# Factorial Moment Generating Function of the Lognormal

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 $$\sigma= \sigma'c$$.

• Why is this question getting so many views? – Zacky Aug 17 '19 at 23:12
• @Zacky Nero = Taleb – JP McCarthy Aug 17 '19 at 23:18
• @JPMcCarthy can you give more details? Who is Taleb? – Zacky Aug 17 '19 at 23:22
• @Zacky This is why twitter.com/nntaleb/status/1162860369225932812 – Arin Chaudhuri Aug 17 '19 at 23:23
• Thanks for clarifying!// @Nero, can you bring some background about your question? – Zacky Aug 17 '19 at 23:28

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.

• Thanks it turns out we need to take -x not x. I corrected above. Will retry with your method. – Nero Aug 18 '19 at 12:26
• 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. – Nero Aug 18 '19 at 12:54
• Great news it is convergent for some values of c – Nero Aug 18 '19 at 13:04