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?

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.  
