Solution Manual for Shreve 2

7/30/2019 Solution Manual for Shreve 2

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Stochastic Calculus for Finance, Volume I and II

by Yan ZengLast updated: August 20, 2007

This is a solution manual for the two-volume textbook Stochastic calculus for nance , by Steven Shreve.If you have any comments or nd any typos/errors, please email me at [email protected].

The current version omits the following problems. Volume I: 1.5, 3.3, 3.4, 5.7; Volume II: 3.9, 7.1, 7.2,7.57.9, 10.8, 10.9, 10.10.

Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through thissolution manual and communicating to me several mistakes/typos.

1 Stochastic Calculus for Finance I: The Binomial Asset Pricing Model

1. The Binomial No-Arbitrage Pricing Model

1.1.

Proof. If we get the up sate, then X 1 = X 1(H ) = 0uS 0 + (1 + r )(X 0 0S 0); if we get the down state,then X 1 = X 1(T ) = 0dS 0 +(1+ r )(X 0 0S 0). If X 1 has a positive probability of being strictly positive,then we must either have X 1(H ) > 0 or X 1(T ) > 0.(i) If X 1(H ) > 0, then 0uS 0 + (1 + r )(X 0 0S 0) > 0. Plug in X 0 = 0, we get u 0 > (1 + r ) 0 .By condition d < 1 + r < u , we conclude 0 > 0. In this case, X 1(T ) = 0dS 0 + (1 + r )(X 0 0S 0) = 0S 0[d (1 + r )] < 0.(ii) If X 1(T ) > 0, then we can similarly deduce 0 < 0 and hence X 1(H ) < 0.So we cannot have X 1 strictly positive with positive probability unless X 1 is strictly negative with positive

probability as well, regardless the choice of the number 0 .Remark: Here the condition X 0 = 0 is not essential, as far as a property denition of arbitrage for

arbitrary X 0 can be given. Indeed, for the one-period binomial model, we can dene arbitrage as a tradingstrategy such that P (X 1 X 0(1 + r )) = 1 and P (X 1 > X 0(1 + r )) > 0. First, this is a generalization of thecase X 0 = 0; second, it is proper because it is comparing the result of an arbitrary investment involvingmoney and stock markets with that of a safe investment involving only money market. This can also be seenby regarding X 0 as borrowed from money market account. Then at time 1, we have to pay back X 0(1 + r )to the money market account. In summary, arbitrage is a trading strategy that beats safe investment.

Accordingly, we revise the proof of Exercise 1.1. as follows. If X 1 has a positive probability of beingstrictly larger than X 0(1 + r ), the either X 1(H ) > X 0(1 + r ) or X 1(T ) > X 0(1 + r ). The rst case yields 0S 0(u 1r ) > 0, i.e. 0 > 0. So X 1(T ) = (1+ r )X 0 + 0S 0(d1r ) < (1 + r )X 0 . The second case canbe similarly analyzed. Hence we cannot have X 1 strictly greater than X 0(1 + r ) with positive probabilityunless X 1 is strictly smaller than X 0(1 + r ) with positive probability as well.

Finally, we comment that the above formulation of arbitrage is equivalent to the one in the textbook.For details, see Shreve [7], Exercise 5.7.

1.2.

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Proof. X 1(u) = 0 8 + 0 354 (4 0 + 1 .200) = 3 0 + 1 .50 , and X 1(d) = 0 254 (4 0 + 1 .200) =3 0 1.50 . That is, X 1(u) = X 1(d). So if there is a positive probability that X 1 is positive, then thereis a positive probability that X 1 is negative.

Remark: Note the above relation X 1(u) = X 1(d) is not a coincidence. In general, let V 1 denote thepayoff of the derivative security at time 1. Suppose X 0 and 0 are chosen in such a way that V 1 can bereplicated: (1 + r )(X 0 0S 0) + 0S 1 = V 1 . Using the notation of the problem, suppose an agent beginswith 0 wealth and at time zero buys 0 shares of stock and 0 options. He then puts his cash position 0S 0 0X 0 in a money market account. At time one, the value of the agents portfolio of stock, optionand money market assets is

X 1 = 0S 1 + 0V 1 (1 + r )( 0S 0 + 0X 0).Plug in the expression of V 1 and sort out terms, we have

X 1 = S 0( 0 + 00)(S 1S 0 (1 + r )) .

Since d < (1 + r ) < u , X 1(u) and X 1(d) have opposite signs. So if the price of the option at time zero is X 0 ,then there will no arbitrage.

1.3.Proof. V 0 = 11+ r

1+ r dud S 1(H ) +u 1ru d S 1(T ) =

S 01+ r

1+ r du d u +u 1rud d = S 0 . This is not surprising, since

this is exactly the cost of replicating S 1 .Remark: This illustrates an important point. The fair price of a stock cannot be determined by the

risk-neutral pricing, as seen below. Suppose S 1(H ) and S 1(T ) are given, we could have two current prices, S 0and S 0 . Correspondingly, we can get u, d and u , d . Because they are determined by S 0 and S 0 , respectively,its not surprising that risk-neutral pricing formula always holds, in both cases. That is,

S 0 =1+ r dud S 1(H ) +

u1ru d S 1(T )1 + r

, S 0 =1+ r du d S 1(H ) +

u 1ru d S 1(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.

X n +1 (T ) = n dS n + (1 + r )(X n n S n )= n S n (d 1 r ) + (1 + r )V n=

V n +1 (H ) V n +1 (T )u d

(d 1 r ) + (1 + r ) pV n +1 (H ) + qV n +1 (T )

1 + r= p(V n +1 (T ) V n +1 (H )) + pV n +1 (H ) + qV n +1 (T )= pV n +1 (T ) + qV n +1 (T )

= V n +1 (T ).

1.6.

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Proof. The banks trader should set up a replicating portfolio whose payoff is the opposite of the optionspayoff. More precisely, we solve the equation

(1 + r )(X 0 0S 0) + 0S 1 = (S 1 K )+ .Then X 0 = 1.20 and 0 = 12 . This means the trader should sell short 0.5 share of stock, put the income2 into a money market account, and then transfer 1.20 into a separate money market account. At time one,the portfolio consisting of a short position in stock and 0 .8(1 + r ) in money market account will cancel outwith the options 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 stockprice goes down at time one, the option will expire without any payoff. The initial money 1.20 we paid attime zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) whichguarantees a sure 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 banks trader only needs to set up the reverse of thereplicating trading strategy described in Example 1.2.4. More precisely, he should short sell 0.1733 share of stock, invest the income 0.6933 into money market account, and transfer 1.376 into a separate money marketaccount. The portfolio consisting a short position in stock and 0.6933-1.376 in money market account willreplicate the opposite of the options payoff. After they cancel out, we end up with 1 .376(1 + r )3 in theseparate money market account.

1.8. (i)

Proof. vn (s, y ) = 25 (vn +1 (2s, y + 2 s) + vn +1 (s2 , y +

s2 )).

(ii)

Proof. 1.696.

(iii)Proof.

n (s, y ) =vn +1 (us,y + us ) vn +1 (ds,y + ds)

(u d)s.

1.9. (i)

Proof. Similar to Theorem 1.2.2, but replace r , u and d everywhere with r n , un and dn . More precisely, set pn = 1+ r n dnu n dn and q n = 1 pn . Then

V n = pn V n +1 (H ) + q n V n +1 (T )

1 + r n.

(ii)

Proof. n = V n +1(H )V n +1 (T )S n +1 (H )S n +1 (T ) =

V n +1 (H )V n +1 (T )(u n dn )S n .

(iii)

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Proof. un = S n +1 (H )S n =S n +10

S n = 1+10S n and dn =

S n +1 (T )S n =

S n 10S n = 1 10S n . So the risk-neutral probabilitiesat time n are pn = 1dnu n dn =

12 and q n =

12 . Risk-neutral pricing implies the price of this call at time zero is

9.375.

2. Probability Theory on Coin Toss Space

2.1. (i)Proof. P (Ac) + P (A) = A c

P () + AP () =

P () = 1.

(ii)

Proof. By induction, it suffices to work on the case N = 2. When A1 and A2 are disjoint, P (A1A2) =A 1A 2

P () = A 1P () + A 2

P () = P (A1) + P (A2). When A1 and A2 are arbitrary, usingthe result when they are disjoint, we have P (A1A2) = P ((A1 A2)A2) = P (A1 A2) + P (A2) P (A1) + P (A2).2.2. (i)

Proof. P (S 3 = 32) = p3 = 18 , P (S 3 = 8) = 3 p2q = 38 , P (S 3 = 2) = 3 pq

2 = 38 , and P (S 3 = 0 .5) = q 3 = 18 .

(ii)

Proof. E [S 1] = 8P (S 1 = 8) + 2 P (S 1 = 2) = 8 p + 2 q = 5, E [S 2] = 16 p2 + 4 2 pq + 1 q 2 = 6 .25, andE [S 3] = 32 18 + 8 38 + 2 38 + 0 .5 18 = 7 .8125. So the average rates of growth of the stock price under P are, respectively: r 0 = 54 1 = 0 .25, r 1 = 6.255 1 = 0 .25 and r 2 = 7.81256.25 1 = 0 .25.

(iii)

Proof. P (S 3 = 32) = ( 23 )3 = 827 , P (S 3 = 8) = 3 ( 23 )2 13 = 49 , P (S 3 = 2) = 2 19 = 29 , and P (S 3 = 0 .5) = 127 .Accordingly, E [S 1] = 6, E [S 2] = 9 and E [S 3] = 13.5. So the average rates of growth of the stock price

under P are, respectively: r 0 = 64 1 = 0 .5, r 1 = 96 1 = 0 .5, and r 2 = 13 .59 1 = 0 .5.2.3.

Proof. Apply conditional Jensens inequality.

2.4. (i)

Proof. E n [M n +1 ] = M n + E n [X n +1 ] = M n + E [X n +1 ] = M n .

(ii)

Proof. E n [S n +1S n ] = E n [eX n +1 2

e + e ] =2

e + e E [eX n +1 ] = 1.

2.5. (i)

Proof. 2I n = 2n 1j =0 M j (M j +1 M j ) = 2

n 1j =0 M j M j +1 n 1j =1 M 2j

n 1j =1 M 2j = 2n 1j =0 M j M j +1 +

M 2n

n 1j =0

M 2j +1

n 1j =0

M 2j

= M 2n

n 1j =0

(M j +1

M j )2 = M 2

n n 1j =0

X 2j +1

= M 2n

n.

(ii)

Proof. E n [f (I n +1 )] = E n [f (I n + M n (M n +1 M n ))] = E n [f (I n + M n X n +1 )] = 12 [f (I n + M n )+ f (I n M n )] =g(I n ), where g(x) = 12 [f (x + 2x + n) + f (x 2x + n)], since 2I n + n = |M n |.2.6.

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Proof. E n [I n +1 I n ] = E n [ n (M n +1 M n )] = n E n [M n +1 M n ] = 0.2.7.

Proof. We denote by X n the result of n-th coin toss, where Head is represented by X = 1 and Tail isrepresented by X = 1. We also suppose P (X = 1) = P (X = 1) = 12 . Dene S 1 = X 1 and S n +1 =S n + bn (X 1 ,

, X n )X n +1 , where bn (

) is a bounded function on

{1, 1

}n , to be determined later on. Clearly

(S n )n 1 is an adapted stochastic process, and we can show it is a martingale. Indeed, E n [S n +1 S n ] =bn (X 1 , , X n )E n [X n +1 ] = 0.For any arbitrary function f , E n [f (S n +1 )] = 12 [f (S n + bn (X 1 , , X n ))+ f (S n bn (X 1 , , X n ))]. Thenintuitively, E n [f (S n +1 ] cannot be solely dependent upon S n when bn s are properly chosen. Therefore ingeneral, ( S n )n 1 cannot be a Markov process.

Remark : If X n is regarded as the gain/loss of n-th bet in a gambling game, then S n would be the wealthat time n. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results.

2.8. (i)

Proof. Note M n = E n [M N ] and M n = E n [M N ].

(ii)

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

(iii)

Proof. V n(1+ r ) n = E nV N

(1+ r ) N , so V 0 ,V 1

1+ r , ,V N 1

(1+ r ) N 1 ,V N

(1+ r ) N is a martingale under P .

(iv)

Proof. Combine (ii) and (iii), then use (i).

2.9. (i)

Proof. u0 = S 1 (H )S 0 = 2, d0 =S 1 (H )

S 0 =12 , u1(H ) =

S 2 (HH )S 1 (H ) = 1 .5, d1(H ) =

S 2 (HT )S 1 (H ) = 1, u1(T ) =

S 2 (T H )S 1 (T ) = 4

and d1(T ) = S 2 (T T )S 1 (T ) = 1.

