答案
Xの期待値は
\begin{align}
\mathbb{E}_{X\sim f_X}[X]&= \int_\alpha^\infty\frac{\beta\alpha^\beta}{x^\beta}dx\\
&= \beta\alpha^\beta\int_\alpha^\infty x^{-\beta}dx\\
&= \beta\alpha^\beta\left[-\frac{1}{\beta-1}x^{-(\beta-1)}\right]_\alpha^\infty\\
&= \frac{\alpha\beta}{\beta-1},\ \ (1<\beta)
\end{align}
と，期待値は1<betaの時のみ発散せずに解が求まる．分散は．
\begin{align}
\mathbb{E}_{X\sim f_X}[X^2]&= \int_\alpha^\infty\frac{\beta\alpha^\beta}{x^{\beta-1}}dx\\
&= \beta\alpha^\beta\int_\alpha^\infty x^{-(\beta-1)}dx\\
&= \beta\alpha^\beta\left[-\frac{1}{\beta-2}x^{-(\beta-2)}\right]_\alpha^\infty\\
&= \frac{\alpha^2\beta}{\beta-2},\ \ (2<\beta)\\
Var(X)&= \frac{\alpha^2\beta}{(\beta-2)(\beta-1)^2}
\end{align}
今，Y=log Xなる変換を行うと，
\begin{align}
F_Y(y)&= P[Y\leq y]\\
&= P[\log X\leq y]\\
&= P[X\leq e^y]\\
&= \int_\alpha^{e^y}f_X(x)dx\\
\therefore f_Y(y)&= f_X(e^y)\frac{d e^y}{dy}\\
&= \frac{\beta\alpha^\beta e^y}{\exp\{y(\beta+1)\}}\\
&= \beta\alpha^\beta\exp(-\beta y)
\end{align}
ここで，
\beta=\frac{1}{\beta},\ \alpha=e^y
を代入すると，
f_Y(y)=\frac{1}{\sigma}\exp(-\frac{y-\mu}{\sigma})
参考文献
- 『現代数理統計学の基礎』(久保川達也 著)