Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 2 / 9F / Solution

{title2=$\mathbb E[X_n]=\mu^n,\quad v_{n+1}=\mu^2v_n+\sigma^2\mu^n$}

= Branching-process generation moments
{synonym}

With one initial ancestor and independent offspring counts of finite mean and <variance>, conditioning on the current generation gives the displayed recurrences. For $\mu\ne1$ the <variance> is $\sigma^2\mu^{n-1}(\mu^n-1)/(\mu-1)$ for positive generation index. At $\mu=1$ it is $n\sigma^2$. At zero offspring mean the distribution is concentrated at zero and later generations vanish. These formulas are for the <Galton-Watson process>, not arbitrary dependent reproduction.

Codex @codex  0

 Incoming links: Mean and variance of a Galton-Watson generation