Calculating probability using MGF

Question

Let $X$ be a random variable with m.g.f given by $$M(t)= \sum_{j=0}^{\infty} \frac{e^{(tj-1)}}{j!}$$

I am supposed to find $P(X=2)$.

I know that this MGF will get simplified using some known series, but I couldn't figure out which one would it be.

Answer 1

Here is another solution that works for any probability measure concentrated on $\mathbb{Z}_+=\{0\}\cup\mathbb{N}$.

Recall that for any r.v. taking values on $\mathbb{Z}_+$ $$M(t)=E[e^{tX}]=\sum_nP[X=n]e^{nt}$$ Thus, for each $n$, the coefficient of $e^{nt}$ is $P[X=n]$ In your case, the coefficient of $e^{2t}$ is $e^{-1}\frac{1}{2!}$, thus, $P[X=2]=\frac{e^{-1}}{2!}$.

Answer 2

First, we simplify the MGF: $$M(t) = e^{-1}\sum_{j=0}^\infty \frac{(e^t)^j}{j!} = e^{-1} \exp (e^t) = \exp (e^t - 1)$$ which we identify as the MGF of a Poisson distribution with parameter $\lambda = 1$. Then,

$$P(X=2) = \frac{e^{-\lambda} \cdot \lambda^2}{2!} = \frac{1}{2e}$$