 2 2 2 1 d+ (t) σForS (t, T )e−d+ (t)/2 √ e−d+ (t)/2 dt + p = ForS (t, T ) √ e−d+ (t)/2 dd+ (t) − dt 2π 2(T − t) 2π 2π(T − t) " # 2 d+ (t) e−d− (t)/2 σ −d2− (t)/2 −d2− (t)/2 √ √ e e dt − −K dd+ (t) + p dt 2π 2(T − t) 2π 2π(T − t) " # 2 2 Kσe−d− (t)/2 Kd+ (t) ForS (t, T )d+ (t) −d2+ (t)/2 σForS (t, T )e−d+ (t)/2 −d2− (t)/2 √ p √ e = e − p − + dt 2(T − t) 2π 2(T − t) 2π 2π(T − t) 2π(T − t) 2 2 1 +√ ForS (t, T )e−d+ (t)/2 − Ke−d− (t)/2 dd+ (t) 2π = 0. 2 2 The last “=” comes from (iii), which implies e−d− (t)/2 = ForSK(t,T ) e−d+ (t)/2 .

10. Term-Structure Models 10.1. (i) Proof. Using the notation I1 (t), I2 (t), I3 (t) and I4 (t) introduced in the problem, we can write Y1 (t) and Y2 (t) as Y1 (t) = e−λ1 t Y1 (0) + e−λ1 t I1 (t) and ( −λ t λ21 21 (e−λ1 t − e−λ2 t )Y1 (0) + e−λ2 t Y2 (0) + λ1λ−λ e 1 I1 (t) − e−λ2 t I2 (t) − e−λ2 t I3 (t), if λ1 6= λ2 ; 2 Y2 (t) = λ1 −λ2 −λ t −λ21 te 1 Y1 (0) + e−λ1 t Y2 (0) − λ21 te−λ1 t I1 (t) − e−λ1 t I4 (t) + e−λ1 t I3 (t), if λ1 = λ2 .

75

e 1 (t)] = Since all the Ik (t)’s (k = 1, · · · , 4) are normally distributed with zero mean, we can conclude E[Y −λ1 t e Y1 (0) and ( λ21 −λ1 t − e−λ2 t )Y1 (0) + e−λ2 t Y2 (0), if λ1 = 6 λ2 ; e 2 (t)] = λ1 −λ2 (e E[Y −λ1 t −λ1 t −λ21 te Y1 (0) + e Y2 (0), if λ1 = λ2 .

(ii) Proof. The calculation relies on the following fact: if Xt and Yt are both martingales, then Xt Yt − [X, Y ]t is e t Yt ] = E{[X, e also a martingale. In particular, E[X Y ]t }. Thus t

Z

2λ1 u

e

e2λ1 t − 1 e du = , E[I1 (t)I2 (t)] = 2λ1

t

e(λ1 +λ2 )t − 1 , λ1 + λ2 0 0 Z t 1 e2λ1 t − 1 2λ1 u 2λ1 t e e te E[I1 (t)I3 (t)] = 0, E[I1 (t)I4 (t)] = ue du = − 2λ1 2λ1 0

e 12 (t)] = E[I

Z

e(λ1 +λ2 )u du =

and e 42 (t)] = E[I

Z

t

u2 e2λ1 u du =

0

t2 e2λ1 t te2λ1 t e2λ1 t − 1 − + . 2λ1 2λ21 4λ31

(iii) Proof. Following the hint, we have Z e 1 (s)I2 (t)] = E[J e 1 (t)I2 (t)] = E[I

t

e(λ1 +λ2 )u 1{u≤s} du =

0

e(λ1 +λ2 )s − 1 . λ1 + λ2

10.2. (i) RT

Proof. Assume B(t, T ) = E[e− t Rs ds |Ft ] = f (t, Y1 (t), Y2 (t)). Then d(Dt B(t, T )) = Dt [−Rt f (t, Y1 (t), Y2 (t))dt+ df (t, Y1 (t), Y2 (t))]. By Itˆ o’s formula, df (t, Y1 (t), Y2 (t))

