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Unformatted text preview: e generation to the next. The individuals could be photons in a photomultiplier, particles in a cloud chamber, microorganisms, insects, or branches in a data
structure.
Let Xn be the number of individuals in generation n of some population. Each of these
Xn individuals, independently of each other, produces a random number of oﬀspring, and
these oﬀspring collectively make up generation n + 1. More precisely, a branching process
is a Markov chain in which the state Xn at time n models the number of individuals in
generation n. Denote the individuals of generation n as {1, 2, ..., Xn } and let Yk,n be the
number of oﬀspring of individual k. The random variables Yk,n are deﬁned to be IID over
k and n, with a PMF pj = Pr {Yk,n = j }. The state at time n + 1, namely the number of
individuals in generation n + 1, is
Xn+1 = Xn
X Yk,n . (5.21) k=1 Assume a given distribution (perhaps deterministic) for the initial state X0 . The transition
probability, Pij = Pr {Xn+1 = j  Xn = i}, is just the probability that Y1,n +Y2,n +· · ·+Yi,n = 208 CHAPTER 5. COUNTABLESTATE MARKOV CHAINS j . The zero state (i e., the state in which there are no individuals) is a trapping state (i.e.,
P00 = 1) since no future oﬀspring can arise in this case.
One of the most important issues about a branching process is the probability that the
population dies out eventually. Naturally, if p0 (the probability that an individual has no
oﬀspring) is zero, then each generation must be at least as large as the generation before,
and the population cannot die out unless X0 = 0. We assume in what follows that p0 > 0
and X0 > 0. Recall that Fij (n) was deﬁned as the probability, given X0 = i, that state j is
entered between times 1 and n. From (5.4), this satisﬁes the iterative relation
X
Fij (n) = Pij +
Pik Fkj (n − 1), n > 1; Fij (1) = Pij .
(5.22)
k6=j The probability that the process dies out by time n or before, given X0 = i, is thus Fi0 (n).
For the nth generation to die out, starting with an initial population of i individuals, the
descendants of each of those i individuals must die out. Since each individual generates
descendants independently, we have Fi0 (n) = [F10 (n)]i for all i and n. Because of this
relationship, it is suﬃcient to ﬁnd F10 (n), which can be determined from (5.22). Observe
that P1k is just pk , the probability that an individual will have k oﬀspring. Thus, (5.22)
becomes
F10 (n) = p0 + 1
X
k=1 k pk [F10 (n − 1)] = 1
X
k=0 pk [F10 (n − 1)]k . (5.23) P
Let h(z ) = k pk z k be the z transform of the number of an individual’s oﬀspring. Then
(5.23) can be written as
F10 (n) = h(F10 (n − 1)). (5.24) This iteration starts with F10 (1) = p0 . Figure 5.3 shows a graphical construction for
evaluating F10 (n). Having found F10 (n) as an ordinate on the graph for a given value of n,
we ﬁnd the same value as an abscissa by drawing a horizontal line over to the straight line of
slope 1; we then draw a vertical line back to the curve h(z ) to ﬁnd h(F10 (n)) = F10 (n + 1).
For the two ﬁgures shown, it can be seen that F10 (1) is equal to the smallest root of
the equation h(z ) − z = 0. We next show that these two ﬁgures are representative of all
possibilities. Since h(z ) is a z transform, we know P h(1) = 1, so that z = 1 is one
that
root of h(z ) − z = 0. Also, h0 (1) = Y , where Y = k kpk is the expected number of an
individual’s oﬀspring. If Y > 1, as in Figure 5.3a, then h(z ) − z is negative for z slightly
smaller than 1. Also, for z = 0, h(z ) − z = h(0) = p0 > 0. Since h00 (z ) ≥ 0, there is exactly
one root of h(z ) − z = 0 for 0 < z < 1, and that root is equal to F10 (1). By the same
type of analysis, it can be seen that if Y ≤ 1, as in Figure 5.3b, then there is no root of
h(z ) − z = 0 for z < 1, and F10 (1) = 1.
As we saw earlier, Fi0 (1) = [F10 (1)]i , so that for any initial population size, there is a
probability strictly between 0 and 1 that successive generations eventually die out for Y > 1,
and probability 1 that successive generations eventually die out for Y ≤ 1. Since state 0 is
accessible from all i, but F0i (1) = 0, it follows from Lemma 5.3 that all states other than
state 0 are transient. 5.2. BRANCHING PROCESSES 1 209 °
°
° h(z )
°
°
F10 (1)
°
F10 (3)
° F10 (2)
° 10 (1)
p0
F
°
° z 1 F10 (1) °
°
°
°
F (3)
° F (2) 10
h(z )
10
°
p0
°F10 (1)
°
°
° Figure 5.3: Graphical construction to ﬁnd the probability that a population dies out.
Here F10 (n) is the probability that a population starting with one member at generation
0 dies out by generation n or before. Thus F10 (1) is the probability that the population
ever dies out. We next evaluate the expected number of individuals in a given generation. Conditional on
Xn−1 = i, (5.21) shows that the expected value of Xn is iY . Taking the expectation over
Xn−1 , we have
E [Xn ] = Y E [Xn−1 ] . (5.25) Iterating this equation, we get
n E [Xn ] = Y E [X0 ] . (5.26) Thus, if Y > 1, the expected number of in...
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This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 R.Srikant

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