Stat 150 Stochastic Processes
Spring 2009
Lecture 14: Branching Processes and Random Walks
Lecturer:
Jim Pitman
Common setting:
p
0
,p
1
,p
2
,...
probability distribution of
X
on
{
0
,
1
,
2
,...
}
.
•
Two surprisingly related problems:
Problem 1
: Use
X
as oﬀspring variable of a branching process
Z
0
,Z
1
,Z
2
,...
.
Consider the random variable Σ :=
∑
∞
n
=0
Z
n
, the
total progeny
of the branching
process. Describe the distribution of Σ given
Z
0
=
k
.
Problem 2
: Use
X

1 as the increment variable of a random walk
S
(

)
n
,
n
=
0
,
1
,
2
,...
on integer states. Let
T
0
:= inf
{
n
:
S
(

)
n
}
= 0. Describe the distribu
tion of
T
0
given
S
(

)
0
=
k
.
•
Theorem:
•
These two problems have the same solution.
•
Moreover, the solution is
P
(Σ =
n

Z
0
=
k
) =
P
(
T
0
=
n

S
(

)
0
=
k
)
=
k
n
P
(
S
n
=
n

k
)
where
S
n
:=
X
1
+
···
+
X
n
is the sum of
n
independent copies
X
i
d
=
X
.
•
Notes
:
•
P
(Σ
<
∞
Z
0
=
k
) =
P
(
T
0
<
∞
S
(

)
0
=
k
);
Here (Σ
<
∞
) is the event of extinction of the branching process, which given
k
= 1 has probability which is the least root
s
∈
[0
,
1] of
s
=
φ
(
s
), and for
general
k
is the
k
th power of the probability for
k
= 1.
•
. The formula is very
explicit for any distribution of
X
for which there is a simple for the distribution
of
S
n
. e.g. for
X
d
= Poisson(
λ
),
S
n
d
= Poisson(
nλ
), so
P
(
S
n
=
n

k
) =
e

nλ
(
nλ
)
n

k
(
n

k
)!
•
Approach both problems with the technique of probability generating func
tions.
. This will show both solutions are the same. Then we can pick which one
1
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2
to work with to establish the formula.
Branching process problem:
Introduce G.F.
G
(
s
) :=
∑
∞
n
=0
s
n
P
(Σ =
n

Z
0
=
1). Notice that Σ given
Z
0
=
k
is the sum of
k
independent copies of Σ given
Z
0
= 1,
[
G
(
s
)]
k
=
∞
X
n
=0
s
n
P
(Σ =
n

Z
0
=
k
)
.
First step analysis
: Start from
Z
0
= 1, condition on
Z
1
=
k
for
k
= 0
,
1
,
2
,...
to see that
(Σ

Z
0
= 1
,Z
1
=
k
)
d
= (1 + Σ

Z
0
=
k
)
which gives
G
(
s
) =
∞
X
k
=0
p
k
s
[
G
(
s
)]
k
where
s
[
G
(
s
)]
k
is the GF of (1 + Σ

Z
0
=
k
)
=
sφ
(
G
(
s
))
=
⇒
G
(
s
)
φ
(
G
(
s
))
=
s
Deﬁne
F
(
z
) :=
z
φ
(
z
)
, then
F
(
G
(
s
)) =
s
. So
G
is the functional inverse of
F
(
z
).
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 Spring '08
 Evans
 Probability, Probability distribution, Stochastic process, random walks, Sn

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