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随机微分方程习题解答.pdf
Stochastic Differential Equations, Sixth Edition
Solution of Exercise Problems
Yan Zeng
July 16, 2006
This is a solution manual for the SDE book by Øksendal, Stochastic Differential Equations, Sixth Edition.
It is complementary to the books own solution, and can be downloaded at www.math.fsu.edu/˜zeng. If you
have any comments or find any typos/errors, please email me at yz44@cornell.edu.
This version omits the problems from the chapters on applications, namely, Chapter 6, 10, 11 and 12. I
hope I will find time at some point to work out these problems.
∞
k=0
∞
k=0
1
k!
(− t
2
)ku2k.
ik
k! E[Bk
t ]uk =
E[eiuBt] =
E[B2k
t
] =
1
k!(− t
2)k
(−1)k
(2k)!
=
(2k)!
k! · 2k tk.
Ex[|Bt − Bs|4] =
Ex[(B(i)
t − B(i)
s )2(B(j)
t − B(j)
s )2]
· (t − s)2 + n(n − 1)(t − s)2
s )4] +
i=j
t − B(i)
Ex[(B(i)
n
= n · 4!
2! · 4
= n(n + 2)(t − s)2.
i=1
2.8. b)
Proof.
So
d)
Proof.
2.11.
Proof. Prove that the increments are independent and stationary, with Gaussian distribution. Note for
Gaussian random variables, uncorrelatedness=independence.
2.15.
Proof. Since Bt − Bs ⊥ Fs := σ(Bu : u ≤ s), U(Bt − Bs) ⊥ Fs. Note U(Bt − Bs) d= N(0, t − s).
3.2.
1
Proof. WLOG, we assume t = 1, then
(B3
j/n − B3
(j−1)/n)
B3
1 =
=
=
j=1
n
n
n
n
j=1
j=1
+
j=1
:= I + II + III
n
[(Bj/n − B(j−1)/n)3 + 3B(j−1)/nBj/n(Bj/n − B(j−1)/n)]
(Bj/n − B(j−1)/n)3 +
3B2
(j−1)/n(Bj/n − B(j−1)/n)
j=1
3B(j−1)/n(Bj/n − B(j−1)/n)2
By Problem EP1-1 and the continuity of Brownian motion.
I ≤ [
To argue II → 3 1
n
0 B2
j=1 B2
(j−1)/n1{(j−1)/n
By looking at a subsequence, we only need to prove the L2-convergence. Indeed,
2
E
E
j=1
n
n
n
n
n
j=1
j=1
j=1
j=1
=
=
=
=
B(j−1)/n[(Bj/n − B(j−1)/n)2 − 1
n
]
(j−1)/n[(Bj/n − B(j−1)/n)2 − 1
B2
(Bj/n − B(j−1)/n)4 − 2
E
n
]2
j − 1
n
j − 1
1
n2 − 2
n
2(j − 1)
n3 → 0
(3
1
n2 +
1
n2 )
(Bj/n − B(j−1)/n)2 +
n
1
n2
(Btj+1 − Btj )
(Btj+1 − B tj +tj+1
2
) + Btj (B tj +tj+1
2
− Btj )
+
j
(B tj +tj+1
2
− Btj )2.
0 BtdBt. For the second term, we note
as n → ∞. This completes our proof.
3.9.
2
2
j
j
B tj +tj+1
B tj +tj+1
The first term converges in L2(P ) to T
Proof. We first note that
=
=
=
tj+1 − tj
=
B tj +tj+1
B tj +tj+1
B tj +tj+1
= E
tj+1−tj
2 ·
B2
j, k
E
E
E
j
j
j
2
2
2
2
2
− Btj
− Btj
2
2
2 − t
2 −
2 − tj+1 − tj
2
2
j
tj+1 − tj
2
2
− Btj
− tj+1 − tj
2
2
2 − tk+1 − tk
2
− Btk
B tk+tk+1
2
since E[(B2
j
≤ T
2
max
1≤j≤n
t − t)2] = E[B4
|tj+1 − tj| → 0,
t − 2tB2
t + t2] = 3E[B2
(Btj+1 − Btj ) →
T
t ]2 − 2t2 + t2 = 2t2. So
1
2 B2
BtdBt + T
2
=
T
0
B tj +tj+1
2
j
in L2(P ).
