# Find distribution function through moment generating function

Suppose that the moment generating function $$M_X(t)$$ of a random variable $$X$$ is given by

$$M_X(t)=\frac{e^t+e^{-t}}{6} + \frac 23$$

I need to find the distribution function $$F_X(x)$$.

Until now, I have been given (in my lecture notes) that I can express $$E(X)$$= $$M_X^{(1)}(0)$$ . But I can't use this here for finding the distribution function $$F_X(x)$$?(Or at least I have no idea how to do it) Could you please tell me how to proceed?

## 2 Answers

Hint: From the moment generating function we can determine the distribution of $$X$$, which is $$P(X=1)=P(X=-1)=\frac16$$, $$P(X=0)=\frac23$$. I believe that you can move on now.

A moment generating function of the form $$M(e^t)$$ can be easily converted into a probability generating function of the original, integer-valued random variable, just by replacing $$e^t$$ by z. This yields $$P(z)=\frac{z}6+\frac{z^{-1}}6+\frac{2}3z^0$$ from which one can just read off the probabilities of $$X=1$$, $$X=-1$$ and $$X=0$$; the coefficient of $$z^k$$ is the probability of $$X=k$$. This of course agrees with the previous answer. In general, when X is a continuous-type random variable, finding its probability generating function based on $$M(t)$$ requires a knowledge of complex calculus and Fourier transform.