How to calculate probability from transform function?

Question

In my book I found an interesting question, given the transform function (moment generating function) of random variable $X$:$$M_X(s)=\frac{1}{2^4}\left (\left (e^s+1\right )^2\left (e^s+3\right )\right ).$$I want to compute $\mathbb{P}(X>2)$, which is equal to $1-\mathbb{P}(X\leq 2)$.

I know that I can use $M_X(s)$ to calculate expected value, but how to use it to calculate probability function.

Plus the question doesn't say anything if $X$ is continuous or discrete...

Answer 1

Using $$M_{\text{Bernoulli}(p)}(t)=p\left(\frac{1-p}p+e^t\right)$$ and the fact that summing independent random variables multiplies their mgfs, we find that

• $M_{\text{Bernoulli}(1/2)}(t)=\frac12(e^t+1)$
• $M_{\text{Bernoulli}(1/4)}(t)=\frac14(e^t+3)$

so $X$ is the sum of two $\text{Bernoulli}(1/2)$ variables and one $\text{Bernoulli}(1/4)$ variable, all independent. Its realisation is greater than $2$ iff it is $3$, so $\mathbb P(X>2)=\frac12×\frac12×\frac14=\frac1{16}$.