So p0 = 1+ r 0 d0u 0 d0 =12 , q 0 =

12 , p1(H ) =

1+ r 1 (H )d1 (H )u 1 (H )d1 (H ) =12 , q 1(H ) =

12 , p1(T ) =

1+ r 1 (T )d1 (T )u 1 (T )d1 (T ) =16 , and

q 1(T ) = 56 .Therefore P (HH ) = p0 p1(H ) = 14 , P (HT ) = p0q 1(H ) =

14 , P (T H ) = q 0 p1(T ) =

112 and P (T T ) =

q 0q 1(T ) = 512 .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 modications (i.e. P would be constructed according to conditional probabili-ties P (n +1 = H

|1 ,

, n ) := pn and P (n +1 = T

|1 ,

, n ) := q n . Cf. notes on page 39.). So

the time-zero value of an option that pays off V 2 at time two is given by the risk-neutral pricing formulaV 0 = E V 2(1+ r 0 )(1+ r 1 ) .

(ii)

Proof. V 2(HH ) = 5, V 2(HT ) = 1, V 2(T H ) = 1 and V 2(T T ) = 0. So V 1(H ) = p1 (H )V 2 (HH )+ q1 (H )V 2 (HT )1+ r 1 (H ) =

2.4, V 1(T ) = p1 (T )V 2 (T H )+ q1 (T )V 2 (T T )1+ r 1 (T ) =19 , and V 0 =

p0 V 1 (H )+ q0 V 1 (T )1+ r 0 1.

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(iii)

Proof. 0 = V 1 (H )V 1 (T )S 1 (H )S 1 (T ) =2.419

82 = 0 .4 1

54 0.3815.(iv)

Proof. 1(H ) = V 2 (HH )V 2 (HT )S 2 (HH )S 2 (HT )

= 51128

= 1.

2.10. (i)

Proof. E n [ X n +1(1+ r ) n +1 ] = E n [ n Y n +1 S n(1+ r ) n +1 +

(1+ r )( X n n S n )(1+ r ) n +1 ] = n S n(1+ r ) n +1 E n [Y n +1 ] + X n n S n(1+ r ) n = n S n(1+ r ) n +1 (u p +dq ) + X n n S n(1+ r )n = n S n + X n n S n(1+ r ) n = X n(1+ r ) n .

(ii)

Proof. From (2.8.2), we have

n uS n + (1 + r )(X n n S n ) = X n +1 (H ) n dS n + (1 + r )(X n n S n ) = X n +1 (T ).

So n =X n +1 (H )

X n +1 (T )

uS n dS n and X n = E n [X n +11+ r ]. To make the portfolio replicate the payoff at time N , we

must have X N = V N . So X n = E n [ X N (1+ r ) N n ] = E n [V N

(1+ r ) N n ]. Since (X n )0n N is the value process of theunique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linearequations), the no-arbitrage price of V N at time n is V n = X n = E n [ V N (1+ r ) N n ].

(iii)

Proof.

E n [S n +1

(1 + r )n +1] =

1(1 + r )n +1

E n [(1 An +1 )Y n +1 S n ]=

S n(1 + r )n +1

[ p(1 An +1 (H ))u + q (1 An +1 (T ))d]< S n

(1 + r )n +1[ pu + qd]

=S n

(1 + r )n.

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

(1+ r ) n +1 (1a)( pu+ qd) = S n(1+ r ) n (1a). So E n [ S n +1(1+ r ) n +1 (1a )n +1 ] =S n(1+ r ) n (1a ) n .

2.11. (i)

Proof. F N + P N = S N K + ( K S N )+ = ( S N K )+ = C N .(ii)

Proof. C n = E n [ C N (1+ r ) N n ] = E n [F N

(1+ r ) N n ] + E n [P N

(1+ r ) N n ] = F n + P n .

(iii)

Proof. F 0 = E [ F N (1+ r ) N ] =1

(1+ r ) N E [S N K ] = S 0 K (1+ r ) N .(iv)

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Proof. At time zero, the trader has F 0 = S 0 in money market account and one share of stock. At time N ,the trader has a wealth of ( F 0 S 0)(1 + r )N + S N = K + S N = F N .

(v)

Proof. By (ii), C 0 = F 0 + P 0 . Since F 0 = S 0 (1+ r )N S 0

(1+ r ) N = 0, C 0 = P 0 .

(vi)

Proof. By (ii), C n = P n if and only if F n = 0. Note F n = E n [ S N K (1+ r ) N n ] = S n (1+ r )N S 0

(1+ r ) N n = S n S 0(1 + r )n .So F n is not necessarily zero and C n = P n is not necessarily true for n 1.2.12.

Proof. First, the no-arbitrage price of the chooser option at time m must be max( C, P ), where

C = E (S N K )+(1 + r )N m , and P = E

(K S N )+(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 optionat time m. Both of them have maturity date N and strike price K . Suppose the market is liquid, then thechooser option is equivalent to receiving a payoff of max( C, P ) at time m. Therefore, its current no-arbitrageprice should be E [max( C,P )(1+ r ) m ].

By the put-call parity, C = S m K (1+ r ) N m + P . So max( C, P ) = P + ( S m K (1+ r ) N m )+ . Therefore, thetime-zero price of a chooser option is

E P

(1 + r )m+ E

(S m K (1+ r ) N m )+(1 + r )m

= E (K S N )+

(1 + r )N + E

(S m K (1+ r ) N m )+(1 + r )m

.

The rst term stands for the time-zero price of a put, expiring at time N and having strike price K , and thesecond term stands for the time-zero price of a call, expiring at time m and having strike price K (1+ r )N m .

If we feel unconvinced by the above argument that the chooser options no-arbitrage price is E [max( C,P )(1+ r ) m ],

due to the economical argument involved (like the chooser option is equivalent to receiving a payoff of max( C, P ) at time m), then we have the following mathematically rigorous argument. First, we canconstruct a portfolio 0 , , m 1 , whose payoff at time m is max( C, P ). Fix , if C () > P (), wecan construct a portfolio m , , N 1 whose payoff at time N is (S N K )+ ; if C () < P (), we canconstruct a portfolio m , , N 1 whose payoff at time N is (K S N )+ . By dening ( m k N 1)

k () = k () if C () > P () k () if C () < P (),

we get a portfolio ( n )0n N 1 whose payoff is the same as that of the chooser option. So the no-arbitrageprice process of the chooser option must be equal to the value process of the replicating portfolio. Inparticular, V 0 = X 0 = E [ X m(1+ r ) m ] = E [

max( C,P )(1+ r ) m ].

2.13. (i)Proof. Note under both actual probability P and risk-neutral probability P , coin tosses n s are i.i.d.. Sowithout loss of generality, we work on P . For any function g, E n [g(S n +1 , Y n +1 )] = E n [g( S n +1S n S n , Y n +S n +1

S n S n )] = pg(uS n , Y n + uS n ) + qg(dS n , Y n + dS n ), which is a function of ( S n , Y n ). So (S n , Y n )0n N isMarkov under P .

(ii)

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Proof. Set vN (s, y ) = f ( yN +1 ). Then vN (S N , Y N ) = f (N n =0 S nN +1 ) = V N . Suppose vn +1 is given, then

V n = E n [V n +11+ r ] = E n [

vn +1 (S n +1 ,Y n +1 )1+ r ] =

11+ r [ pvn +1 (uS n , Y n + uS n ) + qvn +1 (dS n , Y n + dS n )] = vn (S n , Y n ),

wherevn (s, y ) =

vn +1 (us,y + us ) + vn +1 (ds,y + ds)1 + r

.

2.14. (i)

Proof. For n M , (S n , Y n ) = ( S n , 0). Since coin tosses n s are i.i.d. under P , (S n , Y n )0n M is Markovunder P . More precisely, for any function h, E n [h(S n +1 )] = ph(uS n ) + h(dS n ), for n = 0 , 1, , M 1.

For any function g of two variables, we have E M [g(S M +1 , Y M +1 )] = E M [g(S M +1 , S M +1 )] = pg(uS M , uS M )+qg(dS M , dS M ). And for n M +1, E n [g(S n +1 , Y n +1 )] = E n [g( S n +1S n S n , Y n + S n +1S n S n )] = pg(uS n , Y n + uS n )+qg(dS n , Y n + dS n ), so (S n , Y n )0n N is Markov under P .

(ii)

Proof. Set vN (s, y ) = f ( yN M ). Then vN (S N , Y N ) = f (N K = M +1 S k

N M ) = V N . Suppose vn +1 is already given.a) If n > M , then E

n[v

n +1(S

n +1, Y

n +1)] = pv

n +1(uS

n, Y

n+ uS

n) + qv

n +1(dS

n, Y

n+ dS

n). So v

n(s, y ) =

pvn +1 (us,y + us ) + qvn +1 (ds,y + ds).b) If n = M , then E M [vM +1 (S M +1 , Y M +1 )] = pvM +1 (uS M , uS M ) + vn +1 (dS M , dS M ). So vM (s) =

pvM +1 (us, us ) + qvM +1 (ds, ds ).c) If n < M , then E n [vn +1 (S n +1 )] = pvn +1 (uS n ) + qvn +1 (dS n ). So vn (s) = pvn +1 (us ) + qvn +1 (ds).

3. State Prices

3.1.

Proof. Note Z () := P ()P ()

= 1Z () . Apply Theorem 3.1.1 with P , P , Z replaced by P , P , Z , we get theanalogous of properties (i)-(iii) of Theorem 3.1.1.

3.2. (i)

Proof. P () = P () =

Z ()P () = E [Z ] = 1.

(ii)

Proof. E [Y ] = Y ()P () =

Y ()Z ()P () = E [Y Z ].

(iii)

Proof. P (A) = AZ ()P (). Since P (A) = 0, P () = 0 for any A. So P (A) = 0.

(iv)

Proof. If P (A) = AZ ()P () = 0, by P (Z > 0) = 1, we conclude P () = 0 for any A. SoP (A) =

AP () = 0.

(v)

Proof. P (A) = 1 P (Ac) = 0 P (A

c) = 0 P (A) = 1.

(vi)

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Proof. Pick 0 such that P (0) > 0, dene Z () =0, if = 01

P (0 ) , if = 0 .Then P (Z 0) = 1 and E [Z ] =

1P (0 ) P (0) = 1.

Clearly P ( \ {0}) = E [Z 1\{0 }] = = 0 Z ()P () = 0. But P ( \ {0}) = 1 P (0) > 0 if P (0) < 1. Hence in the case 0 < P (0) < 1, P and P are not equivalent. If P (0) = 1, then E [Z ] = 1 if

and only if Z (0) = 1. In this case P (0) = Z (0)P (0) = 1. And P and P have to be equivalent.In summary, if we can nd 0 such that 0 < P (0) < 1, then Z as constructed above would induce aprobability P that is not equivalent to P .

3.5. (i)

Proof. Z (HH ) = 916 , Z (HT ) =98 , Z (T H ) =

38 and Z (T T ) =

154 .

(ii)

Proof. Z 1(H ) = E 1[Z 2](H ) = Z 2(HH )P (2 = H |1 = H ) + Z 2(HT )P (2 = T |1 = H ) = 34 . Z 1(T ) =E 1[Z 2](T ) = Z 2(T H )P (2 = H |1 = T ) + Z 2(T T )P (2 = T |1 = T ) = 32 .(iii)

Proof.

V 1(H ) =[Z 2(HH )V 2(HH )P (2 = H |1 = H ) + Z 2(HT )V 2(HT )P (2 = T |1 = T )]

Z 1(H )(1 + r 1(H ))= 2 .4,

V 1(T ) =[Z 2(T H )V 2(T H )P (2 = H |1 = T ) + Z 2(T T )V 2(T T )P (2 = T |1 = T )]

Z 1(T )(1 + r 1(T ))=

19

,

and

V 0 =Z 2(HH )V 2(HH )

(1 + 14 )(1 +14 )

P (HH ) +Z 2(HT )V 2(HT )(1 + 14 )(1 +

14 )

P (HT ) +Z 2(T H )V 2(T H )(1 + 14 )(1 +

12 )

P (T H ) + 0 1.

3.6.

Proof. U (x) = 1x , so I (x) =1x . (3.3.26) gives E [

Z (1+ r ) N

(1+ r ) N

Z ] = X 0 . So =1

X 0 . By (3.3.25), we

have X N = (1+ r )N

Z =X 0Z (1 + r )

N . Hence X n = E n [ X N (1+ r ) N n ] = E n [X 0 (1+ r ) n

Z ] = X 0(1 + r )n E n [ 1Z ] =

X 0(1 + r )n 1Z n E n [Z 1Z ] = X 0n , where the second to last = comes from Lemma 3.2.6.3.7.

Proof. U (x) = x p1 and so I (x) = x1

p 1 . By (3.3.26), we have E [ Z (1+ r ) N (Z

(1+ r ) N )1

p 1 ] = X 0 . Solve it for ,we get

=X 0

E Z p

p

1

(1+ r )Np

p 1

p1

=X p10 (1 + r )Np

(E [Z

pp

1

]) p

1

.

