Suppose $X,Y$ are discrete integer-valued random variables (possibly dependent), where $Y=0,1,$ or $2$.
I want to compute the joint probability $P(X>Y)$ given below:
$$\begin{matrix} &X&0&1&2\\ Y&&\\ 0&&p_{0,0}&p_{0,1}&p_{0,2}\\ 1&&p_{1,0}&p_{1,1}&p_{1,2}\\ 2&&p_{2,0}&p_{2,1}&p_{2,2}.\\ \end{matrix}$$
Here $$p_{i,j}=P(Y=i\cap X=j).$$
Now,
$$P(X>Y)=p_{0,1}+p_{0,2}+p_{1,2}.$$
If I have the pmfs, and cdfs for $X,Y$ how can they be used to evaluate the $p_{i,j}$'s; what additional information is needed to compute the joint distribution?
I've tried $p_{i,j}=P(X=i, Y=j)=P(X=i)P(Y=j\vert X=i)$, but I don't know how to compute the conditional probability since $X,Y$ have different distributions.
If I understand correctly, you would like to get $P(X > Y)$ from the marginal probability distributions for $X$ and $Y$.
I don't think this can be done! Consider these two scenarios:
$$ \begin{matrix} &X&0&1&2\\ Y&&\\ 0&&\tfrac 1 3 & 0 & 0 \\ 1&&0 & \tfrac 1 3 & 0 \\ 2&&0 & 0 & \tfrac 1 3\\ \end{matrix} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{matrix} &X&0&1&2\\ Y&&\\ 0&&\tfrac 1 9 & \tfrac 1 9 & \tfrac 1 9\\ 1&&\tfrac 1 9 & \tfrac 1 9 & \tfrac 1 9\\ 2&&\tfrac 1 9 & \tfrac 1 9 & \tfrac 1 9\\ \end{matrix}$$
In both cases, we have $$ P(X = 0) = P(X = 1) = P(X = 2) = P(Y = 0) = P(Y = 1) = P(Y = 2) = \tfrac 1 3.$$ So the two cases cannot be distinguished from the marginal probability distributions alone. Intuitively, the problem is that the marginal distributions do not tell you anything about the extent to which the two random variables are correlated.
Anyway, in the first case we have $P(X > Y) = 0$, whereas in the second case we have $P(X > Y) = \frac 1 3$. So $P(X > Y)$ is not determined by the information you are given.