Math 554
Introduction to Stochastic Processes
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Solutions to Homework #7
5.4 Let Fn = σ(X1 , . . . , Xn ). Then, using Markov property of the random walk Sn ,
1
tSn+1
E
e
F
n
[m(t)]n+1
etSn
etSn
etSn
tXn+1
tXn+1
=
=
E e
Fn =
E e
= Mn .
[m(t)]n+1
[m(t)]n+1
[m(t)]n
E(Mn+1 |Fn ) =
5.5 Let ξn,k denote the number of progeny of k-th particle living in generation n. Then
P n
Xn+1 = X
k=1 ξn,k , and
E(Mn+1 |Fn ) =
1
µn+1
E(Xn+1 |Fn ) =
1
µn+1
· Xn E(ξn,1 ) =
Xn
= Mn .
µn
5.7
(a) Let ρ :=
1−p
.
p
Then
E ρSn+1 Fn = ρSn E ρXn+1 = ρSn pρ + (1 − p)ρ−1 = ρSn ,
and hence Mn = ρSn is a martingale.
(b) Since |Sn | ≤ N, the martingale is bounded and hence E(MT |F0 ) = M0 by the optional
stopping theorem. Therefore,
ρa = Pa (ST = 0) + ρN Pa (St = N ) = Pa (ST = 0) + ρN 1 − Pa (St = 0) .
Thus Pa (ST = 0) =
ρa −ρN
.
1−ρN
5.8
(a) We have
E(Mn+1 |Fn ) = E(Sn |Fn ) + E(Xn+1 |Fn ) + (1 − 2p)(n + 1)
= Sn + E(Xn+1 ) + (1 − 2p)(n + 1)
= Sn + p · 1 − (1 − p) · (−1) + (1 − 2p)(n + 1) = Mn ,
and hence Mn = Sn + (1 − 2p)n is a martingale.
1
(b) Since STn ∈ [0, N ], we have:
E(Zn2 ) ≤ E (N + (1 − 2p)T )2 ≤ 2 N 2 + (1 − 2p)2 E(T 2 ) .
To conclude the proof it thus suffices to show that E(T 2 ) < ∞. Toward this end, use
the result in Exercise 1.7 and write
Z ∞
2
E(T ) =
P(T 2 > x)dx
(substitute x = t2 )
Z ∞
Z0 ∞
2
2
2
2tP(T > t)dt.
P(T > t )dt =
=
0
0
Since P(T > t) decays exponentially when t → ∞, this establishes the result.
(c) It follows from the optional stopping theorem that
E(MT |F0 ) = M0 = a,
and hence
N · P(ST = N ) + (1 − 2p)E(T ) = a.
Thus
E(T ) =
1 1 − ρa a−N
.
1 − 2p
1 − ρN
2