Revision as of 03:30, 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>Z_1</math>, <math>Z_2</math>, \ldots, <math>Z_N</math> describe a branching process in which each parent has <math>j</math> offspring with probability <math>p_j</math>. Find the probability <math>d</math> that the process eventually d...")
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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]Z_1[/math], [math]Z_2[/math], \ldots, [math]Z_N[/math] describe a branching process in

which each parent has [math]j[/math] offspring with probability [math]p_j[/math]. Find the probability [math]d[/math] that the process eventually dies out if

  • [math]p_0 = 1/2[/math], [math]p_1 = 1/4[/math], and [math]p_2 = 1/4[/math].
  • [math]p_0 = 1/3[/math], [math]p_1 = 1/3[/math], and [math]p_2 = 1/3[/math].
  • [math]p_0 = 1/3[/math], [math]p_1 = 0[/math], and [math]p_2 = 2/3[/math].
  • [math]p_j = 1/2^{j + 1}[/math], for [math]j = 0[/math], 1, 2, \ldots.
  • [math]p_j = (1/3)(2/3)^j[/math], for [math]j = 0[/math], 1, 2, \ldots.
  • [math]p_j = e^{-2} 2^j/j![/math], for [math]j = 0[/math], 1, 2, \ldots\ (estimate [math]d[/math] numerically).