So by (3.3.25), X N = ( Z (1+ r ) N )1

p 1 = 1

p 1 Z 1

p 1(1+ r )

N p 1

= X 0 (1+ r )Np

p 1

E [Z p

p 1 ]Z

1p 1

(1+ r )N

p 1= (1+ r )

N X 0 Z 1

p 1

E [Z p

p 1 ].

3.8. (i)

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Proof. ddx (U (x) yx) = U (x) y. So x = I (y) is an extreme point of U (x) yx. Because d2

dx 2 (U (x) yx) =U (x) 0 (U is concave), x = I (y) is a maximum point. Therefore U (x) y(x) U (I (y)) yI (y) for everyx.(ii)

Proof. Following the hint of the problem, we have

E [U (X N )] E [X N Z

(1 + r )N ] E [U (I (

Z (1 + r )N

))] E [Z

(1 + r )N I (

Z (1 + r )N

)],

i.e. E [U (X N )] X 0 E [U (X N )] E [ (1+ r ) N X N ] = E [U (X N )] X 0 . So E [U (X N )] E [U (X N )].3.9. (i)

Proof. X n = E n [ X N (1+ r ) N n ]. So if X N 0, then X n 0 for all n.(ii)

Proof. a) If 0 x < and 0 < y 1 , then U (x) yx = yx 0 and U (I (y)) yI (y) = U ( ) y =1 y 0. So U (x) yx U (I (y)) yI (y).

b) If 0 x < and y > 1 , then U (x) yx = yx 0 and U (I (y)) yI (y) = U (0) y 0 = 0. SoU (x) yx U (I (y)) yI (y).c) If x and 0 < y 1 , then U (x) yx = 1 yx and U (I (y)) yI (y) = U ( ) y = 1 y 1yx.

So U (x) yx U (I (y)) yI (y).d) If x and y > 1 , then U (x) yx = 1 yx < 0 and U (I (y)) yI (y) = U (0) y 0 = 0. So

U (x) yx U (I (y)) yI (y).(iii)

Proof. Using (ii) and set x = X N , y = Z (1+ r ) N , where X N is a random variable satisfying E [X N

(1+ r ) N ] = X 0 ,we have

E [U (X N )] E [Z

(1 + r )N X N ] E [U (X N )] E [

Z (1 + r )N

X N ].

That is, E [U (X N )] X 0 E [U (X N )] X 0 . So E [U (X N )] E [U (X N )].(iv)

Proof. Plug pm and m into (3.6.4), we have

X 0 =2N

m =1 pm m I ( m ) =

2N

m =1 pm m 1{ m 1 }.

So X 0 =2N m =1 pm m 1{ m 1 }. Suppose there is a solution to (3.6.4), note

X 0 > 0, we then can conclude

{m : m 1 } = . Let K = max {m : m 1 }, then K 1 < K +1 . So K < K +1 andX 0 =

K m =1 pm m (Note, however, that K could be 2

N . In this case, K +1 is interpreted as . Also, notewe are looking for positive solution > 0). Conversely, suppose there exists some K so that K < K +1 andK

m =1 m pm =X 0 . Then we can nd > 0, such that K holds if and only if b1 > a 1 , b2 < a 2 or b1 < a 1 , b2 > a 2 . By induction, we can show

V S n = max gS (S n ), pV S

n +1+ V S

n +11 + r

max gP (S n ) + gC (S n ), pV P n +1 + V P n +1

1 + r+

pV C n +1 + V C n +11 + r

max gP (S n ), pV P n +1 + V P n +1

1 + r+ max gC (S n ),

pV C n +1 + V C n +11 + r

= V P n + V C

n .

As to when < holds, suppose m = max {n : V S n < V P n + V C n }. Then clearly m N 1 and it is possiblethat {n : V S n < V P n + V C n }= . When this set is not empty, m is characterized as m = max {n : gP (S n ) < pV P n +1 + qV P n +11+ r and gC (S n ) >

pV Cn +1 + qV C

n +11+ r or gP (S n ) >

pV P n +1 + qV P

n +11+ r and gC (S n ) G n for 0 n N 1. So thetime-zero value is S 0 K (1+ r ) N and the optimal exercise time is N .5. Random Walk

5.1. (i)

Proof. E [ 2 ] = E [ ( 2 1 )+ 1 ] = E [ ( 2 1 ) ]E [ 1 ] = E [ 1 ]2 .

(ii)

Proof. If we dene M (m )n = M n + m M m (m = 1 , 2, ), then ( M (m ) )m as random functions are i.i.d. withdistributions the same as that of M . So m +1 m = inf {n : M (m )n = 1 }are i.i.d. with distributions thesame as that of 1 . Therefore

E [ m ] = E [ ( m m 1 )+( m 1 m 2 )+ + 1 ] = E [ 1 ]m .

(iii)

Proof. Yes, since the argument of (ii) still works for asymmetric random walk.

5.2. (i)

Proof. f () = pe qe , so f () > 0 if and only if > 12 (ln q ln p). Since 12 (ln q ln p) < 0,f () > f (0) = 1 for all > 0.(ii)

Proof. E n [S n +1

S n ] = E n [eX n +1 1

f ( ) ] = pe 1

f ( ) + qe 1f ( ) = 1.

(iii)

Proof. By optional stopping theorem, E [S n 1 ] = E [S 0] = 1. Note S n 1 = eM n 1 ( 1f ( ) )

n 1 e 1 ,by bounded convergence theorem, E [1{ 1 < }S 1 ] = E [limn S n 1 ] = lim n E [S n 1 ] = 1, that is,E [1{ 1 < }e

( 1f ( ) ) 1 ] = 1. So e = E [1{ 1 < }(

1f ( ) )

1 ]. Let 0, again by bounded convergence theorem,1 = E [1{ 1 < }(

1f (0) )

1 ] = P ( 1 < ).(iv)

Proof. Set = 1f ( ) =1

pe + qe , then as varies from 0 to , can take all the values in (0 , 1). Write in terms of , we have e = 1 14 pq 22 p (note 4 pq2 < 4( p+ q2 )2 12 = 1). We want to choose > 0, so weshould take = ln( 1+

14 pq 22 p ). Therefore E [

1 ] = 2 p1+ 14 pq 2 =

1 14 pq 22q .

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(v)

Proof. E [ 1 ] = E [

1 ] = E [ 1 1 1], and

1 1 4 pq22q= 1

2q (1 1 4 pq2)1

=12q

[12

(1 4 pq2)12 (4 pq 2 )1 + (1 1 4 pq2)(1) 2].

So E [ 1] = lim 1

E [ 1 ] = 12q [12 (1 4 pq )

12 (8 pq ) (1 1 4 pq )] = 12 p1 .

5.3. (i)

Proof. Solve the equation pe + qe = 1 and a positive solution is ln 1+ 14 pq2 p = ln1 p p = ln q ln p. Set

0 = ln q ln p, then f (0) = 1 and f () > 0 for > 0 . So f () > 1 for all > 0 .(ii)

Proof. As in Exercise 5.2, S n = eM n ( 1f ( ) )n is a martingale, and 1 = E [S 0] = E [S n 1 ] = E [e

M n 1 ( 1f ( ) ) 1n ].

Suppose > 0 , then by bounded convergence theorem,

1 = E [ limn

eM n 1 (1

f ())n 1 ] = E [1{ 1 < }e

(1

f ()) 1 ].

Let 0 , we get P ( 1 < ) = e 0 = pq < 1.(iii)

Proof. From (ii), we can see E [1{ 1 < }(1

f ( ) ) 1 ] = e , for > 0 . Set = 1f ( ) , then e =

1 14 pq 22 p . We

need to choose the root so that e > e 0 = q p , so = ln(1+ 14 pq 2

2 p ), then E [ 1 1{ 1 < }] =

1 14 pq 22q .

(iv)

Proof. E [ 11{ 1 < }] =

E [ 1 1{ 1 < }]| =1 = 12q [ 4 pq 14 pq (1 1 4 pq )] =

12q [

4 pq2q1 1 + 2 q 1] = p

q1

2q1 .

5.4. (i)

Proof. E [ 2 ] = k =1 P ( 2 = 2 k) 2k = k=1 ( 2 )2k P ( 2 = 2 k)4k . So P ( 2 = 2 k) =(2 k )!

4k (k +1)! k! .

(ii)

Proof. P ( 2 = 2) = 14 . For k 2, P ( 2 = 2 k) = P ( 2 2k) P ( 2 2k 2).P ( 2 2k) = P (M 2k = 2) + P (M 2k 4) + P ( 2 2k, M 2k 0)

= P (M 2k = 2) + 2 P (M 2k 4)= P (M 2k = 2) + P (M 2k 4) + P (M 2k 4)= 1 P (M 2k = 2) P (M 2k = 0) .

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Similarly, P ( 2 2k 2) = 1 P (M 2k2 = 2) P (M 2k2 = 0). SoP ( 2 = 2 k) = P (M 2k2 = 2) + P (M 2k2 = 0) P (M 2k = 2) P (M 2k = 0)

= (12

)2k2 (2k 2)!k!(k 2)!

+(2k 2)!

(k 1)!(k 1)! (

12

)2k(2k)!

(k + 1)!( k 1)!+

(2k)!k!k!

=(2k)!

4k (k + 1)! k!4

2k(2k 1) (k + 1) k(k 1) +4

2k(2k 1) (k + 1) k2

k (k + 1)=

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

2(k2 1)2k 1

+2(k2 + k)

2k 1 4k2 12k 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 reectedpath that resides at time 2 m b. So

P (M n m, M n = b) = P (M n = 2 m b) = ( 12)n n!(m + n b2 )!( n + b2 m)!.

(ii)

Proof.

P (M n m, M n = b) =n!

(m + n b2 )!( n + b2 m)! pm +

n b2 q n + b

2 m .

5.6.

Proof. On the innite coin-toss space, we dene M n = {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 perpetualAmerican put as in Section 5.4 can be dened as sup M E [1{< }

(K S ) +(1+ r ) ]. For an American put withthe same strike price K that expires at time n, its time-zero value V (n ) is max M n E [1{< }

(K S ) +(1+ r ) ].Clearly ( V (n ) )n 0 is nondecreasing and V

(n ) V for every n. So limn V (n ) exists and lim n V (n ) V .For any given M , we dene

(n ) = , if = n, if < , then (n ) is also a stopping time, (n )M n

and lim n (n ) = . By bounded convergence theorem,

E 1{< }(K S )+

(1 + r ) = lim

n E 1{ ( n ) < }

(K S ( n ) )+(1 + r ) ( n ) limn V

(n ) .

Take sup at the left hand side of the inequality, we get V limn V (n ) . Therefore V = lim n V (n ) .Remark: In the above proof, rigorously speaking, we should use ( K S ) in the places of ( K S )+ . Sothis needs some justication.

5.8. (i)

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Proof. v(S n ) = S n S n K = g(S n ). Under risk-neutral probabilities, 1(1+ r ) n v(S n ) = S n(1+ r ) n is a martingaleby Theorem 2.4.4.

(ii)

Proof. If the purchaser chooses to exercises the call at time n, then the discounted risk-neutral expectationof her payoff is E S n K

(1+ r )n = S 0

K

(1+ r )n . Since limn

S 0

K

(1+ r )n = S 0 , the value of the call at time

zero is at least sup n S 0 K (1+ r ) n = S 0 .(iii)

Proof. max g(s), pv(us )+ qv (ds )1+ r = max {s K, pu + qv1+ r s}= max {s K, s }= s = v(s), so equation (5.4.16) issatised. Clearly v(s) = s also satises the boundary condition (5.4.18).

(iv)

Proof. Suppose is an optimal exercise time, then E S K (1+ r ) 1{< } S 0 . Then P ( < ) = 0 andE K (1+ r ) 1{< } > 0. So E

S K (1+ r ) 1{< } < E S

(1+ r ) 1{< } . SinceS n

(1+ r ) n n 0is a martingale

under risk-neutral measure, by Fatous lemma, E S (1+ r ) 1{< } liminf n E S n(1+ r ) n 1{< } =liminf n E

S n(1+ r ) n = liminf n E [S 0] = S 0 . Combined, we have S 0 E S K (1+ r ) 1{< } < S 0 .Contradiction. So there is no optimal time to exercise the perpetual American call. Simultaneously, we haveshown E S K (1+ r ) 1{< } < S 0 for any stopping time . Combined with ( ii ), we conclude S 0 is the leastupper bound for all the prices acceptable to the buyer.

5.9. (i)

Proof. Suppose v(s) = s p, then we have s p = 25 2 ps p + 25

s p2p . So 1 =

2p +15 +

21 p5 . Solve it for p, we get p = 1

or p = 1.(ii)

Proof. Since limsv(s) = lim s(As +Bs ) = 0, we must have A = 0.

(iii)

Proof. f B (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 B . By drawing graphs of 4 s and Bs , we should choose B = 4 and sB = 2 + 4 B = 2.(v)

Proof. To have continuous derivative, we must have 1 = Bs 2B . Plug B = s

2B back into s

2B 4sB + B = 0,we get sB = 2. This gives B = 4.

