Let $X$ be a random variable from a normal distribution $N(\mu, \sigma)$. How do we calculate the expectation $E[e^{k\cdot e^{-X}}]$, where $k<0$?


I think we can use the moment generating function


$
E[e^{k\cdot e^{-X}}] = \sum\_{t=0}^{\infty}\frac{k^{t}E[e^{-tX}]}{t!}
$, where $E[e^{-tX}]=e^{-t\mu+\frac{1}{2}\sigma^{2}t^{2}}$


But is there a more "compact" solution beyond the one above?



The expected value of a double exponential function of normal random variable
For simplicity I consider first the case $\mu=0,\sigma=1$. (The more general formula follows at the end.)


A [saddlepoint approximation](https://en.wikipedia.org/wiki/Saddlepoint_approximation_method) of the integral
$$I(k)=(2\pi)^{-1/2}\int\_{-\infty}^\infty e^{-x^2/2}e^{ke^{-x}}\,dx$$
gives for large $-k$ the expression
$$I\_\infty(k)=[1+W(-k)]^{-1/2}e^{-W(-k)-\tfrac{1}{2}W(-k)^2},$$
with $W(x)$ the [Lambert W-function.](https://en.wikipedia.org/wiki/Lambert_W_function)  

This approximation is highly accurate already for moderately large $|k|$. See the plot, where I compare the two (blue for $I\_\infty$, gold for $I$) --- they are nearly indistinguishable.


![](https://i.stack.imgur.com/8Xhdv.png)
In the more general case of arbitrary $\mu,\sigma$ one has
$$I\_\infty(k)=[1+W(-k')]^{-1/2}\exp\left[-\sigma^{-2}W(-k')-\tfrac{1}{2}\sigma^{-2}W(-k')^2\right],$$
with $k'=k\sigma^2 e^{-\mu}$.


