If $X$ is normal, is $\exp(X)$ still normal? How to find its mean and variance?
$X$ is a random variable for normal distribution: $X\sim N(\mu, \sigma^2)$.
What is the mean and variance of $\exp\{x\}$?
My attempt:
$$E[\exp\{x\}]=\exp \{E[x]\} \text{, by the invariance property?}$$
$$\operatorname{var}(\exp\{x\})=\exp\{\operatorname{var}(x)\}, \text{ similarly}$$
This looks too easy, probably not right.
Should I look at exp{x} as a whole. use moment generating function?
But normal pdf requires $\exp\{x^2\}$. I'm stuck.
2 Answers
$\begingroup$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 $$Var(\exp(X)) = (\exp(\sigma^2) -1)\exp(2\mu + \sigma^2)$$
$\endgroup$ 6 $\begingroup$$\mathbb P\{e^X\leqslant 0\}=0$ so $e^X$ cannot be normal. And in general $\mathbb E[f(X)]\neq f(\mathbb E(X))$.
We can only consider the case $\mu=0$ (why?) and we are reduced to compute the integrals $$(\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.$$
$\endgroup$ 1