6. Interest-Rate-Dependent Assets

6.2.

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Proof. X k = S k E k [Dm (S m K )]D1k S nB n,m Bk,m for n k m. Then

E k1[Dk X k ] = E k1[Dk S k E k [Dm (S m K )] S n

Bn,mBk,m Dk ]

= Dk1S k1 E k1[Dm (S m K )] S n

Bn,mE k1[E k [Dm ]]

= Dk1[S k1 E k1[Dm (S m K )]D1k1 S n

Bn,mBk1,m ]

= Dk1X k1 .

6.3.

Proof.

1Dn

E n [Dm +1 Rm ] =1

DnE n [Dm (1 + Rm )1Rm ] = E n [

Dm Dm +1Dn

] = Bn,m Bn,m +1 .

6.4.(i)

Proof. D1V 1 = E 1[D3V 3] = E 1[D2V 2] = D2E 1[V 2]. So V 1 = D 2D 1 E 1[V 2] =1

1+ R 1 E 1[V 2]. In particular,V 1(H ) = 11+ R 1 (H ) V 2(HH )P (w2 = H |w1 = H ) = 421 , V 1(T ) = 0.

(ii)

Proof. Let X 0 = 221 . Suppose we buy 0 shares of the maturity two bond, then at time one, the value of our portfolio is X 1 = (1 + R0)(X 0 B0,2) + 0B1,2 . To replicate the value V 1 , we must have

V 1(H ) = (1 + R0)(X 0 0B0,2) + 0B1,2(H )V 1(T ) = (1 + R0)(X 0 0B0,2) + 0B1,2(T ).So 0 = V 1 (H )V 1 (T )B 1 , 2 (H )B 1 , 2 (T ) =

43 . The hedging strategy is therefore to borrow

43 B0,2 221 = 2021 and buy 43

share of the maturity two bond. The reason why we do not invest in the maturity three bond is thatB1,3(H ) = B1,3(T )(= 47 ) and the portfolio will therefore have the same value at time one regardless theoutcome of rst coin toss. This makes impossible the replication of V 1 , since V 1(H ) = V 1(T ).

(iii)

Proof. Suppose we buy 1 share of the maturity three bond at time one, then to replicate V 2 at timetwo, we must have V 2 = (1 + R1)(X 1 1B1,3) + 1B2,3 . So 1(H ) = V 2 (HH )V 2 (HT )B 2 , 3 (HH )B 2 , 3 (HT ) =

23 , and

1(T ) = V 2 (T H )V 2 (T T )B 2 , 3 (T H )B 2 , 3 (T T ) = 0. So the hedging strategy is as follows. If the outcome of rst coin toss isT , then we do nothing. If the outcome of rst coin toss is H , then short 23 shares of the maturity threebond 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 samevalue regardless outcomes of coin tosses. This makes impossible the replication of V 2 .

6.5. (i)

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Proof. Suppose 1 n m, thenE m +1n 1 [F n,m ] = E n 1[B

1n,m +1 (Bn,m Bn,m +1 )Z n,m +1 Z 1n 1,m +1 ]

= E n 1Bn,m

Bn,m +1 1Bn,m +1 Dn

Bn 1,m +1 Dn 1

=Dn

Bn 1,m +1 Dn 1 E n 1[D1

n E n [Dm ]D1

n E n [Dm +1 ]]

=E n 1[Dm Dm +1 ]

Bn 1,m 1 Dn 1=

Bn 1,m Bn 1,m +1Bn 1,m +1

= F n 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

the strike price K = For n,m =S n

B n,m . At time n + 1, the contract has the value E n +1 [Dm (S m K )] =S n +1 KB n +1 ,m = S n +1 S n B n +1 ,m

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

ow of S n +1 S n B n +1 ,m

B n,m

(ii)

Proof. By (i), the cash ow generated at time n + 1 is

(1 + r )m n 1 S n +1 S n Bn +1 ,m

Bn,m

= (1 + r )m n 1 S n +1 S n

(1+ r ) m n 11

(1+ r ) m n

= (1 + r )m n 1S n +1 (1 + r )m n S n= (1 + r )m E n 1 [

S m(1 + r )m

] + (1 + r )m E n [S m

(1 + r )m]

= Fut n +1 ,m F ut n,m .

6.7.

Proof.

n +1 (0) = E [Dn +1 V n +1 (0)]

= E [Dn

1 + r n (0)1{

# H (1

n +1 )=0

}]

= E [Dn

1 + r n (0)1{# H (1 n )=0 }1{n +1 = T }]

=12

E [Dn

1 + r n (0)]

=n (0)

2(1 + r n (0)).

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For k = 1 , 2, , n ,n +1 (k) = E

Dn1 + r n (# H (1 n ))

1{# H (1 n +1 )= k}

= E Dn

1 + r n (k)1{# H (1 n )= k}1{n +1 = T } + E

Dn1 + r n (k

1)

1{# H (1 n )= k}1{n +1 = H }

=12

E [Dn V n (k)]1 + r n (k)

+12

E [Dn V n (k 1)]1 + r n (k 1)

=n (k)

2(1 + r n (k))+

n (k 1)2(1 + r n (k 1))

.

Finally,

n +1 (n + 1) = E [Dn +1 V n +1 (n + 1)] = E Dn

1 + r n (n)1{# H (1 n )= n }1{n +1 = H } =

n (n)2(1 + r n (n))

.

Remark : In the above proof, we have used the independence of n +1 and ( 1 , , n ). This is guaranteedby the assumption that p = q = 12 (note if and only if E [ |] = constant ). In case the binomial modelhas stochastic up- and down-factor un and dn , we can use the fact that P (n +1 = H |1 , , n ) = pn andP (n +1 = T |1 , , n ) = q n , where pn = 1+ r n dnu n dn and q n =

u1r nu n dn (cf. solution of Exercise 2.9 andnotes on page 39). Then for any X F n = (1 , , n ), we have E [Xf (n +1 )] = E [X E [f (n +1 )|F n ]] =E [X ( pn f (H ) + q n f (T ))].

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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 dene a mapping from A to as follows: (12 ) = 135 . Then is one-to-one andonto. So the cardinality of A is the same as that of , which means in particular that A is uncountablyinnite.

(ii)

Proof. Let An = { = 12 : 1 = 2 , , 2n 1 = 2n }. Then An A as n . So

P (A) = limn P (An ) = limn [P (1 = 2) P (2n 1 = 2n )] = limn ( p2

+ (1 p)2)

n.

Since p2 + (1 p)2 < 1 for 0 < p < 1, we have lim n ( 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 nite, then AB is also nite. SoP (AB ) = 0 = P (A) + P (B ). If at least one of them is innite, then AB is also innite. So P (AB ) =

= P (A) + P (B ). Similarly, we can prove P (N n =1 An ) =N n =1 P (An ), even if An s are not disjoint.

To see countable additivity property doesnt hold for P , let An = {1n }. Then A = n =1 An is an inniteset and therefore P (A) = . However, P (An ) = 0 for each n. So P (A) = 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 uniformlydistributed on [0 , 1]. For the strictly increasing and continuous function N (x) = x 1 2 e 22 d , we letZ = N 1(X ). Then P (Z a) = P (X N (a)) = N (a) for any real number a, i.e. Z is a standard normalrandom variable on the coin-toss space ( , F , P ).

(ii)

Proof. Dene X n =ni=1

Y i2i , where

Y i () =1, if i = H 0, if i = T .

Then X n () X () for every where X is dened as in (i). So Z n = N 1(X n ) Z = N 1(X ) forevery . Clearly Z n depends only on the rst n coin tosses and {Z n }n 1 is the desired sequence.1.5.

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Proof. First, by the information given by the problem, we have

0 1[0,X ()) (x)dxdP () = 0 1[0,X ()) (x)dP ()dx.The left side of this equation equals to

X ()

0dxdP () = X ()dP () = E {X }.

The right side of the equation equals to

0 1{x


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1.8. (i)

Proof. By (1.9.1), |Y n | = etX e s n Xts n = |Xe

X | = Xe X Xe 2tX . The last inequality is by X 0 and thefact that is between t and sn , and hence smaller than 2 t for n sufficiently large. So by the DominatedConvergence Theorem, (t) = lim n E {Y n }= E {limn Y n }= E {Xe tX }.

(ii)Proof. Since E [etX

+1{X 0}] + E [e

tX 1{X< 0}] = E [etX ] < for every tR , E [et |X |] = E [etX

+1{X 0}] +

E [e(t )X 1{X< 0}] < for every tR . Similarly, we have E [|X |et |X |] < for every tR . So, similar to(i), we have |Y n | = |Xe X | |X |e2t |X | for n sufficiently large, So by the Dominated Convergence Theorem, (t) = lim n E {Y n }= E {limn Y n }= E {Xe tX }.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 of Lebesgue integral, the desired equality also holds for simple functions, i.e. g of theform g(x) = ni =1 1B i (x), where each B i is a Borel subset of R . Since any nonnegative, Borel-measurablefunction g is the limit of an increasing sequence of simple functions, the desired equality can be proved bythe 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,P (i =1 Ai ) equals to

1i =1 A i ZdP = limn 1ni =1 A i ZdP = limn 1ni =1 A i ZdP = limn ni=1 A i ZdP = i =1 P (Ai ).Meanwhile, P () = 2 P ([ 12 , 1]) = 1. So P is a probability measure.

(ii)

Proof. If P (A) = 0, then P (A) = A ZdP = 2

A[ 12 ,1] dP = 2 P (A [12 , 1]) = 0.

(iii)

Proof. Let A = [0 , 12 ).

1.11.

Proof.

E {euY }= E {euY Z }= E {euX + u eX 22 }= eu

22 E {e(u )X }= eu

22 e

( u ) 22 = eu 22 .

1.12.

Proof. First, Z = eY 2

2 = e(X + )2

2 = e 22 + X = Z 1 . Second, for any A F , P (A) = A Zd P = (1A Z )ZdP = 1A dP = P (A). So P = P . In particular, X is standard normal under P , since its standardnormal under P .

1.13. (i)

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Proof. 1 P (X B (x, )) =1 x + 2x2 1 2 eu 22 du is approximately 1 1 2 ex 22 = 1 2 eX

2 ( )2 .

(ii)

Proof. Similar to (i).

(iii)

Proof. {X B (x, )}= {X B (y , )}= {X + B (y, )}= {Y B (y, )}.(iv)

Proof. By (i)-(iii), P (A)P (A) is approximately

2 eY 2 ( )

2

2 eX 2 ( )

2

= eY 2 ( ) X 2 ( )2 = e

( X ( )+ ) 2 X 2 ( )2 = eX ()2

2 .

1.14. (i)

Proof.

P () = e( )X dP = 0 e( )x ex dx = 0 ex dx = 1 .(ii)

Proof.

P (X a) = {X a } e( )X dP = a

0

e( )x ex dx = a0 ex dx = 1 ea .1.15. (i)

Proof. Clearly Z 0. Furthermore, we haveE {Z }= E

h(g(X ))g (X )f (X )

= h(g(x))g (x)f (x) f (x)dx = h(g(x))dg(x) = h(u)du = 1 .(ii)

Proof.

P (Y a) = {g(X )a} h(g(X ))g (X )f (X ) dP = g1 (a )

h(g(x))g (x)f (x)

f (x)dx = g1 (a )

h(g(x))dg(x).

By the change of variable formula, the last equation above equals to ah(u)du. So Y has density h underP .24


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2. Information and Conditioning

2.1.

Proof. For any real number a, we have {X a} F 0 = {, }. So P (X a) is either 0 or 1. Sincelima P (X a) = 1 and lim a P (X a) = 0, we can nd a number x0 such that P (X x0) = 1 andP (X

x) = 0 for any x < x 0 . So

P (X = x0) = limn P (x0

1n

< X x0) = limn (P (X x0) P (X x0 1n

)) = 1 .

2.2. (i)

Proof. (X ) = {, , {HT,TH }, {TT,HH }}.(ii)

Proof. (S 1) = {, , {HH,HT }, {TH,TT }}.(iii)

Proof. P ({HT,TH } {HH,HT }) = P ({HT }) = 14 , P ({HT,TH }) = P ({HT }) + P ({T H }) = 14 + 14 = 12 ,and P ({HH,HT }) = P ({HH }) + P ({HT }) = 14 + 14 = 12 . So we have

P ({HT,TH }{HH,HT }) = P ({HT,TH })P ({HH,HT }).Similarly, we can work on other elements of (X ) and (S 1) and show that P (A B ) = P (A)P (B ) for anyA(X ) and B(S 1). So (X ) and (S 1) are independent under P .

(iv)

Proof. P ({HT,TH } {HH,HT }) = P ({HT }) = 29 , P ({HT,TH }) = 29 + 29 = 49 and P ({HH,HT }) =49 +

29 =

69 . So

P (

{HT,TH

}{HH,HT

}) = P (

{HT,TH

})P (

{HH,HT

}).

