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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.

ddswsd
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1 Answers1

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

Kavi Rama Murthy
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  • Should the substitution be $y=\log x$? – ddswsd Jan 26 '19 at 05:19
  • @ddswsd Yes, sorry for the typo. – Kavi Rama Murthy Jan 26 '19 at 05:24
  • 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 $\sqrt{e}$. – ddswsd Jan 26 '19 at 14:16
  • @ddswsd The integral you have written is $\sqrt e$ as far as I can see. Can you show me how you evaluated it? – Kavi Rama Murthy Jan 26 '19 at 23:26
  • 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! – ddswsd Jan 27 '19 at 00:13
  • Where can I find the derivation of that integral? The textbook I use just states that it equals $1$. I want to be able to understand the derivation. – ddswsd Jan 27 '19 at 02:18
  • $(\int_0^{\infty} e^{-x^{2}/2}, dx)^{2} =\int_0^{\infty} e^{-x^{2}/2}, dx\int_0^{\infty} e^{-y^{2}/2}, dy=\int_{\mathbb R^{2}} e^{-(x^{2}+y^{2})/2}, dxdy$. Use polar coordinates to evaluate this. You will get $2\pi$. – Kavi Rama Murthy Jan 27 '19 at 04:57