3
Proof. According to the result of Exercise 3.9., it suffices to show
f(t
j, ω)∆Bj
→ 0.
f(t
j, ω)∆Bj
3.10.
Indeed, note
3.11.
E
f(tj, ω)∆Bj −
j
j)||∆Bj|]
j
j
f(tj, ω)∆Bj −
≤
≤
≤
E[|f(tj) − f(t
E[|f(tj) − f(t
K|tj − t
√
E
j
j
j
j
j
√
K
√
|tj − t
j| 1+
j|1+
|tj − t
j|
2
2
K max
1≤j≤n
=
≤ T
→ 0.
j)|2]E[|∆Bj|2]
2 |tj − t
j| 1
2
Proof. Assume W is continuous, then by bounded convergence theorem, lims→t E[(W (N )
Since Ws and Wt are independent and identically distributed, so are W (N )
and W (N )
s
t
E[(W (N )
t − W (N )
s
)2] = E[(W (N )
t
)2] − 2E[W (N )
t
]E[W (N )
s
] + E[(W (N )
s
)2] = 2E[(W (N )
t
)2] = 0.
s
t − W (N )
. Hence
)2] − 2E[W (N )
t
]2.
Since the RHS=2V ar(W (N )
N → ∞ and apply dominated convergence theorem to E[W (N )
3.18.
t
t
) is independent of s, we must have RHS=0, i.e. W (N )
= E[W (N )
], we get Wt = 0. Therefore W· ≡ 0.
t
t
] a.s. Let
Proof. If t > s, then
Mt
Ms
E
|Fs
= E
eσ(Bt−Bs)− 1
2 σ2(t−s)|Fs
= E[eσBt−s]
2 σ2(t−s)
1
= 1
e
The second equality is due to the fact Bt − Bs is independent of Fs.
4.4.
Proof. For part a), set g(t, x) = ex and use Theorem 4.12. For part b), it comes from the fundamental
property of Itˆo integral, i.e. Itˆo integral preserves martingale property for integrands in V.
Comments: The power of Itˆo formula is that it gives martingales, which vanish under expectation.
4.5.
4
Proof.
Therefore,
t
0
Bk
t =
kBk−1
s
dBs +
t
0
Bk−2
s
ds
1
2 k(k − 1)
t
βk(t) = k(k − 1)
2
0
βk−2(s)ds
This gives E[B4
t ] and E[B6
t ]. For part b), prove by induction.
4.6. (b)
Proof. Apply Theorem 4.12 with g(t, x) = ex and Xt = ct +n
j=1 αjBj. Noten
j=1 αjBj is a BM, up to a
constant coefficient.
4.7. (a)
Proof. v ≡ In×n.
(b)
Proof. Use integration by parts formula (Exercise 4.3.), we have
0 + 2 t
So Mt = X 2
t
t
X 2
t = X 2
0 + 2
XsdX +
0
0
|vs|2ds = X 2
0 + 2
t
t
0
E
0 XsvsdBs. Let C be a bound for |v|, then
t
≤ C 2E
|Xs|2ds
= C 2
|Xsvs|2ds
s
= C 2
E
0
0
|vu|2du
0
ds ≤ C 4t2
2 .
t
0
t
0
|vs|2ds.
2
ds
vudBu
XsvsdBs +
t
0
E
s
0
So Mt is a martingale.
4.12.
Proof. Let Yt = t
0 u(s, ω)ds. Then Y is a continuous {F (n)
|Ytk+1 − Ytk|2 ≤ lim
∆tk→0
Y t = lim
∆tk→0
tk≤t
t }-martingale with finite variation. On one hand,
(total variation of Y on [0, t]) · max
|Ytk+1 − Ytk| = 0.
tk
On the other hand, integration by parts formula yields
t
0
t = 2
Y 2
YsdYs + Y t.
