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Math 542: Stochastic Processes

0
α（s，Xs）ds和
t
0
σ（s，Xs）dWs定义明确。对于第一个积分，我们有
t
0
α（s，Xs）ds =
t
0
[µ-Xs] ds =
t
0
[xe-αs-µe-αs + Zs] ds，（1）
Zs =σe-αsR s
0
Ë
αudWu。观察到（1）中整数符号下的所有三个过程都是

t
0
σ（s，Xs）dWs =
t
0
σdWs=σWt，

d（f（t）It）= Itdf（t）+ f（t）dIt。

dYt = e
αtdWt，t≥0，Y0 = 0。

t
0
Ë
αudWu，因此
Xt =xe-αt+ µ（1 − e
-αt）+σe-αtYt，t≥0。

xe−αt + µ（1 − e
-αt）+σe-αty。我们获得
f（t，Yt）− f（0，Y0）= Z t
0
σdWs+
t
0
-αe-αs
x − µ +σYs
</ s> </ s> </ s>
ds
=
t
0
σdWs+
t
0
α
</ s> </ s> </ s>
µ-

Ë
−αsx + µ（1 − e
-αs）+σe-αsYs
</ s> </ s> </ s>
</ s> </ s> </ s>
ds =
t
0
σdWs+
t
0
α
</ s> </ s> </ s>
µ-Xs
</ s> </ s> </ s>
ds。

Xt = x +
t
0
σdWs+
t
0
α
</ s> </ s> </ s>
µ-Xs
</ s> </ s> </ s>
ds，

Exercise 5 We have
Q(−∞, ∞) = EP
m(X) = Z ∞
−∞
m(x)dP(x) = Z ∞

m(x)f(x)dx =
Z ∞
−∞
(2π)
− 1
2 exp

(x − q)
2
2

dx = 1,
where the last equality follows since the last integral is equal to P robP
(X˜ ∈ (−∞, ∞)) which is
equal to 1, since X˜ P∼ N(q, 1). [Of course one can just evaluate the last inegral to get 1.]
Exercise 6 It is enough to observe that Xt, the strong solution, is a function of Wft [what is the
form of this function?], which, in turn, is a function of Wt, for every t ∈ [0, T].
Exercise 7 Use Itˆo formula.
Exercise 8 Use the fact that Xt = e
−γt2
Nt
.
Exercise 9 Use Itˆo formula.
2) α = 0
3) The answer depends on whether you want X to be MTG under the measure P or under the
measure Q. If under the measure P, then we must have µ + σ
2/2 = 0. If under measure Q then we
must have σ 6= 0.
4) (a) The process X is a GBM: Xt = e
1
2
t+Wt
, t ∈ [0, T]. Thus,
EXT = e
T
, V ar(XT ) = EX2
T − (EXT )
2 = e
2T
(e
T − 1).
Since EXT = e
T 6= EX0 = 1, the process X is not a martingale w.r.t. natural filtration of the SBM
W. Another way to see that X is not a MTG is the following:
Xt = 1 + Z t
0
Xs ds +
Z t
0
Xs dWs.
Thus, for t ≥ s ≥ 0 we have
E(Xt|Fs) = why? = Xs + E(
Z t
s
Xu du|Fs) 6= Xs.
(b) The process Y is a (O-U) process: Yt = e
−2t
R t
0
e
2sdWs, t ≥ 0. Thus,
Cov(Yt, Ys) = E[YtYs] = E[(Yt − Ys)Ys] + V ar(Ys)
= e
−2(t+s)E[
Z t
s
e
2u
dWu
Z s
0
e
2u
dWu] + e
−2(t+s)E
 Z s
0
e
2u
dWu
2
=
1
2
(e
−2(t−s) − e
−2(t+s)
).
The limiting distribution of the process Y is normal N(0, 1/4). Thus, the limiting mean is 0, and
the limiting variance is 1/4.
(c) So, what is this distribution

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