# Is $f_{X,Y}(X,Y)$ the same as $f_{X,Y|X}(X,Y|X)$

This might be a very simple question to answer, but for some reason I have just been banging my head against the wall for a little while now. To me, it makes some intuitive sense that $f_{X,Y}(X,Y)$ is same as $f_{X,Y|X}(X,Y|X)$, but I cannot for the life of me figure out how to show it.

However at the same time, if those two joint distributions were the same, wouldn't that imply that $X$ and $Y|X$ are independent? Which definitely seems off to me.

• Tell us this: (1) Is it possible that $f_{X,Y}(1,0) \ne 0$? (Obviously yes.) (2) Is it possible that $f_{X,Y|X}(1,0|X=0) \ne 0$?
– whuber
Jun 27, 2013 at 15:31
• @whuber, Hmmm, no, (2) doesn't seem possible, seeing that X is not 0, it is 1. Which seems like it would indicate that the two distributions are not the same. But wouldn't $f_{X,Y|X}(1,0|X=1) = f_{X,Y}(1,0)$? Jun 27, 2013 at 15:49
• (Let me try this again, because my previous comment was flagged as offensive and, although it was tongue-in-cheek, absolutely no offense was intended. I apologize for any misunderstanding.) You have observed a contradiction. The correct response is to drop your now obviously erroneous assumption that the conditional and joint densities are always the same, and seek to understand why they can differ. A good way to approach this is to concoct simple counterexamples. Discrete distributions are good for this. Maybe one that has only two possible values of $X$ and $Y$ will do?
– whuber
Jun 27, 2013 at 16:34

Try a discrete distiribution $\Pr(X=0,Y=0) = 0.2, \Pr(X=0,Y=1) = 0.4, \Pr(X=1,Y=0) = 0.3, \Pr(X=1,Y=1) = 0.1$.
Then consider $\Pr(X=1,Y=0|X=1) = 0.75, \Pr(X=1,Y=1|X=1) = 0.25$.
• Thank you for the answer, it's definitely helpful. Though I am still a tad confused as to how $Pr(X=1, Y=0|X=1)$ is different from $Pr(Y=0|X=1)$. If they are the same, which I think your answer indicates, then of course, $Pr(X = x, Y = y)$ is not the same as $Pr(Y = y|X = x)$. It just seems that $Pr(X=1, Y=0|X=1)$ should take into account the distribution of X, even if in the conditional statement X is set to 1. Jun 27, 2013 at 16:33
• $P(A|B) = P(A \cap B)/P(B)$ Which would let you calculate everything off the joint distribution. Based on Henry's example that would lead to $Pr(X=1, Y=0|X=1) = Pr(X=1|(Y=0|X=1))Pr(Y=1|X=0)$ Ahhhh, okay I see now. I was getting caught up on the joint distribution of (X, Y|X) and not seeing how the conditional takes precedence in calculations. Thanks, my confusion has lifted. Jun 27, 2013 at 16:54