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# 確率論 ~ベイズの定理~

$B_1,\;B_2,\;...$ が以下の条件を満たすとき

"互いに排反" \; かつ \; \bigcup _{n=1}^{\infty} B_n = Ω \\
(この\;\{B_n\}\;を\;Ω\;の分割という)


A=A\capΩ\\
\begin{align}
=A\cap\bigl(\bigcup_{n=1}^{\infty} B_n\bigr)\\
=\bigcup_{n=1}^{\infty}\;(\;A\cap B_n\,)
\end{align}


このことから

P(A) = P\bigl(\bigcup_{n=1}^{\infty}\;(\;A\cap B_n\,)\bigr) \\
=\sum_{n=1}^{\infty}P(\;A\cap B_n\,) \\
=\sum_{n=1}^{\infty}P(B_n)\,P(\,A\,|\,B_n\,)


ベイズの定理

$A \in F$ を $P(A)\neq 0$ なる事象、{ $B_n$ } を $Ω$ の分割とする

P(\,B_i\,|\,A\,) = \frac{P(\;A\cap B_i\,)}{P(A)}\\
=\frac{P(\,A\,|\,B_i\,)\,P(B_i)}{\sum_{n=1}^{\infty}P(\,A\,|\,B_n\,)\,P(B_n)}

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