How to use probability mass functions to find probability mass function and moment generating function of Z? Let $X$ and $Y$ be discrete random variables such that 
$p_{(x)}(x) = \frac{1}{3}, x = -1,0,1$
$p_{(y)}(y) = \frac{1}{2}, y = 2,4$
Let's say $Z = X + Y$.
I am trying to find the probability mass function of $Z$ using what I know about $X$ and $Y$, their functions above. Here's my intuition, I think 
$p_{(z)}(z) = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$. But how do I know the values $z$ can take on? Is it the all values of $x$ and $y$ together? So, $,-1,0,1,2,4$. This seems reasonable to think. Also, I want to determine the moment generating function of $Z$. I figured there should be a way to use it's probability mass function once found. But is it possible to find a moment generating function from it's probability mass function, I can't see how? Typically, you're given both but with one missing, I can't see any idea on how to solve anything without it. Any ideas? 
 A: We will assume that $X$ and $Y$ are independent. This is not mentioned explicitly, but the problem cannot be solved if we do not have information about possible dependencies between $X$ and $Y$.
The moment generating function of $X$ is $E(e^{tX})$. We have $e^{tX}=e^{-t}$, $1$, or $e^t$, each with probability $\frac{1}{3}$. Thus the moment generating function $M_X(t)$ of $X$ is given by
$$M_X(t)=\frac{1}{3}e^{-t}+\frac{1}{3}+\frac{1}{3}e^t.$$
The moment generating function of $Y$ is found in a similar way. 
For the moment generating function of $Z$, use the fact that if $X$ and $Y$ are independent, then 
$$M_{X+Y}(t)=M_X(t)M_Y(t).$$
Added: We could also first find the probability mass function of $Z$, and then find the mgf. Let us start on finding the mass function of $Z$, in order to clear up possible misconceptions.
The random variable $X+Y$ takes on possible values $-1+2,-1+4,0+2,0+4,1+2, 1+4$, or more simply $1,2,3,4,5$. To find the pobability mass function of $Z$, we calculate $\Pr(Z=z)$ for each of these possible values.
For example, we have $Z=1$ precisely if $X=-1$ and $Y=2$. This has probability $\frac{1}{3}\cdot \frac{1}{2}$.
The other probabilities are computed in a similar way. But note that we can have $Z=3$ in two ways, $X=-1$ and $Y=4$ or $X=1$ and $Y=2$. Thus $\Pr(Z=3)=\frac{1}{3}\cdot\frac{1}{2}+\frac{1}{3}\cdot \frac{1}{2}$.
You should end up with the conclusion that $\Pr(Z=z)=\frac{1}{6}$ for $z=1,2,4,5$, and $\Pr(Z=z)=\frac{2}{6}$ for $z=3$.
