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 yz44@cornell.edu.
The current version omits the following problems. Volume I: 1.5, 3.3, 3.4, 5.7; Volume II: 3.9, 7.1, 7.2,
7.5–7.9, 10.8, 10.9, 10.10.
Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this
solution manual and communicating to me several mistakes/typos.
1
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 are chosen in such a way that V1 can be
payoff of the derivative security at time 1. Suppose X̄0 and ∆
¯
¯
replicated: (1 + r)(X̄0 − ∆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 S0 − Γ0 X̄0 in a money market account. At time one, the value of the agent’s portfolio of stock, option
and money market assets is
X1 = ∆0 S1 + Γ0 V1 − (1 + r)(∆0 S0 + Γ0 X̄0 ).
Plug in the expression of V1 and sort out terms, we have
¯ 0 Γ0 )(
X1 = S0 (∆0 + ∆
S1
− (1 + r)).
S0
Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X̄0 ,
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
p̃Vn+1 (H) + q̃Vn+1 (T )
Vn+1 (H) − Vn+1 (T )
(d − 1 − r) + (1 + r)
=
u−d
1+r
= p̃(Vn+1 (T ) − Vn+1 (H)) + p̃Vn+1 (H) + q̃Vn+1 (T )
=
= p̃Vn+1 (T ) + q̃Vn+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 p̃n = u1−d
= 12 and q̃n = 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)
Proof. P̃ (A) =
P
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
sn X
−e
= |XeθX | = XeθX ≤ Xe2tX . The last inequality is by X ≥ 0 and the
Proof. By (1.9.1), |Yn | = e t−s
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
Proof. First, Ẑ = eθY − 2 = eθ(X+θ)− 2 = e 2 +θX = Z −1 . Second, for any A ∈ F, P̂ (A) = A ẐdPe =
R
R
(1A Ẑ)ZdP = 1A dP = P (A). So P = P̂ . In particular, X is standard normal under P̂ , 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 (ω̄)
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 (ω̄)
2
X 2 (ω̄)
− 2
= e−
Y 2 (ω̄)−X 2 (ω̄)
2
= e−
(X(ω̄)+θ)2 −X 2 (ω̄)
2
= e−θX(ω̄)−
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
Z t0 +
(Θi (u) − Θi (t0 ))du|Ft0 ≤ E
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 Itô’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 Lévy’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 Itô’s formula, dWt2 = 2Wt dWt + dt.
(iv)
Proof. By Itô’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 Itô’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 Itô’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 Itô’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 Lévy’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 Itô’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 Itô’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 Itô’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
. Itô’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 Itô’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−
Let X̄u = e−
Ru
t
bv dv
Xu , āu = e−
Ru
Ru
t
t
bv dv
bv dv
) = e−
Ru
t
bv dv
(au du + (γu + σu Xu )dWu ).
au and γ̄u = e−
Ru
t
bv dv
γu , then X̄ satisfies the SDE
dX̄u = āu du + (γ̄u + σu X̄u )dWu = (āu du + γ̄u dWu ) + σu X̄u dWu .
To deal with the term σu X̄u dWu , we consider X̂u = X̄u e−
dX̂u
=
e−
Ru
t
σv dWv
Ru
t
σv dWv
[(āu du + γ̄u dWu ) + σu X̄u dWu ] + X̄u e−
+(γ̄u + σu X̄u )(−σu )e−
Ru
t
σv dWv
Ru
t
. Then
σv dWv
1 Ru
(−σu )dWu + e− t σv dWv σu2 du
2
du
1
= âu du + γ̂u dWu + σu X̂u dWu − σu X̂u dWu + X̂u σu2 du − σu (γ̂u + σu X̂u )du
2
1
2
= (âu − σu γ̂u − X̂u σu )du + γ̂u dWu ,
2
where âu = āu e−
Ru
t
σv dWv
and γ̂u = γ̄u e−
Ru
t
σv dWv
. Finally, use the integrating factor e
Ru
t
1 2
2 σv dv
, we have
Ru 2 Ru 2
Ru 2
1
1
1
1
d X̂u e 2 t σv dv = e 2 t σv dv (dX̂u + X̂u · σu2 du) = e 2 t σv dv [(âu − σu γ̂u )du + γ̂u dWu ].
