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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.

Thanks in advance!

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    $\begingroup$ Tell us this: (1) Is it possible that $f_{X,Y}(1,0) \ne 0$? (Obviously yes.) (2) Is it possible that $f_{X,Y|X}(1,0|X=0) \ne 0$? $\endgroup$
    – whuber
    Jun 27, 2013 at 15:31
  • $\begingroup$ @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 $f_{X,Y|X}(1,0|X=1) = f_{X,Y}(1,0)$? $\endgroup$
    – Teague
    Jun 27, 2013 at 15:49
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    $\begingroup$ (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? $\endgroup$
    – whuber
    Jun 27, 2013 at 16:34

1 Answer 1

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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 $\Pr(X=1,Y=0|X=1) = 0.75, \Pr(X=1,Y=1|X=1) = 0.25$.

Now try to translate this into a similar example for your original question.

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  • $\begingroup$ 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. $\endgroup$
    – Teague
    Jun 27, 2013 at 16:33
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    $\begingroup$ Teague, what definition of conditional probability are you using? One that shows its connection to the joint and marginal probabilities will be useful here, because it makes it possible to calculate everything in sight based on the joint distribution. I believe that's where Henry is trying to lead you. $\endgroup$
    – whuber
    Jun 27, 2013 at 16:35
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    $\begingroup$ $P(A|B) = P(A \cap B)/P(B)$ 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. $\endgroup$
    – Teague
    Jun 27, 2013 at 16:54

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