Exercise 2.6 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 2.6 解答例

\( A \perp B \mid C \) means that

\begin{equation} \label{2.6condition} \text{Pr}(A\cap B \mid C) = \text{Pr}(A \mid C) \text{Pr}(B \mid C) \end{equation}

holds.

\begin{equation} \label{2.6condition2} \text{Pr}(B \mid C) = \text{Pr}(A \cap B \mid C) + \text{Pr}(A^c \cap B \mid C) \end{equation}

From this equation, we can rewrite \(\text{Pr}(A^c \cap B \mid C)\) as follows.

\begin{equation} \label{2.6ans1} \begin{aligned} \text{Pr}(A^c \cap B \mid C) &= \text{Pr}(B \mid C) - \text{Pr}(A \cap B \mid C) \qquad (\because \eqref{2.6condition2})\\ &= \text{Pr}(B \mid C) - \text{Pr}(A \mid C) \text{Pr}(B \mid C) \qquad (\because \eqref{2.6condition})\\ &= \text{Pr}(B \mid C) \left(1 - \text{Pr}(A \mid C)\right) \\ &= \text{Pr}(A^c \mid C) \text{Pr}(B \mid C) \\ \end{aligned} \end{equation}

This means \( A^c \perp B \mid C \).

Similarly, \(\text{Pr}(A \cap B^c \mid C)\) can be rewritten as follows.

\begin{equation} \label{2.6ans2} \begin{aligned} \text{Pr}(A \cap B^c \mid C) &= \text{Pr}(A \mid C) - \text{Pr}(A \cap B \mid C) \\ &= \text{Pr}(A \mid C) - \text{Pr}(A \mid C) \text{Pr}(B \mid C) \qquad (\because \eqref{2.6condition})\\ &= \text{Pr}(A \mid C) \left(1 - \text{Pr}(B \mid C)\right) \\ &= \text{Pr}(A \mid C) \text{Pr}(B^c \mid C). \\ \end{aligned} \end{equation}

This means \( A \perp B^c \mid C \).

Furthermore, \(\text{Pr}(A^c \cap B^c \mid C)\) can also be calculated in the same way.

\begin{align*} \text{Pr}(A^c \cap B^c \mid C) &= \text{Pr}(A^c \mid C) - \text{Pr}(A^c \cap B \mid C) \\ &= \text{Pr}(A^c \mid C) - \text{Pr}(A^c \mid C) \text{Pr}(B \mid C) \qquad (\because \eqref{2.6ans1})\\ &= \text{Pr}(A^c \mid C) \left(1 - \text{Pr}(B \mid C)\right) \\ &= \text{Pr}(A^c \mid C) \text{Pr}(B^c \mid C) \end{align*}

This means \( A^c \perp B^c \mid C \).