Homework-1 (due date: 08.11.11) To hand in problems 1–5; other ...

M11 Stochastic Processes III/IV – MATH 3251/4091 HW1:sol s
Homework-1 (due date: 08.11.11)
To hand in problems 1–5; other problems are optional
Problem 1 Let the r.v. Z1 have mean EZ1 = m, variance VarZ1 = σ2 , and
generating function ϕ(s). Let Zn be the size of the nth generation of the related
branching process; we write ϕn(s) for the corresponding generating function.
� �
a) Using ϕn(s) ≡ ϕn−1 ϕ(s) or otherwise, deduce that EZn = mn ;
b) Using ϕn(s) ≡ ϕ � ϕn−1(s) � or otherwise, deduce that
�
σ
Var(Zn) =
2mn−1 mn−1 m−1 , m �= 1 ,
σ2n , m = 1 .
Solution. Thinking of Zn as a random sum of random variables, we have Zn ∼ SN
with N ∼ Z1 and X ∼ Zn−1; therefore EZn = EZ1 · EZn−1 = mEZn−1 and
Var(Zn) = E(Z1)Var(Zn−1) + Var(Z1) ` ´ 2
EZn−1 = mVar(Zn−1) + σ 2 m 2(n−1) ;
a straightforward induction now gives the result (do this carefully!).
Alternatively, differentiate the equalities and use the fact that if X ∼ ϕX(s), then
EX = ϕ ′ X(1) and Var(X) = ϕ ′′ X(1)+ϕ ′ X(1)− ` ϕ ′ X(1) ´ 2 ; next, carefully apply induction.
Problem 2 In a branching process with generating function
ϕ(s) = as 2 + bs + c
where a > 0, b > 0, c > 0, ϕ(1) = 1, compute the extinction probability ρ and give
the condition for sure extinction. Can you interpret your results?
Solution. Since ϕ(1) = 1 implies a + b + c = 1, the fixed point equation becomes
(as − c)(s − 1) = 0, giving the extinction probability ρ = min(c/a, 1).
Similarly, sure extinction happens when ϕ ′ (1) ≡ 2a + b ≤ 1, which is equivalent to
a ≤ c. (Can you interpret these results?)
Problem 3 Let (Zn)n≥0 be a supercritical branching process 1 with offspring distribution
Poi(λ), λ > 1. Let T0 = min{n ≥ 0 : Zn = 0} be its extinction time,
and let ρ = P(T0 < ∞) > 0 be its extinction probability. Define ( � Zn)≥0 as Zn
conditioned on extinction, ie., � Zn = � Zn | T0 < ∞ � .
a) Show that the transition probabilities ˆpxy of ( � Zn)n≥0 and the transition probabilities
pxy of the original process (Zn)n≥0 are related via ˆpxy = pxyρy−x , x, y ≥ 0;
� b � � �
Z1
Z1
b) Deduce that the generating functions �ϕ(s) ≡ E1 s and ϕ(s) ≡ E1 s satisfy
the identity �ϕ(s) = 1ϕ(ρs),
0 ≤ s ≤ 1;
ρ
c) Use the fixed point equation ρ = eλ(ρ−1) to show that �ϕ(s) = eλρ(s−1) , ie., that
the offspring distribution for ( � Zn)n≥0 is just Poi(λρ).
1 ie., with the expected offspring size m larger than one;
O.H. http://maths.dur.ac.uk/stats/courses/StochProc34/ 1

M11 Stochastic Processes III/IV – MATH 3251/4091 HW1:sol s
Solution. a) If we start with x individuals at time zero, Px(T0 < ∞) = ρ x . Then the
Markov property for (Zn)n≥0 implies 2
ˆpxy = Px(Z1 = y, T0 < ∞)
Px(T0 < ∞)
b) The result in a) implies
ˆ Z1
b
bϕ(s) ≡ E1 s ˜ = X
y≥0
c) Using the fixed-point equation, we get
bϕ(s) = ϕ(ρs)
ρ
= Px(Z1 = y)Py(T0 < ∞)
Px(T0 < ∞)
ˆp1ys y = 1
ρ
X
y≥0
= pxyρ y
p1y(ρs) y ≡ 1
ϕ(ρs) .
ρ
≡ exp{λ(ρs − 1)}
exp{λ(ρ − 1)} = eλρ(s−1) ,
ρ x .
which is the generating function of Poi(λρ), as requested. Now use the uniqueness
result (state it!).
Problem 4 Let (Zn)n≥0 be a supercritical branching process 1 with offspring distribution
{pk}k≥0, generating function ϕ(s) and extinction probability ρ ∈ [0, 1).
a) If Z0 = 1, let ˜pk be the probability that conditioned on survival the first generation
has exactly k individuals with an infinite line of descent. Show that
˜pk = 1
1 − ρ
∞�
n=k
pn
� �
n
(1 − ρ)
k
k ρ n−k .
b) Let ( � Zn)n≥0 count only those individuals in (Zn)n≥0, who conditioned on survival
have an infinite line of descent. Show that ( � Zn)n≥0 is a branching process with
offspring generating function
�ϕ(s) = 1
�
ϕ
1 − ρ
� (1 − ρ)s + ρ � �
− ρ .
Solution. a) Use the partition theorem for probabilities: for a single individual
(Z0 = 1), conditional on survival (prob. 1 − ρ) to produce exactly k individuals in the
next generation with an infinite line of descent, one has to produce n ≥ k individuals
of which exactly n −k die out (total prob. ρ n−k ) and the remaining k individuals have
an infinite line of descent (total prob. (1 − ρ) k ); hence the result.
b) In view of the previous part, we just need to compute the generating function
eϕ(s) = X
s
k≥1
k ˜pk = 1
!
X X n
pn (1 − ρ)
1 − ρ k
k≥1 n≥k
k ρ n−k s k
X nX
!
(1 − ρ) k ρ n−k s k .
= 1
1 − ρ
n≥1
pn
k=1
n
k
2 You might also wish to show that ( b Zn) ≥0 is a Markov chain! In this case the following
observation might be useful: if events B, C and D are such that for every event A one
has P ` A ∩ B | C ´ = P ` A ∩ B | D ´ , then P(A | B ∩ C) = P(A | B ∩ D); next apply it to
A = {Zn=1 = zn+1}, B = {T0 < ∞}, C = {Zn = zn} and D = {Zn = zn, . . . , Z0 = z0}.
O.H. http://maths.dur.ac.uk/stats/courses/StochProc34/ 2

