MA547 Homework 7
Rt
11. From Xt = e−αt x + e−αt
eαs dWs , X0 = x, we get
Z t
−αt
−αt
eαs dWs + e−αt eαt dWt
dXt = −αe xdt − αe dt
0
Z t
eαs dWs )dt + dWt
= −α(e−αt x + e−αt
0
0
= −αXt dt + dWt .
Now EXt = E[e−αt x] = e−αt x, and
E[Xt2 ]
−αt
−αt
Z
t
eαs dWs ]2
0
Z t
−2αt 2
−2αt
=e
x +0+e
E[( eαs dWs )2 ]
0
Z t
e2αs ds
= e−2αt x2 + e−2αt
= E[e
x+e
0
1
(1 − e−2αt ),
= e−2αt x2 +
2α
so
V [Xt ] = E[Xt2 ] − [EXt ]2 =
1
(1 − e−2αt ).
2α
√
√
12. If α = 0, we have drt = −βrt dt + σ rt dWt . Let f (rt ) = rt , then
1
df (rt ) = fr drt + frr drt drt
2
√
1
1 −3/2 2
= √ (−βrt dt + σ rt dWt ) − rt
σ rt dt
2 rt
8
√
β rt
σ2
σ
= (−
− √ )dt + dWt
2
8 rt
2
= (−
βf
σ2
σ
−
)dt + dWt .
2
8f
2
1
From
du
σ2
(t) = −βu(t) − u(t)2 , u(0) = θ,
dt
2
we get
du(T − t)
σ2
= βu(T − t) + u(T − t)2 .
dt
2
Let f (t, rt ) = e−u(T −t)rt , then
1
df (t, rt ) = ft dt + fr drt + frr drt drt
2
1
= fu u0 dt + fr drt + frr drt drt
2
σ2
= −rt e−u(T −t)rt (βu(T − t) + u(T − t)2 )dt
2
√
− u(T − t)e−u(T −t)rt (−βrt dt + σ rt dWt )
1
+ u(T − t)2 e−u(T −t)rt σ 2 rt dt
2
√
= −u(T − t)e−u(T −t)rt σ rt dWt .
There is no drift term, so f (t, rt ) = e−u(T −t)rt is a martingale, that is
MrT (θ) = E[e−u(T −t)rt ] = constant = e−u(T )r0 = E[e−θrT ].
First solve the ODE for u (Hint: divide both sides by u2 , let y =
get y 0 = ay + b), we get
θe−βt
,
1 + γθ − γθe−βt
σ2
u(0) = θ, γ =
.
2β
u(t) =
So
MrT (θ) = E[e−θrT ] = e−u(T )r0 = exp(−
θe−βT
r0 ).
1 + γθ − γθe−βT
Now
E[rT ] = −Mr0 T (θ)|θ=0 = e−βT r0 ,
E[rT2 ] = Mr00T (θ)|θ=0 = 2γ(1 − e−βT )e−βT r0 ,
V ar[rt ] = E[rT2 ] − E[rT ]2 =
2
σ2
(1 − e−βT )e−βT r0 − e−2βT r02 .
β
1
u,
Finally
P (rT = 0) = P (0rT = 1) = E[0rT ]
= E[(e−∞ )rT ] = MrT (∞)
r0 e−βT
= exp(−
)
γ(1 − e−βT )
2βr0 e−βT
.
= exp − 2
σ (1 − e−βT )
14. Assume that St = f (t, Wt ). From Ito’s formula, we have
1
dSt = ft dt + fx dWt + fxx dt.
2
Since dSt = σSt dWt , we have
1
ft + fxx = 0
2
fx = σf,
From the above, we get
St = eσWt b (t) ,
where b(t) satisfies
1
b0 (t) + b (t) σ 2 = 0.
