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I have the following function

$$Y = a e^X$$

where $X$ is a normally distributed random variable. I would like to compute the variance of $Y$. So far I did the following

\begin{eqnarray} \text{Var}(Y) &= \text{Var}(ae^X)\\ &= a^2 \text{Var}(e^X)\\ &\approx a^2 e^{2E(X)} \text{Var}(X) \end{eqnarray}

Where I used information I have found on this site on variance propagation. I would like to know if this calculation is correct and if there are better and more correct ways to compute $\text{Var}(Y)$? So far, this is only an approximation, if what I did is right.

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  • $\begingroup$ Your approximation does not make sense because $Var(Y)$ is a number, but your approximation is a function of $X$. So you approximate it as the wrong type of thing ("type error"). In general, for the expectation of a function of $X$ you can use $E[g(X)] = \int_{-\infty}^{\infty} g(x)f_X(x)dx$. And $Var(g(X))=E[g(X)^2]-E[g(X)]^2$. $\endgroup$
    – Michael
    Aug 20, 2021 at 14:49
  • $\begingroup$ @Michael Yes, you are absolutely right. I think it should be $\text{Var}(Y) \approx a^2e^{2E(X)}\text{Var}(X)$, right? Would that make more sense? Does this also hold for random variables that are not normally distributed? $\endgroup$
    – Gilfoyle
    Aug 20, 2021 at 20:30
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    $\begingroup$ I have never used or seen such a variance approximation formula. I would not worry about memorizing it. Nevertheless your modified approximation makes sense, I can see it being justified (for any random variable $X$ with $m=E[X]$ and with $X$ "typically close" to $m$) by $$ Var(a e^X) = a^2Var(e^X) = a^2Var(e^me^{X-m}) = a^2e^{2m}Var(e^{X-m}) \approx a^2 e^{2m}Var(1+(X-m)) = a^2e^{2m}Var(X)$$ Of course this approximation may be "infinitely bad" in the sense that $E[X]$ and $Var(X)$ is finite but $E[e^X]$ and $Var(e^X)$ can be infinite. $\endgroup$
    – Michael
    Aug 20, 2021 at 21:48

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If $X\sim N(\mu;\sigma^2)$ then $Y=e^X\sim\text{LogNormal}$ thus you can calculate $\mathbb{V}[Y]$ exactly

Lognormal distribution

$$\mathbb{V}[Y]=a^2\cdot e^{2\mu+\sigma^2}(e^{\sigma^2}-1)$$

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  • $\begingroup$ Cool, I didn't know that. What if $X$ is a random variable where the distribution is not know? Does my approach work then? $\endgroup$
    – Gilfoyle
    Aug 20, 2021 at 14:32

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