3.16(標準) 正規分布4

1.答案

Z=|X|^kの分布関数は，

\begin{align}
F_Z(z)&= P[|X|^k\leq z]\\
&= P[|X|\leq z^{1/k}]\ \ \ \ \ \because 0<k\\
&= P[-z^{1/k}\leq X\leq z^{1/k}]\\
&= \int_{-\infty}^{z^{1/k}}f_X(x)dx-\int_{-\infty}^{-z^{1/k}}f_X(x)dx
\end{align}

ゆえ，確率密度関数は

\begin{align}
f_Z(z)&= f_X(z^{1/k})(\frac{1}{k}z^{\frac{1}{k}-1})-f_X(-z^{1/k})(-\frac{1}{k}z^{\frac{1}{k}-1})\\
&= 2f_X(z^{1/k})(\frac{1}{k}z^{\frac{1}{k}-1})-f_X(-z^{1/k})(-\frac{1}{k}z^{\frac{1}{k}-1})\\
&= 2f_X(z^{1/k})(\frac{1}{k}z^{\frac{1}{k}-1})\\
&= 2\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^{2/k}}{2})\frac{1}{k}z^{\frac{1}{k}-1}\\
&= \frac{2}{k\sqrt{2\pi}}z^{\frac{1}{k}-1}\exp(-\frac{z^{2/k}}{2})
\end{align}

2.答案

Z=|X|^2mと置くと，

f_Z(z)=\frac{1}{m\sqrt{2\pi}}z^{\frac{1}{2m}-1}\exp(-\frac{z^{1/m}}{2})

なので，

\begin{align}
\mathbb{E}_{X\sim f_X}[X^{2m}]&= \mathbb{E}_{X\sim f_X}[|X|^{2m}]\\
&= \mathbb{E}_{Z\sim f_Z}[Z]\\
&= \int_0^\infty\frac{1}{m\sqrt{2\pi}}z^{\frac{1}{2m}}\exp(-\frac{1}{2}z^{\frac{1}{m}})dz\tag{1}
\end{align}

ここで，t=z^1/mと置換すると，

\begin{align}
(1)&= \int_0^\infty\frac{1}{m\sqrt{2\pi}}t^{1/2}\exp(-\frac{1}{2}t)mt^{m-1}dt\\
&= \int_0^\infty\frac{1}{\sqrt{2\pi}}t^{(m+\frac{1}{2})-1}\exp(-\frac{1}{2}t)dt\\
&= \frac{(\frac{1}{2})^{1/2}}{\Gamma(\frac{1}{2})}\frac{\Gamma(\frac{1}{2}+m)}{(\frac{1}{2})^{\frac{1}{2}+m}}\int_0^\infty\frac{(\frac{1}{2})^{\frac{1}{2}+m}}{\Gamma(\frac{1}{2}+m)}t^{(m+\frac{1}{2})-1}\exp(-\frac{1}{2}t)dt\\
&= \frac{2^m\Gamma(\frac{1}{2}+m)}{\Gamma(\frac{1}{2})}
\end{align}

次に，Z=|X|^{2m+1}と置くと，

f_Z(z)=\frac{2}{(2m+1)\sqrt{2\pi}}z^{\frac{1}{2m+1}-1}\exp(-\frac{z^{\frac{2}{2m+1}}}{2})

であるから

\begin{align}
\mathbb{E}_{X\sim f_X}[|X|^{2m+1}]&= \mathbb{E}_{Z\sim f_Z}[Z]\\
&= \int_0^\infty\frac{2}{(2m+1)\sqrt{2\pi}}z^{\frac{1}{2m+1}}\exp(-\frac{1}{2}z^{\frac{2}{2m+1}})dz\tag{2}
\end{align}

ここで，

t=z^{\frac{2}{2m+1}}

と置換すると

\begin{align}
(2)&= \int_0^\infty\frac{1}{\sqrt{2\pi}}t^{(m+1)-1}\exp(-\frac{1}{2}t)dt\\
&= \frac{(\frac{1}{2})^{1/2}}{\Gamma(\frac{1}{2})}\frac{\Gamma(m+1)}{(\frac{1}{2})^{m+1}}\int_0^\infty\frac{(\frac{1}{2})^{m+1}}{\Gamma(m+1)}t^{(m+1)-1}\exp(-\frac{1}{2}t)dt\\
&= \frac{2^{m+\frac{1}{2}}\Gamma(m+1)}{\Gamma(\frac{1}{2}}
\end{align}

参考文献

• 『現代数理統計学の基礎』(久保川達也　著)