exercise:9b763b8e9e: 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>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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Let <math>Z_1</math>, <math>Z_2</math>, ..., <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> | 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 | that the process eventually dies out if | ||
<ul><li> <math>p_0 = 1/2</math>, <math>p_1 = 1/4</math>, and <math>p_2 = 1/4</math>. | <ul style="list-style-type:lower-alpha"><li> <math>p_0 = 1/2</math>, <math>p_1 = 1/4</math>, and <math>p_2 = 1/4</math>. | ||
</li> | </li> | ||
<li> <math>p_0 = 1/3</math>, <math>p_1 = 1/3</math>, and <math>p_2 = 1/3</math>. | <li> <math>p_0 = 1/3</math>, <math>p_1 = 1/3</math>, and <math>p_2 = 1/3</math>. | ||
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<li> <math>p_0 = 1/3</math>, <math>p_1 = 0</math>, and <math>p_2 = 2/3</math>. | <li> <math>p_0 = 1/3</math>, <math>p_1 = 0</math>, and <math>p_2 = 2/3</math>. | ||
</li> | </li> | ||
<li> <math>p_j = 1/2^{j + 1}</math>, for <math>j = 0</math>, 1, 2, | <li> <math>p_j = 1/2^{j + 1}</math>, for <math>j = 0</math>, 1, 2, .... | ||
</li> | </li> | ||
<li> <math>p_j = (1/3)(2/3)^j</math>, for <math>j = 0</math>, 1, 2, | <li> <math>p_j = (1/3)(2/3)^j</math>, for <math>j = 0</math>, 1, 2, .... | ||
</li> | </li> | ||
<li> <math>p_j = e^{-2} 2^j/j!</math>, for <math>j = 0</math>, 1, 2, | <li> <math>p_j = e^{-2} 2^j/j!</math>, for <math>j = 0</math>, 1, 2, ... (estimate <math>d</math> | ||
numerically). | numerically). | ||
</li> | </li> | ||
</ul> | </ul> | ||
Latest revision as of 00:48, 15 June 2024
Let [math]Z_1[/math], [math]Z_2[/math], ..., [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, ....
- [math]p_j = (1/3)(2/3)^j[/math], for [math]j = 0[/math], 1, 2, ....
- [math]p_j = e^{-2} 2^j/j![/math], for [math]j = 0[/math], 1, 2, ... (estimate [math]d[/math] numerically).