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Let $A$ (or $X$) be

$\log A \sim N(\mu,\sigma^2)$, (lognormal distribution)

I have to show

$$E[A] = \exp[\mu + (\sigma^2/2)]\mbox{ and }E[A^2] = \exp[2\mu + 2\sigma^2].$$

Do I have to use mgf of the normal dist. ?

It is easy to show E[$A^2$] since it is the second order derivative of the mgf.

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2 Answers

up vote 1 down vote accepted

One may start with $\log A=\mu+\sigma Z$ and $Z$ standard gaussian. Thus $A^x=\mathrm e^{x\mu}\cdot\mathrm e^{x\sigma Z}$ for every real number $x$, hence $\mathbb E(A^x)=\mathrm e^{x\mu}\cdot\psi(x\sigma)$ where $\psi(t)=\mathbb E(\mathrm e^{t Z})$. Furthermore, $$ \psi(t)=\int_{-\infty}^{+\infty}\mathrm e^{tz}\cdot\mathrm e^{-z^2/2}\frac{\mathrm dz}{\sqrt{2\pi}}=\mathrm e^{t^2/2}\int_{-\infty}^{+\infty}\mathrm e^{-(z-t)^2/2}\frac{\mathrm dz}{\sqrt{2\pi}}=\mathrm e^{t^2/2}. $$ Hence, for every real number $x$, $$ \mathbb E(A^x)=\mathrm e^{x\mu+x^2\sigma^2/2}. $$

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Yes, the mgf would be an easy way to do this. Just replace $t = 1$ and you get the mean.

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But replacing with t= 1 instead of t = 0 like in normal dist, gives E[A] = ($\mu$ + $\sigma^2$)exp[$\mu$ + ($\sigma^2$/2)] –  Ricardian Oct 17 '12 at 19:36
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