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If $Y_1,...,Y_n$ are normal with $\mu$ and $\sigma^2$, and $X_i = e^{Y_i}$, then $X_i$ are log-normal.

I want to show that $E(X_i) = e^{\mu+\frac{1}{2}\sigma^2}$.

The explanation is that $E(X_i) = E(e^{Y_i}) = M_{Y_i}(1) = e^{\mu+\frac{1}{2}\sigma^2}$, but I dont understand exactly why this is true... can someone help? I understand the "algebra", but I thought that the moment generating function had to be differentiated?

In the text the moment generating function of $Y_i$ is $M_Y(t)=e^{\mu t+\frac{1}{2}\sigma^2 t^2}$, so this is given.

insipidintegrator
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1 Answers1

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By definition $M_{Y_i}(t)=Ee^{tY_i}$. Putting $t=1$ in this gives $EX_i=Ee^{Y_i}=M_{Y_i}(1)=e^{\mu (1)+\frac 1 2\sigma ^{2} (1^{2})}$

geetha290krm
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  • Ok, so you just take the definition... but why do you differentiate the moment generating function, in other contexts? I thought that if you wanted to find $E(X)$ you would have to differentiate $M_X$ once... but, thanks! –  Jun 25 '22 at 05:10
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    To find moments of $Y_i$ from its MGF you keep differentiating it. But here what you need is simply the value of the MGF at $t=1$. You want to find $Ee^{Y_i}$ and not $EY_i$. @NumberMan – geetha290krm Jun 25 '22 at 05:13