$y = 0,1,2,3$
$x = 0,1$
I have already made a joint distribution table.
I understand that I can use the equation $f(x|y) = f(x,y)/h(y)$ when I'm given a particular value for $Y$. In this case, would I go through using all the values of $y$ and $x$ ? Like, compute $(0,0), (0,1), (0,2)$, etc. using the corresponding $h(y)$ values?
A following question I have is "Compute $E(X|y)$ for all possible values $y$ of $Y$". How is this different?
Any nudge in the right direction is greatly appreciated. Thanks!
We have $$E[X|Y=y] = \sum_x xP(X=x|Y=y) = \sum_x x\frac{P(X=x,Y=y)}{P(Y=y)}.$$ You could compute $E[X|Y]$ using the last equality, meaning for each $y$ compute $$E[X|Y=y]=\sum_x x\frac{P(X=x,Y=y)}{P(Y=y)}.$$
For example, say I have a joint distribution table \begin{array}{r|ccc|l} P(X = x,Y=y)&0&1&2&P(X=x)\\\hline 0&.1&.1&.3&.5\\\hline 1&.1&.2&.2&.5\\\hline P(Y=y)&.2&.3&.5&1 \end{array} To compute $E[X|Y = 2]$, we have $$E[X|Y=2] = \sum_x x\frac{P(X=x,Y=y)}{P(Y=y)} = 0\left(\frac{.1}{.3}\right)+1\left(\frac{.2}{.3}\right) = \frac{2}{3}.$$
So, in your case, you will have to compute 4 expectations, one for each value of $Y$.