=

[ft (t, Y1 (t), Y2 (t)) + fy1 (t, Y1 (t), Y2 (t))(µ − λ1 Y1 (t)) + fy2 (t, Y1 (t), Y2 (t))(−λ2 )Y2 (t)] 1 +fy1 y2 (t, Y1 (t), Y2 (t))σ21 Y1 (t) + fy1 y1 (t, Y1 (t), Y2 (t))Y1 (t) 2 1 2 + fy2 y2 (t, Y1 (t), Y2 (t))(σ21 Y1 (t) + α + βY1 (t))]dt + martingale part. 2

Since Dt B(t, T ) is a martingale, we must have ∂ ∂ ∂ 1 ∂2 ∂2 ∂2 2 + (µ − λ1 y1 ) 2σ21 y1 f = 0. −(δ0 + δ1 y1 + δ2 y2 ) + − λ 2 y2 + + y1 2 + (σ21 y1 + α + βy1 ) 2 ∂t ∂y1 ∂y2 2 ∂y1 ∂y2 ∂y1 ∂y2

(ii)

76

Proof. If we suppose f (t, y1 , y2 ) = e−y1 C1 (T −t)y2 C2 (T −t)−A(T −t) , then A0 (T − t)]f , and

2

∂ f ∂y22

∂ ∂y1 f

= −C1 (T − t)f ,

∂f ∂y2

= −C2 (T − t)f ,

2

∂ f ∂y1 ∂y2

∂ ∂t f

= [y1 C10 (T − t) + y2 C20 (T − t) +

= C1 (T − t)C2 (T − t)f ,

∂2f ∂y12

= C12 (T − t)f ,

= C22 (T − t)f . So the PDE in part (i) becomes

−(δ0 +δ1 y1 +δ2 y2 )+y1 C10 +y2 C20 +A0 −(µ−λ1 y1 )C1 +λ2 y2 C2 +

1 2 2σ21 y1 C1 C2 + y1 C12 + (σ21 y1 + α + βy1 )C22 = 0. 2

Sorting out the LHS according to the independent variables y1 and y2 , we get  1 1 2 2 2 0  −δ1 + C1 + λ1 C1 + σ21 C1 C2 + 2 C1 + 2 (σ21 + β)C2 = 0 0 −δ2 + C2 + λ2 C2 = 0   −δ0 + A0 − µC1 + 12 αC22 = 0. In other words, we can obtain the ODEs for C1 , C2 and A as follows  1 1 2 2 2 0  C1 = −λ1 C1 − σ21 C1 C2 − 2 C1 − 2 (σ21 + β)C2 + δ1 different from (10.7.4), check! C20 = −λ2 C2 + δ2   0 A = µC1 − 21 αC22 + δ0 .

10.3. (i) Proof. d(Dt B(t, T )) = Dt [−Rt f (t, T, Y1 (t), Y2 (t))dt + df (t, T, Y1 (t), Y2 (t))] and df (t, T, Y1 (t), Y2 (t)) =

[ft (t, T, Y1 (t), Y2 (t)) + fy1 (t, T, Y1 (t), Y2 (t))(−λ1 Y1 (t)) + fy2 (t, T, Y1 (t), Y2 (t))(−λ21 Y1 (t) − λ2 Y2 (t)) 1 1 + fy1 y1 (t, T, Y1 (t), Y2 (t)) + fy2 y2 (t, T, Y1 (t), Y2 (t))]dt + martingale part. 2 2

Since Dt B(t, T ) is a martingale under risk-neutral measure, we have the following PDE: 1 ∂ ∂ ∂ ∂ 1 ∂2 − λ1 y1 + f (t, T, y1 , y2 ) = 0. −(δ0 (t) + δ1 y1 + δ2 y2 ) + − (λ21 y1 + λ2 y2 ) + ∂t ∂y1 ∂y2 2 ∂y12 2 ∂y22 Suppose f (t, T, y1 , y2 ) = e−y1 C1 (t,T )−y2 C2 (t,T )−A(t,T ) , then  d d ft (t, T, y1 , y2 ) = −y1 dt C1 (t, T ) − y2 dt C2 (t, T ) −      fy1 (t, T, y1 , y2 ) = −C1 (t, T )f (t, T, y1 , y2 ),    f (t, T, y , y ) = −C (t, T )f (t, T, y , y ), y2 1 2 2 1 2  f (t, T, y , y ) = C (t, T )C (t, T )f (t, T, y1 , y2 ), y1 y2 1 2 1 2    2  fy1 y1 (t, T, y1 , y2 ) = C1 (t, T )f (t, T, y1 , y2 ),    fy2 y2 (t, T, y1 , y2 ) = C22 (t, T )f (t, T, y1 , y2 ).

