Revision as of 02:31, 9 June 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> Let <math>N</math> be the expected total number of offspring in a branching process. Let <math>m</math> be the mean number of offspring of a single parent. Show that <math display="block"> N = 1 + \left(\sum p_k \cdot k\right) N = 1 + mN </math...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
BBy Bot
Jun 09'24

Exercise

[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Let [math]N[/math] be the expected total number of offspring in a branching

process. Let [math]m[/math] be the mean number of offspring of a single parent. Show that

[[math]] N = 1 + \left(\sum p_k \cdot k\right) N = 1 + mN [[/math]]

and hence that [math]N[/math] is finite if and only if [math]m \lt 1[/math] and in that case [math]N = 1/(1 - m)[/math].