A branching process \(Z_{n}\) with reproduction law \(X\) counts the number of individuals at each generation
\[Z_n = \sum_{i=1}^{Z_{n-1}} X_{i},\]
where each \(X_i\) are independent with same distribution as \(X\) and \(Z_{0} = 1\).
Proposition. Denote \(\mu = EX\), \(\sigma^2 = \mbox{Var}(X)\), and \(\eta = P(\text{eventual extinction})\). Then,
Note: if it’s given that the number of individuals at any generation is \(k\), then the probability of extinction is \(\eta^{k}\) as each individual can be thought as independent branching processes, so the probability of extinction is the probability that each process is eventually goes extinct.
Let \(Z_{n}\) be a branching process with reproduction law \(X\) and let \(\mu = EX\), \(\sigma^2 = \mbox{Var}(X)\). The expectation of \(Z_{n}\) is
\[E[Z_{n}] = E[E[Z_{n} \mid Z_{n-1}]] = E[Z_{n-1}\mu] = \mu E[Z_{n-1}] = \cdots = \mu^{n} E[Z_{0}] = \mu^{n}.\]
The variance of \(Z_{n}\) is
\[\mbox{Var}(Z_{n}) = \begin{cases} \sigma^2 \mu^{n-1}\left(\frac{1 - \mu^n}{1 - \mu}\right), &\text{if } \mu\neq 1 \\ n\sigma^2, &\text{if } \mu = 1 \end{cases}\]
Remark: the proof for this is more cumbersome, but can be done with conditional variance (see textbook).
\[\mbox{Var}(Z_n) = E[\mbox{Var}(Z_n\mid Z_{n-1})] + \mbox{Var}(E[Z_{n}\mid Z_{n-1}])\]
Definition. The generating function of \(X\) is \(G_X(s) = E[s^{X}]\).
Proposition. Let \(X_{1},\ldots, X_{N}\) be i.i.d. and \(N\) be an integer-valued random variable. Define \(Y = X_1 + \cdots + X_{N}\) and let \(Z_{n}\) be defined as above. Then, the following holds:
Additional properties: