How to find probability from moment generating Function?

Question

Given:

$$M_X(t)=\frac13\exp(e^t-1) + \frac23\exp(2(e^t-1))$$

How can I find $P(X=0)?$

I know that: $M_X(t)=E[e^{tx}]$ so: $M_X(1)=E[e^{x}] = \frac13\exp(e-1) + \frac23\exp(2(e-1))$

Answer 1

From the moment generating function of $X$, one can see that, for some random variables $U$ and $V$,

$$X=\begin{cases}U &, \text{ with probability } \frac13 \\ V &, \text{ with probability } \frac23 \end{cases}$$

Here $U$ has moment generating function $E(e^{tU})=\exp(e^t-1)$ and $V$ has moment generating function $E(e^{tV})=\exp\left\{2(e^t-1)\right\}$. Identify the distributions of $U$ and $V$ and you are essentially done.

Answer 2

If $\xi\sim\mathsf{Pois}(\mu)$ then $$ \mathbb E[e^{t\xi}] = \sum_{n=0}^\infty e^{-\mu}\frac{(\mu t)^n}{n!} = e^{\mu(t-1)},\ t\in\mathbb R. $$ Here we see that $$ M_X(t) = \frac13 M_{X_1}(t) + \frac23 M_{X_2}(t) $$ where \begin{align} X_1&\sim \mathsf{Pois}(1)\\ X_2&\sim \mathsf{Pois}(2). \end{align} It follows that $X=BX_1 + (1-B)X_2$ where $B\sim\mathsf{Ber}\left(\frac13\right)$ and $B$, $X_1$, and $X_2$ are independent. For each nonnegative integer $n$, we compute \begin{align} \mathbb P(X=n) &= \mathbb P(X=n, B=0) + \mathbb P(X=n, B=1)\\ &= \mathbb P(X=n\mid B=0)\mathbb P(B=0) + P(X=n\mid B=1)\mathbb P(B=1)\\ &= \frac13\cdot\frac{e^{-1}}{n!} + \frac23\cdot\frac{e^{-2}2^n}{n!}\\ &= \frac{2^{n+1}+e}{3 e^2 n!}. \end{align} To verify we may of course compute $$ \sum_{n=0}^{\infty } \frac{\left(2^{n+1}+e\right) }{3 e^2 n!} e^{tn}= \frac13\exp(e^t-1) + \frac23\exp(2(e^t-1)) = M_X(t). $$