演習問題 2.6

(a)

$A \perp B \mid C$ は

\[\begin{align} \Pr(A \cap B \mid C) = \Pr(A \mid C)\Pr(B \mid C) \end{align} \tag{1}\]

と表せる

\[\begin{align} \Pr(B \mid C) = \Pr(A \cap B \mid C) + \Pr(A^c \cap B \mid C) \end{align} \tag{2}\]

より

\[\begin{align} \Pr(A^c \cap B \mid C) &= \Pr(B \mid C) - \Pr(A \cap B \mid C) \quad \text{(∵ (2))} \\ &= \Pr(B \mid C) - \Pr(A \mid C)\Pr(B \mid C) \quad \text{(∵ (1))} \\ &= \Pr(B \mid C)(1 - \Pr(A \mid C)) \\ &= \Pr(A^c \mid C)\Pr(B \mid C) \end{align} \tag{3}\]

以上より $A^c \perp B \mid C$

(b)

同様に

\[\begin{align} \Pr(A \cap B^c \mid C) &= \Pr(A \mid C) - \Pr(A \cap B \mid C) \\ &= \Pr(A \mid C) - \Pr(A \mid C)\Pr(B \mid C) \quad \text{(∵ (1))} \\ &= \Pr(A \mid C)(1 - \Pr(B \mid C)) \\ &= \Pr(A \mid C)\Pr(B^c \mid C) \end{align} \tag{4}\]

以上より $A \perp B^c \mid C$

(c)

さらに同様に

\[\begin{align} \Pr(A^c \cap B^c \mid C) &= \Pr(A^c \mid C) - \Pr(A^c \cap B \mid C) \\ &= \Pr(A^c \mid C) - \Pr(A^c \mid C)\Pr(B \mid C) \quad \text{(∵ (3))} \\ &= \Pr(A^c \mid C)(1 - \Pr(B \mid C)) \\ &= \Pr(A^c \mid C)\Pr(B^c \mid C) \end{align} \tag{5}\]

以上より $A^c \perp B^c \mid C$

(d)

$C$ $A$ $B$ $P(A,B,C)$
0 0 0 1/4
0 0 1 0
0 1 0 0
0 1 1 1/4
1 0 0 1/8
1 0 1 1/8
1 1 0 1/8
1 1 1 1/8

このとき、 $A \perp B \mid C = 1$ かつ $A \not\perp B \mid C = 0$ が成り立つ

演習問題 3.6

(a)

\[\begin{align} p(y \mid \phi) = c(\phi) h(y) e^{\phi t(y)} \end{align} \tag{6}\]

$\int p(y \mid \phi)dy = 1$ の左辺を $\phi$ で微分して

\[\begin{align} \frac{d}{d\phi} \int p(y \mid \phi)dy &= \int \frac{d}{d\phi} p(y \mid \phi)dy \\ &= \int \frac{d}{d\phi} c(\phi) h(y) e^{\phi t(y)}dy \\ &= \int h(y) \Big(c(\phi) \frac{de^{\phi t(y)}}{d\phi} + \frac{dc(\phi)}{d\phi} e^{\phi t(y)}\Big)dy \\ &= \int h(y) \Big(c(\phi) t(y) e^{\phi t(y)} + c'(\phi) e^{\phi t(y)}\Big)dy \\ &= \int h(y) c(\phi) t(y) e^{\phi t(y)}dy + \int h(y) c'(\phi) e^{\phi t(y)}dy \\ &= E[t(Y) \mid \phi] + \frac{c'(\phi)}{c(\phi)} \int h(y) c(\phi) e^{\phi t(y)}dy \\ &= E[t(Y) \mid \phi] + \frac{c'(\phi)}{c(\phi)} \int p(y \mid \phi)dy \\ &= E[t(Y) \mid \phi] + \frac{c'(\phi)}{c(\phi)} \end{align} \tag{7}\]

右辺は $\phi$ で微分すると0より $E[t(Y) \mid \phi] = - \frac{c’(\phi)}{c(\phi)}$

(b)

\[\begin{align} p(\phi) \propto c(\phi)^{n_0} e^{n_0 t_0 \phi} \end{align} \tag{8}\]

より、比例定数を $\alpha$ として

\[\begin{align} p(\phi) = \alpha c(\phi)^{n_0} e^{n_0 t_0 \phi} \end{align} \tag{9}\]

