How to prove $E[e^{e^y}]=\infty$? y is a normal random variable

Question

The question is, given $Y\sim N(\mu,\sigma^2)$, how to prove$E[e^{e^Y}]=\infty$?

I tried to look Y as some kind of Ito's process and apply Ito's formula to it but it doesn't make sense. Next I tried to use variable change $u=e^y$ and I still can't prove it. Is there anyway to do that?

Answer 1

Well, let's explore the expectation directly:

$$E\left[e^{e^y}\right] = \int_{-\infty}^\infty e^{e^x} e^{-(x-\mu)^2/(2\sigma^2)}\, dx = \int_{-\infty}^\infty \exp\left( e^x-\frac{(x-\mu)^2}{2\sigma^2}\right)\, dx.$$

Now take $x \to +\infty$; it is clear that $e^x \to \infty$ much faster than $-x^2 \to -\infty$, so it dominates in the exponential. Therefore, $\int_a^\infty \exp\left( e^x-\frac{(x-\mu)^2}{2\sigma^2}\right)\, dx = \infty$. This is enough.

Answer 2

We have to check that the integral $$\int_0^{+\infty}\exp\left(e^y-(y-\mu)^2/(2\sigma^2)\right)\mathrm dy$$ is divergent. There is a huge problem at $+\infty$ because $\lim_{y \to +\infty}e^y-(y-\mu)^2/(2\sigma^2)=+\infty$.