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$X$ is a random variable for normal distribution: $X\sim N(\mu, \sigma^2)$.

What is the mean and variance of $e^{X}$?

My attempt:
$$E[e^{X}]=e^{E[x]} \text{, by the invariance property?}$$ $$\operatorname{var}(e^{x})=e^{\operatorname{var}(x)}, \text{ similarly}$$

This looks too easy, probably not right.

Should I look at $e^{X}$ as a whole. use moment generating function?
But normal pdf requires $e^{x^2}$. I'm stuck.

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  • $\begingroup$ No, $\exp(X)$ is certainly not normal (for example, all its values are positive). Also, usually $E(\exp(X)) \ne \exp(E\; X)$and $var(\exp(X)) \ne \exp(var \;X)$ $\endgroup$
    – GEdgar
    Commented Feb 2, 2014 at 17:59
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    $\begingroup$ There is no such "invariance property" applying to any but affine functions. The exponential function is not affine. Maybe you're confusing this with an invariance property that applies to maximum-likelihood estimation. $\endgroup$
    – Michael Hardy
    Commented Feb 2, 2014 at 18:03
  • $\begingroup$ @MichaelHardy So, how do I know when to use the invariance property (of MLE)? I suppose that's only for MLE. $\endgroup$
    – user13985
    Commented Feb 2, 2014 at 18:10
  • $\begingroup$ You can apply it to MLEs. $\endgroup$
    – Michael Hardy
    Commented Feb 2, 2014 at 18:14

2 Answers 2

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If $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$ then $\exp(X)$ has a log-normal distribution; it is not symmetric and it cannot take negative values so it cannot be normal.

In fact $$E[\exp(X)]= \exp(\mu + \sigma^2/2)$$ and $$Var(\exp(X)) = (\exp(\sigma^2) -1)\exp(2\mu + \sigma^2)$$

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  • $\begingroup$ Now I am confused about negative values. It seems strange to me why normal could take negative values. They are both exp to some power. Why is one exp different from the other exp? $\endgroup$
    – user13985
    Commented Feb 2, 2014 at 18:05
  • $\begingroup$ The exponential function may appear in the normal distribution's density, but a normal random variable $X$ is not $\exp(Y)$ for some other random variable $Y$ $\endgroup$
    – Henry
    Commented Feb 2, 2014 at 18:07
  • $\begingroup$ then, why does one spit out negative numbers, the other doesn't? It just doesn't seem intuitive. Is the negative sign in front of the normal pdf the culprit? $\endgroup$
    – user13985
    Commented Feb 2, 2014 at 18:16
  • $\begingroup$ Why is it called lognormal when you are actually raising it to the power, namely, exp{x}. Shouldn't it be called exponential-normal? $\endgroup$
    – user13985
    Commented Feb 2, 2014 at 19:03
  • $\begingroup$ Does lognormal actually have MGF? Wiki says yes. Another post here asks why it doesn't. I'm confused. $\endgroup$
    – user13985
    Commented Feb 2, 2014 at 19:20
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$\mathbb P\{e^X\leqslant 0\}=0$ so $e^X$ cannot be normal. And in general $\mathbb E[f(X)]\neq f(\mathbb E(X))$.

We can only consider the case $\mu=0$ (why?) and we are reduced to compute the integrals $$(\sqrt{2\pi}\sigma)^{-1}\int_{-\infty}^{+\infty}\exp\left(t-\frac{t^2}{2\sigma^2}\right)\mathrm dt, \mbox{ and }$$ $$(\sqrt{2\pi}\sigma)^{-1}\int_{-\infty}^{+\infty}\exp\left(2t-\frac{t^2}{2\sigma^2}\right)\mathrm dt.$$

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  • $\begingroup$ That integration looks doable. $\endgroup$
    – user13985
    Commented Feb 2, 2014 at 18:20

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