Grinstead-Snell-Probability · Problem 4.1.48

Conditional Probability

Mathematics · conditional-distribution-function

Problem

Let Ω\Omega be the sample space of an experiment. Let EE be an event with P(E)>0P(E) > 0 and define mE(ω)m_E(\omega) by mE(ω)=m(ω∣E)m_E(\omega) = m(\omega|E). Prove that mE(ω)m_E(\omega) is a distribution function on EE, that is, that mE(ω)≥0m_E(\omega) \geq 0 and that ∑ω∈ΩmE(ω)=1\sum_{\omega \in \Omega} m_E(\omega) = 1. The function mEm_E is called the conditional distribution given EE.

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