### Hints for the Exercises 1 Throughout these exercises, let (Xt)t≥0 be

```Hints for the Exercises 1
Throughout these exercises, let (Xt )t≥0 be a Markov process with countable state space S =
{xi : i ∈ N}, i.e.:
• X takes only values in S and
• ∀τ > 0, t ≥ 0, x ∈ S : P (Xt+τ = x|Xs , s ≤ t) = P (Xt+τ = xi |Xt )
First, we consider S = N0 and assume that pn (t) := P (Xt = n) satisfies the following transition
rule:
p˙n (t) = −λpn (t) + λpn−1 (t), p0 (0) = 1, pi (0) = 0 ∀i > 0.
(1)
In this case, X is called a Poisson counter or Poisson process with rate λ.
(i) Write for arbitrary n ∈ N
 

 
1
p0
p˙0
0
 p1 
 p˙1 
 
 
 
 ..  = A ·  ..  , p(0) =  .. 
.
.
.

pn
p˙n
(2)
0
with the correct martrix A ∈ Rn×n . Solve this ODE and recall that its solution is given
by
p(t) = exp(At) · p(0),
(3)
where exp(·) denotes the matrix exponential.
(ii) P
In the discrete case, expectation is particularly easy: E[Xt ] =
∞
n=0 npn (t).
P∞
n=0 nP (Xt
= n) =
(iii) Compute the moment-generating function Mt of Xt for each t ≥ 0 and recall that
E[Xtk ] =
∂k
M
(u)
.
t
∂uk
u=0
Then compute the first two or three moments, for instance.
Now assume S = Z and let X be bi-directional Poisson counter, i.e.:
p˙i (t) = λpi−1 (t) − 2λpi (t) + λpi+1 (t), i ∈ Z, p0 (0) = 1, pi (0) = 0 ∀i 6= 0.
(4)
Compute the probability-generating function of Xt , namely
gt (z) := E[z Xt ] =
∞
X
z i pi (t), z ∈ C, |z| ≤ 1, z 6= 0.
i=−∞
Obviously, pi (t) is the ith Taylor coefficient in the Taylor expansion of gt . In order to obtain this,
show that
∂
gt (z) = λ(z − 2 + z −1 )gt (z)
∂t
holds for every |z| ≤ 1, z 6= 0 and solve this differential equation. Finally, use the binomial
formula during the expansion to compute pi (t).
1
Consider now stochastic differential equations driven by Poisson counters:
Z t
Z t
g(Xs , s)dNs
f (Xs , s)ds +
Xt = X0 +
(5)
0
0
or, in short,
dX = f (X, t)dt + g(X, t)dN.
(6)
Note that (5) has to be understood in the pathwise sense, i.e., it holds for every ω ∈ Ω whereas
in the Brownian case, a pathwise definition of the integral is not possible.
Definition 0.1 A right-continuous process X is the solution of (5) in the Itˆo sense if X satisfies
the following two properties:
• On intervals where N is constant, we have X˙ t = f (Xt , t), and
• if N jumps at time t, we have that ∆Xt := Xt − Xt− = g(Xt− , t). Here, Xt− := lims↑t Xs .
(iv) Itˆo’s differential rule has the following form: if ψ is a C 1 -function, then
Z
t
ψ(Xt ) − ψ(X0 ) =
Z
0
0
t
[ψ(Xs + g(Xs , s)) − ψ(Xs )]dNs
ψ (Xs )f (Xs , s)ds +
(7)
0
To show this, prove that ψ(X) satisfies Definition 0.1 if the right-hand side of (7) is given.
(v) For this exercise, one has to assume the following:
• There
exists a differentiable function ρ : R+ × R → R+ such that P (Xt ∈ A) =
R
A ρ(t, x)dx for every t ≥ 0 and Borel set A.
• The function g˜t : R → R, g˜t (x) := x + g(x, t), is bijective for every t ≥ 0.
• f and g are both differentiable.
To start with, use (5) to show that the evolution of the expectation of Xt is given by
d
E[Xt ] = E[f (Xt , t)] + E[g(Xt , t)] · λ.
dt
d
Now take some arbitray C 1 -function ψ and use
R Itˆo’s formula to compute dt E[ψ(Xt )].
Compare this with the fact that E[ψ(Xt )] = ψ(x)ρ(t, x)dx. Integration by parts and
∂
using the substitution dz = (1 + ∂x
g(x, t))dx yield the rest.
Now, we consider finite-state jump processes X, i.e. S = {x1 , . . . , xn } and the probabilities pi (t)
satisfy (2) with some arbitrary matrix A ∈ Rn×n . A is called the infinitesimal generator or
intensity matrix of X. Due to the non-negativity and the conservation of probability, A must
satisfy the following conditions:
Pn
•
i=1 aij = 0 for all j = 1, . . . , n and
• aij ≥ 0 for all i, j = 1, . . . , n.
(vi) In order to compute E[Xt Xt+τ ], observe that P (Xt+τ = xj |Xt = xi ) = (exp(Aτ ))ij for
every i, j = 1, . . . , n.
2
√
(vii) Let (Yt )t≥0 be a bi-directional Poisson counter of rate λ/2 and let Xλ (t) := Yt / λ. Then,
we can find Poisson counters N1 and N2 such that Xλ = λ−1/2 (N1 − N2 ). Use Itˆ
o’s
differential rule to prove
1 p
1 p
p
p
p
dXλ =
Xλ + √
− Xλ dN1 +
Xλ − √
− Xλ dN2 .
λ
λ
Now compute
p
d
dt E[Xλ (t)]
for odd and even p separately. Deduce that for even p, we have
lim E[Xλp (t)]
λ→∞
1
=
2
Z
t
p(p − 1) lim E[Xλp−2 (s)]ds
λ→∞
0
and solve these equations. Observe that these moments correspond to a N (0, t) distribud
tion. Finally show that dτ
E[Xλ (t)Xλ (τ )] = 0 for τ ≥ t and check the defining properties
of Brownian motion.
(viii) Let ψ be a C 2 function and X be a solution of
Z
t
Z
t
g(Xs , s)dWs
f (Xs , s)ds +
Xt = X0 +
(8)
0
0
where (Wt )t≥0 is a Brownian motion. Our aim is to derive Itˆo’s formula for Brownian
motion, i.e.:
1
dψ(X) = ψ 0 (X)f (X, t)dt + ψ 0 (X)g(X, t)dW + ψ 00 (X)g 2 (x, t)dt
2
(9)
To this end, first take Xλ as the solution of
Z t
Z t
√
Xλ (t) = Xλ (0) +
f (Xλ (s), s)ds +
g(Xλ (s), s)(dN1 − dN2 )/ λ
0
0
with two independent Poisson counters N1 , N2 of rate λ/2. Use (7) to compute dψ(Xλ )
and use Taylor’s formula to derive
√
dψ(Xλ ) = ψ 0 (Xλ )f (Xλ , t)dt + ψ 0 (Xλ )g(Xλ , t)(dN1 − dN2 )/ λ+
1
+ ψ 00 (X)g 2 (X, t)(dN1 + dN2 )/λ + O(λ−3/2 )dN1
2
Show that the variance of Zλ (t) = (N1 (t) + N2 (t))/λ is just t/λ and conclude that Zλ (t)
tends to Z(t) = t for λ → ∞, which finishes the proof.
3
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