How to find the sum of 2 discrete random variables Let X and Y be independent random discrete variables and 
$X = \begin{pmatrix}-1&0&1\\\frac13&\frac13&\frac13\end{pmatrix}$ and $Y = \begin{pmatrix}0&1\\\frac13&\frac23\end{pmatrix}$
Then what is $X + Y$? As a discrete random variable.. I understood that I have to add the first row just like in matrix addition but what happens when they don't have the same size and how to add the probabilities? Also what is the difference between independent variables and incompatible variables?
The missing entries were:
$X + Y = \begin{pmatrix}-1&0&1&2\\\frac19&\frac39&\frac39&\frac29\end{pmatrix}$
 A: "I understood that I have to add the first row just like in matrix addition". This is not true even when they have the same size.
Note that $X+Y\in\{-1,0,1,2\}$: 
i) $X+Y=-1$ iff $(X,Y)=(-1,0)$, therefore, since $X$ and $Y$ are  independent ,
 $$p(X+Y=-1)=p(\{X=-1\} \cap \{Y=0\})=p(X=-1)\cdot p(Y=0)=\frac{1}{3}\cdot \frac{1}{3}=\frac{1}{9}.$$
ii) $X+Y=0$ iff $(X,Y)\in\{(-1,1),(0,0)\}$;
iii) $X+Y=1$ iff $(X,Y)\in\{(0,1),(1,0)\}$;
iv) $X+Y=2$ iff $(X,Y)=(1,1)$.
Can you take it from here and fill the missing entries?
$$X+Y = \begin{pmatrix}-1&0&1&0\\\frac19&?&?&?\end{pmatrix}$$
A: Hint: Can you enumerate all the cases? For example, if $X+Y=1$ it means either $X=0$, $Y=1$ (probability $\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$) or $X=1$, $Y=0$ (probability $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$) $\Longrightarrow P\left(X+Y=1\right)=\frac{2}{9}+\frac{1}{9}=\frac{1}{3}$.
A: Adding an $x$ and an $y$, you can obtain one of $-1,0,1$ or $0,1,2$, with probabilities equal to the product of respective probabilities. You see that there is an overlap on $0$ and $1$; just cumulate the corresponding probabilities.