d dt A(t, T )

f (t, T, y1 , y2 ),

So the PDE becomes d d d −(δ0 (t) + δ1 y1 + δ2 y2 ) + −y1 C1 (t, T ) − y2 C2 (t, T ) − A(t, T ) + λ1 y1 C1 (t, T ) dt dt dt 1 2 1 2 +(λ21 y1 + λ2 y2 )C2 (t, T ) + C1 (t, T ) + C2 (t, T ) = 0. 2 2

77

Sorting out the terms according to independent variables y1 and y2 , we get  1 2 1 2 d  −δ0 (t) − dt A(t, T ) + 2 C1 (t, T ) + 2 C2 (t, T ) = 0 d −δ1 − dt C1 (t, T ) + λ1 C1 (t, T ) + λ21 C2 (t, T ) = 0   d −δ2 − dt C2 (t, T ) + λ2 C2 (t, T ) = 0. That is  d   dt C1 (t, T ) = λ1 C1 (t, T ) + λ21 C2 (t, T ) − δ1 d dt C2 (t, T ) = λ2 C2 (t, T ) − δ2  d 1 2 1 2 dt A(t, T ) = 2 C1 (t, T ) + 2 C2 (t, T ) − δ0 (t).

(ii) d −λ2 t Proof. For C2 , we note dt [e C2 (t, T )] = −e−λ2 t δ2 from the ODE in (i). Integrate from t to T , we have R T 0 − e−λ2 t C2 (t, T ) = −δ2 t e−λ2 s ds = λδ22 (e−λ2 T − e−λ2 t ). So C2 (t, T ) = λδ22 (1 − e−λ2 (T −t) ). For C1 , we note

λ21 δ2 −λ1 t d −λ1 t (e C1 (t, T )) = (λ21 C2 (t, T ) − δ1 )e−λ1 t = (e − e−λ2 T +(λ2 −λ1 )t ) − δ1 e−λ1 t . dt λ2 Integrate from t to T , we get −e−λ1 t C1 (t, T ) ( δ2 −λ1 T (e − e−λ1 t ) − − λλ21 2 λ1 = λ21 δ2 −λ1 T − e−λ1 t ) − − λ2 λ1 (e

λ21 δ2 −λ2 T e(λ2 −λ1 )T −e(λ2 −λ1 )t + λδ11 (e−λ1 T λ2 e λ2 −λ1 λ21 δ2 −λ2 T (T − t) + λδ11 (e−λ1 T − e−λ1 T ) λ2 e

− e−λ1 T )

if λ1 6= λ2 if λ1 = λ2 .

So ( C1 (t, T ) =

λ21 δ2 −λ1 (T −t) λ2 λ1 (e λ21 δ2 −λ1 (T −t) λ2 λ1 (e

− 1) + − 1) +

λ21 δ2 e−λ1 (T −t) −e−λ2 (T −t) − λδ11 (e−λ1 (T −t) − 1) λ2 λ2 −λ1 λ21 δ2 −λ2 T +λ1 t (T − t) − λδ11 (e−λ1 (T −t) − 1) λ2 e

if λ1 6= λ2 . if λ1 = λ2 .

(iii) Proof. From the ODE

d dt A(t, T )

= 12 (C12 (t, T ) + C22 (t, T )) − δ0 (t), we get Z T 1 A(t, T ) = δ0 (s) − (C12 (s, T ) + C22 (s, T )) ds. 2 t

