Finding the Law of X

Question

I am woking on a problem and am unsure how to approach it.

It is finding the law of X (the distribution that correspond to) for MGF $2e^t\over3-e^t$

I am thinking that this is perhaps an exponential distribution

X ~ Exp(lambda) or perhaps X ~ Poi(lambda)

Do we need to break this apart into:

2e$^t$ and $1\over3-e^t$

Answer 1

Use the formula for a geometric series namely $(1-x)^{-1}=\sum_{k=0}^\infty x^k$ if $|x|<1$, to write that $$ Ee^{tX}=\frac{2e^t}{3-e^t}=2\frac{e^t/3}{1-(e^t/3)}=\sum_{k=1}^\infty2\left(\frac{e^t}{3}\right)^k=\sum_{k=1}^\infty2\times 3^{-k}e^{tk} $$ provided that $e^t<3$. In any case it follows that $$ P(X=k)=2\times 3^{-k}\quad (k=1,2,\dotsc) $$