The joint probability function of the random variables $X$ and $Y$ is given by
$p(x,y)=\Biggl\{\frac 1{e^2y!(x-y)!}$ if $x=1,2...$ and $y=0,1,2,...,x$
and $p(x,y)=0$, $otherwise$.
Prove that $E[Y| X=x]=x/2$
I have doubts on how to attack this problem, I feel that it has to do something with the binomial variable, although I am not sure how to properly use the expected conditional value? Any contribution would be very helpful. Thanks, I'm new to the probabilities course!
Bayes Formula tells us $$\mathbb P(Y=y|X=x) = \frac{\mathbb P(X=x,Y=y)}{\mathbb P(X=x)}.$$
We already have the numerator in the hypothesis. $$\mathbb P(X=x)= \sum_{y=0}^xp(x,y) = \frac{1}{x!e^2}\sum_{y=0}^x {x \choose y} = \frac{2^x}{x!e^2}.$$
Now $$\mathbb E(Y|X=x) = \sum_{y=0}^x yP(Y=y|X=x) = \frac{1}{2^x}\sum_{y=0}^x y{x \choose y} = \frac{1}{2^x}\cdot x \cdot 2^{x-1} = \frac x 2.$$