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April 1st, 2017, 09:29 AM   #1
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Computing the expectation of a log-normal variable using its pdf


Someone help me understand what is wrong in the log-normal pdf. Say we have this guy $\displaystyle Y=e^X$, for which pdf is:

$\displaystyle f(Y)=\frac{1}{Y\sigma\sqrt{2\pi}}\exp\left(-\frac{(\ln(Y)-\mu _X)^2}{2\sigma ^2}\right) = \frac{f(X)}{Y}$.

Generally speaking, the expectation of a random variable given by $\displaystyle E[Y]=\int{Y\cdot f(Y)dY}$. So:

$\displaystyle \int{Y\cdot f(Y)dY}=\int{Y\frac{f(X)}{Y}dY}=\int{f(X)dX}=1$

This is wrong because the expectation of our lognormal random variable is $\displaystyle E[Y]=\exp\left(E[X]+\frac{\sigma^2}{2}\right)$.

So, why does E[Y] using $\displaystyle \int{Y\cdot f(Y)dY}$ doesn't give the E[Y]?

I know the way to obtain the correct expectation, I should simply take the integral with the normal pdf, $\displaystyle E[Y]=\int{\exp(X)\cdot f(X)dX}=\exp\left(E[X]+\frac{\sigma^2}{2}\right)$, but it still sounds reasonable that it should succeed both ways.

Is there something wrong with the lognormal pdf? Since anyway it doesn't seem right that you would divide something that integrates to one (that is f(X)) by something that is mostly larger than one (that is Y) and still get f(Y) integrate to one.

Help anyone?

Last edited by skipjack; April 1st, 2017 at 09:59 AM.
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April 1st, 2017, 02:48 PM   #2
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It's not $f(X)$

it's $f(\ln(X))$

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