3.8 Branching Processes
Each individual in a population produces, by the end of its life, \(j\) offspring with probability \(p_j\), \(j\geq 0\), independently of all the others. The initial individuals form generation \(0\), their offspring generation \(1\), and so on. Let \(X_n\) be the size of generation \(n\); then \(\{X_n\}\) is a Markov chain on \(\{0, 1, 2, \dots \}\). State \(0\) is absorbing: a population that has died out stays extinct. If \(p_0>0\), every other state \(i\) leads to \(0\) in one step with probability \(p_0^i>0\), never to return, so all states \(i\geq 1\) are transient; the population either dies out or grows without bound.
Let \(\mu = \sum _jjp_j\) and \(\sigma ^2\) be the mean and variance of the number of offspring of one individual, and suppose \(X_0 = 1\).
Proof. Given \(X_{n-1}\), generation \(n\) is \(X_n = Z_1 + \cdots + Z_{X_{n-1}}\), where \(Z_r\) is the number of offspring of the \(r\)th member of generation \(n - 1\). These are independent with mean \(\mu \), so \(E(X_n\mid X_{n-1}) = \mu X_{n-1}\) and, by lemma 1.3.1, \(E(X_n) = \mu E(X_{n-1})\). Since \(E(X_0) = 1\), \(E(X_n) = \mu ^n\). □
If \(\mu <1\) the population dies out with probability one, since \[P(X_n\geq 1) = \sum _{j\geq 1}P(X_n = j)\leq \sum _{j\geq 1}jP(X_n = j) = E(X_n) = \mu ^n\to 0 .\]
Theorem 3.8.2 (Extinction probability). Let \(\pi _0\) be the probability that the population, started from one individual, eventually dies out, and let \(G(s) = \sum _jp_js^j\) be the offspring probability generating function. Then \(\pi _0\) is the smallest non-negative root of \(s = G(s)\). If \(p_0 + p_1<1\), then \(\pi _0 = 1\) when \(\mu \leq 1\) and \(\pi _0<1\) when \(\mu >1\).
Sketch of proof. Condition on \(X_1\). Given \(X_1 = j\), the population dies out exactly when each of the \(j\) independent family lines started by the first generation dies out, which has probability \(\pi _0^j\). So \[\pi _0 = \sum _jP(\text {extinction}\mid X_1 = j)p_j = \sum _j\pi _0^jp_j = G(\pi _0) .\] The function \(G\) is increasing and convex on \([0, 1]\) with \(G(1) = 1\) and \(G'(1) = \mu \). If \(\mu \leq 1\) the graph of \(G\) lies above the line \(y = s\) on \([0, 1)\), so \(s = 1\) is the only root; if \(\mu >1\) there is a second root in \([0, 1)\). That \(\pi _0\) is the smaller root follows from \(P(X_n = 0) = G_n(0)\), where \(G_n\) is the \(n\)-fold composition of \(G\), and \(G_n(0)\) increases to the smallest fixed point. □
Example 3.8.3. Find the extinction probability when (a) \(p_0 = \frac 12\), \(p_1 = \frac 14\), \(p_2 = \frac 14\) and (b) \(p_0 = 0.2\), \(p_1 = 0.3\), \(p_2 = 0.5\).
Solution.
- (a)
- \(\mu = \frac 14 + 2\cdot \frac 14 = \frac 34<1\), so \(\pi _0 = 1\). Indeed \(\pi _0 = \frac 12 + \frac 14\pi _0 + \frac 14\pi _0^2\) gives \(\pi _0^2 - 3\pi _0 + 2 = (\pi _0 - 1)(\pi _0 - 2) = 0\), whose smallest non-negative root is \(1\).
- (b)
- \(\mu = 0.3 + 2(0.5) = 1.3>1\). The equation \(\pi _0 = 0.2 + 0.3\pi _0 + 0.5\pi _0^2\) gives \(\pi _0^2 - 1.4\pi _0 + 0.4 = (\pi _0 - 1)(\pi _0 - 0.4) = 0\), so \(\pi _0 = 0.4\). Even though each individual has on average more than one offspring, the line dies out with probability \(0.4\), usually in the first few generations. Starting from \(k\) individuals the extinction probability is \(0.4^k\). Problem 6.1.3 is another case.
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