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We know that $X$ and $Y$ are conditionally independent given $Z$ iff $\mathbb{P}(X,Y|Z) = \mathbb{P}(X|Z) \mathbb{P}(Y|Z)$.

The following condition is weaker than conditional independence: \begin{equation} \mathbb{E}_Z(\mathbb{P}(X|Z) \mathbb{P}(Y|Z)) = \mathbb{E}_Z(\mathbb{P}(X,Y|Z)), \end{equation}

in the sense that if $X$ and $Y$ are conditionally independent given $Z$, then $\mathbb{E}_Z(\mathbb{P}(X|Z) \mathbb{P}(Y|Z)) = \mathbb{E}_Z(\mathbb{P}(X,Y|Z))$, but the converse is apparently not necessarily true.

I am trying to find an example where this reverse direction does not hold. I.e., what is an example of a case when we have $\mathbb{E}_Z(\mathbb{P}(X|Z) \mathbb{P}(Y|Z)) = \mathbb{E}_Z(\mathbb{P}(X,Y|Z))$ but $X$ and $Y$ are conditionally dependent given $Z$?

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  • $\begingroup$ What, exactly do you think $\Bbb E_Z(\Bbb P(X\mid Z)\Bbb P(Y\mid Z))$ means? $\endgroup$ – Graham Kemp Jan 31 '17 at 5:41
  • $\begingroup$ $\mathbb{E}_Z (\mathbb{P}(X|Z) \mathbb{P}(Y|Z)) = \int_{\mathcal{Z}) \mathbb{P}(X|z) \mathbb{P}(Y|z) d\mathbb{P}_z$. Is this incorrect? $\endgroup$ – Stephanie Jan 31 '17 at 5:46
  • $\begingroup$ So you are mixing up notations for events and continuous random variables? $\endgroup$ – Graham Kemp Jan 31 '17 at 5:50
  • $\begingroup$ If we think of $\mathbb{P}(X|Z)$ and $\mathbb{P}(Y|Z)$ as random variables w.r.t. $Z$ , then we can think about their expectation with respect to $Z$. Right? Please correct me if I'm wrong. I am not sure what you are trying to get at $\endgroup$ – Stephanie Jan 31 '17 at 5:56
  • $\begingroup$ Are $X$ and $Y$ events and $Z$ a random variable? It seems like we should be able to come up with probabilities for a simple discrete probability space that make the expectations equal but have the probability equation for conditional independence not hold. $\endgroup$ – Jonathan Hahn Feb 1 '17 at 19:05

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