方針
標準正規分布に従う確率変数Xの確率密度関数は，
f_X(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})
である．
1.答案
\begin{align}
\mathbb{E}_{X\sim\mathcal{N}(0,1)}[X^2]&= \int_{-\infty}^\infty\frac{x^2}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}0dx\\
&= \int_{-\infty}^\infty\frac{x}{\sqrt{2\pi}}\{-\exp(\frac{x^2}{2})\}'dx\\
&= \left[-\frac{x}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})\right]_{-\infty}^\infty+\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})dx\\
&= 1
\end{align}
2.答案
Y=X^2の分布関数は，
\begin{align}
F_Y(y)&= P[Y\leq y]\\
&= P[X^2\leq y]\\
&= P[-\sqrt{y}\leq X\leq\sqrt{y}]\\
&= \int_{-\infty}^{\sqrt{y}}f_X(x)dx-\int_{-\infty}^{-\sqrt{y}}f_X(x)dx
\end{align}
より，
\begin{align}
f_Y(y)&= f_X(\sqrt{y})\frac{1}{2\sqrt{y}}-f_X(-\sqrt{y})(-\frac{1}{2\sqrt{y}})\\
&= \frac{1}{\sqrt{y}}f_X(\sqrt{y})\\
&= \frac{1}{\sqrt{2\pi y}}\exp(-\frac{y}{2})
\end{align}
である．従って，
\begin{align}
\mathbb{E}_{Y\sim f_Y}[Y]&= \int_0^\infty\frac{1}{\sqrt{2\pi}}y^{1/2}\exp(-\frac{1}{2}y)dy\\
&= \frac{1}{\sqrt{2\pi}}\frac{\Gamma(\frac{3}{2})}{(\frac{1}{2})^{3/2}}\int_0^\infty\frac{(\frac{1}{2})^{3/2}}{\Gamma(\frac{3}{2})}y^{\frac{3}{2}-1}\exp(-\frac{1}{2}y)dy\\
&= 1
\end{align}
3.答案
Y=|X|の分布関数は，
\begin{align}
F_Y(y)&= P[Y\leq y]\\
&= P[|X|\leq y]\\
&= P[-y\leq X\leq y]\\
&= \int_{-\infty}^yf_X(x)dx-\int_{-\infty}^{-y}f_X(x)dx
\end{align}
確率密度関数は，
\begin{align}
f_Y(y)&= f_X(y)-f_X(-y)(-1)\\
&= 2f_X(y)\\
&= \frac{2}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})
\end{align}
より，
\begin{align}
\mathbb{E}_{Y\sim f_Y}[Y]&= \int_0^\infty\frac{2y}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})dy\\
&= \left[-\frac{2}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})\right]_0^\infty\\
&= \sqrt{\frac{2}{\pi}}\\
\mathbb{E}_{Y\sim f_Y}[Y^2]&= \int_0^\infty\frac{2y^2}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})dy\\
&= \int_0^\infty\frac{2y}{\sqrt{2\pi}}\{-\exp(-\frac{y^2}{2})\}'dy\\
&= \left[-\frac{2y}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})\right]_0^\infty+\int_0^\infty\frac{2}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})dy\\
&= \int_0^\infty\frac{1}{\sqrt{2\pi}}t^{-\frac{1}{2}}\exp(-\frac{1}{2}t)dt,\ \ \ \ (t=y^2)\\
&= \frac{1}{\sqrt{2\pi}}\frac{\Gamma(\frac{1}{2})}{(\frac{1}{2})^{1/2}}\int_0^\infty\frac{(\frac{1}{2})^{1/2}}{\Gamma(\frac{1}{2})}t^{\frac{1}{2}-1}\exp(-\frac{1}{2}t)dt\\
&= 1\\
Var(Y)&= 1-\frac{2}{\sqrt{\pi}}
\end{align}
参考文献
- 『現代数理統計学の基礎』(久保川達也 著)