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Let $X,Y,Z$ be random variables jointly distributed on the same probability space then we know the conditional probability $P(X=a,Y=b\mid Z)$ is a random variable which is $Z$ measurable. Can we say

$$P(X=a,Y=b\mid Z)=P(X=a\mid Y=b,Z)P(Y=b\mid Z)?$$

Moreover is it true to say

$$\operatorname E[P(x=a,Y=b \mid Z)]=P(x=a,Y=b)?$$

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closed as off-topic by R W, Mark Wildon, kodlu, Chris Godsil, Pace Nielsen Feb 11 at 17:46

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ That last is one of the related things that are sometimes called the law of total probability. It can be stated by saying that the prior expected value of the posterior probability is equal to the prior probability. $$\operatorname E( \Pr(A\mid Z)) = \Pr(A). $$ $\endgroup$ – Michael Hardy Feb 8 at 20:49
  • $\begingroup$ Thank you Michael. What can we say about the first comment? $\endgroup$ – user131465 Feb 8 at 21:06
  • $\begingroup$ The problem with your first question is: How do you define $P(X = a ~|~ Y = b,Z)$? If you define it as $P(X = a, Y = b|Z) / P(Y = b|Z)$, then the answer is of course yes. I don't think this is what you want. Otherwise it seems more simple and clear to work with conditional expectations with respect to some sub-$\sigma$-algebras $\cal{G} \subset \cal{H}$. $\endgroup$ – Dieter Kadelka Feb 8 at 23:29

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