14 Branching processes

Transcription

1 4 BRANCHING PROCESSES 6 4 Branching processes In this chapter we will consider a rom model for population growth in the absence of spatial or any other resource constraints. So, consider a population of individuals which evolves according to the following rule: in every generation n = 0,,2,..., each individual produces a rom number of offspring in the next generation, independently of other individuals. The probability mass function for offspring is often called the offspring distribution is given by p i = P(number of offspring = i), for i = 0,,2,... We will assume that p 0 < p < to eliminate the trivial cases. This model was introduced by F. Galton, in late 800s, to study the disappearance of family names; in this case p i is the probability that a man has i sons. We will start with a single individual in generation 0, generate the resulting rom family tree. This tree is either finite (when some generation produces no offspring at all) or infinite in the former case, we say that the branching process dies out, in the latter that it survives. We can look at this process as a Markov chain, where X n is the number of individuals at generation n. Let s start with the following observations. If X n reaches 0, it stays there, so 0 is an absorbing state. Of p 0 > 0, P(X n+ = 0 X n = k) > 0 for all k. Therefore, by Proposition 3.5, all states other than 0 are transient if p 0 > 0; the population must either die out or increase to infinity. If p 0 = 0, then the population cannot decrease, increases each generation with probability at least p, therefore must increase to infinity. It is possible to write down the transition probabilities for this chain, but they have a rather complicated explicit form, as P(X n+ = i X n = k) = P(W + W W k = i), where W,...,W k are independent rom variables, each with the offspring distribution. This suggest the use of the moment generating functions, which we will indeed do. Recall that we are assuming that X 0 =. Let δ n = P(X n = 0) be the probability that the population is extinct by generation (which we also think of as time) n. The probability π 0 that the brcing process dies out is then the limit of these probabilities: π 0 = P(the process dies out) = P(X n = 0 for some n) = lim n P(X n = 0) = lim n δ n.

2 4 BRANCHING PROCESSES 62 Note that π 0 = 0 if p 0 = 0. Our main task will be to compute π 0 for general probabilities p k. We start, however, with computing expectation variance of the population at generation n. Let µ σ 2 be the expectation variance of the offspring distribution, that is, µ = EX = kp k, σ 2 = Var(X ). Let m n = E(X n ) v n = Var(X n ). Now, X n+ is the sum of a rom number, which equals X n, of independent rom variables, each with the offspring distribution. Thus we have, by Theorem., m n+ = m n µ, v n+ = m n σ 2 + v n µ 2. Together with initial conditions m 0 =, v 0 = 0, the two recursive equations determine m n v n. We can very quickly solve the first recursion to get m n = µ n, so v n+ = µ n σ 2 + v n µ 2. This recursion has a general solution v n = Aµ n + Bµ 2n. The constant A must satisfy so that, when µ, Aµ n+ = σ 2 µ n + Aµ n+2, A = σ 2 µ( µ). From v 0 = 0 we get A + B = 0 the solution is given in the next theorem. Theorem 4.. Expectation m n variance v n of the n th generation count. We have m n = µ n v n = { σ 2 µ( µ) µn nσ 2 σ2 µ( µ) µ2n if µ, otherwise. We can immediately conclude that µ < implies π 0 =, as P(X n 0) = P(X n ) EX n = µ n 0. That is, if the individuals have less than one offspring on the average, the branching process dies out.

3 4 BRANCHING PROCESSES 63 Let now φ be the moment generating function of the offspring distribution. It is more convenient to replace e t in our original definition with s, so that φ(s) = φ X (s) = E(s X ) = p k s k. In combinatorics, this would be exactly the generating function of the sequence p k. The moment generating function of X n then is φ Xn (s) = E[s Xn ] = P(X n = k)s k. We will assume that 0 s, observe that for such s this power series converges. Let s try to get a recursive equation for φ Xn, by conditioning on the population count at generation n : φ Xn (s) = E[s Xn ] = E[s Xn X n = k]p(x n = k) = = = So φ Xn is the n th iterate of φ, we can also write E[s W +...+W k ]P(X n = k) (E(s W )E(s W 2 ) E(s W k )P(X n = k) φ(s) k P(X n = k) = φ Xn (φ(s)). φ X2 (s) = φ(φ(s)), φ X3 (s) = φ(φ(φ(s))),... φ Xn (s) = φ(φ Xn (s)). We will take a closer look at the properties of φ next. Clearly, φ(0) = p 0 > 0, Moreover, φ() = p k =. φ (s) = kp k s k 0,

