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Suppose we are give $X$ which is normally distributed with mean $\mu$ and variance $\sigma^2$. How do I show that

$$E[e^{e^X}]=\infty$$

Of course I wanted to find a lower bound, which also explodes to conclude. However I did not find the right one.

Thanks for your help

hulik

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up vote 2 down vote accepted

The problem comes from the divergence of $\int_{\Bbb R}e^{e^x-x^2/2}dx$, which can be seen noticing that $e^x-x^2/2\geqslant x\geqslant 0$ for $x\geqslant 0$.

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