(iv) Proof. We want to find δ0 so that f (0, T, Y1 (0), Y2 (0)) = e−Y1 (0)C1 (0,T )−Y2 (0)C2 (0,T )−A(0,T ) = B(0, T ) for all T > 0. Take logarithm on both sides and plug in the expression of A(t, T ), we get Z T 1 2 (C1 (s, T ) + C22 (s, T )) − δ0 (s) ds. log B(0, T ) = −Y1 (0)C1 (0, T ) − Y2 (0)C2 (0, T ) + 2 0 Taking derivative w.r.t. T, we have ∂ ∂ ∂ 1 1 log B(0, T ) = −Y1 (0) C1 (0, T ) − Y2 (0) C2 (0, T ) + C12 (T, T ) + C22 (T, T ) − δ0 (T ). ∂T ∂T ∂T 2 2

78

So δ0 (T )

∂ ∂ ∂ = −Y1 (0) C1 (0, T ) − Y2 (0) C2 (0, T ) − log B(0, T ) ∂Th ∂T i ∂T ( δ2 −λ2 T ∂ − Y2 (0)δ2 e−λ2 T − ∂T log B(0, T ) if λ1 = 6 λ2 −Y1 (0) δ1 e−λ1 T − λ21 λ2 e = −λ T ∂ −λ2 T −λ2 T 1 −Y1 (0) δ1 e − λ21 δ2 e T − Y2 (0)δ2 e − ∂T log B(0, T ) if λ1 = λ2 .

10.4. (i) Proof. bt dX

= dXt + Ke−Kt

t

Z

eKu Θ(u)dudt − Θ(t)dt Z t bt + Ke−Kt = −KXt dt + ΣdB eKu Θ(u)dudt 0

0

bt dt + ΣdB bt . = −K X

(ii) Proof. ft = CΣB et = W

1 σ1 − √ρ 2 σ1 1−ρ

0 √1 σ2

!

1−ρ2

σ1 0

0 e Bt = σ2

f is a martingale with hW f 1 it = hB e 1 it = t, hW f 2 it = h− √ ρ So W

1−ρ2

ρ2 +1−2ρ2 t 1−ρ2

f1, W f 2 it = hB e1, − √ ρ = t, and hW

1−ρ2

e1 + √ 1 B

1−ρ2

1 −√ ρ

e1 + √ 1 B

e 2 it = B

√1

1−ρ2

et . B

1−ρ2

e 2 it = B

1−ρ2 ρt −√ 2 + 1−ρ

!

0

√ ρt

ρ2 t 1−ρ2

1−ρ2

ρ t + 1−ρ 2 −2 1−ρ2 ρt =

f is a = 0. Therefore W

bt = −CK X et dt+CΣdB et = −CKC −1 Yt dt+dW ft = −ΛYt dt+dW ft , two-dimensional BM. Moreover, dYt = CdX where   ! 1 √1 2 0 0 1 λ1 0 σ  σ2 ρ1−ρ  · Λ = CKC −1 = − √1ρ √1 −1 λ2 |C| σ √1−ρ2 σ11 σ1 1−ρ2 σ2 1−ρ2 1 ! λ1 0 σ1 0 σ1 p = ρλ √λ2 − √ 1 2 − √1 2 ρσ2 σ2 1 − ρ2 σ1 1−ρ σ2 1−ρ σ2 1−ρ2 ! λ1 0 ρσ2 (λ2 −λ1 )−σ1 = √ 2 λ2 . σ2

1−ρ

(iii) Proof. Xt

= = =

Z t Z t bt + e−Kt X eKu Θ(u)du = C −1 Yt + e−Kt eKu Θ(u)du 0 0 Z t σ1 0 Y (t) 1 −Kt Ku p +e e Θ(u)du Y2 (t) ρσ2 σ2 1 − ρ2 0 Z t σ1 Y1p (t) −Kt +e eKu Θ(u)du. ρσ2 Y1 (t) + σ2 1 − ρ2 Y2 (t) 0 79

p Rt So Rt = X2 (t) = ρσ2 Y1 (t)+σ2 1 − ρ2 Y2 (t)+δ0 (t), where δ0 (t) is the second coordinate of e−Kt 0 eKu Θ(u)du p and can be derived explicitly by Lemma 10.2.3. Then δ1 = ρσ2 and δ2 = σ2 1 − ρ2 . 10.5. Proof. We note C(t, T ) and A(t, T ) are dependent only on T − t. So C(t, t + τ¯) and A(t, t + τ¯) aare constants when τ¯ is fixed. So d Lt dt