M11 Stochastic Processes III/IV – MATH 3251/4091 HW1:sol s
Notice that the internal sum is just ` (1 − ρ)s + ρ ´ n − ρ n (and that this expression
vanishes for n = 0), so that
eϕ(s) = 1
1 − ρ
as required.
X
n≥0
pn
“ `(1 − ρ)s + ρ ´ n − ρ n ”
= 1
“
ϕ
1 − ρ
` (1 − ρ)s + ρ ´ ”
− ϕ(ρ) ,
Problem 5 For n ≥ 1 let X (n)
k , k = 1, . . ., n, be independent Bernoulli random
variables with
P(X (n)
k = 1) = 1 − P(X(n)
k = 0) = p(n)
k .
Assume that as n → ∞
and that
n�
k=1
p (n)
k
δ (n) def
= max
1≤k≤n p(n)
k → 0
≡ E
n�
k=1
X (n)
k
→ λ ∈ (0, ∞) .
Using generating functions or otherwise, show that the distribution of �n k=1 X(n)
k
converges to that of a Poisson(λ) random variable. This result is known as the law
of rare events.
Solution. By indepedence, the generating function of the sum is Q n
k=1
` (n) ´
1+(s−1)p k ,
so it is enough to show that its logarithm converges to λ(s−1) for every fixed s, |s| ≤ 1.
By the elementary inequality 3 |y − log(1 + y)| ≤ y 2 , valid for all |y| ≤ 1/2, we get 4
˛
nX
k=1
log ` 1 + (s − 1)p (n) ´
k − (s − 1)
nX
k=1
p (n)
k
˛ ≤ (s − 1) 2
nX ` (n) ´ 2
p ,
and it remains to observe that the last sum is bounded above by δ (n) · Pn k=1 p(n)
k → 0
as n → ∞. Similarly, the assumptions above imply (s − 1) Pn k=1 p(n)
k → (s − 1)λ as
n → ∞ for every fixed s, |s| ≤ 1. The result follows.
optional
Problem 6 In a sequence of independent Bernoulli experiments with success probability
p ∈ (0, 1), let D be the first moment of two consecutive successful outcomes.
Show that dn = P(D = n) satisfy d2 = p 2 , d3 = qp 2 , and, by conditioning on the
value of the first outcome, that
dn = q dn−1 + pq dn−2 , n > 3.
Use these relations to derive the generating function GD(s). Compute E[D].
R
3 y`
´ R 1 y
Indeed, since y − log(1 + y) = 0 1 − dx = 1+x 0
k=1
k
x
dx if only y > −1, taking the
1+x
absolute value on both sides with |y| ≤ 1/2 and integrating, we get |y − log(1 + y)| ≤ y 2 .
4 as long as |1 − s|δ (n) ≤ 1/2;
O.H. http://maths.dur.ac.uk/stats/courses/StochProc34/ 3

Z
Z
M11 Stochastic Processes III/IV – MATH 3251/4091 HW1:sol s
Solution. The first failure (prob. q = 1 − p) is followed by a valid string of n − 1
attemps (prob. dn−1); similarly, the first success if followed by a failure at the second
attempt (prob. pq) and then by a valid string of n − 2 attempts (prob. dn−2). Hence
the recursion. A standard computation thus gives
GD(s) =
p 2 s 2
d
, and E[D] =
1 − qs − pqs2 ds
˛
˛ s=1
GD(s) = 1
p
1
+ .
p2 Thus the average time in a die-tossing experiment until a double-six appeares is 42.
Problem 7 A biased coin showing ‘heads’ with probability p ∈ (0, 1) is flipped
repeatedly. Let Cw be the first moment when the word w appears in the observed
sequence of results. Find the generating function of Cw and the expectation E � �
Cw
for each of the following words: HH, HT, TH and TT.
Solution. A standard method gives 5
GHH(s) =
p 2 s 2
1 − qs − pqs 2 = p2 s 2 (1 − ps)
1 − s + p 2 qs 3 , GHT(s) =
For the other two cases, interchange H ↔ T and p ↔ q.
pqs 2
(1 − ps)(1 − qs) .
5Notice that (1 − qs − pqs2 ) −1 is the generating function of all finite strings ending with
T, which never have two H in a row! One can use a similar observation to immediately write
the result for HHH, HHHH etc. (can you?!)
O.H. http://maths.dur.ac.uk/stats/courses/StochProc34/ 4