If $X$ is normal, is $\exp(X)$ still normal? How to find its mean and variance?

Question

$X$ 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.

Answer 1

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 $$Var(\exp(X)) = (\exp(\sigma^2) -1)\exp(2\mu + \sigma^2)$$

Answer 2

$\mathbb P\{e^X\leqslant 0\}=0$ 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 $$(\sqrt{2\pi}\sigma)^{-1}\int_{-\infty}^{+\infty}\exp\left(t-\frac{t^2}{2\sigma^2}\right)\mathrm dt, \mbox{ and }$$ $$(\sqrt{2\pi}\sigma)^{-1}\int_{-\infty}^{+\infty}\exp\left(2t-\frac{t^2}{2\sigma^2}\right)\mathrm dt.$$