= − = =

B(t, t + τ¯)[−C(t, t + τ¯)R0 (t) − A(t, t + τ¯)] τ¯B(t, t + τ¯)

1 [C(t, t + τ¯)R0 (t) + A(t, t + τ¯)] τ¯ 1 [C(0, τ¯)R0 (t) + A(0, τ¯)]. τ¯

Hence L(t2 )−L(t1 ) = τ1¯ C(0, τ¯)[R(t2 )−R(t1 )]+ τ1¯ A(0, τ¯)(t2 −t1 ). Since L(t2 )−L(t1 ) is a linear transformation, it is easy to verify that their correlation is 1. 10.6. (i) h f1 (t)) = δ1 ( δ0 − Proof. If δ2 = 0, then dRt = δ1 dY1 (t) = δ1 (−λ1 Y1 (t)dt + dW δ1

Rt δ1 )λ1 dt

i f1 (t) = (δ0 λ1 − + dW

f1 (t). So a = δ0 λ1 and b = λ1 . λ1 Rt )dt + δ1 dW (ii) Proof. dRt

= δ1 dY1 (t) + δ2 dY2 (t) f1 (t) − δ2 λ21 Y1 (t)dt − δ2 λ2 Y2 (t)dt + δ2 dW f2 (t) = −δ1 λ1 Y1 (t)dt + λ1 dW f1 (t) + δ2 dW f2 (t) = −Y1 (t)(δ1 λ1 + δ2 λ21 )dt − δ2 λ2 Y2 (t)dt + δ1 dW f1 (t) + δ2 dW f2 (t) = −Y1 (t)λ2 δ1 dt − δ2 λ2 Y2 (t)dt + δ1 dW f1 (t) + δ2 dW f2 (t) = −λ2 (Y1 (t)δ1 + Y2 (t)δ2 )dt + δ1 dW " # q δ1 δ2 2 2 f f = −λ2 (Rt − δ0 )dt + δ1 + δ2 p 2 dW1 (t) + p 2 dW2 (t) . δ1 + δ22 δ1 + δ22

So a = λ2 δ0 , b = λ2 , σ =

p

et = √ δ1 δ12 + δ22 and B 2

δ1 +δ22

f1 (t) + √ δ2 W 2

δ1 +δ22

f2 (t). W

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)) = de−Y1 (t)C1 (T −t)−Y2 (t)C2 (T −t)−A(T −t) f1 (t) − C2 (T − t)dW f2 (t)] = dt term + B(t, T )[−C1 (T − t)dW ! f1 (t) dW = dt term + B(t, T )(−C1 (T − t), −C2 (T − t)) f2 (t) . dW

f T (t) = So the volatility vector of B(t, T ) under Pe is (−C1 (T − t), −C2 (T − t)). By (9.2.5), W j fj (t) (j = 1, 2) form a two-dimensional PeT −BM. u)du + W (ii) 80

Rt 0

Cj (T −

Proof. Under the T-forward measure, the numeraire is B(t, T ). By risk-neutral pricing, at time zero the risk-neutral price V0 of the option satisfies V0 1 T −C1 (T¯ −T )Y1 (T )−C2 (T¯ −T )Y2 (T )−A(T¯ −T ) + e =E (e − K) . B(0, T ) B(T, T ) Note B(T, T ) = 1, we get (10.7.19). (iii) Proof. We can rewrite (10.2.4) and (10.2.5) as ( f T (t) − C1 (T − t)dt dY1 (t) = −λ1 Y1 (t)dt + dW 1 f T (t) − C2 (T − t)dt. dY2 (t) = −λ21 Y1 (t)dt − λ2 Y2 (t)dt + dW 2 Then ( Rt R f T (s) − t C1 (T − s)eλ1 (s−t) ds Y1 (t) = Y1 (0)e−λ1 t + 0 eλ1 (s−t) dW 1 Rt R0t R f2 (s) − t C2 (T − s)eλ2 (s−t) ds. Y2 (t) = Y0 e−λ2 t − λ21 0 Y1 (s)eλ2 (s−t) ds + 0 eλ2 (s−t) dW 0 So (Y1 , Y2 ) is jointly Gaussian and X is therefore Gaussian. (iv) ft , then Proof. First, we recall the Black-Scholes formula for call options: if dSt = µSt dt + σSt dW e −µT (S0 eσWT +(µ− 12 σ2 )T − K)+ ] = S0 N (d+ ) − Ke−µT N (d− ) E[e with d± = and

