exercise:A9cd331806: Difference between revisions
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(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...") |
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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 | |||
process. Let <math>m</math> be the mean number of offspring of a single parent. Show that | |||
<math display="block"> | <math display="block"> | ||
Latest revision as of 00:55, 15 June 2024
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].