随机微分方程 oksendal 答案

7.9. a)
d 2 Proof. By Theorem 7.3.3, A restricted to C0 (R) is rx dx + α 2x
2 (r − α 2 2 2
)t+αBt , and E [f (Xt )] = xγ e(r− calculated by definition. Indeed, Xt = xe So E x [f (Xt )] − f (x) α2 lim = (rγ + γ (γ − 1))xγ t↓0 t 2
t k Bt = 0
1 k−1 kBs dBs + k (k − 1) 2 k (k − 1) 2
t
t k−2 Bs ds 0
Therefore, βk (t) = βk−2 (s)ds
0
4 6 This gives E [Bt ] and E [Bt ]. For part b), prove by induction.
n n n n n
2 2 2 2 so E (B( j −1)/n − Bt ) = 3(t − (j − 1)/n) + 4(t − (j − 1)/n)(j − 1)/n, and
j n j −1 n
2 2 E (B 2 j −1 − Bt ) dt =
n
2j + 1 n3
Hence E
1 (Bt 0
1
3.2.
Problems in Oksendal’s book
Proof. WLOG, we assume t = 1, then
n 3 B1
=
j =1 n
3 3 (Bj/n − B( j −1)/n )
=
j =1 n
[(Bj/n − B(j −1)/n )3 + 3B(j −1)/n Bj/n (Bj/n − B(j −1)/n )]
3
Proof. First, we check by integration-by-parts formula,
t
dYt = Set Xt = (1 − t)
−a + b −
0
dBs 1−s
dt + (1 − t)
dBt b − Yt = dt + dBt 1−t 1−t
t dBs , 0 1−s
d2 dx2 . x
For f (x) = xγ , Af can be
α2 2
+ α2 γ )γt
2
.
So f ∈ DA and Af (x) = (rγ + b)
α2 2 γ (γ
− 1))xγ .
2 Proof. We choose ρ such that 0 < ρ < x < R. We choose f0 ∈ C0 (R) such that f0 = f on (ρ, R). Define τ(ρ,R) = inf {t > 0 : Xt ∈ (ρ, R)}. Then by Dynkin’s formula, and the fact 2 Af0 (x) = Af (x) = γ1 xγ1 (r + α 2 (γ1 − 1)) = 0 on (ρ, R), we get
− Bt )2 dt =
1 0
( n)
n 2j −1 j =1 n3
→ 0 as n → ∞.
To argue III → 3
n
Bt dt as n → ∞, it suffices to prove
n 2
B(j −1)/n (Bj/n − B(j −1)/n ) −
j =1 j =1
B(j −1)/n (
n
I≤[
j =1
(Bj/n − B(j −1)/n )2 ] max |Bj/n − B(j −1)/n | → 0 a.s.
1≤j ≤n 1 1 (Bt 0
2 To argue II → 3 0 Bt dBt as n → ∞, it suffices to show E [ ( n) n 2 where Bt = j =1 B(j −1)/n 1{(j −1)/n<t≤j/n} . Indeed, 1 n j/n (j −1)/n
− Bt )2 dt] → 0,
( n)
E[
0
|Bt − Bt |2 dt] =
j =1
( n)
2 2 2 E (B( j −1)/n − Bt ) dt
2 2 is equal to − B2 We note (Bt j −1 )
n
(Bt − B j−1 )4 + 4(Bt − B j−1 )3 B j−1 + 4(Bt − B j−1 )2 B 2 j −1
then Xt is centered Gaussian, with variance
t
2 E [Xt ] = (1 − t)2 0 2
ds = (1 − t) − (1 − t)2 (1 − s)2
So Xt converges in L to 0 as t → 1. Since Xt is continuous a.s. for t ∈ [0, 1), we conclude 0 is the unique a.s. limit of Xt as t → 1. 7.8 Proof. {τ1 ∧ τ2 ≤ t} = {τ1 ≤ t} ∪ {τ2 ≤ t} ∈ Nt And since {τi ≥ t} = {τi < t}c ∈ Nt , {τ1 ∨ τ2 ≥ t} = {τ1 ≥ t} ∪ {τ2 ≥ t} ∈ Nt
n j =1
αj Bj . Note
n j =1
αj Bj
(iv)
1 Proof. dXt = dt is obvvious. Set f (t, x) = et x, then 2 2 dXt = df (t, Bt ) = et Bt dt + et dBt = Xt dt + et dBt
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 )| ≤ 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. 5.11. 2|ξ | |x − y | + |1{x>0} x − 1{y>0} y | 1 + ξ2
2
c) Proof. We consider ρ > 0 such that ρ < x < R. τ(ρ,R) is the first exit time of X from (ρ, R). 2 Choose f0 ∈ C0 (R) such that f0 = f on (ρ, R). By Dynkin’s formula with f (x) = log x and 2 the fact Af0 (x) = Af (x) = r − α 2 for x ∈ (ρ, R), we get E x [f0 (Xτ(ρ,R) ∧k )] = f0 (x) + (r − Since r >
n
=
j =1ห้องสมุดไป่ตู้n
2 2 E B( j −1)/n [(Bj/n − B(j −1)/n ) −
1 2 ] n
=
j =1 n
j−1 2 1 E (Bj/n − B(j −1)/n )4 − (Bj/n − B(j −1)/n )2 + 2 n n n j−1 1 1 1 (3 2 − 2 2 + 2 ) n n n n 2(j − 1) →0 n3
α2 2 ,
α2 x )E [τ(ρ,R) ∧ k ] 2
Xt → ∞ a.s. as t → ∞. So τ(ρ,R) < ∞ a.s.. Let k ↑ ∞, we get E x [τ(ρ,R) ] = f0 (R)p(ρ) + f0 (ρ)(1 − p(ρ)) − f0 (x) 2 r− α 2
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 Ito integral, i.e. Ito integral preserves martingale property for integrands in V . Comments: The power of Ito formula is that it gives martingales, which vanish under expectation. 4.5. Proof.