Generically, suppose that we have a system of particles that can generate or split into other particles of the same type. Here are some typical examples:
The particles are biological organisms that reproduce.
The particles are neutrons in a chain reaction.
The particles are electrons in an electron multiplier.
We assume that each particle, at the end of its life, is replaced by a random number of new particles that we will refer to as children of the original particle. Our basic assumption is that the particles act independently, each with the same offspring distribution on
.
Let
denote the common probability density function of the number of offspring of a particle. We will also let
denote the convolution power of degree
of
; this is the probability density function of the total number of children of
particles.
We will not yet try to study the evolution of the system in real time, but rather in generational time. Specifically, the particles that we start with are in generation 0, and recursively, the children of a particle in generation
are in generation
.
Let
denote the number of particles in generation
.
One way to define the process mathematically is to start with an array of independent random variables
where
and
,
each with probability density function
. We interpret
as the number of children of the
particle in generation
(if this particle exists). Note that we have more random variables than we need, but this causes no harm, and we know that we can construct a probability space that supports such an array of random variables. We can now define our state variables recursively by
Show that
is a Markov chain on
with transition matrix
P given by
Pxyfxy, xy2
Note that the descendants of each initial particle form a branching chain, and these chains are independent. Thus, the branching chain starting with
x particles is equivalent to
x independent copies of the branching chain starting with 1 particle. This features turns out to be very important in the analysis of the chain. Note also that 0 is an absorbing state that corresponds to extinction. Computing the probability of extinction is one of the fundamental problems in branching chains; we will essentially solve this problem in the next section.
Explicitly find the one-step transition matrix
P for each of the following offspring distributions:
Consider the branching chain with offspring probability density function given by
f01p,
f2p
where
p01
is a parameter. Thus, each particle either dies or splits into two new particles. Find the probability density function of
X1X2X3
starting with 1 particle.
Extinction and Explosion
Let
m denote the mean of the offspring distribution. Show that
Xn1mXn
for
n
XnmnX0
for
n
Xn0
as
n
(extinction in the mean) if
m1.
XnX0
for each
n
(stability in the mean) if
m1.
Xn
as
n
(explosion in the mean) if
m1
and
X00.
Recall that state 0 is absorbing (there are no particles), and hence
nXn0T0
is the extinction event (where as usual,
T0
is the time of the first return to 0). We are primarily concerned with the probability of extinction, as a function of the initial state. First, however, we will make some simple observations and eliminate some trivial cases.
Suppose that
f11,
so that each particle is replaced by a single new particle. Show that
Every state is absorbing.
The equivalence classes are the singleton sets.
XnX0
for every
n
Suppose now that
f00
so that it is possible for a particle to simply die without offspring. Show that
Every state leads to 0.
Every positive state is transient.
With probability 1 either
Xn0
for some
n (extinction) or
Xn
as
n
(explosion).
Suppose now that
f00
and
f11,
so that every particle is replaced by at least one particle, and possibly more than one. Show that explosion is certain, starting with at least one particle. That is,
Xn as nX0x1
for every
x.
Suppose now that
f00
and
f0f11,
so that it is possible for a particle to die without offspring, and a particle is not replaced by more than one particle. Show that with probability 1,
Xn0
for some
n,
so extinction is certain.
Thus, the interesting case is when
f00
and
f0f11,
so that it is possible for a particle to die without offsrping and also for the particle to be replaced by more than one new particles. We will assume these conditions for the remainder of our discussion. By Exercise 6, all states lead to 0 (extinction). We will denote the probability of extinction, starting with one particle, by
qT0X01nXn0X01
Show that
is a transient equivalence class, and the probability of extinction starting with
x particles is
qxT0X0xnXn0X0x
Condition on the first state to establish the following equation:
qx0fxqx
Thus the extinction probability
q starting with 1 particle is a fixed point of the probability generating functionΦtx0fxtx
of the offspring distribution. Moreover, from the general discussion of hitting probabilities in the section on recurrence and transience,
q is the smallest such number in the interval
01.
Show that
Φ0f0
Φ11
Φt0
for
t01
so
Φ in increasing on
01
2Φt0
for
t01
so
Φ in concave upward
01
Φ1m
Based on the previous exercise and the graphs below, conclude that
If
m1
then
q1,
so extinction is certain.
If
m1
then
0q1,
so there is a positive probability of extinction and a positive probability of explosion.
Find the mean
m of each of the following offspring probability density functions. Find the extinction probability
q starting with 1 particle.
fn1ppn
for
n
where
p01
is a parameter (the geometric distribution on
).
f01p,
f2p
where
p01
is a parameter. Thus, each particle either dies or splits into two new particles.