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I am having a bit of trouble with one of the problems I am working right now.

The problem is:

"What is $E\left \{ E\left \{ E\left \{ Z \mid X , Y \right \} \right \} \right \}$ ?"

During an office hours the professor explained that it would be necessary to use a "probability chain rule" and that you would need to use the following relationship: $f\left ( x,y,z \right )= f\left ( x \mid y,z \right)f\left( y\mid z \right )f\left(z \right )$

I am having some trouble seeing how they relate, and also what the actual problem is looking for as well.

Naturally, I am not looking for the answer, but rather a path to the answer. I know I have already been given that path by the professor, but it still does not make much sense to me.

Any help anyone could provide would be very much appreciated!

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The notation $$\mathbb E\left \{ \mathbb E\left \{ \mathbb E\left \{ Z \mid X , Y \right \} \right \} \right \}$$ is incorrect, it should be $$\mathbb E [ \overbrace{\mathbb E \{ \underbrace{\mathbb E( Z \mid X , Y )}_\text{function of $(X,Y)$} \mid X \}}^\text{function of $X$}]$$ or $$\mathbb E [ \overbrace{\mathbb E \{ \underbrace{\mathbb E( Z \mid X , Y )}_\text{function of $(X,Y)$} \mid Y \}}^\text{function of $Y$}]$$

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  • $\begingroup$ Xi'an, thank you for the feedback. I wish the notation was incorrect, but all I am given is exactly what is in the original question. This question, I think, has been used for years by the same professor, so I believe he would have changed it by now if the notation was incorrect for the answer he is looking for. $\endgroup$
    – pflykyle
    Commented Jun 29, 2020 at 12:27
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    $\begingroup$ The inner expectation provides a function of $(X,Y)$, but the following two expectations are not uniquely defined and are missing a supplementary conditioning. That someone has used the notation for several years is not a mathematical argument for validity! $\endgroup$
    – Xi'an
    Commented Jun 29, 2020 at 13:28
  • $\begingroup$ Xi'an, I agree totally! However, that is all the information I have to work with and I have to enter these problems with some sort of notional assumption that the notation is correct, or is at least as intended. I just do not know enough about the material to determine if the notation is incorrect. When I have an answer to this from the professor I will post it. $\endgroup$
    – pflykyle
    Commented Jun 29, 2020 at 16:37
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    $\begingroup$ I interpret the notation differently: the inner expectation is a function of $(X,Y),$ whence the next expectation is the expectation of this bivariate variable--a vector--and finally the outer expectation operator is either nonsensical or should be understood as treating that vector as a constant random variable (making it superfluous). $\endgroup$
    – whuber ♦
    Commented Nov 23, 2023 at 19:01
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    $\begingroup$ You could improve the notation by adding the variables which the expectation takes into account as an index $\endgroup$
    – Ggjj11
    Commented Dec 31, 2023 at 7:13

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