Variance of $Y = a \cdot e^X$ if $X$ is Gussian random variable

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.

• 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 $Var(g(X))=E[g(X)^2]-E[g(X)]^2$. Aug 20, 2021 at 14:49
• @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? Aug 20, 2021 at 20:30
• 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. Aug 20, 2021 at 21:48

If $$X\sim N(\mu;\sigma^2)$$ then $$Y=e^X\sim\text{LogNormal}$$ thus you can calculate $$\mathbb{V}[Y]$$ exactly
$$\mathbb{V}[Y]=a^2\cdot e^{2\mu+\sigma^2}(e^{\sigma^2}-1)$$
• Cool, I didn't know that. What if $X$ is a random variable where the distribution is not know? Does my approach work then? Aug 20, 2021 at 14:32