I am working on a problem and am a little bit confused.
I need to find the distribution that corresponds to the MGF:
$2e^t\over3-e^t$
Do we need to separate this into something like:
2e$^t$ and $1\over3-e^t$
It seems like we might be able to separate this into X ~ Exp() or Poi()
The MGF of a Poisson r.v. features a double exponential, and that of an exponential r.v. is a rational function (no exponential), so it looks unlikely at first glance these two would be candidates (also, unclear what you mean by "separate" here.)
However, eyeballing the Wikipedia table showing a list of standard MGFs, you can spot one that looks quite similar, that of the geometric distribution. Namely, if $X$ follows a Geometric distribution with parameter $p$, then the MGF of X is $$ M_x(t) = \frac{p e^t}{1-(1-p)e^t}, \quad t < \log\frac{1}{1-p} \tag{1} $$ So let's see. What you have is $$ M_X(t) = \frac{2e^t}{3-e^t} = \frac{\frac{2}{3}e^t}{1-\frac{1}{3}e^t} = \frac{\frac{2}{3}e^t}{1-\left(1-\frac{2}{3}\right)e^t} \tag{2} $$ which should allow you to conclude.