1) I'm trying to prove that two R.V.s X & Y are related iff Y & X are related. Assuming they are discretely distributed.

So basically from what I've learned is that two R.V.s are related if the joint pdf changes as Y changes. So basically if f(X|Y=yi) changes when i changes. So from that definition this is what I came up with.

if I have 2 pdf's and assuming X and Y are related then

f(X=x1|Y=y1) = P(X=x1 n Y=y1)/P(Y=y1)

should not be the same as:

f(X=x1|Y=y2) = P(X=x1 n Y=y2)/P(Y=y2)

However if Y and X are not related then:

f(Y=y1|X=x1) = P(Y=y1 n X=x1)/P(X=x1)

should be the same as:

f(Y=y1|X=x2) = P(Y=y1 n X=x2)/P(X=x2)

But P(X=x1 n Y=y1) = P(Y=y1 n X=x1) thus:

f(Y=y1|X=x1)*P(X=x1)/P(Y=y1) = f(X=x1|Y=y1)

So we can see that f(X=x1|Y=y1) depends on f(Y=y1|X=x1), so if X and Y are related, it should mean that Y and X are related as well? Since f(Y=y1|X=x1) = f(Y=y1|X=x2), I can put in any f(Y|X=xi) in there, and the left side should remain unchanged, but it is contradictory to the right side since it has to change if X and Y are related. I'm not sure if that's the right way to prove it.

2) Prove Var(Y) = Var(E(Y|X)) + E(Var(Y|X)).

earlier I proved that Var(Y) = Var(E(Y|X)) + Var(epsilon). The way i did this was pretty long because i decomposed E(Y|X) into Y - epsilon and then expanded that, but in the end i did end up with Var(Y).

I was taking a similar approach to this proof but after expanding I'm stuck at evaluating E(epsilon*Y), any ideas how I can get past this?