Hence (X ) and (S 1) are not independent under P .

(v)

Proof. Because S 1 and X are not independent under the probability measure P , knowing the value of X will affect our opinion on the distribution of S 1 .

2.3.

Proof. We note ( V, W ) are jointly Gaussian, so to prove their independence it suffices to show they areuncorrelated. Indeed, E {V W }= E {X 2 sin cos + XY cos2 XY sin2 + Y 2 sin cos }= sin cos +0 + 0 + sin cos = 0.2.4. (i)

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Proof.

E {euX + vY } = E {euX + vXZ }= E {euX + vXZ |Z = 1 }P (Z = 1) + E {euX + vXZ |Z = 1}P (Z = 1)=

12

E {euX + vX }+12

E {euX vX }= 1

2[e ( u + v ) 22 + e ( u v ) 22 ]

= eu 2 + v 2

2euv + euv

2.

(ii)

Proof. Let u = 0.

(iii)

Proof. E {euX }= eu 22 and E {evY }= e

v 22 . So E {euX + vY } = E {euX }E {evY }. Therefore X and Y cannot

be independent.2.5.

Proof. The density f X (x) of X can be obtained by

f X (x) = f X,Y (x, y )dy = {y|x |} 2|x|+ y 2 e (2 |x |+ y ) 22 dy = {|x |} 2 e 22 d = 1 2 ex 22 .The density f Y (y) of Y can be obtained by

f Y (y) = f XY (x, y )dx= 1{|

x |y}2

|x

|+ y

2 e(2 |x |+ y ) 22

dx

= 0(y) 2x + y 2 e (2 x + y ) 22 dx + 0y

2x + y 2 e

( 2 x + y ) 22 dx

= 0(y) 2x + y 2 e (2 x + y ) 22 dx + 0(y)

2x + y 2 e

(2 x + y ) 22 d(x)

= 2 |y | 2 e 22 d( 2)=

1 2 e

y 22 .

So both X and Y are standard normal random variables. Since f X,Y (x, y ) = f X (x)f Y (y), X and Y are not

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independent. However, if we set F (t) = t u 2 2 eu 22 du, we haveE {XY } = xyf X,Y (x, y )dxdy

=

xy1{y|x |}2|x|+ y 2 e

(2 |x |+ y ) 22 dxdy

= xdx |x | y 2|x|+ y 2 e (2 |x |+ y ) 22 dy= xdx |x | ( 2|x|) 2 e 22 d = xdx ( |x | 2 2 e 22 d 2|x|e

x 22

2 )

= 0 x x 2 2 e 22 ddx + 0x x 2 2 e 22 ddx=

0xF (x)dx +

0

xF (x)dx.

So E {XY }= 0 xF (x)dx 0 uF (u)du = 0.2.6. (i)

Proof. (X ) = {, , {a, b}, {c, d}}.(ii)

Proof.

E {Y |X }={a,b,c,d }

E {Y 1{X = }}P (X = )

1{X = }.

(iii)

Proof.

E {Z |X }= X + E {Y |X }= X +{a,b,c,d }

E {Y 1{X = }}P (X = )

1{X = }.

(iv)

Proof. E {Z |X }E {Y |X }= E {Z Y |X }= E {X |X }= X .2.7.

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Proof. Let = E {Y X }and = E {Y X |G}. Note is G-measurable, we haveV ar(Y X ) = E {(Y X )2}

= E {[(Y E {Y |G}) + ( E {Y |G}X )]2}= V ar(Err ) + 2 E {(Y E {Y |G}) }+ E { 2}= V ar(Err ) + 2 E

{Y

E

{Y

|G}

}+ E

{ 2

}= V ar(Err ) + E { 2} V ar(Err ).

2.8.

Proof. It suffices to prove the more general case. For any (X )-measurable random variable , E {Y 2 }=E {(Y E {Y |X }) }= E {Y E {Y |X } }= E {Y }E {Y }= 0.

2.9. (i)

Proof. Consider the dice-toss space similar to the coin-toss space. Then a typical element in this spaceis an innite sequence 123 , with i {1, 2, , 6} (i N ). We dene X () = 1 and f (x) =1{odd integers }(x). Then its easy to see(X ) = {, , { : 1 = 1 }, , { : 1 = 6 }}

and (f (X )) equals to

{, , { : 1 = 1 }{ : 1 = 3 }{ : 1 = 5 }, { : 1 = 2 }{ : 1 = 4 }{ : 1 = 6 }}.So {, } (f (X ))(X ), and each of these containment is strict.

(ii)

Proof. No. (f (X ))(X ) is always true.2.10.

Proof.

A g(X )dP = E {g(X )1B (X )}= g(x)1B (x)f X (x)dx= yf X,Y (x, y )f X (x) dy1B (x)f X (x)dx=

y1B (x)f X,Y (x, y )dxdy

= E {Y 1B (X )}= E {Y I A}= A Y dP.

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2.11. (i)

Proof. We can nd a sequence {W n }n 1 of (X )-measurable simple functions such that W n W . Each W ncan be written in the form K ni =1 a

ni 1A ni , where A

ni s belong to (X ) and are disjoint. So each Ani can be

written as {X B ni }for some Borel subset B ni of R , i.e. W n =K ni =1 a

ni 1{XB ni } =

K ni =1 a

ni 1B ni (X ) = gn (X ),

where gn (x) =K ni =1 a

ni 1B ni (x). Dene g = limsup gn , then g is a Borel function. By taking upper limits on

both sides of W n = gn (X ), we get W = g(X ).

(ii)

Proof. Note E {Y |X } is (X )-measurable. By (i), we can nd a Borel function g such that E {Y |X }=g(X ).3. Brownian Motion

3.1.

Proof. We have F t F u 1 and W u 2 W u 1 is independent of F u 1 . So in particular, W u 2 W u 1 is independentof F t .

3.2.Proof. E [W 2t W 2s |F s ] = E [(W t W s )2 + 2 W t W s 2W 2s |F s ] = t s + 2 W s E [W t W s |F s ] = t s.3.3.

Proof. (3) (u) = 2 4ue12

2 u 2 +( 2 + 4u2)2ue12

2 u 2 = e12

2 u 2 (34u+ 4u2), and (4) (u) = 2ue12

2 u 2 (34u+4u2) + e

12

2 u 2 (34 + 2 4u). So E [(X )4] = (4) (0) = 3 4 .3.4. (i)

Proof. Assume there exists A F , such that P (A) > 0 and for every A, limnn 1j =0 |W t j +1 W t j |() 0, Z t m em . By bounded convergence theorem,E [1{ m < }Z m ] = E [ limt

Z t m ] = limtE [Z t m ] = 1,

since on the event { m = }, Z t m em 12

2 t 0 as t . Therefore, E [em ( + 22 ) m ] = 1. Let

0, by bounded convergence theorem, we have P ( m < ) = 1. Let + 2

2 = , we get

E [e m ] = em = em m 2 + 2 .

(iv)Proof. We note for > 0, E [ m e m ] < since xex is bounded on [0 , ). So by an argument similarto Exercise 1.8, E [e m ] is differentiable and

E [e m ] = E [ m e m ] = em m 2 + 2 m 2 + 2 .

Let 0, by monotone increasing theorem, E [ m ] = m < for > 0.(v)

Proof. By > 2 > 0, we get + 2

2 > 0. Then Z t m em and on the event { m = }, Z t m em (

22 + ) t

0 as t

. Therefore,

E [em ( + 2

2 ) m 1{ m < }] = E [ limtZ t m ] = limt

E [Z t m ] = 1.

Let 2, then we get P ( m < ) = e2m = e2| |m < 1. Set = + 2

2 . So we get

E [e m ] = E [e m 1{ m < }] = em = em m 2 + 2 .

3.8. (i)

Proof.

n (u) = E [eu1

n M nt,n ] = (E [eu

n X 1 ,n ])nt = ( eu

n pn + eu

n q n )nt

= eu n

rn + 1 e

n

e n e

n+ e

u n rn 1 + e n

e n e

n

nt

.

(ii)

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Proof.

1x 2

(u) = euxrx 2 + 1 ex

ex ex eux rx

2 + 1 exex ex

tx 2

.

So,

ln 1x 2 (u) =t

x2 ln(rx 2 + 1)( eux

eux ) + e(u )x

e(u )x

ex ex=

tx2

ln(rx 2 + 1) sinh ux + sinh( u)x

sinh x

=t

x2ln

(rx 2 + 1) sinh ux + sinh x cosh ux cosh x sinh uxsinh x

=t

x2ln cosh ux +

(rx 2 + 1 cosh x) sinh uxsinh x

.

(iii)

Proof.

cosh ux +(rx 2 + 1 cosh x) sinh ux

sinh x

= 1 +u2x2

2+ O(x4) +

(rx 2 + 1 1 2 x 22 + O(x

4))( ux + O(x3))x + O(x3)

= 1 +u2x2

2+

(r 2

2 )ux3 + O(x5)

x + O(x3)+ O(x4)

= 1 +u2x2

2+

(r 2

2 )ux3(1 + O(x2))

x (1 + O(x2))+ O(x4)

= 1 +u2x2

2+

rux 2

1

2ux 2 + O(x4).

(iv)

Proof.

ln 1x 2

=t

x2ln(1 +

u2x2

2+

ru

x2 ux 2

2+ O(x4)) =

tx2

(u2x2

2+

ru

x2 ux 2

2+ O(x4)) .

So limx0 ln 1x 2 (u) = t(u 22 +

ru u2 ), and E [e

u 1 n M nt,n ] = n (u) 12 tu 2 + t( r 2 )u. By the one-to-onecorrespondence between distribution and moment generating function, ( 1 n M nt,n )n converges to a Gaussianrandom variable with mean t( r

2 ) and variance t. Hence (

n M nt,n )n converges to a Gaussian randomvariable with mean t(r 22 ) and variance 2 t.4. Stochastic Calculus

4.1.

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Proof. Fix t and for any s < t , we assume s[tm , t m +1 ) for some m.Case 1. m = k. Then I (t)I (s) = t k (M t M t k ) t k (M sM t k ) = t k (M tM s ). So E [I (t)I (s)|F t ] = t k E [M t M s |F s ] = 0.Case 2. m < k . Then tm s < t m +1 tk t < t k+1 . So

I (t)

I (s) =

k1

j = m

t j (M t j +1

M t j ) + t k (M s

M t k )

t m (M s

M t m )

=k1

j = m +1 t j (M t j +1 M t j ) + t k (M t M t k ) + t m (M t m +1 M s ).

Hence

E [I (t) I (s)|F s ]=

k1

j = m +1

E [ t j E [M t j +1 M t j |F t j ]|F s ] + E [ t k E [M t M t k |F t k ]|F s ] + t m E [M t m +1 M s |F s ]= 0 .

Combined, we conclude I (t) is a martingale.

4.2. (i)

Proof. We follow the simplication in the hint and consider I (tk ) I (t l ) with t l < t k . Then I (tk ) I (t l ) =k1j = l t j (W t j +1 W t j ). Since t is a non-random process and W t j +1 W t j F t j F t l for j l, we musthave I (tk ) I (t l ) F t l .(ii)

Proof. We use the notation in (i) and it is clear I (tk ) I (t l ) is normal since it is a linear combination of independent normal random variables. Furthermore, E [I (tk ) I (t l )] =

k1j = l t j E [W t j +1 W t j ] = 0 andV ar(I (tk ) I (t l )) =

k1j = l 2t j V ar(W t j +1 W t j ) =k1j = l 2t j (t j +1 t j ) = t kt l 2u du.(iii)

Proof. E [I (t) I (s)|F s ] = E [I (t) I (s)] = 0, for s < t .(iv)

Proof. For s < t ,

E [I 2(t) t0 2u du (I 2(s) s0 2u du)|F s ]= E [I 2(t) I 2(s) ts 2u du|F s ]= E [(I (t) I (s))2 + 2 I (t)I (s) 2I 2(s)|F s ] ts 2u du= E [(I (t) I (s))2] + 2I (s)E [I (t) I (s)|F s ]

t

s 2u du

= ts 2u du + 0 ts 2u du= 0 .

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4.3.

Proof. I (t) I (s) = 0(W t 1 W 0) + t 1 (W t 2 W t 1 ) 0(W t 1 W 0) = t 1 (W t 2 W t 1 ) = W s (W t W s ).(i) I (t) I (s) is not independent of F s , since W s F s .(ii) E [(I (t) I (s))4] = E [W 4s ]E [(W t W s )4] = 3s 3(t s) = 9 s(t s) and 3 E [(I (t) I (s))2] =3E [W 2s ]E [(W t W s )2] = 3s(t s). Since E [(I (t) I (s))4] = 3 E [(I (t) I (s)) 2], I (t) I (s) is not normallydistributed.(iii) E [I (t) I (s)|F s ] = W s E [W t W s |F s ] = 0.(iv)

E [I 2(t) t0 2u du (I 2(s) s0 2u du)|F s ]= E [(I (t) I (s))2 + 2 I (t)I (s) 2I 2(s) ts W 2u du|F s ]= E [W 2s (W t W s )2 + 2 W s (W t W s ) W 2s (t s)|F s ]= W 2s E [(W t W s )2] + 2W s E [W t W s |F s ]W 2s (t s)= W 2s (t s) W 2s (t s)= 0 .

