Timeline for Is $f_{X,Y}(X,Y)$ the same as $f_{X,Y|X}(X,Y|X)$
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2013 at 16:55 | vote | accept | Teague | ||
| Jun 27, 2013 at 16:54 | comment | added | Teague | $P(A|B) = P(A \cap B)/P(B)$ 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 16:35 | comment | added | whuber♦ | 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. | |
| Jun 27, 2013 at 16:33 | comment | added | Teague | 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 16:16 | history | answered | Henry | CC BY-SA 3.0 |