I would like to ask this question that I received from my lecturer. The question was to prove the given mgf is actually a poisson distribution.
Here's the mgf; $M_x(t)=8^{(e^t)-1}$
The pmf of the poisson distribution; $P(X=x)={(\frac{3}{2})}^x {\frac{(\ln 2)^x}{x!{2^{(3-x)}}}}$
What I know is: the mean and variance of the Poisson are the same (= $\lambda$), but I can't figure out how to do it.
Thank you.
The MGF of a poisson is
$$M_X(t)=e^{\theta(e^t-1)}$$
You can rewrite your expression in the following way:
$$8^{e^t-1}=e^{\log8^{(e^t-1)}}=e^{\log 8(e^t-1)}$$
Which is the MGF of a poisson $\text{Poi}(\log8)$ that is
$$\mathbb{P}[X=x]=\frac{\left(\frac{3}{2} \right)^x(\log 2)^x}{x!\cdot2^{3-x}}=\frac{e^{-\log8}(\log 8)^x}{x!}\cdot\mathbb{1}_{\{0;1;2;3;\dots\}}(x)$$