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Course: M362K Stochastic Processes I Term: Spring 2008 Instructor: Gordan Zitkovic
Lecture 7
Branching processes
7.1. A bit of history
In the mid 19th century several aristocratic families in Victorian Eng- land realized that their family names could become extinct. Was it just unfounded paranoia, or did something real prompt them to come to this conclusion? They decided to ask around, and Sir Francis Gal- ton (a “polymath, anthropologist, eugenicist, tropical explorer, geogra- pher, inventor, meteorologist, proto-geneticist, psychometrician and statistician” and half-cousin of Charles Darwin) posed the following question (1873, Educational Times ):
How many male children (on average) must each generation of a family have in order for the family name to continue in perpetuity?
The first complete answer came from Reverend Henry William Wat- son soon after, and the two wrote a joint paper entitled One the prob- ability of extinction of families in 1874. By the end of this lecture, you will be able to give a precise answer to Galton’s question.
Sir Francis Galton
7.2. A mathematical model
The model proposed by Watson was the following:
(1) A population starts with one individual at time n = 0: Z 0 = 1. (2) After one unit of time (at time n = 1) the sole individual produces Z 1 identical clones of itself and dies. Z 1 is an N 0 -valued random variable. (3) (a) If Z 1 happens to be equal to 0 the population is dead and nothing happens at any future time n ≥ 2. (b) If Z 1 > 0 , a unit of time later, each of Z 1 individuals gives birth to a random number of children and dies. The first one has Z 1 ; 1 children, the second one Z 1 ; 2 children, etc. The last, Z 1 th one, gives birth to Z 1 ;Z 1 children. We assume that the distribution of the number of children is the same for each individual in every generation and independent of either the number of individuals in the generation and of the number of children the others have. This distribution, shared by all Zn;i and Z 1 , is called the offspring distribution. The total number of individuals in the second generation is now
Z 2 =
∑^ Z^1
k=
Z 1 ;k:
(c) The third, fourth, etc. generations are produced in the same way. If it ever happens that Zn = 0, for some n, then Zm = 0 for all m ≥ n - the population is extinct. Otherwise,
Zn+1 =
∑^ Zn
k=
Zn;k:
Definition 7.1. A stochastic process with the properties described in (1), (2) and (3) above is called a (simple) branching process.
The mechanism that produces the next generation from the present one can differ from application to application. It is the offspring distribution alone that determines the evolution of a branching process. With this new formalism, we can pose Galton’s question more precisely:
Under what conditions on the offspring distribution will the process { Zn } n ∈ N 0 never go extinct, i.e., when does P[Zn ≥ 1 for all n ∈ N 0 ] = 1 (^) (7.1)
hold?
7.3. Construction and simulation of branching processes
Before we answer Galton’s question, let us figure out how to simulate a branching process, for a given offspring distribution { pn } n ∈ N 0 (pk = P[Z 1 = k]). The distribution { pn } n ∈ N 0 is N 0 -valued - we have learned how to simulate such distributions in Lecture 3. We can, therefore, assume that a transformation function F is known, i.e., that the random variable η = F(γ) is N 0 -valued with pmf { pn } n ∈ N 0 , where γ ∼ U[0; 1]. Some time ago we assumed that a probability space with a sequence { γn } n ∈ N 0 of independent U[0; 1] random variables is given. We think of { γn } n ∈ N 0 as a sequence of random numbers produced by a computer. Let us first apply the function F to each member of { γn } n ∈ N 0 to obtain an independent sequence { ηn } n ∈ N 0 of N 0 -valued random variables with pmf { pn } n ∈ N 0. In the case of a simple random walk, we would be done at this point - an accumulation of the first n elements of { ηn } n ∈ N 0 would give you the value Xn of the random walk at time n. Branching processes are a bit more complicated; the increment Zn+1 − Zn depends on Zn: the more individuals in a generation, the more offspring they will produce. In other words, we need a black box with two inputs - “randomness” and Zn - which will produce Zn+1. What do we mean by “randomness”? Ideally, we would need exactly Zn (unused) elements of { ηn } n ∈ N 0 to simulate the number of children for each of Zn members of generation n. This is exactly how one would do it in practice: given the size Zn of generation n, one would draw Zn simulations from the distribution { pn } n ∈ N 0 , and sum up the results to get Zn+1. Mathematically, it is easier to be more wasteful. The sequence { ηn } n ∈ N 0 can
be rearranged into a double sequence^1 { Zn;i } n ∈ N 0 ;i ∈ N. In words, instead of one sequence of independent random variables with pmf { pn } n ∈ N 0 , we have a sequence of sequences. Such an abundance allows us to feed the whole “row” { Zn;i } i ∈ N into the black box which produces Zn+1 from Zn. You can think of Zn;i as the number of children the ith^ individual in the nth^ generation would have had he been born. The black box uses only the first Zn elements of { Zn;i } i ∈ N and discards the rest:
Z 0 = 1; Zn+1 =
∑^ Zn
i=
Zn;i;
where all { Zn;i } n ∈ N 0 ;i ∈ N are independent of each other and have the same distribution with pmf { pn } n ∈ N 0. Once we learn a bit more about the probabilistic structure of { Zn } n ∈ N 0 , we will describe another way to simulate it.
(^1) Can you find a one-to-one and onto mapping from N into N × N?