4.4.

Proof. (Cf. ksendal [3], Exercise 3.9.) We rst note that

j

B t j + t j +12

(B t j +1 B t j )

=j

B t j + t j +12

(B t j +1 B t j + t j +12 ) + B t j (B t j + t j +12 B t j ) +j

(B t j + t j +12 B t j )2 .

The rst term converges in L2(P ) to

T

0 B t dB t . For the second term, we note

E j

B t j + t j +12 B t j

2

t2

2

= E j

B t j + t j +12 B t j

2

j

t j +1 t j2

2

=j, k

E B t j + t j +12 B t j

2

t j +1 t j

2B t k + t k +1

2 B t k2

tk+1 tk

2

=j

E B 2t j +1 t j2

t j +1 t j2

2

=j

2 t j +1 t j

2

2

T 2

max1j n |t j +1 t j | 0,

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since E [(B 2t t)2] = E [B 4t 2tB 2t + t2] = 3E [B 2t ]2 2t2 + t2 = 2 t2 . So

j

B t j + t j +12

(B t j +1 B t j ) T 0 B t dB t + T 2 = 12B 2T in L2(P ).4.5. (i)

Proof.

d ln S t =dS tS t

12

d S tS 2t

=2S t dS t d S t

2S 2t=

2S t ( t S t dt + t S t dW t ) 2t S 2t dt2S 2t

= t dW t + ( t 12

2t )dt.

(ii)

Proof.

ln S t = ln S 0 +

t

0

s dW s +

t

0

( s 12

2s )ds.

So S t = S 0 exp{ t0 s dW s + t0 ( s 12 2s )ds}.4.6.Proof. Without loss of generality, we assume p = 1. Since ( x p) = px p1 , (x p) = p( p 1)x p2 , we have

d(S pt ) = pS p1t dS t +

12 p( p1)S p2t d S t

= pS p1t (S t dt + S t dW t ) +12 p( p1)S p2t 2S 2t dt

= S pt [ pdt + pdW t +12 p( p 1)2dt]

= S pt p[dW t + ( +

p

1

2 2)dt].

4.7. (i)

Proof. dW 4t = 4 W 3t dW t +12 4 3W 2t d W t = 4 W 3t dW t + 6 W 2t dt . So W 4T = 4 T 0 W 3t dW t + 6 T 0 W 2t dt .(ii)

Proof. E [W 4T ] = 6 T 0 tdt = 3 T 2 .(iii)Proof. dW

6t = 6 W

5t dW t +

12 6 5W

4t dt. So W

6T = 6

T

0 W 5t dW t + 15

T

0 W 4t dt . Hence E [W

6T ] = 15

T

0 3t2

dt =15T 3 .

4.8.

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Proof. d(et R t ) = e t R t dt + et dR t = et (dt + dW t ) . Hence

et R t = R0 + t0 es (ds + dW s ) = R0 + (et 1) + t0 es dW s ,and R t = R0et + (1 et ) +

t

0 e ( ts ) dW s .

4.9. (i)

Proof.

Ke r (T t ) N (d) = Ke r (T t ) e

d 22

2= Ke r (T t ) e

( d + T t ) 22

2= Ke r (T t ) e T td + e

2 ( T t )2 N (d+ )

= Ke r (T t ) xK

e( r + 22 )( T t ) e

2 ( T t )2 N (d+ )

= xN (d+ ).

(ii)

Proof.

cx = N (d+ ) + xN (d+ )

xd+ (T t, x ) Ke r (T t ) N (d)

x

d(T t, x )= N (d+ ) + xN (d+ )

x

d+ (T t, x ) xN (d+ )

xd+ (T t, x )

= N (d+ ).

(iii)

Proof.

ct = xN (d+ )

xd+ (T t, x ) rKe r (T t ) N (d) Ke r (T t ) N (d)

t

d(T t, x )= xN (d+ )

t

d+ (T t, x ) rKe r (T t ) N (d) xN (d+ ) t

d+ (T t, x ) +

2 T t= rKe r (T t ) N (d)

x2 T t

N (d+ ).

(iv)

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Proof.

ct + rxc x +12

2x2cxx

= rKe r (T t ) N (d) x

2 T tN (d+ ) + rxN (d+ ) +

12

2x2N (d+ )

xd+ (T t, x )

= rc x2 T tN (d+ ) + 12

2x2N (d+ ) 1 T tx= rc.

(v)

Proof. For x > K , d+ (T t, x ) > 0 and lim tT d+ (T t, x ) = lim 0 d+ (, x ) = . limtT d(T t, x ) =lim 0 d(, x ) = lim 0

1 ln xK + 1 (r + 12 2) = . Similarly, lim tT d = for x

(0, K ). Also its clear that lim tT d = 0 for x = K . So

limtT

c(t, x ) = xN (limtT

d+ )

KN (lim

tT d

) =x K, if x > K 0, if x K

= ( x

K )+ .

(vi)

Proof. It is easy to see lim x0 d = . So for t[0, T ], limx0 c(t, x ) = lim x0 xN (lim xd+ (T t, x )) Ke r (T t ) N (lim x0 d(T t, x )) = 0.(vii)

Proof. For t[0, T ], it is clear lim xd = . Note

limx

x(N (d+ )

1) = lim

x

N (d+ ) x d+

x2 = lim

x

N (d+ ) 1 T t

x1 .

By the expression of d+ , we get x = K exp{ T td+ (T t)( r + 12 2)}. So we have

limx

x(N (d+ ) 1) = limxN (d+ ) x

T t= lim

d+ e

d 2+2

2 Ke T td + (T t )( r + 12 2 )

T t= 0 .

Therefore

limx

[c(t, x ) (x er (T t ) K )]= lim

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

= limx

[x(N (d+ )

1) + Ke r (T t ) (1

N (d

))]

= limx

x(N (d+ ) 1) + Ke r (T t ) (1 N ( limxd))= 0 .

4.10. (i)

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Proof. We show (4 .10.16) + (4 .10.9) (4.10.16) + (4 .10.15), i.e. assuming X has the representationX t = t S t + t M t , continuous-time self-nancing condition has two equivalent formulations, (4 .10.9) or(4.10.15). Indeed, dX t = t dS t + t dM t +( S t d t + dS t d t + M t dt + dM t dt ). So dX t = t dS t + t dM t S t d t + dS t d t + M t dt + dM t dt = 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 X t = t S t + t M t . We let c(t, S t ) denote the price of call option at time tand set t = cx (t, S t ). Finally, we assume the portfolio is self-nancing. The problem is to show

rN t dt = ct (t, S t ) +12

2S 2t cxx (t, S t ) dt,

where N t = c(t, S t ) t S t .Indeed, by the self-nancing property and t = cx (t, S t ), we have c(t, S t ) = X t (by the calculations inSubsection 4.5.1-4.5.3). This uniquely determines t as

t =X t t S t

M t=

c(t, S t ) cx (t, S t )S tM t

=N tM t

.

Moreover,

dN t = ct (t, S t )dt + cx (t, S t )dS t +12

cxx (t, S t )d S t t d( t S t )= ct (t, S t ) +

12

cxx (t, S t )2S 2t dt + [ cx (t, S t )dS t d(X t t M t )]

= ct (t, S t ) +12

cxx (t, S t )2S 2t dt + M t dt + dM t dt + [cx (t, S t )dS t + t dM t dX t ].By self-nancing property, cx (t, S t )dt + t dM t = t dS t + t dM t = dX t , so

ct (t, S t ) +1

2cxx (t, S t )2S 2t dt = dN t

M t dt

dM t dt = t dM t = t rM t dt = rN t dt.

4.11.

Proof. First, we note c(t, x ) solves the Black-Scholes-Merton PDE with volatility 1 :

t

+ rx

x+

12

x221 2

x 2 r c(t, x ) = 0 .So

ct (t, S t ) + rS t cx (t, S t ) +12

21 S 2t cxx (t, S t ) rc (t, S t ) = 0 ,

and

dc(t, S t ) = ct (t, S t )dt + cx (t, S t )(S t dt + 2S t dW t ) +12

cxx (t, S t )22 S 2t dt

= ct (t, S t ) + cx (t, S t )S t +12

22 S 2t cxx (t, S t ) dt + 2S t cx (t, S t )dW t

= rc (t, S t ) + ( r )cx (t, S t )S t +12

S 2t (22 21 )cxx (t, S t ) dt + 2S t cx (t, S t )dW t .

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Therefore

dX t = rc (t, S t ) + ( r )cx (t, S t )S t +12

S 2t (22 21 )xx (t, S t ) + rX t rc (t, S t ) + rS t cx (t, S t )

12

(22 21 )S 2t cxx (t, S t ) cx (t, S t )S t dt + [2S t cx (t, S t ) cx (t, S t )2S t ]dW t= rX t dt.

This implies X t = X 0ert . By X 0 , we conclude X t = 0 for all t[0, T ].

4.12. (i)

Proof. By (4.5.29), c(t, x ) p(t, x ) = x er (T t ) K . So px (t, x ) = cx (t, x ) 1 = N (d+ (T t, x )) 1, pxx (t, x ) = cxx (t, x ) = 1x T t N (d+ (T t, x )) and pt (t, x ) = ct (t, x ) + re r (T t ) K

= rKe r (T t ) N (d(T t, x )) x

2 T tN (d+ (T t, x )) + rKe r (T t )

= rKe r (T t ) N (

d

(T

t, x ))

x

2 T tN (d+ (T

t, x )) .

(ii)

Proof. For an agent hedging a short position in the put, since t = px (t, x ) < 0, he should short theunderlying stock and put p(t, S t ) px (t, S t )S t (> 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 ) satises the Black-Scholes-Mertonpartial differential equation. Indeed,

t+

1

22x2

2

x2 + rx

x r f (t, x ) =

rKe r (T t ) + 1

22x2

0 + rx

1

r (x

Ke r (T t ) ) = 0 .

Remark: The Black-Scholes-Merton PDE has many solutions. Proper boundary conditions are the keyto uniqueness. For more details, see Wilmott [8].

4.13.

Proof. We suppose ( W 1 , W 2) is a pair of local martingale dened by SDE

dW 1(t) = dB1(t)dW 2(t) = (t)dB1(t) + (t)dB2(t).

(1)

We want to nd (t) and (t) such that

(dW 2(t))2 = [ 2(t) + 2(t) + 2 (t)(t) (t)]dt = dtdW 1(t)dW 2(t) = [ (t) + (t)(t)]dt = 0 . (2)

Solve the equation for (t) and (t), we have (t) = 1 12 ( t ) and (t) = ( t ) 12 ( t ) . So

W 1(t) = B1(t)W 2(t) = t0 (s ) 12 ( s ) dB1(s) + t0 1 12 ( s ) dB2(s) (3)

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is a pair of independent BMs. Equivalently, we have

B1(t) = W 1(t)B2(t) = t0 (s)dW 1(s) + t0 1 2(s)dW 2(s). (4)

4.14. (i)

Proof. Clearly Z j F t j +1 . MoreoverE [Z j |F t j ] = f (W t j )E [(W t j +1 W t j )2 (t j +1 t j )|F t j ] = f (W t j )(E [W 2t j +1 t j ](t j +1 t j )) = 0 ,

since W t j +1 W t j is independent of F t j and W t N (0, t ). Finally, we haveE [Z 2j |F t j ] = [f (W t j )]2E [(W t j +1 W t j )4 2(t j +1 t j )(W t j +1 W t j )2 + ( t j +1 t j )2|F t j ]

= [ f (W t j )]2(E [W 4t j +1 t j ]2(t j +1 t j )E [W 2t j +1 t j ] + ( t j +1 t j )2)

= [ f (W t j )]2[3(t j +1 t j )2 2(t j +1 t j )2 + ( t j +1 t j )2]

= 2[ f (W t j )]2(t j +1

t j )2 ,

where we used the independence of Browian motion increment and the fact that E [X 4] = 3E [X 2]2 if X isGaussian with mean 0.

(ii)

Proof. E [ n 1j =0 Z j ] = E [n 1j =0 E [Z j |F t j ]] = 0 by part (i).

(iii)

Proof.

V ar[n 1

j =0

Z j ] = E [(n 1

j =0

Z j )2]

= E [n 1

j =0

Z 2j + 20i


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Proof. B i is a local martingale with

(dB i (t))2 =d

j =1

ij (t)i (t)

dW j (t)

2

=d

j =1

2ij (t)2i (t)

dt = dt.

