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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 $\sigma= \sigma'c$.

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    $\begingroup$ Why is this question getting so many views? $\endgroup$ – カカロット Aug 17 at 23:12
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    $\begingroup$ @Zacky Nero = Taleb $\endgroup$ – JP McCarthy Aug 17 at 23:18
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    $\begingroup$ @JPMcCarthy can you give more details? Who is Taleb? $\endgroup$ – カカロット Aug 17 at 23:22
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    $\begingroup$ @Zacky This is why twitter.com/nntaleb/status/1162860369225932812 $\endgroup$ – Arin Chaudhuri Aug 17 at 23:23
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    $\begingroup$ Thanks for clarifying!// @Nero, can you bring some background about your question? $\endgroup$ – カカロット Aug 17 at 23:28
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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.

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  • $\begingroup$ Thanks it turns out we need to take -x not x. I corrected above. Will retry with your method. $\endgroup$ – Nero Aug 18 at 12:26
  • $\begingroup$ So it looks like we need to put $(-1)^n$ inside the summation so we get the moments of the flipped Lognormal but I am not sure of the convergence. $\endgroup$ – Nero Aug 18 at 12:54
  • $\begingroup$ Great news it is convergent for some values of c $\endgroup$ – Nero Aug 18 at 13:04

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