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

$$\begin{align} MSE[\hat{\theta}] &= E[(\hat{\theta} - \theta)^2] \\ &= E[{\hat{\theta}}^2 - 2\theta\hat{\theta} + \theta^2] \\ &= E[{\hat{\theta}}^2] - 2\theta E[\hat{\theta}] + \theta^2 \\ &= E[(e^{\bar{X}})^2] - 2\theta E[e^{\bar{X}}] + \theta^2 \\ &= E[e^{2\bar{X}}] - 2\theta E[e^{\bar{X}}] + \theta^2 \\ \end{align}$$

$$\begin{equation} E[2\bar{X}] = 2E[\bar{X}] = 2\mu \end{equation}$$

$$\begin{equation} V[2\bar{X}] = 4V[\bar{X}] = \frac{4}{n} \end{equation}$$

$$\begin{equation} M_{2\bar{X}}(t) = E[e^{2t\bar{X}}] = \exp\left[2\mu t + \frac{2}{n}t^2\right] \end{equation}$$

$$\begin{equation} M_{2\bar{X}}(1) = E[e^{2\bar{X}}] = \exp\left[2\mu + \frac{2}{n}\right] \end{equation}$$

$$\begin{align} MSE[\hat{\theta}] &= \exp\left[2\mu + \frac{2}{n}\right] - 2e^{\mu}\exp\left[\mu + \frac{1}{2n}\right] + e^{2\mu} \\ &= \exp\left[2\mu + \frac{2}{n}\right] - 2\exp\left[2\mu + \frac{1}{2n}\right] + \exp\left[2\mu\right] \\ &= e^{2\mu}\left(e^{\frac{2}{n}} - 2e^{\frac{1}{2n}} + 1\right) \end{align}$$

$$\begin{align} \lim_{n \to \infty}MSE[\hat{\theta}] &= e^{2\mu}\left(e^{0} - 2e^{0} + 1\right) \\ &= 0 \end{align}$$