答案
Xの期待値は，
\begin{align}
\mathbb{E}_{X\sim f_X}[X]&= \int_0^\infty\frac{1}{\sqrt{2\pi}}\exp\{-\frac{(\log x)^2}{2}\}dx\tag{1}
\end{align}
t=log xと置換すると，
\begin{align}
(1)&= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp(-\frac{t^2}{2}+t)dt\\
&= \int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}\exp\{-\frac{(t-1)^2}{2}+\frac{1}{2}\}dt\\
&= e^{1/2}
\end{align}
分散は，
\begin{align}
\mathbb{E}_{X\sim f_X}[X^2]&= \int_0^\infty\frac{x}{\sqrt{2\pi}}\exp\{-\frac{(\log x)^2}{2}\}dx\tag{2}
\end{align}
t=log xと置換すると，
\begin{align}
(2)&= \int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}\exp(-\frac{t^2}{2}+2t)dt\\
&= \int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}\exp\{-\frac{(t-2)^2}{2}+2\}dt\\
&= e^2\\
Var(X)&= e^2-e=e
\end{align}
次に，Y=logXなる変換を考える．
\begin{align}
P[Y\leq y]&= P[\log X\leq y]\\
&= P[X\leq e^y]\\
&= \int_0^{e^y}f_X(x)dx
\end{align}
なので，
\begin{align}
f_Y(y)&= f_X(e^y)\frac{de^y}{dy}\\
&= \frac{1}{\sqrt{2\pi}}\exp\{\frac{-(\log e^y)^2}{2}+y\}\\
&= \frac{1}{\sqrt{2\pi}}\exp(-\frac{y^2}{2})
\end{align}
参考文献
- 『現代数理統計学の基礎』(久保川達也 著)