So B i is a Brownian motion.(ii)

Proof.

dB i (t)dBk (t) =d

j =1

ij (t)i (t)

dW j (t)d

l=1

kl (t)k (t)

dW l (t)

=1j, l d

ij (t)kl (t)i (t)k (t)

dW j (t)dW l (t)

=d

j =1

ij (t)kj (t)i (t)k (t)

dt

= ik (t)dt.

4.16.

Proof. To nd the m independent Brownion motion W 1(t), , W m (t), we need to nd A(t) = ( a ij (t)) sothat(dB1(t), , dB m (t)) tr = A(t)(dW 1(t), , dW m (t)) tr ,

or equivalently

(dW 1(t), , dW m (t)) tr = A(t)1(dB1(t), , dB m (t)) tr ,and

(dW 1(t), , dW m (t)) tr (dW 1(t), , dW m (t))= A(t)1(dB1(t), , dB m (t)) tr (dB1(t), , dB m (t))( A(t)1)tr dt= I m m dt,

where I m m is the m m unit matrix. By the condition dB i (t)dBk (t) = ik (t)dt , we get(dB1(t), , dB m (t)) tr (dB1(t), , dB m (t)) = C (t).

So A(t)1C (t)(A(t)1) tr = I m m , which gives C (t) = A(t)A(t)tr . This motivates us to dene 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 X i :

X i (t) = X i (0) + t0 i (u)du + t0 i (u)dB i (u), i = 1 , 2.41


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The goal is to show

lim0

C ( )

V 1( )V 2( ) = (t0).First, for i = 1 , 2, we haveM i ( ) = E [X i (t0 + )

X i (t0)

|F t 0 ]

= E t 0 +

t 0i (u)du +

t 0 +

t 0i (u)dB i (u)|F t 0

= i (t0) + E t 0 +t 0 ( i (u) i (t0))du|F t 0 .By Conditional Jensens Inequality,

E t 0 +t 0 ( i (u) i (t0))du|F t 0 E t 0 +t 0 | i (u) i (t0)|du|F t 0Since 1

t 0 +

t 0 |i (u) i (t0)|du 2M and lim 0 1

t 0 +

t 0 | i (u) i (t0)|du = 0 by the continuity of i ,the Dominated Convergence Theorem under Conditional Expectation implies

lim0

1E t 0 +t 0 |i (u) i (t0)|du|F t 0 = E lim0 1 t 0 +t 0 | i (u) i (t0)|du|F t 0 = 0 .

So M i ( ) = i (t0) + o( ). This proves (iii).To calculate the variance and covariance, we note Y i (t) = t0 i (u)dB i (u) is a martingale and by It osformula Y i (t)Y j (t) t0 i (u)j (u)du is a martingale ( i = 1 , 2). So

E [(X i (t0 + ) X i (t0))( X j (t0 + ) X j (t0)) |F t 0 ]= E Y i (t0 + ) Y i (t0) + t 0 +t 0 i (u)du Y j (t0 + ) Y j (t0) + t 0 +t 0 j (u)du |F t 0= E [(Y i (t0 + )

Y i (t0)) ( Y j (t0 + )

Y j (t0))

|F t

0] + E

t 0 +

t 0i (u)du

t 0 +

t 0j (u)du

|F t

0

+ E (Y i (t0 + ) Y i (t0)) t 0 +t 0 j (u)du|F t 0 + E (Y j (t0 + ) Y j (t0)) t 0 +t 0 i (u)du|F t 0= I + II + III + IV.

I = E [Y i (t0 + )Y j (t0 + ) Y i (t0)Y j (t0)|F t 0 ] = E t 0 +t 0 i (u)j (u)ij (t)dt|F t 0 .By an argument similar to that involved in the proof of part (iii), we conclude I = i (t0)j (t0)ij (t0) + o( )and

II = E t 0 +t 0 ( i (u) i (t0))du t 0 +t 0 j (u)du|F t 0 + i (t0) E t 0 +t 0 j (u)du|F t 0= o( ) + ( M i ( ) o( ))M j ( )= M i ( )M j ( ) + o( ).

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By Cauchys inequality under conditional expectation (note E [XY |F ] denes an inner product on L2()),

III E |Y i (t0 + ) Y i (t0)| t 0 +t 0 |j (u)|du|F t 0 M

E [(Y i (t0 + ) Y i (t0))2|F t 0 ]

M E [Y i (t0 + )

2

Y i (t0)2

|F t 0 ] M E [ t 0 +t 0 i (u)2du|F t 0 ] M M = o( )

Similarly, IV = o( ). In summary, we have

E [(X i (t0 + ) X i (t0))( X j (t0 + ) X j (t0)) |F t 0 ] = M i ( )M j ( ) + i (t0)j (t0)ij (t0) + o( ) + o( ).This proves part (iv) and (v). Finally,

lim0C ( )

V 1( )V 2( ) = lim0(t0)1(t0)2(t0) + o( )

(21 (t0) + o( ))( 22 (t0) + o( )) = (t0).

This proves part (vi). Part (i) and (ii) are consequences of general cases.

4.18. (i)

Proof.d(ert t ) = ( deW t

12

2 t ) = eW t 12

2 t dW t = (ert t )dW t ,where for the second =, we used the fact that eW t 12 2 t solves dX t = X t dW t . Since d(ert t ) =re rt t dt + ert d t , we get d t = t dW t r t dt .

(ii)

Proof.

d( t X t ) = t dX t + X t d t + dX t d t= t (rX t dt + t ( r )S t dt + t S t dW t ) + X t ( t dW t r t dt)

+( rX t dt + t ( r )S t dt + t S t dW t )( t dW t r t dt)= t ( t ( r )S t dt + t S t dW t ) X t t dW t t S t t dt= t t S t dW t X t t dW t .

So t X t is a martingale.

(iii)

Proof. By part (ii), X 0 = 0X 0 = E [ T X t ] = E [ T V T ]. (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. B t is a local martingale with [ B ]t = t0 sign (W s )2ds = t. So by Levys theorem, B t is a Brownianmotion.43


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(ii)

Proof. d(B t W t ) = B t dW t + sign (W t )W t dW t + sign (W t )dt . Integrate both sides of the resulting equationand the expectation, we get

E [B t W t ] =

t

0

E [sign (W s )]ds =

t

0

E [1{W s 0}1{W s < 0}]ds =12

t 12

t = 0 .

(iii)

Proof. By It os formula, dW 2t = 2 W t dW t + dt .

(iv)

Proof. By It os formula,

d(B t W 2t ) = B t dW 2t + W

2t dB t + dB t dW

2t

= B t (2W t dW t + dt) + W 2t sign (W t )dW t + sign (W t )dW t (2W t dW t + dt)= 2 B t W t dW t + B t dt + sign (W t )W 2t dW t + 2 sign (W t )W t dt.

So

E [B t W 2t ] = E [ t0 B s ds] + 2E [ t0 sign (W s )W s ds]= t0 E [B s ]ds + 2 t0 E [sign (W s )W s ]ds= 2 t0 (E [W s 1{W s 0}]E [W s 1{W s < 0}])ds= 4 t0 0 x ex

22 s

2s dxds

= 4 t

0 s2 ds= 0 = E [B t ] E [W 2t ].Since E [B t W 2t ] = E [B t ] E [W 2t ], B t and W t are not independent.4.20. (i)

Proof. f (x) =x K, if x K 0, if x < K.

So f (x) =1, if x > K undened , if x = K 0, if x < K

and f (x) =0, if x = K undened , if x = K.

(ii)

Proof. E [f (W T )] = K (x K ) ex 22 T

2T dx = T 2 eK 22 T K ( K T ) where is the distribution function of standard normal random variable. If we suppose T 0 f (W t )dt = 0, the expectation of RHS of (4.10.42) isequal to 0. So (4.10.42) cannot hold.(iii)

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Proof. This is trivial to check.

(iv)

Proof. If x = K , limn f n (x) =1

8n = 0; if x > K , for n large enough, x K + 12n , so limn f n (x) =limn (x K ) = x K ; if x < K , for n large enough, x K 12n , so lim n f n (x) = lim n 0 = 0. Insummary, lim n

f n (x) = ( x

K )+ . Similarly, we can show

limn

f n (x) =0, if x < K 12 , if x = K 1, if x > K.

(5)

(v)

Proof. Fix , so that W t () < K for any t [0, T ]. Since W t () can obtain its maximum on [0 , T ], thereexists n0 , so that for any n n0 , max 0tT W t () < K 12n . So

LK (T )() = limn

n

T

01(K

1

2 n ,K +1

2 n ) (W t ())dt = 0 .

(vi)

Proof. Take expectation on both sides of the formula (4.10.45), we have

E [LK (T )] = E [(W T 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 [(S T K )+ ] > 0. SoX T = ( S T K )+ .5. Risk-Neutral Pricing

5.1. (i)

Proof.

df (X t ) = f (X t )dt +12

f (X t )d X t

= f (X t )(dX t + 12d X t )

= f (X t ) t dW t + ( t R t 12

2t )dt +12

2t dt

= f (X t )( t R t )dt + f (X t )t dW t .This is formula (5.2.20).

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(ii)

Proof. d(D t S t ) = S t dD t + D t dS t + dD t dS t = S t R t D t dt + D t t S t dt + D t t S t dW t = D t S t ( t R t )dt +D t S t t dW t . This is formula (5.2.20).5.2.

Proof. By Lemma 5.2.2., E [DT V T |F t ] = E D T V T Z T

Z t |F t . Therefore (5.2.30) is equivalent to D t V t Z t =E [DT V T Z T |F t ].5.3. (i)

Proof.

cx (0, x) =d

dxE [erT (xe W T +( r 12 2 )T K )+ ]

= E erT ddx

h(xe W T +( r 12 2 )T )

= E erT e W T +( r 12 2 )T 1{xe W T +( r

12

2 ) T >K }= e

1

2 2 T

E e W T

1{W T > 1 (ln Kx ( r 12 2 )T )}= e12 2 T E e T

W T T 1{

W T T T > 1 T (ln Kx ( r 12 2 )T ) T }= e12 2 T 1 2 ez 22 e T z 1{z T > d+ (T,x )}dz= 1 2 e ( z T ) 22 1{z T > d+ (T,x )}dz= N (d+ (T, x)) .

(ii)

Proof. If we set Z T = e W T 12

2 T and Z t = E [ Z T |F t ], then Z is a P -martingale, Z t > 0 and E [ Z T ] =E [e W T 12 2 T ] = 1. So if we dene P by d P = Z T dP on F T , then P is a probability measure equivalent toP , and cx (0, x) = E [ Z T 1{S T >K }] = P (S T > K ).Moreover, by Girsanovs Theorem, W t = W t + t0 ()du = W t t is a P -Brownian motion (set = in Theorem 5.4.1.)(iii)

Proof. S T = xe W T +( r 12

2 )T = xe W T +( r + 12 2 )T . So

P (S T > K ) = P (xe W T +( r + 12 2 )T > K ) = P W

T T > d+ (T, x) = N (d+ (T, x)) .

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5.4. First, a few typos. In the SDE for S , (t)dW (t) (t)S (t)dW (t). In the rst equation forc(0, S (0)), E E . In the second equation for c(0, S (0)), the variable for BSM should be

BSM T, S (0); K,1T T 0 r (t)dt, 1T T 0 2(t)dt .

(i)

Proof. d ln S t = dS tS t 12S 2t d S t = r t dt + t dW t 12 2t dt . So S T = S 0 exp{ T 0 (r t 12 2t )dt + T 0 t dW t}. LetX = T 0 (r t 12 2t )dt + T 0 t dW t . The rst term in the expression of X is a number and the second termis a Gaussian random variable N (0, T 0 2t dt), since both r and ar deterministic. Therefore, S T = S 0eX ,with X N ( T 0 (r t 12 2t )dt, T 0 2t dt),.(ii)Proof. For the standard BSM model with constant volatility and interest rate R, under the risk-neutralmeasure, we have S T = S 0eY , where Y = ( R 12 2)T + W T N ((R 12 2)T, 2T ), and E [(S 0eY K )+ ] =eRT BSM (T, S 0 ; K,R, ). Note R = 1T (E [Y ] +

12 V ar(Y )) and =

1T V ar(Y ), we can get

E [(S 0eY K )+ ] = eE [Y ]+12 V ar (Y ) BSM T, S 0 ; K,

1T

E [Y ] +12

V ar(Y ) , 1T V ar(Y ) .So for the model in this problem,

c(0, S 0) = e T 0 r t dt E [(S 0eX K )+ ]= e T 0 r t dt eE [X ]+ 12 V ar (X ) BSM T, S 0 ; K, 1T E [X ] +

12

V ar(X ) , 1T V ar(X )= BSM T, S 0 ; K,

1

T T

0

r t dt,

1

T T

0

2t dt .

5.5. (i)

Proof. Let f (x) = 1x , then f (x) = 1x 2 and f (x) = 2x 3 . Note dZ t = Z t t dW t , so

d1

Z t= f (Z t )dZ t +

12

f (Z t )dZ t dZ t = 1

Z 2t(Z t ) t dW t +

12

2Z 3t

Z 2t 2t dt =

tZ t

dW t +2tZ t

dt.

