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Could anyone derive or explain why the formula $p ( x , y | z ) = p ( x | z ) p ( y | x , z )$ is true?

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closed as off-topic by Did, user91500, Najib Idrissi, Aloizio Macedo, user223391 Oct 3 '15 at 17:08

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  • $\begingroup$ Have you tried write the definition of $P(A|B)$ and the check the formula? $\endgroup$ – Nikita Evseev Oct 2 '15 at 10:55
  • $\begingroup$ No indication about what you tried, how can we help? $\endgroup$ – Did Oct 2 '15 at 11:54
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    $\begingroup$ $$ P(x,y|z) = \frac{P(x\cap y)(\cap z)}{P(z)} $$ $$ P(x,y|z) P(z) = P(x\cap y)(\cap z) $$ $$ \frac {P(x,y|z) P(z)}{P(z)} = P(x\cap y)(\cap z) $$ $$ P(x,y|z) = P(x\cap y)(\cap z) $$ $\endgroup$ – Siemens Oct 2 '15 at 13:05
  • $\begingroup$ @Siemens Indeed one can run in circles as you do in your comment, for a long time (but note that identities 1 and 2 hold provided one replaces the absurd P(x∩y)(∩z) by P(x∩y∩z), while 3 and 4 are squarely wrong). Is this all you tried? So you never even considered p(x|z)p(y|x,z)? $\endgroup$ – Did Oct 3 '15 at 8:18
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Take the defintion (for $P(B) \neq 0$): $$P(A|B) := \frac{P(A\cap B)}{P(B)}$$ Just use that for every term and you will see that the equation holds.

Edit
I'll make it clearer and use your notation:
$$p(x,y|z)=\frac{p(x,y,z)}{p(z)}$$ and $$p(y|x,z)=\left(\frac{p(y,x,z)}{p(x,z)}= \right)\frac{p(x,y,z)}{p(x,z)}.$$ And this holds by definition.
There is no computation done here.

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  • $\begingroup$ i know this formula but when i use i can not prove that is true. $\endgroup$ – Siemens Oct 2 '15 at 10:53
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    $\begingroup$ @Siemens Hint: $(x\cap y)\cap z = (x\cap z)\cap y$ $\endgroup$ – Graham Kemp Oct 2 '15 at 10:57
  • $\begingroup$ for example P(x,y|z) = P(x|z) * P(y|z) then P(x)P(z|x) P(y) P(z|X) i stuck here $\endgroup$ – Siemens Oct 2 '15 at 10:59
  • $\begingroup$ i i can wrote this formula correctly for you guys to see what I've done $\endgroup$ – Siemens Oct 2 '15 at 11:01
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    $\begingroup$ @Siemens No. $P(x,y\mid z) = \dfrac{P(x,y,z)}{P(z)}$. $\endgroup$ – Graham Kemp Oct 2 '15 at 11:04
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To get you started. We use: $\mathsf P(\alpha\mid \beta)\;\mathsf P(\beta)=\mathsf P(\alpha\cap \beta)$

$\begin{align} \mathsf P ( x \cap y \mid z ) & = \frac{\mathsf P((x\cap y)\cap z)}{\mathsf P(z)} & \text{if } \mathsf P(z)\neq 0 \\[1ex] & = \frac{\mathsf P((x\cap z)\cap y)}{\mathsf P(z)} \end{align}$

Can you take it from here?

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  • $\begingroup$ i i can wrote this formula correctly for you guys to see what I've done. I already did this step. $\endgroup$ – Siemens Oct 2 '15 at 11:04
  • $\begingroup$ $$ P(x,y|z) = \frac{P(x\cap y)(\cap z)}{P(z)} $$ $$ P(x,y|z) P(z) = P(x\cap y)(\cap z) $$ $$ \frac {P(x,y|z) P(z)}{P(z)} = P(x\cap y)(\cap z) $$ $$ P(x,y|z) = P(x\cap y)(\cap z) $$ $\endgroup$ – Siemens Oct 2 '15 at 13:05
  • $\begingroup$ @Siemens No; try again. Aside from misplacing the bracket in the RHS of first line, you divided by $P(z)$ on the LHS of the third line without also doing so to the RHS. $\endgroup$ – Graham Kemp Oct 3 '15 at 15:48

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