1 √ σ T

(log

S0 K

d

+ (µ ± 12 σ 2 )T ). Let T = 1, S0 = 1 and ξ = σW1 + (µ − 12 σ 2 ), then ξ = N (µ − 21 σ 2 , σ 2 ) e ξ − K)+ ] = eµ N (d+ ) − KN (d− ), E[(e

d where d± = σ1 (− log K + (µ ± 21 σ 2 )) (different from the problem. Check!). Since under PeT , X = N (µ − 12 σ 2 , σ 2 ), we have

e T [(eX − K)+ ] = B(0, T )(eµ N (d+ ) − KN (d− )). B(0, T )E

10.11. Proof. On each payment date Tj , the payoff of this swap contract is δ(K − L(Tj−1 , Tj−1 )). Its no-arbitrage price at time 0 is δ(KB(0, Tj ) − B(0, Tj )L(0, Tj−1 )) by Theorem 10.4. So the value of the swap is n+1 X

δ[KB(0, Tj ) − B(0, Tj )L(0, Tj−1 )] = δK

j=1

n+1 X j=1

10.12.

81

B(0, Tj ) − δ

n+1 X j=1

B(0, Tj )L(0, Tj−1 ).

Proof. Since L(T, T ) =

1−B(T,T +δ) δB(T,T +δ)

∈ FT , we have

e E[D(T + δ)L(T, T )]

e E[D(T e = E[ + δ)L(T, T )|FT ]] 1 − B(T, T + δ) e e = E E[D(T + δ)|FT ] δB(T, T + δ) 1 − B(T, T + δ) e = E D(T )B(T, T + δ) δB(T, T + δ) D(T ) − D(T )B(T, T + δ) e = E δ B(0, T ) − B(0, T + δ) = δ = B(0, T + δ)L(0, T ).

11. Introduction to Jump Processes 11.1. (i) Proof. First, Mt2 = Nt2 − 2λtNt + λ2 t2 . So E[Mt2 ] < ∞. f (x) = x2 is a convex function. So by conditional Jensen’s inequality, E[f (Mt )|Fs ] ≥ f (E[Mt |Fs ]) = f (Ms ), ∀s ≤ t. So Mt2 is a submartingale. (ii) Proof. We note Mt has independent and stationary increment. So ∀s ≤ t, E[Mt2 − Ms2 |Fs ] = E[(Mt − 2 Ms )2 |Fs ] + E[(Mt − Ms ) · 2Ms |Fs ] = E[Mt−s ] + 2Ms E[Mt−s ] = V ar(Nt−s ) + 0 = λ(t − s). That is, 2 2 E[Mt − λt|Fs ] = Ms − λs. 11.2. Proof. P (Ns+t = k|Ns = k) = P (Ns+t − Ns = 0|Ns = k) = P (Nt = 0) = e−λt = 1 − λt + O(t2 ). Similarly, 1 −λt = λt(1 − λt + O(t2 )) = λt + O(t2 ), and we have P (Ns+t = k + 1|Ns = k) = P (Nt = 1) = (λt) 1! e P∞ (λt)k −λt P (Ns+t ≥ k + 2|N2 = k) = P (Nt ≥ 2) = k=2 k! e = O(t2 ). 11.3. Proof. For any t ≤ u, we have Su E Ft = St

E[(σ + 1)Nt −Nu e−λσ(t−u) |Ft ]

=

e−λσ(t−u) E[(σ + 1)Nt−u ]

=

e−λσ(t−u) E[eNt−u log(σ+1) ]

=

e−λσ(t−u) eλ(t−u)(e

=

−λσ(t−u) λσ(t−u)

e

=

1.

e

So St = E[Su |Ft ] and S is a martingale. 11.4.

