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?
For simplicity I consider first the case $\mu=0,\sigma=1$. (The more general formula follows at the end.)
A saddlepoint approximation 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.
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.
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 $k'=k\sigma^2 e^{-\mu}$.
