Consider a random variable $X$ with the log-normal pdf $f(x) ={1\over \sqrt{2π}}x^{−1}exp^{{−0.5 (logx)^2}}$, $x>0$.

Question

Consider a random variable $X$ with the log-normal pdf $f(x) ={1\over \sqrt{2π}}x^{−1}exp^{{−0.5 (logx)^2}}$, $x>0$.

a) Find the mean and the variance of $X$.

I know that $\int_0^\infty f(x) dx=1$, but this says "consider the random variable $X$ with the log-normal pdf" so would I actually be integrating $\int_0^\infty x f(x) dx$ for the mean? The reason I ask (and I'm skeptical) is because the solution to that integral is disgusting and has $i$ in the solution and that doesn't seem to be applicable for what I'm doing. Any help is greatly appreciated.

Answer 1

hint: it is $\int_0^{\infty} xf(x)\, dx$. To evaluate this make the substitution $y=\log\, x$. And then use the following: $-\frac 12 y^{2}+y=-\frac 1 2 (y-1)^{2} +\frac 1 2$. You should get the answer as $\sqrt e$. [ You have to use the fact $\frac 1 {\sqrt {2\pi}} \int_{-\infty}^{\infty} e^{-\frac 1 2 y^{2}}\, dy =1$]