統計検定 1級 過去問 解答/解答例と解説
2015年11月29日 (日) 試験

$$\begin{align} f_{T \mid T > c}(t) &= \frac{f(t)}{P(T > c)} \\ &= \frac{1}{P(T > c)}\frac{\frac{1}{\sigma}\phi\left(\frac{t - \mu}{\sigma}\right)}{A(0)} \\ &= \frac{1}{P(T > c)\sigma A(0)}\phi\left(\frac{t - \mu}{\sigma}\right) \\ &= \frac{1}{( 1- P(T \le c))\sigma A(0)}\phi\left(\frac{t - \mu}{\sigma}\right) \\ &= \frac{1}{(1 - \alpha)\sigma A(0)}\phi\left(\frac{t - \mu}{\sigma}\right) \\ \end{align}$$

$$\begin{align} E[T \mid T > c] &= \int_{c}^{\infty} t\,f_{T \mid T > c}(t)\,dt \\ &= \int_{c}^{\infty} t\,\frac{1}{(1 - \alpha)\sigma A(0)}\phi\left(\frac{t - \mu}{\sigma}\right)\,dt \\ &= \frac{1}{\sigma(1 - \alpha)\,A(0)}\int_{c}^{\infty} t\,\phi\left(\frac{t - \mu}{\sigma}\right)\,dt \\ \end{align}$$

$$\begin{align} E[T \mid T > c] &= \frac{1}{\sigma(1 - \alpha)\,A(0)}\int_{\frac{c - \mu}{\sigma}}^{\infty} (\mu + \sigma v)\,\phi(v) \cdot \sigma\,dv \\ &= \frac{1}{(1 - \alpha)\,A(0)}\int_{\frac{c - \mu}{\sigma}}^{\infty} (\mu + \sigma v)\,\phi(v)\,dv \\ &= \frac{1}{(1 - \alpha)\,A(0)}\left\{\mu \int_{\frac{c - \mu}{\sigma}}^{\infty}\phi(v)\,dv + \sigma\int_{\frac{c - \mu}{\sigma}}^{\infty}v\,\phi(v)\,dv\right\} \\ &= \frac{1}{(1 - \alpha)\,A(0)}\left\{\mu\Bigl[\Phi(v)\Bigr]_{\frac{c - \mu}{\sigma}}^{\infty} + \sigma\int_{\frac{c - \mu}{\sigma}}^{\infty}v\,\phi(v)\,dv\right\} \\ &= \frac{1}{(1 - \alpha)\,A(0)}\left\{\mu\left(\Phi(\infty) - \Phi\left(\frac{c - \mu}{\sigma}\right)\right) + \sigma\int_{\frac{c - \mu}{\sigma}}^{\infty}v\,\phi(v)\,dv\right\} \\ &= \frac{1}{(1 - \alpha)\,A(0)}\left\{\mu\left(1 - \Phi\left(\frac{c - \mu}{\sigma}\right)\right) + \sigma\int_{\frac{c - \mu}{\sigma}}^{\infty}v\,\phi(v)\,dv\right\} \\ \end{align}$$

$$\begin{equation} \Phi\left(\frac{c - \mu}{\sigma}\right) = 1 - (1 - \alpha)A(0) \end{equation}$$

$$\begin{align} \int_{a}^{\infty}v\,\phi(v)\,dv &= \int_{a}^{\infty}v\,\frac{1}{\sqrt{2\pi}}e^{-\frac{v^2}{2}}\,dv \\ &= \frac{1}{\sqrt{2\pi}}\int_{a}^{\infty}v\,e^{-\frac{v^2}{2}}\,dv \\ &= \frac{1}{\sqrt{2\pi}}\int_{a}^{\infty}\frac{d}{dv}\left(-e^{-\frac{v^2}{2}}\right)\,dv \\ &= \frac{1}{\sqrt{2\pi}}\left[-e^{-\frac{v^2}{2}}\right]_{a}^{\infty} \\ &= \frac{1}{\sqrt{2\pi}}\left\{0 + e^{-\frac{a^2}{2}}\right\} \\ &= \frac{1}{\sqrt{2\pi}}e^{-\frac{a^2}{2}} \\ &= \phi(a) \end{align}$$

$$\begin{equation} \int_{\frac{c - \mu}{\sigma}}^{\infty}v\,\phi(v)\,dv = \phi\left(\frac{c - \mu}{\sigma}\right) \end{equation}$$

$$\begin{align} E[T \mid T > c] &= \frac{1}{(1 - \alpha)\,A(0)}\left\{(1 - \alpha)A(0)\mu + \sigma\phi\left(\frac{c - \mu}{\sigma}\right)\right\} \\ &= \mu + \frac{\sigma\phi\left(\frac{c - \mu}{\sigma}\right)}{(1 - \alpha)\,A(0)} \end{align}$$

$$\begin{align} E[T \mid T > c] &= \mu + \frac{\sigma\phi\left(\frac{c - \mu}{\sigma}\right)}{(1 - \alpha)\,A(0)} \\ &= \mu + \frac{\sigma\phi\left(\frac{\mu + \sigma\Phi^{-1}\left(1 - (1 - \alpha)A(0)\right) - \mu}{\sigma}\right)}{(1 - \alpha)\,A(0)} \\ &= \mu + \frac{\sigma\phi\left(\Phi^{-1}\left(1 - (1 - \alpha)A(0)\right)\right)}{(1 - \alpha)\,A(0)} \\ \end{align}$$