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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.

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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 $\frac 1 {\sqrt {2\pi}} \int_{-\infty}^{\infty} e^{-\frac 1 2 y^{2}}\, dy =1$]

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  • $\begingroup$ Should the substitution be $y=\log x$? $\endgroup$ – ddswsd Jan 26 at 5:19
  • $\begingroup$ @ddswsd Yes, sorry for the typo. $\endgroup$ – Kavi Rama Murthy Jan 26 at 5:24
  • $\begingroup$ If I let $y= \log x$, then the integral becomes ${1\over \sqrt{2\pi}}\int_{- \infty} ^{\infty} e^{-0.5y^{2}+y} dy$. I don't think that equals $\sqrt{e}$. $\endgroup$ – ddswsd Jan 26 at 14:16
  • $\begingroup$ @ddswsd The integral you have written is $\sqrt e$ as far as I can see. Can you show me how you evaluated it? $\endgroup$ – Kavi Rama Murthy Jan 26 at 23:26
  • $\begingroup$ Interesting...I put that integral into wolfram alpha and used -0.5 and 0.5 instead of writing them as fractions and the answer they gave wasn't $\sqrt{e}$. Then I changed the decimals to fractions and the answer is $\sqrt{e}$. I see why you chose to complete the square. Really clever! Thank you! $\endgroup$ – ddswsd Jan 27 at 0:13

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