exercise:A4b1ca13a5: Difference between revisions
From Stochiki
(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>\Omega</math> be the sample space of an experiment. Let <math>E</math> be an event with <math>P(E) > 0</math> and define <math>m_E(\omega)</math> by <math>m_E(\omega) = m(\omega|E)</math>. Prove that <math>m_E(\omega)</math> is a dis...") |
No edit summary |
||
| Line 1: | Line 1: | ||
Let <math>\Omega</math> be the sample space of an experiment. Let <math>E</math> be an event with <math>P(E) > 0</math> and define <math>m_E(\omega)</math> by | |||
event with <math>P(E) > 0</math> and define <math>m_E(\omega)</math> by | |||
<math>m_E(\omega) = m(\omega|E)</math>. Prove that <math>m_E(\omega)</math> is a distribution function on | <math>m_E(\omega) = m(\omega|E)</math>. Prove that <math>m_E(\omega)</math> is a distribution function on | ||
<math>E</math>, that is, that <math>m_E(\omega) \geq 0</math> and that <math>\sum_{\omega\in\Omega} | <math>E</math>, that is, that <math>m_E(\omega) \geq 0</math> and that <math>\sum_{\omega\in\Omega} | ||
m_E(\omega) = 1</math>. The function <math>m_E</math> is called the ''conditional distribution given <math>E</math>.'' | m_E(\omega) = 1</math>. The function <math>m_E</math> is called the ''conditional distribution given <math>E</math>.'' | ||
Latest revision as of 00:12, 13 June 2024
Let [math]\Omega[/math] be the sample space of an experiment. Let [math]E[/math] be an event with [math]P(E) \gt 0[/math] and define [math]m_E(\omega)[/math] by [math]m_E(\omega) = m(\omega|E)[/math]. Prove that [math]m_E(\omega)[/math] is a distribution function on [math]E[/math], that is, that [math]m_E(\omega) \geq 0[/math] and that [math]\sum_{\omega\in\Omega} m_E(\omega) = 1[/math]. The function [math]m_E[/math] is called the conditional distribution given [math]E[/math].