2
1
1
2
2
Solve it to get b (t) = Ke− 2 σ t . So St = KeσWt − 2 σ t . Suppose initially
1 2
we have S0 , then St = S0 eσWt − 2 σ t . Now we have
Z t
Z t
Z t
Su du =
E[Su ]du =
S0 du = S0 t.
E[Yt ] = E
0
0
Also notice that St = S0 +
E[Yt2 ]
Z
=E
t
Rt
0
σSs dWs , so
2
Z t Z
Su du = E[
S0 +
0
Notice that
Z tZ u
0
Z tZ
σSs dWs du =
0
0
0
t
u
σSs dWs
Z
s
2
Rt
E[Yt2 ] = E[ S0 t + 0 (t − s) σSs dWs ]
3
t
(t − s) σSs dWs ,
0
we can get
2
du ]
0
σSs dudWs =
0
R
2
Rt
t
= E[S02 t2 + 2S0 t 0 (t − s) σSs dWs + 0 (t − s) σSs dWs ]
R
2
t
= S02 t2 + 0 + E[ 0 (t − s) σSs dWs ].
Using Ito’s isometry, we can get
R
2
t
E[ 0 (t − s) σSs dWs ]
Rt
= E[ 0 (t − s)2 σ 2 Ss2 ds]
Rt
= 0 (t − s)2 σ 2 E[Ss2 ]ds
Rt
2
= 0 (t − s)2 σ 2 S02 eσ s ds
2 0
Rt
= S02 0 (t − s)2 eσ s ds
Rt 2
= S02 −t2 + 2 0 eσ s (t − s) ds integration by parts
Rt 2 = S02 −t2 − σ2t2 + σ22 0 eσ s ds
2
= S02 −t2 − σ2t2 + σ24 eσ t − σ24 .
2
So E[Yt2 ] = −S02 σ2t2 + S02 σ24 eσ t − S02 σ24 .
Now we have
V ar[Yt ] = E[Yt2 ] − E[Yt ]2
2
= −S02 t2 − S02 σ2t2 + S02 σ24 eσ t − S02 σ24
4
2
2S 2
= σ40 eσ t − 1 − σ 2 t − σ2 t2 .
µ−ν
16. dXt = µdt + σdWt = νdt + σ µ−ν
σ dt + dWt . We want
σ t + Wt
to be a Brownian motion under the new measure P? . Let θ = µ−ν
σ ,
1
Rt
1 2
2
then notice that E[e 2 0 θ ds ] = e 2 θ t < ∞, so Novikov’s condition is
satisfied.
Let
Rt
R
1 t 2
dP?
Lt =
= e− 0 θdWs − 2 0 θ ds
dP
which is
µ−ν
1 µ−ν 2
e− σ Wt − 2 ( σ ) t
By Girsanov’s theory, W̃t = µ−ν
σ t + Wt is a Brownian motion under
P? , and Xt = νt + σ W̃t is a Brownian motion with drift ν under P? .
18 Notice that there is no drift, so dXt = σdWt , i.e., Xt = X0 + σWt .
By Feynman-Kac representation formula, the solution to
∂F
1 ∂2F
(t, x) + σ 2 2 (t, x) = 0
∂t
2 ∂x
4
with the terminal value condition F (T, x) = x4 = Φ (x) is F (t, x) =
E[Φ (XT ) |Xt = x]. Notice that Xt ∼ N X0 , σ 2 t and Xt − Xs ∼
N 0, σ 2 (t − s) , which is independent of Xs , so we have
F (t, x) = E[XT4 |Xt = x] = E[(XT − Xt + Xt )4 |Xt = x]
= E[(XT − Xt )4 + 4x (XT − Xt )3
+ 6x2 (XT − Xt )2 + 4x3 (XT − Xt ) + x4 |Xt = x]
= E[(XT − Xt )4 ] + 0 + 6x2 E[(XT − Xt )2 ] + 0 + x4
Since E[(XT − Xt )2 ] = σ 2 (T − t) and from previous homework we
know E[(XT − Xt )4 ] = 3σ 4 (T − t)2 , so the solution is
F (t, x) = 3σ 4 (T − t)2 + 6x2 σ 2 (T − t) + x4
RT
RT
Since 0 σFx (t, x) dt = 0 12σ 3 x (T − t) + 4σx3 dt < ∞, so the
condition for Feynman-Kac representation formula is satisfied, and
the above F (t, x) is the desired solution.
