統計検定 1級 過去問 解答/解答例と解説
2016年11月27日 (日) 試験

$$\begin{align} \frac{d}{d\theta}\ell(\theta) &= \frac{d}{d\theta}\log L(\theta) \\ &= \frac{1}{\theta}\sum_{i=1}^{n}(x_i - \log \theta) \\ &= \frac{n}{\theta}(\bar{x} - \log \theta) \\ &= \frac{n}{\theta}(\bar{x} - \mu) \end{align}$$

$$\begin{align} \left\{\frac{d}{d\theta}\log L(\theta)\right\}^2 &= \left\{\frac{n}{\theta}(\bar{x} - \mu)\right\}^2 \\ &= \frac{n^2}{\theta^2}\left\{\bar{x}^2 - 2\mu \bar{x} + \mu^2\right\} \\ \end{align}$$

$$\begin{align} I(\theta) &= E\left[\left\{\frac{d}{d\theta}\log L(\theta)\right\}^2\right] \\ &= E\left[\frac{n^2}{\theta^2}\left\{\bar{X}^2 - 2\mu \bar{X} + \mu^2\right\}\right] \\ &= \frac{n^2}{\theta^2} E\left[\bar{X}^2 - 2\mu \bar{X} + \mu^2\right] \\ &= \frac{n^2}{\theta^2} \left\{E[\bar{X}^2] - 2\mu E[\bar{X}] + \mu^2\right\} \\ &= \frac{n^2}{\theta^2} \left\{V[\bar{X}] + (E[\bar{X}])^2 - 2\mu E[\bar{X}] + \mu^2\right\} \\ &= \frac{n^2}{\theta^2} \left\{\frac{1}{n} + \mu^2 - 2\mu^2 + \mu^2\right\} \\ &= \frac{n}{\theta^2} \end{align}$$

$$\begin{equation} \frac{1}{I(\theta)} = \frac{\theta^2}{n} \end{equation}$$

$$\begin{align} V\left[\tilde{\theta}\right] &= V\left[\exp\left[\bar{X} - \frac{1}{2n}\right]\right] \\ &= V\left[e^{\bar{X}} e^{-\frac{1}{2n}}\right] \\ &= e^{-\frac{1}{n}} V[e^{\bar{X}}] \\ &= e^{-\frac{1}{n}}\left\{E[e^{2\bar{X}}] - (E[e^{\bar{X}}])^2\right\} \\ &= e^{-\frac{1}{n}}\left\{\exp\left[2\mu + \frac{2}{n}\right] - \exp\left[\mu + \frac{1}{2n}\right]^2\right\} \\ &= e^{-\frac{1}{n}}\left\{\exp\left[2\mu + \frac{2}{n}\right] - \exp\left[2\mu + \frac{1}{n}\right]\right\} \\ &= e^{-\frac{1}{n}}e^{2\mu + \frac{1}{n}} \left\{e^{\frac{1}{n}} - 1\right\} \\ &= e^{2\mu}(e^{\frac{1}{n}} - 1) \\ &= \theta^2(e^{\frac{1}{n}} - 1) \\ \end{align}$$

$$\begin{align} V\left[\tilde{\theta}\right] - \frac{1}{I(\theta)} &= \theta^2(e^{\frac{1}{n}} - 1) - \frac{\theta^2}{n} \\ &= \theta^2\left(e^{\frac{1}{n}} - 1 - \frac{1}{n}\right) \\ \end{align}$$

$$\begin{equation} - \frac{1}{n^2}e^{\frac{1}{n}} + \frac{1}{n^2} = \frac{1}{n^2}(1 - e^{\frac{1}{n}}) < 0 \end{equation}$$

$$\begin{align} \lim_{n \to 1}\theta^2\left(e^{\frac{1}{n}} - 1 - \frac{1}{n}\right) &= \theta^2\left(e - 1 - 1\right) \\ &= \theta^2\left(e - 2\right) > 0~~~~(\because~e \approx 2.72) \end{align}$$

$$\begin{align} \lim_{n \to \infty}\theta^2\left(e^{\frac{1}{n}} - 1 - \frac{1}{n}\right) &= \theta^2\left(e^0 - 1 - 0\right) \\ &= 0 \end{align}$$

$$\begin{equation} V\left[\tilde{\theta}\right] - \frac{1}{I(\theta)} > 0 \end{equation}$$

$$\begin{equation} V\left[\tilde{\theta}\right] > \frac{1}{I(\theta)} \end{equation}$$