82

log(σ+1)

−1)

(by (11.3.4))

Proof. The problem is ambiguous in that the relation between N1 and N2 is not clearly stated. According to page 524, paragraph 2, we would guess the condition should be that N1 and N2 are independent. Suppose N1 and N2 are independent. Define M1 (t) = N1 (t) − λ1 t and M2 (t) = N2 (t) − λ2 t. Then by independence E[M1 (t)M2 (t)] = E[M1 (t)]E[M2 (t)] = 0. Meanwhile, by Itˆo’s product formula, M1 (t)M2 (t) = Rt Rt Rt Rt M1 (s−)dM2 (s) + 0 M2 (s−)dM1 (s) + [M1 , M2 ]t . Both 0 M1 (s−)dM2 (s) and 0 P M2 (s−)dM1 (s) are mar0 tingales. So taking expectation on both sides, we get 0 = 0 + E{[M , M ] } = E[ 1 2 t 0<s≤t ∆N1 (s)∆N2 (s)]. P P Since 0<s≤t ∆N1 (s)∆N2 (s) ≥ 0 a.s., we conclude 0<s≤t ∆N (s)∆N (s) = 0 a.s. By letting t = 1, 2, · · · , 1 2 P we can find a set Ω0 of probability 1, so that ∀ω ∈ Ω0 , 0<s≤t ∆N1 (s)∆N2 (s) = 0 for all t > 0. Therefore N1 and N2 can have no simultaneous jump. 11.5. Proof. We shall prove the whole path of N1 is independent of the whole path of N2 , following the scheme suggested by page 489, paragraph 1. Fix s ≥ 0, we consider Xt = u1 (N1 (t) − N1 (s)) + u2 (N2 (t) − N2 (s)) − λ1 (eu1 − 1)(t − s) − λ2 (eu2 − 1)(t − s), t > s. Then by Itˆ o’s formula for jump process, we have Z t Z X 1 t Xu c Xu Xs Xt e dXu + e dXuc dXuc + (eXu − eXu− ) = e −e 2 s s s
0
Since ∆Xt = u1 ∆N1 (t)+u2 ∆N2 (t) and N1 , N2 have no simultaneous jump, eXu −eXu− = eXu− (e∆Xu −1) = eXu− [(eu1 − 1)∆N1 (u) + (eu2 − 1)∆N2 (u)]. So eXt − 1 Z t X = eXu− [−λ1 (eu1 − 1) − λ2 (eu2 − 1)]du + eXu− [(eu1 − 1)∆N1 (u) + (eu2 − 1)∆N2 (u)] s

Z =

s
eXu− [(eu1 − 1)d(N1 (u) − λ1 u) − (eu2 − 1)d(N2 (u) − λ2 u)].

s

This shows (eXt )t≥s is a martingale w.r.t. (Ft )t≥s . So E[eXt ] ≡ 1, i.e. E[eu1 (N1 (t)−N1 (s))+u2 (N2 (t)−N2 (s)) ] = eλ1 (e

u1 −1)(t−s)

eλ2 (e

u2 −1)(t−s)

= E[eu1 (N1 (t)−N1 (s)) ]E[eu2 (N2 (t)−N2 (s)) ].

This shows N1 (t) − N1 (s) is independent of N2 (t) − N2 (s). Now, suppose we have 0 ≤ t1 < t2 < t3 < · · · < tn , then the vector (N1 (t1 ), · · · , N1 (tn )) is independent of (N2 (t1 ), · · · , N2 (tn )) if and only if (N1 (t1 ), N1 (t2 ) − N1 (t1 ), · · · , N1 (tn ) − N1 (tn−1 )) is independent of (N2 (t1 ), N2 (t2 ) − N2 (t1 ), · · · , N2 (tn ) − N2 (tn−1 )). Let t0 = 0, then E[e = = =

E[e E[e

Pn

i=1

ui (N1 (ti )−N1 (ti−1 ))+

Pn−1 i=1

Pn−1 i=1

Pn−1

E[e

i=1

Pn

j=1

vj (N2 (tj )−N2 (tj−1 ))

]

ui (N1 (ti )−N1 (ti−1 ))+

Pn−1 j=1

vj (N2 (tj )−N2 (tj−1 ))

E[eun (N1 (tn )−N1 (tn−1 ))+vn (N2 (tn )−N2 (tn−1 )) |Ftn−1 ]]

ui (N1 (ti )−N1 (ti−1 ))+

Pn−1 j=1

vj (N2 (tj )−N2 (tj−1 ))