(ii)

Proof. By Lemma 5.2.2., for s, t 0 with s < t , M s = E [M t |F s ] = E Z t M tZ s |F s . That is, E [Z t M t |F s ] =Z s M s . So M = Z M is a P -martingale.

(iii)

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Proof.

dM t = d M t 1

Z t=

1Z t

dM t + M t d1

Z t+ dM t d

1Z t

=tZ t

dW t +M t t

Z tdW t +

M t 2tZ t

dt +t t

Z tdt.

(iv)Proof. In part (iii), we have

dM t =tZ t

dW t +M t t

Z tdW t +

M t 2tZ t

dt +t t

Z tdt =

tZ t

(dW t + t dt) +M t t

Z t(dW t + t dt).

Let t = t + M t tZ t , then dM t = t dW t . This proves Corollary 5.3.2.

5.6.

Proof. By Theorem 4.6.5, it suffices to show W i (t) is an F t -martingale under P and [W i , W j ](t) = t ij(i, j = 1 , 2). Indeed, for i = 1 , 2, W i (t) is an F t -martingale under P if and only if W i (t)Z t is an F t -martingaleunder P , since

E [W i (t)|F s ] = E W i (t)Z tZ s |F s .

By It os product formula, we have

d(W i (t)Z t ) = W i (t)dZ t + Z t dW i (t) + dZ t dW i (t)

= W i (t)(Z t )( t) dW t + Z t (dW i (t) + i (t)dt) + ( Z t t dW t )(dW i (t) + i (t)dt)= W i (t)(Z t )

d

j =1

j (t)dW j (t) + Z t (dW i (t) + i (t)dt) Z t i (t)dt

= W i (t)(Z t )d

j =1

j (t)dW j (t) + Z t dW i (t)

This shows W i (t)Z t is an F t -martingale under P . So W i (t) is an F t -martingale under P . Moreover,[W i , W j ](t) = W i + 0 i (s)ds,W j + 0 j (s)ds (t) = [ W i , W j ](t) = t ij .

Combined, this proves the two-dimensional Girsanovs Theorem.

5.7. (i)

Proof. Let a be any strictly positive number. We dene X 2(t) = ( a + X 1(t))D (t)1 . Then

P X 2(T ) X 2(0)D (T )

= P (a + X 1(T ) a) = P (X 1(T ) 0) = 1 ,

and P X 2(T ) > X 2 (0)D (T ) = P (X 1(T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem.Remark : The intuition is that we invest the positive starting fund a into the money market account,

and construct portfolio X 1 from zero cost. Their sum should be able to beat the return of money marketaccount.

(ii)

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Proof. We dene X 1(t) = X 2(t)D (t) X 2(0). Then X 1(0) = 0,

P (X 1(T ) 0) = P X 2(T ) X 2(0)D(T )

= 1 , P (X 1(T ) > 0) = P X 2(T ) >X 2(0)D (T )

> 0.

5.8. The basic idea is that for any positive P -martingale M , dM t = M t 1M t dM t . By Martingale Repre-sentation Theorem, dM t = t dW t for some adapted process t . So dM t = M t ( tM t )dW t , i.e. any positivemartingale must be the exponential of an integral w.r.t. Brownian motion. Taking into account discountingfactor and apply It os product rule, we can show every strictly positive asset is a generalized geometricBrownian motion.

(i)

Proof. V t D t = E [e T 0 R u du V T |F t ] = E [DT V T |F t ]. So (D t V t ) t0 is a P -martingale. By Martingale Represen-tation Theorem, there exists an adapted process t , 0 t T , such that D t V t = t0 s dW s , or equivalently,V t = D1t t0 s dW s . Differentiate both sides of the equation, we get dV t = R t D1t t0 s dW s dt + D1t t dW t ,i.e. dV t = R t V t dt + tD t dW t .

(ii)Proof. We prove the following more general lemma.

Lemma 1. Let X be an almost surely positive random variable (i.e. X > 0 a.s.) dened on the probability space (, G, P ). Let F be a sub -algebra of G, then Y = E [X |F ] > 0 a.s.Proof. By the property of conditional expectation Y t 0 a.s. Let A = {Y = 0 }, we shall show P (A) = 0. In-deed, note A F , 0 = E [Y I A ] = E [E [X |F ]I A ] = E [XI A ] = E [X 1A{X 1}] + n =1 E [X 1A{1n >X 1n +1 }] P (A{X 1})+ n =1 1n +1 P (A{1n > X 1n +1 }). So P (A{X 1}) = 0 and P (A{1n > X 1n +1 }) = 0,n 1. This in turn implies P (A) = P (A {X > 0}) = P (A {X 1}) + n =1 P (A {1n > X 1n +1 }) =0.

By the above lemma, it is clear that for each t [0, T ], V t = E [e T

t R u du V T |F t ] > 0 a.s.. Moreover,by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3.4)), we have thefollowing stronger result: for a.s. , V t () > 0 for any t[0, T ].

(iii)

Proof. By (ii), V > 0 a.s., so dV t = V t 1V t dV t = V t1

V t R t V t dt + tD t dW t = V t R t dt + V t

tV t D t dW t = R t V t dt +

t V t dW t , where t = tV t D t . 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 = 1 T (ln xK + ( r 12 2)T ). Let f (y) = 1 2 ey 22 ,

then f (y) = yf (y),

cK (0,T,x ,K ) = xf (d+ ) d+y erT N (d) Ke rT f (d) dy= xf (d+ ) 1

T K erT N (d) + e

rT f (d)1

T ,

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and

cKK (0,T,x ,K )

= xf (d+ )1

T K 2 x

T K f (d+ )(d+ )d+y e

rT f (d)dy

+erT T (d)f (d)

dy

=x

T K 2 f (d+ ) +xd+

T K f (d+ ) 1

K T erT

f (d) 1

K T erT d

T f (d) 1

K T = f (d+ )

xK 2 T [1

d+ T ] +

erT f (d)K T [1 +

d T ]

=erT

K 2T f (d)d+

xK 22T

f (d+ )d.

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) + ( er (T t 0 ) K S (t0))+ .(ii)

Proof. By the risk-neutral pricing formula, V (0) = E [ert 0 V (t0)] = E [ert 0 C (t0)+( erT K ert 0 S (t0)+ ] =C (0) + E [ert 0 (er (T t 0 ) K S (t0))+ ]. The rst term is the value of a call expiring at time T with strikeprice K and the second term is the value of a put expiring at time t0 with strike price er (T t 0 ) K .

5.11.

Proof. We rst make an analysis which leads to the hint, then we give a formal proof.(Analysis) If we want to construct a portfolio X that exactly replicates the cash ow, we must nd a

solution to the backward SDE

dX t = t dS t + R t (X t

t S t )dt

C t dt

X T = 0 .

Multiply D t on both sides of the rst equation and apply It os product rule, we get d(D t X t ) = t d(D t S t ) C t D t dt . Integrate from 0 to T , we have DT X T D0X 0 = T 0 t d(D t S t ) T 0 C t D t dt. By the terminalcondition, we get X 0 = D10 ( T 0 C t D t dt T 0 t d(D t S t )). X 0 is the theoretical, no-arbitrage price of the cash ow, provided we can nd a trading strategy that solves the BSDE. Note the SDE for S gives d(D t S t ) = ( D t S t )t (t dt + dW t ), where t = t R t t . Take the proper change of measure so thatW t = t0 s ds + W t is a Brownian motion under the new measure P , we get

T 0 C t D t dt = D0X 0 + T 0 t d(D t S t ) = D0X 0 + T 0 t (D t S t )t dW t .This says the random variable

T 0 C t D t dt has a stochastic integral representation D0X 0 + T 0 t D t S t t dW t .

This inspires us to consider the martingale generated by T 0 C t D t dt , so that we can apply Martingale Rep-resentation Theorem and get a formula for by comparison of the integrands.

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(Formal proof) Let M T = T 0 C t D t dt , and M t = E [M T |F t ]. Then by Martingale Representation Theo-rem, we can nd an adapted process t , so that M t = M 0 + t0 t dW t . If we set t = tD t S t t , we can checkX t = D1t (D0X 0 + t0 u d(Du S u ) t0 C u Du du), with X 0 = M 0 = E [ T 0 C t D t dt] solves the SDEdX t = t dS t + R t (X t t S t )dt C t dtX T = 0 .

Indeed, it is easy to see that X satises the rst equation. To check the terminal condition, we noteX T DT = D0X 0 + T 0 t D t S t t dW t T 0 C t D t dt = M 0 + T 0 t dW t M T = 0. So X T = 0. Thus, we havefound a trading strategy , so that the corresponding portfolio X replicates the cash ow and has zeroterminal value. So X 0 = E [ T 0 C t D t dt] is the no-arbitrage price of the cash ow at time zero.Remark : As shown in the analysis, d(D t X t ) = t d(D t S t ) C t D t dt . Integrate from t to T , we get0 D t X t = T t u d(Du S u ) T t C u Du du. Take conditional expectation w.r.t. F t on both sides, we getD t X t = E [ T t C u Du du|F t ]. So X t = D1t E [ T t C u Du du|F t ]. This is the no-arbitrage price of the cashow at time t, and we have justied formula (5.6.10) in the textbook.5.12. (i)

Proof. dB i (t) = dB i (t) + i (t)dt =dj =1

ij

( t ) i ( t ) dW j (t) +

dj =1

ij

( t ) i ( t ) j (t)dt =

dj =1

ij

( t ) i ( t ) dW j (t). So B i is a

martingale. Since dB i (t)dB i (t) =dj =1

ij ( t ) 2

i ( t ) 2 dt = dt, by Levys Theorem, B i is a Brownian motion underP .

(ii)

Proof.

dS i (t) = R(t)S i (t)dt + i (t)S i (t)dB i (t) + ( i (t) R(t))S i (t)dt i (t)S i (t) i (t)dt= R(t)S i (t)dt + i (t)S i (t)dB i (t) +

d

j =1

ij (t) j (t)S i (t)dt S i (t)d

j =1

ij (t) j (t)dt

= R(t)S i (t)dt + i (t)S i (t)dB i (t).

(iii)

Proof. dB i (t)dBk (t) = ( dB i (t) + i (t)dt)(dB j (t) + j (t)dt) = dB i (t)dB j (t) = ik (t)dt.

(iv)

Proof. By It os product rule and martingale property,

E [B i (t)Bk (t)] = E [ t0 B i (s)dBk (s)] + E [ t0 Bk (s)dB i (s)] + E [ t0 dB i (s)dBk (s)]= E [

t

0ik (s)ds] =

t

0ik (s)ds.

Similarly, by part (iii), we can show E [B i (t)Bk (t)] = t0 ik (s)ds.(v)51


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Proof. By It os product formula,

E [B1(t)B2(t)] = E [ t0 sign (W 1(u))du] = t0 [P (W 1(u) 0) P (W 1(u) < 0)]du = 0 .Meanwhile,

E [B1(t)B2(t)] = E [ t

0sign (W 1(u))du

= t0 [P (W 1(u) 0) P (W 1(u) < 0)]du= t0 [P (W 1(u) u) P (W 1(u) < u )]du= t0 2 12 P (W 1(u) < u ) du< 0,

for any t > 0. So E [B1(t)B2(t)] = E [B1(t)B2(t)] for all t > 0.

5.13. (i)

Proof. E [W 1(t)] = E [W 1(t)] = 0 and E [W 2(t)] = E [W 2(t) t0 W 1(u)du] = 0, for all t[0, T ].(ii)Proof.

Cov[W 1(T ), W 2(T )] = E [W 1(T )W 2(T )]

= E T 0 W 1(t)dW 2(t) + T 0 W 2(t)dW 1(t)= E

T

0W 1(t)(dW 2(t)

W 1(t)dt) + E

T

0W 2(t)dW 1(t)

= E T 0 W 1(t)2dt= T 0 tdt=

12

T 2 .

5.14. Equation (5.9.6) can be transformed into d(ert X t ) = t [d(ert S t ) aert dt] = t ert [dS t rS t dt adt ]. So, to make the discounted portfolio value ert X t a martingale, we are motivated to change the measurein such a way that S t r

t0 S u duat is a martingale under the new measure. To do this, we note the SDE for S is dS t = t S t dt + S t dW t . Hence dS t rS t dtadt = [( t r )S t a]dt + S t dW t = S t ( t r )S t aS t dt + dW t .Set t = ( t r )S t aS t and W t = t0 s ds + W t , we can nd an equivalent probability measure P , under whichS satises the SDE dS t = rS t dt + S t dW t + adt and W t is a BM. This is the rational for formula (5.9.7).

This is a good place to pause and think about the meaning of martingale measure. What is to be a martingale? The new measure P should be such that the discounted value process of the replicating

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portfolio is a martingale, no

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