What is p(x=1) of this moment generating function?

Question

So for a MGF like so $M_x(s) = \frac14e^s + \frac34e^{5e^s-5} $

What is P(x=1)? How do I take into account of the 5e^s?

Answer 1

Expanding the last exponential as a series in $\mathrm e^s$, one sees that $$M_X(s)=\frac14\mathrm e^s+\frac34\mathrm e^{-5}\sum_n\frac1{n!}(5\mathrm e^s)^n=\frac14\mathrm e^s+\frac34\mathrm e^{-5}\sum_n\frac{5^n}{n!}\mathrm e^{ns}.$$ Recall that $$M_X(s)=E(\mathrm e^{sX})=\sum_nP(X=n)\mathrm e^{ns},$$ hence $P(X=1)$ is the coefficient of $\mathrm e^s$ in $M_X(s)$, that is, $$P(X=1)=\frac14+\frac34\mathrm e^{-5}\cdot5.$$ Likewise, for every positive integer $n\ne1$, $$P(X=n)=\frac34\mathrm e^{-5}\cdot\frac{5^n}{n!}.$$

Answer 2

It would be of an interest to find the distribution $X$.

According to the MGF, it is the following:

$$ X= B + (1-B) Y, $$ where $B$ is the Bernoulli distribution with $P(B=1) = 1/4$, $P(B=0)=3/4$, and $Y$ is the Poisson distribution with parameter $5$.

Thus $X$ returns the constant random variable $1$ with probability $1/4$, and it returns the Poisson distribution with probability $3/4$.

You can also figure out $P(X=1)$ from this information.