What is P(x=0) and P(x=1) of this moment generating function?

Question

$$M_x(s) = (\frac12 + \frac12e^s)e^{e^s-1}$$

What is P(X = 0) and P(X=1)?

Answer 1

The moment generating function of a discrete random variable is defined as $$M_X(s) := \mathbb{E}[e^{sX}] = \sum_{n=0}^\infty e^{sn}\mathbb{P}(X = n).$$ The probability generating function of a discrete random variable is defined as $$P_X(z) := \mathbb{E}[z^X] = \sum_{n=0}^\infty z^n \mathbb{P}(X=n).$$ Notice that if we substitute $z = e^{s}$ for the probability generating function, we get $\mathbb{E}[(e^s)^X] = \mathbb{E}[e^{sX}]$, which is just the moment generating function. To go backward, substitute $s = \log(z)$ in the moment generating function to get the probability generating function. Thus, $$\left(\frac{1}{2}+\frac{1}{2}e^s\right)e^{e^s-1} \rightarrow \left(\frac{1}{2}+\frac{1}{2}z\right)e^{z-1}.$$ The RHS in the above display is the probability generating function $P_X$, whih yields the desired probabilities via $\mathbb{P}(X=0) = P_X(0)$ and $\mathbb{P}(X=1) = \frac{dP_X}{dz}(1)$. Therefore, $\mathbb{P}(X=0) = \frac{e^{-1}}{2}$, like you said. Others suggested you notice that if $B \sim \mathrm{Ber}(1/2)$ and $P \sim \mathrm{Pois}(1)$, then $$\mathbb{E}[e^{sP}] = \sum_{n=0}^\infty \frac{e^{-1}}{n!}e^{sn} = e^{e^s-1},$$ while $$\mathbb{E}[e^{sB}] = \frac{1}{2}e^{s\cdot 0} + \frac{1}{2}e^{s\cdot 1} = \frac{1}{2}+\frac{1}{2}e^s.$$ By uniqueness of Laplace transforms, $$\mathbb{P}(X = 0) = \mathbb{P}(B = 0, P = 0) = \mathbb{P}(B = 0)\mathbb{P}(P = 0),$$ (why can I factor them?) while $$\mathbb{P}(X = 1) = \mathbb{P}(B = 0, P = 1) + \mathbb{P}(B = 1, P = 0).$$ Forgive the use of $P$ as a random variable.

Answer 2

Hint: decompose $X$ into two separate MGFs

$$\begin{align} &M_{Y}(s) = \dfrac{1}{2}+\dfrac{1}{2}e^{s} \\ &M_{Z}(s) = e^{e^{s}-1} \end{align}$$ In particular, if you have a list of MGFs, you will recognize that $Y$ and $Z$ follow certain distributions.

Since $M_{X}(s) = M_{Y}(s)M_{Z}(s)$, $X = Y+Z$ and you can use this to help you find the probability distribution once you figure out the distributions of $Y$ and $Z$.