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

$$\begin{equation} f_Z(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} \end{equation}$$

$$\begin{equation} P(Z \ge 0) = \frac{1}{2} \end{equation}$$

$$\begin{equation} f_{Z \mid Z \ge 0}(z) = \frac{f_Z(z)}{P(Z \ge 0)} = \frac{2}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} \end{equation}$$

$$\begin{align} E[Z \mid Z \ge 0] &= \int_{0}^{\infty}z\,f_{Z \mid Z \ge 0}(z)\,dz \\ &= \int_{0}^{\infty}z \frac{2}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\,dz \\ &= \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty}z e^{-\frac{z^2}{2}}\,dz \\ &= \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty} \frac{d}{dz}\left(-e^{-\frac{z^2}{2}}\right)\,dz \\ &= \frac{2}{\sqrt{2\pi}} \left[-e^{-\frac{z^2}{2}}\right]_{0}^{\infty} \\ &= \frac{2}{\sqrt{2\pi}} \\ &= \sqrt{\frac{2}{\pi}} \\ \end{align}$$

$$\begin{align} E[Z \mid Z < 0] = -\sqrt{\frac{2}{\pi}} \end{align}$$