Exercise 3.13 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 3.13 解答例

\begin{align*} \log p(Y \mid \theta) &= \log \left[ \frac{\theta^Y e^{-\theta}}{Y!} \right] \\ &= Y \log \theta - \theta - \log Y! \\ \end{align*}

より、 (As the above equation holds,)

\begin{align*} \frac{\partial^2 \log p(Y \mid \theta)}{\partial \theta^2} &= \frac{\partial}{\partial \theta} \left( \frac{Y}{\theta} - 1 \right) \\ &= - \frac{Y}{\theta^2} \\ \end{align*}

よって、 (therefore, we have)

\begin{align*} I(\theta) &= - E \left[ \frac{\partial^2 \log p(Y \mid \theta)}{\partial \theta^2} \middle| \theta \right] \\ &= E \left[ \frac{Y}{\theta^2} \middle| \theta \right] \\ &= \frac{E[Y \mid \theta]}{\theta^2} \\ &= \frac{\theta}{\theta^2} \\ &= \frac{1}{\theta} \\ \end{align*}

したがって、prior distribution は、 (therefore, the prior distribution is)

\begin{align*} p_J(\theta) \propto \times \frac{1}{\sqrt{\theta}} \end{align*}

となるが、 \(\int_{0}^{\infty} \frac{1}{\sqrt{\theta}} d\theta \)は無限大に発散するので、確率分布とはならない。 (The integral diverges to infinity, so it does not become a probability distribution.)

\begin{align*} f(\theta, y) &= \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*} \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 \text{dgamma}(\theta, y + \frac{1}{2}, 1) \end{align*}

Probably yes because we are interested in only the shape of the distribution.