Therefore, PZn (s) = (1 − qn) + qns: Of course, the value of Zn will be equal to 1 if and only if all of the coin-tosses of its ancestors turned out to be heads. The probability of that event is qn. So we didn’t need Proposition 7.2 after all. This example can be interpreted alternatively as follows. Each individual has exactly one child, but its gender is determined at random - male with probability q and female with probability p. Assuming that all females change their last name when they marry, and assuming that all of them marry, Zn is just the number of individuals carrying the family name after n generations. (5) p 0 = p^2 ; p 1 = 2pq; p 2 = q^2 ; pn = 0, n ≥ 3 : In this case each individual has exactly two children and their gender is female with probability p and male with probability q, independently of each other. The generating function P of the offspring distribution { pn } n ∈ N is given by P(s) = (p + qs)^2. Then PZn = (p + q(p + q(: : : p + qs)^2 : : : )^2 )^2 ︸ ︷︷ ︸ n pairs of parentheses
Unlike the example above, it is not so easy to simplify the above expression. Proposition 7.2 can be used to compute the mean and variance of the population size Zn, for n ∈ N.
Proposition 7.4. Let { pn } n ∈ N 0 be a pmf of the offpsring distribution of a branching process { Zn } n ∈ N 0_. If {_ pn } n ∈ N 0 admits an expectation, i.e., if
μ =
∑^ ∞
k=
kpk < ∞ ;
then
E[Zn] = μn: (7.2)
If the variance of { pn } n ∈ N 0 is also finite, i.e., if
σ^2 =
∑^ ∞
k=
(k − μ)^2 pk < ∞ ;
then
Var[Zn] = σ^2 μn(1 + μ + μ^2 + · · · + μn) =
σ^2 μn^1 − μ
n+ 1 − μ ;^ μ^^6 = 1; σ^2 (n + 1); μ = 1
Proof. Since the distribution of Z 1 is just { pn } n ∈ N 0 , it is clear that E[Z 1 ] = μ and Var[Z 1 ] = σ^2. We proceed by induction and assume that the formulas (7.2) and (7.3) hold for n ∈ N. By Proposition 7.2, the generating function PZn+1 is given as a composition PZn+1 (s) = PZn (P(s)). Therefore, if we use the identity E[Zn+1] = P Z ′ n+1 (1), we get
P Z ′ n+1 (1) = P ′ Zn (P(1))P ′ (1) = P ′ Zn (1)P ′ (1) = E[Zn]E[Z 1 ] = μnμ = μn+1:
A similar (but more complicated and less illuminating) argument can be used to establish (7.3).
7.5. Extinction probability
We now turn to the central question (the one posed by Galton). We define extinction to be the following event: E = { ω ∈ Ω : Zn(ω) = 0 for some n ∈ N } :
It is the property of the branching process that Zm = 0 for all m ≥ n whenever Zn = 0. Therefore, we can write E as an increasing union of sets En, where
En = { ω ∈ Ω : Zn(ω) = 0 } :
Therefore, the sequence { P[En] } n ∈ N is non-decreasing and “continuity of probability” (see the very first lecture) implies that P[E] = lim n ∈ N
P[En]:
The number P[E] is called the extinction probability. Using generating functions, and, in particular, the fact that P[En] = P[Zn = 0] = PZn (0) we get
P[E] = lim n ∈ N
PZn (0) = lim n ∈ N
P(P(: : : P(0) : : : ))
n P’s
It is amazing that this probability can be computed, even if the explicit form of the generating function PZn is not known.
Proposition 7.5. The extinction probability p = P[E] is the smallest non-negative solution of the equation
x = P(x); called the extinction equation ,
where P is the generating function of the offspring distribution.
Proof. Let us show first that p = P[E] is a solution of the equation x = P(x). Indeed, P is a continuous function, so P(limn →∞ xn) = limn →∞ P(xn) for every convergent sequence { xn } n ∈ N 0 in [0; 1] with xn → x ∞. Let us take a particular sequence given by
xn = P(P(: : : P(0) : : : )) ︸ ︷︷ ︸ n P’s
Then
(1) p = P[E] = limn ∈ N xn, and (2) P(xn) = xn+1.
Therefore, p = lim n →∞ xn = lim n →∞ xn+1 = lim n →∞ P(xn) = P( lim n →∞ xn) = P(p);
and so p solves the equation P(x) = x. The fact that p = P[E] is the smallest solution of x = P(x) on [0; 1] is a bit trickier to get. Let p ′^ be another solution of x = P(x) on [0; 1]. Since 0 ≤ p ′^ and P is a non-decreasiing function, we have
P(0) ≤ P(p ′ ) = p ′ :
We can apply the function P to both sides of the inequality above to get
P(P(0)) ≤ P(P(p ′ )) = P(p ′ ) = p ′ :
Continuing in the same way we get
P[En] = P(P(: : : P(0) : : : )) ︸ ︷︷ ︸ n P’
≤ p ′ ;
we get p = P[E] = limn ∈ N P[En] ≤ limn ∈ N p ′^ = p ′ , so p is not larger then any other solution p ′^ of x = P(x).
Example 7.6. Let us compute extinction probabilities in the cases from Example 7.3.
(1) p 0 = 1, pn = 0, n ∈ N: No need to use any theorems. P[E] = 1 in this case. (2) p 0 = 0; p 1 = 1; pn = 0, n ≥ 2 : Like above, the situation is clear - P[E] = 0. (3) p 0 = 0; p 1 = 0; : : : ; pk = 1; pn = 0, n ≥ k, for some k ≥ 2 : No extinction here - P[E] = 0.