So Y 2
t
is a local martingale. If (Tn)n is a localizing sequence of stopping times, by Fatou’s lemma,
t ] ≤ lim
So Y· ≡ 0. Take derivative, we conclude u = 0.
4.16. (a)
E[Y 2
n
E[Y 2
t∧Tn
] = E[Y 2
0 ] = 0.
Proof. Use Jensen’s inequality for conditional expectations.
(b)
5
(ii) B3
Proof. (i) Y = 2 T
T = T
3 t
0 sdBs = t
0 BsdBs.
0 B2
(iii)Mt = E[exp(σBT )|Ft] = E[exp(σBT − 1
0 3B2
0 3(B2
0 BsdBs. So Mt = T + 2 t
0 Bsds = 3 T
s dBs + 3 T
s + (T − s)dBs.
t
2 σ2t). Since Z solves the SDE dZt = ZtσdBt, we have
s dBs + 3(BT T − T
t
2 σ2T )|Ft] exp( 1
1
Mt = (1 +
ZsσdBs) exp(
0 sdBs). So Mt = 3 t
0 B2
s dBs + 3T Bt −
2 σ2T ), where Zt = exp(σBt−
2 σ2T ) = Zt exp( 1
1
2 σ2T ) = exp(
1
2 σ2T ) +
0
σ exp(σBs +
1
2 σ2(T − s))dBs.
0
5.1. (ii)
Proof. Set f(t, x) = x/(1 + t), then by Itˆo’s formula, we have
(1 + t)2 dt + dBt
1 + t
dXt = df(t, Bt) = − Bt
= − Xt
1 + t
dt + dBt
1 + t
(iii)
Proof. By Itˆo’s formula, dXt = cos BtdBt − 1
0 : Bs ∈ [− π
2 ]}. Then
2 , π
t
0 Xsds. Let τ = inf{s >
0 cos BsdBs − 1
2
0
2 sin Btdt. So Xt = t
t∧τ
t∧τ
t∧τ
cos BsdBs − 1
t
2
cos Bs1{s≤τ}dBs − 1
t
2
1 − sin2 Bs1{s≤τ}dBs − 1
t∧τ
t∧τ
1 − X 2
2
t
s dBs − 1
2
0
0
0
0
0
0
Xsds
Xsds
t∧τ
0
Xsds.
Xsds
Xt∧τ =
=
=
=
So for t < τ, Xt = t
0
(iv)
1 − X 2
s dBs − 1
2
0 Xsds.
Proof. dX 1
t = dt is obvious. Set f(t, x) = etx, then
dX 2
t = df(t, Bt) = etBtdt + etdBt = X 2
t dt + etdBt
5.3.
Proof. Apply Itˆo’s formula to e−rtXt.
5.5. (a)
Proof. d(e−µtXt) = −µe−µtXtdt + e−µtdXt = σe−µtdBt. So Xt = eµtX0 + t
0 σeµ(t−s)dBs.
(b)
6
Proof. E[Xt] = eµtE[X0] and
So
X 2
t = e2µtX 2
0 + σ2e2µt(
e−µsdBs)2 + 2σe2µtX0
t
0
e−µsdBs.
t
0
t
0
e−2µsds
since t
E[X 2
t ] = e2µtE[X 2
0 ] + σ2e2µt
0 e−µsdBs is a martingale vanishing at time 0
0 ] + σ2e2µt e−2µt − 1
−2µ
0 ] + σ2 e2µt − 1
t ] − (E[Xt])2 = e2µtV ar[X0] + σ2 e2µt−1
2µ .
= e2µtE[X 2
= e2µtE[X 2
2µ
.
So V ar[Xt] = E[X 2
5.6.
Proof. We find the integrating factor Ft by the follows. Suppose Ft satisfies the SDE dFt = θtdt + γtdBt.
