Exercise 2.4 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 2.4 解答例
Answer
a
\begin{align*}
\text{Pr}(H_j \mid E) \text{Pr}(E)
&= \text{Pr}(H_j \mid E \cap (E \text{ or not} E)) \text{Pr}(E \mid E \text{ or not} E) \\
&= \text{Pr}(H_j \cap E \mid E \text{ or not} E)
\qquad (\because \text{P3}) \\
\\
\text{Pr}(E \mid H_j) \text{Pr}(H_j)
&= \text{Pr}(E \mid H_j \cap (E \text{ or not} E)) \text{Pr}(H_j \mid E \text{ or not} E) \\
&= \text{Pr}(E \cap H_j \mid E \text{ or not} E)
\qquad (\because \text{P3}) \\
\\
\therefore \text{Pr}(H_j \mid E) \text{Pr}(E)
&= \text{Pr}(E \mid H_j) \text{Pr}(H_j)
\end{align*}
b
\(\text{Pr}( \bigcup^K_{k=1} H_k ) = 1\) と仮定する。
\begin{align*} \text{Pr}(E) &= \text{Pr}\left(E \cap \bigcup_{k=1}^K H_k\right) \\ &= \text{Pr}\left( \left(E \cap H_1 \right) \cup \left( E \cap \left\{\bigcup_{k=2}^K H_k \right\} \right) \right)\\ &= \text{Pr}\left( E \cap H_1 \right) + \text{Pr}\left( E \cap \left\{ \bigcup_{k=2}^K H_k \right\} \right) \qquad (\because \{H_1, \dots , H_K\} \text{ is a partition and P2}) \end{align*}c
\begin{align*}
\text{Pr}(E)
&= \text{Pr}\left(E \cap H_1 \right)+
\text{Pr}\left( E \cap \left\{ \bigcup_{k=2}^K H_k \right\} \right) \\
&= \text{Pr}\left( E \cap H_1 \right) +
\text{Pr}\left( E \cap H_2 \right) +
\text{Pr}\left( E \cap \left\{ \bigcup_{k=3}^K H_k \right\} \right) \\
&= \cdots \\
&= \text{Pr}\left( E \cap H_1 \right) +
\text{Pr}\left( E \cap H_2 \right) +
\cdots + \text{Pr}\left( E \cap H_K \right) \\
&= \sum_{k=1}^K \text{Pr}\left( E \cap H_k \right)
\end{align*}
d
\begin{align*}
\text{Pr}(H_j \mid E)
&= \frac{\text{Pr}(E \mid H_j) \text{Pr}(H_j)}{\text{Pr}(E)}
\qquad (\because \text{a}) \\
&= \frac{\text{Pr}(E \mid H_j) \text{Pr}(H_j)}{\sum_{k=1}^K \text{Pr}(E \cap H_k)}
\qquad (\because \text{c})\\
&= \frac{\text{Pr}(E \mid H_j) \text{Pr}(H_j)}{\sum_{k=1}^K \text{Pr}(E \mid H_k) \text{Pr}(H_k)}
\qquad (\because \text{P3})\\
\end{align*}
