1Prof. Tomasz R. Bielecki
Math 542: Stochastic Processes
Answers to: HW Topic 3III
Exercises 1-3 Do them yourself.
Exercise 4 Verification of property 2: We need to verify that the two integrals
∫ t
0
α(s,Xs)ds and∫ t
0
σ(s,Xs)dWs are well defined. For the first integral we have∫ t
0
α(s,Xs)ds =
∫ t
0
[µ−Xs]ds =
∫ t
0
[xe−αs − µe−αs + Zs]ds, (1)
where Zs = σe
−αs ∫ s
0
eαudWu. Observe that all three processes under the integral sign in (1) are
continuous in s. Thus, the integral is well defined. For the second (stochastic) integral we have∫ t
0
σ(s,Xs)dWs =
∫ t
0
σdWs = σWt,
so, of course, it is well defined.
Verification of property 3: You may apply Itoˆ formula directly. In the process, however, you will
need to use so called Itoˆ product rule, which says that if f(t) is a differentiable function and It is
an Itoˆ integral, then
d(f(t)It) = Itdf(t) + f(t)dIt.
Or, if you like to work hard you may proceed as follows: consider an auxiliary Ito´ process Y
defined as
dYt = e
αtdWt, t ≥ 0, Y0 = 0.
Observe that we have Yt =
∫ t
0
eαudWu, and thus
Xt = xe
−αt + µ(1− e−αt) + σe−αtYt, t ≥ 0.
We may now apply the second extension of Ito´ formula to the process Y and the function f(t, y) =
xe−αt + µ(1− e−αt) + σe−αty. We obtain
f(t, Yt)− f(0, Y0) =
∫ t
0
σdWs +
∫ t
0
−αe−αs
(
x− µ+ σYs
)
ds
=
∫ t
0
σdWs +
∫ t
0
α
(
µ− (e−αsx+ µ(1− e−αs) + σe−αsYs))ds = ∫ t
0
σdWs +
∫ t
0
α
(
µ−Xs
)
ds.
But f(t, Yt) = Xt and f(0, Y0) = x, and so we finally obtain
Xt = x+
∫ t
0
σdWs +
∫ t
0
α
(
µ−Xs
)
ds,
which demonstrates validity of the Property 3.
2 Intro. to Stochastic Processes - HW Topic 3; Prof. T.R.Bielecki
Exercise 5 We have
Q(−∞,∞) = EPm(X) =
∫ ∞
−∞
m(x)dP(x) =
∫ ∞
∞
m(x)f(x)dx =
∫ ∞
−∞
(2pi)−
1
2 exp
(− (x− q)2
2
)
dx = 1,
where the last equality follows since the last integral is equal to ProbP(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 W˜t [what is the
form of this function?], which, in turn, is a function of Wt, for every t ∈ [0, T ].
Exercise 7 Use Itoˆ formula.
Exercise 8 Use the fact that Xt = e
−γt2Nt .
Exercise 9 Use Itoˆ 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+ σ2/2 = 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 ) = EX
2
T − (EXT )2 = e2T (eT − 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 +
∫ t
0
Xs ds+
∫ t
0
Xs dWs.
Thus, for t ≥ s ≥ 0 we have
E(Xt|Fs) = why? = Xs + E(
∫ t
s
Xu du|Fs) 6= Xs.
(b) The process Y is a (O-U) process: Yt = e
−2t ∫ t
0
e2sdWs, t ≥ 0. Thus,
Cov(Yt, Ys) = E[YtYs] = E[(Yt − Ys)Ys] + V ar(Ys)
= e−2(t+s)E[
∫ t
s
e2udWu
∫ s
0
e2udWu] + e
−2(t+s)E
(∫ s
0
e2udWu
)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?
5) We have Zt = e
t2Nt , t ≥ 0, which can be confirmed using Itoˆ formula. Let s ≤ t. Then
Cov(Yt, Ys) = (ln 2)
2
Cov(Nt, Ns) = (ln 2)
2 (
E ((Nt −Ns)Ns) + EN2s − ENtENs
)
3= λ (ln 2)
2
s.
Thus,
Cov(Yt, Ys) = λ (ln 2)
2
min(s, t).
6) Do it yourself.
7) This can’t be done. You have seen in problem 5 above that Zt = e
t2Nt , t ≥ 0, so that Z is an
increasing process. Therefore, it can’t a MTG under any probability measure!
In class we considered a process V satisfying the following SDE (cf. Example 6.2 of Lecture 11,
with γ = 1):
dVt = Vt−(−dt+ dNt), t ≥ 0, V0 = 1.
Thus, process V is given as,
Vt = e
−t2Nt , t ≥ 0.
This process can be made to be a MTG by change of probability measure, as is shown in Example