20 We need to show that vθ (t, x) satisfies
∂vθ
1
∂ 2 vθ
(t, x) + σ 2 x2 2 (t, x) − rvθ (t, x) = 0.
∂t
2
∂x
By virtue of the definition of vθ , it is not hard to verify that
2
∂vθ
x ∂v
θ
(t, x) =
·
t,
;
∂t
θ ∂t
x
2
θ
∂ 2 vθ
θ3 ∂ 2 v
.
(t, x) =
· 2 t,
2
3
∂x
x ∂x
x
Therefore, we can get
∂vθ
1
∂ 2 vθ
(t, x) + σ 2 x2 2 (t, x)
∂t"
2
∂x
2
2 2 2 #
x ∂v
θ
1 2 θ2
∂ v
θ
=
t,
+ σ
t,
θ ∂t
x
2
x
∂x2
x
2
x
θ
=
· rv t,
θ
x
= rvθ (t, x),
which implies (1).
5
(1)
Let Xt be a solution of
dXt = σXt dWt .
(2)
Then it is easy to verify that
1
XT = Xt eσ(WT −Wt )− 2 σ
2 (T −t)
.
(3)
If the terminal condition for v(t, x) is given by v(t, x) = Φ(x), then by
the Feynman-Kac representation, we should have
v(t, x) = E[e−r(T −t) Φ(XT )|Xt = x].
Therefore, we should have
2
x
θ
x
vθ (t, x) = v t,
= E[e−r(T −t) Φ(XT )|Xt = θ2 /x].
θ
x
θ
(4)
On the other hand, for the terminal condition v(T, x) = Φ(x), the
corresponding terminal condition for vθ (t, x) is
vθ (T, x) =
x
x
v(T, θ2 /x) = Φ(θ2 /x).
θ
θ
Since vθ (t, x) satisfies the same PDE, by virtue of the Feynman-Kac
representation, we should have
2 θ
−r(T −t) XT
vθ (t, x) = E e
Φ
Xt = x .
θ
XT Comparing it with (4), we can get
2 θ
Xt = x = xE[Φ(XT )|Xt = θ2 /x],
E XT Φ
XT which can be proved by virtue of (3).
21 First, it is easy to verify that F (T, x) = Φ(x). By Ito’s formula,
1
dF = Fs ds + Fx dXs + σ 2 Fxx ds
2
where dXs = µ (s, Xs ) ds + σ (s, Xs ) dWs , so we have
1 2
dF = Fs + µ (s, Xs ) Fx + σ (s, Xs ) Fxx ds + σ (s, Xs ) Fx dWs .
2
6
Taking the integral and using F (T, XT ) = Φ(XT ), we can get
Φ (XT ) − F (t, Xt )
Z T
1 2
Fs + µ (s, Xs ) Fx + σ (s, Xs ) Fxx ds
=
2
t
Z T
+
σ (s, Xs ) Fx dWs
t
Take the conditional expectation and get
E[Φ (XT ) |Xt = x] − F (t, x)
Z T
1
=
E[Fs + µ (s, Xs ) Fx + σ 2 (s, Xs ) Fxx |Xt = x]ds + 0.
2
t
Comparing it with
Z
T
E[k (Xs ) |Xt = x]ds,
F (t, x) = E[φ (XT ) |Xt = x] +
t
we know that the following should be satisfied
1
Fs + µ (s, Xs ) Fx + σ 2 (s, Xs ) Fxx = −k (Xs ) .
2
7