]E[eun (N1 (tn )−N1 (tn−1 ))+vn (N2 (tn )−N2 (tn−1 )) ]

ui (N1 (ti )−N1 (ti−1 ))+

Pn−1

vj (N2 (tj )−N2 (tj−1 ))

]E[eun (N1 (tn )−N1 (tn−1 )) ]E[evn (N2 (tn )−N2 (tn−1 )) ],

j=1

where the second equality comes from the independence of Ni (tn ) − Ni (tn−1 ) (i = 1, 2) relative to Ftn−1 and the third equality comes from the result obtained in the above paragraph. Working by induction, we have Pn

=

Pn

E[e i=1 ui (N1 (ti )−N1 (ti−1 ))+ j=1 vj (N2 (tj )−N2 (tj−1 )) ] n n Y Y E[eui (N1 (ti )−N1 (ti−1 )) ] E[evj (N2 (tj )−N2 (tj−1 )) ] i=1

=

j=1 Pn

E[e

i=1 ui (N1 (ti )−N1 (ti−1 ))

83

]E[e

Pn

j=1

vj (N2 (tj )−N2 (tj−1 ))

].

This shows the whole path of N1 is independent of the whole path of N2 . 11.6. Proof. Let Xt = u1 Wt − 12 u21 t + u2 Qt − λt(ϕ(u2 ) − 1) where ϕ is the moment generating function of the jump size Y . Itˆ o’s formula for jump process yields Z Z t X 1 1 t Xs 2 (eXs − eXs− ). e u1 ds + eXt − 1 = eXs (u1 dWs − u21 ds − λ(ϕ(u2 ) − 1)ds) + 2 2 0 0 0<s≤t

Note ∆Xt = u2 ∆Qt = u2 YNt ∆Nt , where Nt is the Poisson process associated with Qt . So eXt − eXt− = PNt u2 Yi (e − 1), eXt− (e∆Xt − 1) = eXt− (eu2 YNt − 1)∆Nt . Consider the compound Poisson process Ht = i=1 then Ht − λE[eu2 YNt − 1]t = Ht − λ(ϕ(u2 ) − 1)t is a martingale, eXt − eXt− = eXt− ∆Ht and eXt − 1

t

Z Z t 1 1 t Xs 2 eXs (u1 dWs − u21 ds − λ(ϕ(u2 ) − 1)ds) + e u1 ds + eXs− dHs 2 2 0 0 0 Z t Z t eXs− d(Hs − λ(ϕ(u2 ) − 1)s). = eXs u1 dWs + Z

=

0

0 1

This shows eXt is a martingale and E[eXt ] ≡ 1. So E[eu1 Wt +u2 Qt ] = e 2 u1 t eλt(ϕ(u2 )−1)t = E[eu1 Wt ]E[eu2 Qt ]. This shows Wt and Qt are independent. 11.7. Proof. E[h(QT )|Ft ] = E[h(QT −Qt +Qt )|Ft ] = E[h(QT −t +x)]|x=Qt = g(t, Qt ), where g(t, x) = E[h(QT −t + x)].

References [1] F. Delbaen and W. Schachermayer. The mathematics of arbitrage. Springer, 2006. [2] J. Hull. Options, futures, and other derivatives. Fourth Edition. Prentice-Hall International Inc., New Jersey, 2000. [3] B. Øksendal. Stochastic differential equations: An introduction with applications. Sixth edition. SpringerVerlag, Berlin, 2003. [4] D. Revuz and M. Yor. Continous martingales and Brownian motion. Third edition. Springer-Verlag, Berline, 1998. [5] A. N. Shiryaev. Essentials of stochastic finance: facts, models, theory. World Scientific, Singapore, 1999. [6] S. Shreve. Stochastic calculus for finance I. The binomial asset pricing model. Springer-Verlag, New York, 2004. [7] S. Shreve. Stochastic calculus for finance II. Continuous-time models. Springer-Verlag, New York, 2004. [8] P. Wilmott. The mathematics of financial derivatives: A student introduction. Cambridge University Press, 1995.

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