このとき

\[\begin{align} \frac{dp(\phi)}{d\phi} &= \frac{d}{d\phi} \alpha c(\phi)^{n_0} e^{n_0 t_0 \phi} \\ &= \alpha \Big(\frac{dc(\phi)^{n_0}}{d\phi}e^{n_0 t_0 \phi} + c(\phi)^{n_0}\frac{d}{d\phi}e^{n_0 t_0 \phi} \Big) \\ &= \alpha \Big(n_0 c(\phi)^{n_0-1}c'(\phi)e^{n_0 t_0 \phi} + c(\phi)^{n_0} n_0 t_0 e^{n_0 t_0 \phi} \Big) \\ &= \frac{c'(\phi)}{c(\phi)}\alpha c(\phi)^{n_0}e^{n_0 t_0 \phi} + n_0 t_0 \alpha c(\phi)^{n_0} e^{n_0 t_0 \phi} \\ &= \frac{c'(\phi)}{c(\phi)} n_0 p(\phi) + n_0 t_0 p(\phi) \end{align} \tag{10}\]

(10)の左辺を $\phi$ で微分して

\[\begin{align} \int \frac{dp(\phi)}{d\phi} d\phi = \frac{d}{d\phi} \int p(\phi) d\phi = \frac{d}{d\phi} 1 = 0 \end{align} \tag{11}\]

(10)の右辺を $\phi$ で微分して

\[\begin{align} n_0\int \frac{c'(\phi)}{c(\phi)} p(\phi) d\phi + n_0 t_0\int p(\phi) d\phi = n_0 E\Big[\frac{c'(\phi)}{c(\phi)} \Big] + n_0 t_0 \end{align} \tag{12}\]

以上より $E\Big[- \frac{c’(\phi)}{c(\phi)} \Big] = t_0$

演習問題 3.12

(a)

\[\begin{align} \log\ p(Y \mid \theta) &= \log\ \Big[ \frac{\theta^Ye^{-\theta}}{Y!} \Big] \\ &= Y \log\ \theta - \theta - \log\ Y! \end{align} \tag{13}\]

より

\[\begin{align} \frac{\partial ^2 \log\ p(Y \mid \theta)}{\partial \theta ^2} &= \frac{\partial}{\partial \theta} \Big( \frac{Y}{\theta} - 1 \Big) \\ &= - \frac{Y}{\theta ^2} \end{align} \tag{14}\]

以上よりフィッシャー情報量は

\[\begin{align} I(\theta) &= -E\Big[ \frac{\partial ^2 \log\ p(Y \mid \theta)}{\partial \theta ^2} \mid \theta \Big] \\ &= -E\Big[ \frac{Y}{\theta ^2} \mid \theta \Big] \\ &= \frac{E[Y \mid \theta]}{\theta ^2} \\ &= \frac{\theta}{\theta ^2} \\ &= \frac{1}{\theta} \end{align} \tag{15}\]

よって $p_j(\theta) \propto \frac{1}{\sqrt{\theta}}$となり、 $\int_0^\infty \frac{1}{\sqrt{\theta}} d\theta$ は無限大に発散するため、確率分布とならない

(b)

\[\begin{align} f(\theta , y) &\propto \sqrt{I(\theta)} \times p(y \mid \theta) \\ &= \sqrt{\frac{1}{\theta}} \times \frac{\theta^y e^{-\theta}}{y!} \\ &= \frac{\theta^{y- \frac{1}{2}} e^{-\theta}}{y!} \end{align} \tag{16}\] \[\begin{align} f(\theta \mid y) &= \frac{\theta^{y- \frac{1}{2}} e^{-\theta}}{y!} \\ &\propto \theta^{y- \frac{1}{2}} e^{-\theta} \\ &= \theta^{y+ \frac{1}{2}-1} e^{-\theta} \\ &\propto \rm{dgamma} (\theta , y+ \frac{1}{2} , 1) \end{align} \tag{17}\]

よって $f(y \mid \theta)$ は $\theta$ に対するガンマ分布に比例する

以上より、 $\frac{f(\theta , y)}{\int f(\theta , y)d\theta}$ は $Y=y$ を与えたもとでの $\theta$ の事後分布と考えることができる