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April 1st, 2017, 10:29 AM  #1 
Newbie Joined: Apr 2017 From: Canada Posts: 1 Thanks: 0  Computing the expectation of a lognormal variable using its pdf
Hello, Someone help me understand what is wrong in the lognormal 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 10:59 AM. 
April 1st, 2017, 03:48 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,660 Thanks: 844 
It's not $f(X)$ it's $f(\ln(X))$ 

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computing, expectation, lognormal, pdf, variable 
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