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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?
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.