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 Lévy’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.
2r
x − σ2 −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 Itô’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 <L∗ } (rK − aSt )dt + e−rt vL
(St )σSt dW
∗
68
Following the reasoning in the proof of Theorem 8.3.5, we only need to show 1{x<L∗ } (rK − ax) ≥ 0 to finish
γK
into the expression and
the solution. This is further equivalent to proving rK − aL∗ ≥ 0. Plug L∗ = γ+1
q
note γ ≥ σ1 σ12 (r − a − 12 σ 2 )2 + σ12 (r − a − 12 σ 2 ) ≥ 0, the inequality is further reduced to r(γ + 1) − aγ ≥ 0.
We prove this inequality as follows.
Assume for some K, r, a and σ (K and σ are assumed to be strictly positive, r and a are assumed to be
γK
≤ K. As shown before, this means
non-negative), rK − aL∗ < 0, then necessarily r < a, since L∗ = γ+1
q
1
1
1 2
1
1 2 2
r(γ + 1) − aγ < 0. Define θ = r−a
,
then
θ
<
0
and
γ
=
(r
−
a
−
σ
)
+
2
σ
σ
2
σ
σ 2 (r − a − 2 σ ) + 2r =
q
1
1
1
(θ − 12 σ)2 + 2r. We have
σ (θ − 2 σ) + σ
r(γ + 1) − aγ < 0
⇐⇒
⇐⇒
(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 Numéraire
To provide an intuition for change of numéraire, we give a summary of results for change of
numéraire in discrete case. This summary is based on Shiryaev [5].
e B̄, S) as in [1] Definition 2.1.1 or [5] page 383. Here B
e and
Consider a model of financial market (B,
e and B̄ are both strictly
B̄ are both one-dimensional while S could be a vector price process. Suppose B
positive, then both of them can be chosen as numéaire.
Several results hold under this model. First, no-arbitrage and completeness properties of market are
independent of the choice of numéraire (see, for example, Shiryaev [5] page 413 Remark and page 481).
e (resp. B̄), there is an equivalent probability
Second, if the market is arbitrage-free,
then
corresponding
to B
e
S
B
S
B̄
,
(resp.
Pe (resp. P̄ ), such that
,
) is a martingale under Pe (resp. P̄ ). Third, if the market is
e
B
B̄
e
B
B̄
both arbitrage-free and complete, we have the relation
dP̄ =
B̄T
1
h i dPe.
e
BT E B̄e0
B
0
Finally, if fT is a European contingent claim with maturity N and the market is both arbitrage-free and
complete, then
fT
et E
e fT |Ft .
B̄t Ē
|Ft = B
eT
B̄T
B
That is, the price of fT is independent of the choice of numéraire.
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 numéraire 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 Lévy’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 Itô’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 + τ̄ ) and A(t, t + τ̄ ) aare constants
when τ̄ is fixed. So
d
Lt
dt
= −
=
=
B(t, t + τ̄ )[−C(t, t + τ̄ )R0 (t) − A(t, t + τ̄ )]
τ̄ B(t, t + τ̄ )
1
[C(t, t + τ̄ )R0 (t) + A(t, t + τ̄ )]
τ̄
1
[C(0, τ̄ )R0 (t) + A(0, τ̄ )].
τ̄
Hence L(t2 )−L(t1 ) = τ̄1 C(0, τ̄ )[R(t2 )−R(t1 )]+ τ̄1 A(0, τ̄ )(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 (T̄ −T )Y1 (T )−C2 (T̄ −T )Y2 (T )−A(T̄ −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 Itô’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 Itô’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<u≤t
Z t
X
=
eXu [−λ1 (eu1 − 1) − λ2 (eu2 − 1)]du +
(eXu − eXu− ).
s
0<u≤t
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<u≤t
t
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 . Itô’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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