Math 554
Introduction to Stochastic Processes
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Fall 2015
Solutions to Homework #5
2.8
In all the case a is the minimal solution of φ(a) = a in (0, 1].
(b) We have:
a = p0 + p1 a + p3 a3 ⇐⇒ 4a3 − 9a + 5 = 0 ⇐⇒ (a − 1)(4a2 + 4a − 5) = 0.
√
Hence a =
6−1
.
2
(d) We have: φ(s) =
1−q
.
1−qs
Hence,
α=
which implies that α =
1−q
⇐⇒ 1 − q − α + qα2 = 0,
1 − qα
1−q
q
if q ≥ 1/2 and α = 1 otherwise.
2.9
In general,
P(Xn+1 = 0 ∩ Xn 6= 0)
P(Xn+1 = 0) − P(Xn = 0)
=
P(Xn 6= 0)
1 − P(Xn = 0)
φ(αn ) − αn
αn+1 − αn
=
.
=
1 − αn
1 − αn
P(Xn+1 = 0|Xn 6= 0) =
(a) We have:
P(X2 = 0) − P(X1 = 0)
φ(p0 ) − p0
p0 + p1 p0 + p3 p30 − p0
=
=
1 − P(X1 = 0)
1 − p0
1 − p0
3
p1 p0 + p3 p 0
=
.
1 − p0
P(X2 = 0|X1 6= 0) =
(b) We have:
φ φ(p0 ) − φ(p0 )
P(X3 = 0) − P(X2 = 0)
P(X3 = 0|X2 =
6 0) =
=
.
1 − P(X2 = 0)
1 − φ(p0 )
Since φ(s) = p0 + p1 s + p3 s3 it is in principle possible to compute it even without
calculators.
1
2.11 Let µ :=
P∞
n=1
npn . Using the notation of the textbook, P(Y1 = 0) = p0 q and
P(Y1 = n) = pn q + pn−1 (1 − q),
because the parent particle counts if it doesn’t die out. Hence,
µ
e := E(Y1 ) =
∞
X
n pn q + pn−1 (1 − q) = 1 + µ − q.
n=1
Therefore, the extinction is certain if q ≥ µ.
2.13
(a) Using the notation of the textbook,
q
P(Y1 = 1) = P(Y1 = 2) = .
2
P(Y1 = 0) = 1 − q,
Hence,
µ := E(Y1 ) =
q
q
3q
+2 = .
2
2
2
Therefore, the extinction is certain if q ≤ 2/3.
(b) Using the notation of the textbook, we must compute 1 − α(4) = 1 − [α(1)]4 . We have:
q
q
q
q
α(1) = 1 − q + α(1) + α(2) = 1 − q + α(1) + [α(1)]2 .
2
2
2
2
Therefore (because α(1) is the minimal root),
α(1) =
and 1 − α(4) = 1 − 16
2(1 − q)
,
q
1−q 4
.
q
2.14
(a) Assume that φ0 (a) ≥ 1. Then, since φ0 (s) is monotone increasing,
Z 1
1 − a = φ(1) − φ(a) =
φ0 (s)ds > φ0 (a)(1 − a) ≥ 1 − a.
a
This contradiction establishes the fact that φ0 (a) < 1.
(b) Denote ρ := φ0 (a). Using the mean value theorem for derivatives and the fact that
αn < α (see p. 56 in the text),
α − αn+1 = φ(α) − φ(αn ) ≤ ρ α − αn ).
Using induction, α − αn ≤ ρn−1 (α − α0 ) = αρn−1 .
2
(c) Let E := limn→∞ Xn = 0 denote the event of extinction. Then,
P(E|Xn 6= 0) =
P(E) − P(E ∩ Xn = 0)
α − αn
α − αn
P(E ∩ Xn 6= 0)
=
=
≤
,
P(Xn 6= 0)
1 − P(Xn = 0)
1 − αn
1−α
where in the last inequality we used the fact that αn < α (see p. 56 in the text). Thus
P(E|Xn 6= 0) ≤
3
αρn−1
.
1−α