Exercise 2.3 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 2.3 解答例

\begin{align*} p(x \mid y, z) &= \frac{p(x, y, z)}{p(y, z)} \\ &= \frac{p(x, y, z)}{\int p(x, y, z) \; dx} \\ &= \frac{f(x, z) g(y, z) h(z)}{\int f(x, z) g(y, z) h(z) \; dx} \qquad (*[1]) \\ &= \frac{f(x, z) g(y, z) h(z)}{g(y, z) h(z) \int f(x, z) \; dx} \\ &= \frac{f(x, z)}{\int f(x, z) \; dx} \end{align*}

*[1]:
仮定より \(p(x, y, z) = Const \times f(x, z) g(y, z) h(z)\) となるが、定数はインテグラルの外に出て打ち消されるから。
By assumption, \(p(x, y, z) = Const \times f(x, z) g(y, z) h(z)\), but the constant cancels out after integration.

\begin{align*} p( y \mid x, z) & = \frac{p(x, y, z)}{p(x, z)} \\ &= \frac{p(x, y, z)}{\int p(x, y, z) \; dy} \\ &= \frac{f(x, z) g(y, z) h(z)}{\int f(x, z) g(y, z) h(z) \; dy} \\ &= \frac{f(x, z) g(y, z) h(z)}{ f(x, z) h(z) \int g(y, z) \; dy} \\ &= \frac{g(y, z)}{\int g(y, z) \; dy} \end{align*}

To show \(p(x, y \mid z) = p(x \mid z) \times p(y \mid z)\):

Since \(p(x,y \mid z) = p(x \mid y, z) p(y \mid z)\) always holds, it suffices to show \(p(x \mid z) = p(x \mid y, z)\).

\begin{align*} p(x \mid z) &= \frac{p(x, z)}{p(z)} \\ &= \frac{\int p(x, y, z) \; dy}{\int \int p(x, y, z) \; dy \; dx} \\ &= \frac{\int f(x, z) g(y, z) h(z) \; dy}{\int \int f(x, z) g(y, z) h(z) \; dy \; dx} \\ &= \frac{f(x, z) h(z) \int g(y, z) \; dy}{h(z) \int \int g(y, z) f(x, z) \; dy \; dx } \\ &= \frac{f(x, z) h(z) \int g(y, z) \; dy}{h(z) \int f(x, z) \left( \int g(y, z) \; dy \right) \; dx } \\ &= \frac{f(x, z) h(z) \int g(y, z) \; dy}{h(z) \left( \int g(y, z) \; dy \right) \left( \int f(x, z) \; dx \right)} \\ &= \frac{f(x, z)}{\int f(x, z) \; dx} \\ &= p(x \mid y, z) \qquad ( \because \text{ a }) \end{align*}

Therefore, \(p(x \mid z) = p(x \mid y, z)\) holds.