Does $E[E(X|Y)|Z] =E[X|Y,Z]$? Also, what about $E[E(X|Y=y)|Z=z] =E[X|Y=y,Z=z]$ I'm confused by the relations. It seems intuitively to be the case. If it is correct, how do I mathematically prove it. I've searched on this site and elsewhere...

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    $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Dec 1 '16 at 16:41
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    $\begingroup$ No, this is not from a course or textbook. Just trying to understand for myself. $\endgroup$ – KUZ Dec 1 '16 at 17:26

Those two conditional expectations differ in general:$$\mathbb{E}[\mathbb{E}(X|Y)|Z] \ne\mathbb{E}[X|Y,Z]$$

As a matter of fact, strictly speaking, they do not even live in the same functional space as the first one is a function of $Z$, measurable wrt $\sigma(Z)$, the $\sigma$ algebra induced by $Z$, while the second one is a function of $(Y,Z)$, hence measurable wrt $\sigma(Y,Z)$, the $\sigma$ algebra induced by $(Y,Z)$,

As a counter-example, consider the setting when

  1. $X$ and $Y$ are independent
  2. $X$ and $Z$ are dependent, with $\mathbb{E}[X|Z]\ne \mathbb{E}[X]$

Then, because of the independence between $X$ and $Y$, $\mathbb{E}(X|Y)=\mathbb{E}[X]$ and therefore$$\mathbb{E}[\mathbb{E}(X|Y)|Z]= \mathbb{E}[X]\ne\mathbb{E}[X|Y,Z]$$

A valid equality is instead$$\mathbb{E}[\mathbb{E}(X|Y,Z)|Z]=\mathbb{E}[X|Z]$$that holds for all dependence relations between the three random variables.

Notations: The difference between the notations $\mathbb{E}[\mathbb{E}(X|Y)|Z]$ and $\mathbb{E}[\mathbb{E}(X|Y=y)|Z=z]$ is that

  1. $\mathbb{E}[\mathbb{E}(X|Y)|Z]$ is a random variable, transform of the random variable $Z$ (and not of the random variable $Y$ since $Y$Y is also conditioned on $Z$);
  2. $\mathbb{E}[\mathbb{E}(X|Y=y)|Z=z]$ is a function of apparently both $y$ and $z$, but in fact only of $y$ (as explained below) which has no clear meaning from a probabilistic point of view. Indeed, for a given value $y$, $\mathbb{E}(X|Y=y)$ is a constant for which taking a conditional expectation conditional on the realisation $Z=z$ makes little sense as it also returns $\mathbb{E}(X|Y=y)$. For instance, if $X$ depends on both $Y$ and $X$ as a random variable, for a given realisation $y$ of $Y$ and $Z$ of $z$, $\mathbb{E}(X|Y=y)$ is a constant that generally differs from $\mathbb{E}(X)$ and from $\mathbb{E}(X|Y=y,Z=z)$. But $\mathbb{E}[\mathbb{E}(X|Y=y)|Z=z]$ is not a realisation of the random variable $\mathbb{E}[\mathbb{E}(X|Y)|Z]$. The correct realisation is $\mathbb{E}[\mathbb{E}(X|Y)|Z=z]$
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  • $\begingroup$ Thank you for this comment, Xi'an. I am unclear about your response on one thing: in your counterexample, if X and Y are independent (therefore $\mathbb{E}(X|Y)= \mathbb{E}[X]$) and if X and Z are dependent, then why is $\mathbb{E}[\mathbb{E}(X|Y)|Z]= \mathbb{E}[\mathbb{E}(X)|Z]= \mathbb{E}[X]$? $\endgroup$ – KUZ Dec 1 '16 at 22:07
  • $\begingroup$ ..because $\mathbb{E}[X]$ is a constant... $\endgroup$ – Xi'an Dec 1 '16 at 22:09

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