4 4 BRANCHING PROCESSES 64 so φ is increasing, with Finally, φ (s) = φ () = µ. k(k )p k s k 2 0, k= so φ is also convex. The crucial observation is that δ n = φ Xn (0), so it is obtained by starting at 0 computing the n th iteration of φ. It is also clear that δ n is a nondecreasing sequence (because X n = 0 implies that X n = 0). We now consider separately two cases: Assume that φ is always above the diagonal, φ(s) s for all s [0,]. This happens exactly when µ = φ (). In this case δ n converges to, so π 0 =. This is the case in the right graph of the figure below. Now assume that φ is not always above the diagonal, which happens when µ >. In this case there exists exactly one s < which solves s = φ(s). As δ n converges to this solution, we conclude that π 0 < is now the smallest solution to s = φ(s). This is the case in the left graph of the figure below. φ φ δ δ 2 δ δ 2 Next theorem is the summary of our findings. Theorem 4.2. Probability that the branching process dies out. If µ, π 0 =. If µ >, then π 0 is the smallest solution on [0,] to s = φ(s). Example 4.. Assume that a branching process is started with X 0 = k instead of X 0 =. How does this change the survival probability? The k individuals all evolve independent family trees, so that the probability of eventual death is π0 k. It follows that also P(the process ever dies out X n = k) = π k 0

5 4 BRANCHING PROCESSES 65 for every n. If µ is barely larger than, the probability π 0 of extinction is quite close to. In the context of family names, this means that the ones with already large number of representatives in the population are at a distinct advantage, as the probability that they die out by chance is much lower than that of those with only a few representatives. Thus common family names become ever more common, especially in societies that have used family names for a long time. The most famous example of this phenomenon is Korea, where three family names (Kim, Lee, Park in common English transcriptions) account for about 45% of the population. Example 4.2. Assume that p k = p k ( p), k = 0,,2,... This means that the offspring distribution is Geometric( p ), minus. Thus µ = p = p p, if p 2, π 0 =. Suppose now that p > 2. Then we have to compute φ(s) = s k p k ( p) = p ps. The equation φ(s) = s has two solutions, s = s = p p. Thus, when p > 2. π 0 = p p. Example 4.3. Assume that the offspring distribution is Binomial(3, 2 ). Compute π 0. As µ = 3 2 >, π 0 is given by φ(s) = s s2 + 8 s3 = s, with solutions s =, 5 2, 5 2. The one that lies in (0,), , is the probability π 0. Problems. For a branching process with offspring distribution given by p 0 = 6,p = 2,p 3 = 3, determine (a) expectation variance of X 9, the population at generation 9, (b) probability that the branching process dies by generation 3, but not by generation 2, (c) the probability that

6 4 BRANCHING PROCESSES 66 the process ever dies out. Then assume you start 5 independent copies of this branching process at the same time (equivalently, change X 0 to 5), (d) compute the probability the that the process ever dies out. 2. Assume that the offspring distribution of a branching process is Poisson with parameter λ. (a) Determine the expected combined population through generation 0. (b) Determine, with the aid of computer if necessary, the probability that the process ever dies out for λ = 2, λ = λ = Assume that the offspring distribution of a branching process is given by p = p 2 = p 3 = 3. Note that p 0 = 0. Solve the following problem for a =,2,3. In a generation n, choose a rom individual let Y n be the proportion of individuals, among X n, from families of size a. (A family consists of individuals that are offspring of the same parent from previous generation.) Compute the limit of Y n as n. Solutions to problems. For (a), compute µ = 3 2, σ2 = = 5 4, plug into the formula. Then compute For (b), φ(s) = s + 3 s3. P(X 3 = 0) P(X 2 = 0) = φ(φ(φ(0))) φ(φ(0)) For (c), we solve φ(s) = s, 0 = 2s 3 3s+ = (s )(2s 2 +2s ), so π 0 = For (d), the answer is π For (a), E(X 0 + X X 0 ) = EX 0 + EX EX 0 = + µ + + µ 0 = µ µ, if µ, if µ =. For (b), if λ then π 0 =, while if λ >, then π 0 is the solution in s (0,) to e λ(s ) = s. This equation cannot be solved analytically, but we can numerically obtained the solution for λ = 2, π 0 = Assuming X n = k, this is the number of families at time n. Each of these has, independently, a members with probability p a. If k is large, which it will be for large n, as the branching process can t die out, then with overwhelming probability the number of children in such families is about ap a k, while X n is about µk. The proportion Y n then is about apa µ, which works out to be 6, 3, 2, for a =, 2, 3.