Then
d(FtYt) = FtdYt + YtdFt + dYtdFt
= Ft(rdt + αYtdBt) + Yt(θtdt + γtdBt) + αγtYtdt
= (rFt + θtYt + αγtYt)dt + (αFtYt + γtYt)dBt.
(1)
Solve the equation system
θt + αγt = 0
αFt + γt = 0,
we get γt = −αFt and θt = α2Ft. So dFt = α2Ftdt − αFtdBt. To find Ft, set Zt = e−α2tFt, then
dZt = −α2e−α2tFtdt + e−α2tdFt = e−α2t(−α)FtdBt = −αZtdBt.
Hence Zt = Z0 exp(−αBt − α2t/2). So
Ft = eα2tF0e−αBt− 1
2 α2t = F0e−αBt+ 1
2 α2t.
Choose F0 = 1 and plug it back into equation (1), we have d(FtYt) = rFtdt. So
Yt = F −1
t
(F0Y0 + r
Fsds) = Y0eαBt− 1
2 α2t + r
eα(Bt−Bs)− 1
2 α2(t−s)ds.
t
0
t
0
t
0
esdBs.
5.7. (a)
Proof. d(etXt) = et(Xtdt + dXt) = et(mdt + σdBt). So
Xt = e−tX0 + m(1 − e−t) + σe−t
(b)
7
Proof. E[Xt] = e−tE[X0] + m(1 − e−t) and
E[X 2
t ] = E[(e−tX0 + m(1 − e−t))2] + σ2e−2tE[
t
0
e2sds]
Hence V ar[Xt] = E[X 2
0 ] + 2m(1 − e−t)e−tE[X0] + m2(1 − e−t)2 +
= e−2tE[X 2
t ] − (E[Xt])2 = e−2tV ar[X0] + 1
2 σ2(1 − e−2t).
1
2 σ2(1 − e−2t).
5.9.
Proof. Let b(t, x) = log(1 + x2) and σ(t, x) = 1{x>0}x, then
|b(t, x)| + |σ(t, x)| ≤ log(1 + x2) + |x|
Note log(1 + x2)/|x| is continuous on R − {0}, has limit 0 as x → 0 and x → ∞. So it’s bounded on R.
Therefore, there exists a constant C, such that
|b(t, x)| + |σ(t, x)| ≤ C(1 + |x|)
Also,
|b(t, x) − b(t, y)| + |σ(t, x) − σ(t, y)| ≤ 2|ξ|
1 + ξ2|x − y| + |1{x>0}x − 1{y>0}y|
for some ξ between x and y. So
|b(t, x) − b(t, y)| + |σ(t, x) − σ(t, y)| ≤ |x − y| + |x − y|
Conditions in Theorem 5.2.1 are satisfied and we have existence and uniqueness of a strong solution.
0 σ(s, Xs)dBs. Since Jensen’s inequality implies (a1 + ··· + an)p ≤
2
t
t
t
0
0
0
b(s, Xs)ds
+ E
σ(s, Xs)dBs
|b(s, Xs)|2ds] + E[
|σ(s, Xs)|2ds]
(1 + |Xs|)2ds] + C 2E[
(1 + |Xs|)2ds])
2
5.10.
Proof. Xt = Z + t
np−1(ap
1 + ··· + ap
E[|Xt|2] ≤ 3
E[|Z|2] + E
0 b(s, Xs)ds + t
t
n) (p ≥ 1, a1,··· , an ≥ 0), we have
t
t
t
t
= 3(E[|Z|2] + 2C 2E[
≤ 3(E[|Z|2] + C 2E[
E[|Z|2] + E[
≤ 3
≤ 3(E[|Z|2] + 4C 2E[
0
0
0
(1 + |Xs|)2ds])
(1 + |Xs|2)ds])
t
0
0
≤ 3E[|Z|2] + 12C 2T + 12C 2
E[|Xs|2]ds
= K1 + K2
0
E[|Xs|2]ds,
t
where K1 = 3E[|Z|2] + 12C 2T and K2 = 12C 2. By Gronwall’s inequality, E[|Xt|2] ≤ K1eK2t.
5.11.
0
8
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