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I am wondering if for the random variables $A,B,C$ and $P$ there exists a property which says

$$E[A\mid B = b,C = c] = E[E[A\mid P]\mid B = b,C = c] \text{?}$$

Unfortunately I lack the mathematical knowledge to prove this statement.

Note that this is similar to the tower property for conditional expectations but instead it is conditional on two random variables:

Proof of the tower property for conditional expectations

Best regards

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1 Answer 1

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The desired property is: $$E[A\mid B=b,C=c]=E\left[ \: E[A\mid P,B=b,C=c] \mid B=b,C=c\right]$$

The idea is that we are living in a world where $B=b,C=c$, and so all probabilities and expectations should be conditioned on this. The above is true because conditional probability distributions given we live in this world are indeed valid probability distributions. So the tower property works the same way for these. Notice that the above is not the same as your guess, since $E[A\mid P]$ is not the same as $E[A\mid P,B=b,C=c]$, and so your guess was close but not quite correct.

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  • $\begingroup$ Thank you for your answer. However, it doesn't answer my question. My question is if $E[A∣B=b,C=c]=E[E[A∣P]∣B=b,C=c]$ is true. Your answer neither proves or disproves the statement i wrote. $\endgroup$
    – Angelos
    Apr 8, 2017 at 15:31
  • $\begingroup$ Interesting response. Based on my answer, what do you think? What if $P$ is independent of $A$? $\endgroup$
    – Michael
    Apr 8, 2017 at 16:57
  • $\begingroup$ Maybe it is just that I misinterpreted your answer and didn't see the connection. But, if I assume P is independent of A then I can prove that the statement you wrote is true. On the other hand, the statement I wrote doesn't hold if A and P is independent. Is it not possible that it anyway holds if A and P are dependent? $\endgroup$
    – Angelos
    Apr 8, 2017 at 18:47
  • $\begingroup$ Certainly you can design special cases where your statement is (accidentally) true (say, when $(B,C)=(b,c)$ with probability 1). However, it is not true in general, as you can verify easily in the case when $A$ and $P$ are independent, and also in cases when they are dependent (you can design your own examples, say, when the variables in question take only 2 values). $\endgroup$
    – Michael
    Apr 8, 2017 at 23:47

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