Looking for the expectation $\mathbb{E}(t^{-x})$, with the random variable X $\approx$ Lognormal Distribution $(\mu,\sigma)$.
More specifically I am looking for the special case $t=2$.
$\textbf{Background}$:
Was interested in the expectation of $2^{-e^{-c x}},\; c>0$, which is a sigmoidal function, with $X \approx$ a normal distribution$(\mu',\sigma')$. Let $z={-e^{-c x}}$, then z is Lognormal with parameters $-c K-c \mu' ,\sigma' c$. Use $\mu=−𝑐𝐾−𝑐𝜇′$ and $\sigma= \sigma'c$.
Best Answer
Let $t\in\mathbb{R}_{>1}$ and $x\in \mathbb{R}$.
Then we have that $t^x=\sum_{n=0}^{\infty}\frac{\log(t)^n}{n!}x^n$.
Now for $X\sim\mathcal{LN}(\mu,\sigma)$, $X=e^Y$, with $Y\sim\mathcal{N}(\mu,\sigma)$ on has that:
$$\mathbb{E}[t^X] = \mathbb{E}[t^{e^Y}] = \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{+\infty}t^{e^y}e^{-\frac{(y-\mu)^2}{2\sigma^2}}\mathrm{d}y = \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{+\infty}\sum_{n=0}^{\infty}\frac{\log(t)^n}{n!}(e^{y})^n e^{-\frac{(y-\mu)^2}{2\sigma^2}}\mathrm{d}y$$ $$\geq \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{+\infty}\sum_{n=0}^{N}\frac{\log(t)^n}{n!}e^{ny} e^{-\frac{(y-\mu)^2}{2\sigma^2}}\mathrm{d}y = \sum_{i=0}^{N} \frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{+\infty}\frac{\log(t)^n}{n!}e^{ny} e^{-\frac{(y-\mu)^2}{2\sigma^2}}\mathrm{d}y$$. $$= \sum_{i=0}^{N}\frac{\log(t)^n}{n!}e^{n\mu+\frac{n^2\sigma^2}{2}}$$ for all $N\in\mathbb{N}$.
(For the inequality we use that all summands are positive and the last integral is a standard computation.)
It is easy to see that the last expression goes to $\infty$ for $N\longrightarrow\infty$, so in particular in your special case the factorial moment generating dunction does not exist.