I am trying to find the distribution that corresponds to this moment-generating function.
$$ M(t) =\frac{1}{3e^{-t}-2} , \quad t < \ln \frac 3 2 $$
I can not even consider where to start. Any push in the right direction would be appreciated! Thanks:)
My best advice for situations like this is to be familiar with the forms of the MGFs of your special distributions.
In particular, consider the following.
Claim: If $X\sim\operatorname{Geom}(\theta),$ then the MGF of $X$ is $$M_X(t)=\frac{\theta e^t}{1-e^t(1-\theta)}=\cfrac1{\frac1\theta e^{-t}-\frac{1-\theta}\theta},$$ where $t<\ln\frac1{1-\theta}.$
Proof: Observe that $$\begin{align}\Bbb E\left[e^{tX}\right] &= \sum_{x=1}^\infty e^{tx}\theta(1-\theta)^{x-1}\\ &= \sum_{x=0}^\infty e^{t(x+1)}\theta(1-\theta)^x\\ &= \theta e^t\sum_{x=0}^\infty e^{tx}(1-\theta)^x\\ &= \theta e^t\sum_{x=0}^\infty\left[e^t(1-\theta)\right]^x.\end{align}$$ Now, $e^t(1-\theta)=|e^t(1-\theta)|,$ so when $e^t(1-\theta)<1$--equivalently, when $t<\ln\frac1{1-\theta}$--the series at the end is a convergent geometric series, so $$M_X(t):=\Bbb E\left[e^{tX}\right]=\theta e^t\cdot\frac{1}{1-e^t(1-\theta)}=\frac{\theta e^t}{1-e^t(1-\theta)},$$ as desired. $\Box$
Can you see what $\theta$ you should use, here?
