- #1
Kuma
- 134
- 0
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.
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.