If the expected value of $X^n$ is $n!$, what is the probability density function of the random variable $X$?

Question

I am working through my homework and this problem has me stumped. I don't know how to come up with a pdf just by being given an expected value? We are learning about exponential distributions, but this detail isn't stated. Is that because simply the fact that the function is $X^n$ implies it's an exponential distribution? I am also wondering if it has to do with the gamma function because I know that $$E[X^s] = \frac{\Gamma(s+1)}{\gamma^s}.$$ Any direction at all would be greatly appreciated as I am thoroughly confused.

Answer 1

Perhaps this might help:

$$\mathbb{E}[e^{tX}]=\mathbb{E}\left[\sum_{n=0}^\infty \frac{(tX)^n}{n!}\right]=\sum_{n=0}^\infty t^n=\frac{1}{1-t}$$

for $|t|<1$.

Answer 2

So $$M_X(t)=\mathsf{E}[e^{tX}]=\sum_{n=0}^{\infty}t^n\frac{\mathsf{E}[X^n]}{n!}=\frac{1}{1-t}$$ The problem boils down to finding a pdf from the above $M_X(t)$.

In doing so, note that $$\begin{align}f_X(x)&=\mathcal{L}^{-1}\{M_X(-s)\}\\&=\mathcal{L}^{-1}\{\frac{1}{1+s}\}\\&=e^{-x}\end{align}$$

Thus, $$X\sim{\tt Exp}(1)$$ That is $X$ is exponentially distributed with parameter $\lambda=1$.