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I'm struggling to understand the correct interpretation of conditional expectation of the form $$ E[X \mid Y, Z ]. $$ I know that $E[X \mid Y]$ is itself a random variable $f(y) = E[X \mid Y=y]$. Does this mean that the above is a random variable $g(Y,Z)$ where $$g(y,z) = E[X \mid Y = y, Z = z]\ ?$$

On the other hand, $E[X \mid Y,Z ]$ is nothing but $E[X \mid \sigma(Y,Z)]$. Clearly $\sigma(Y) \subseteq \sigma(Y,Z)$, so $E[X\mid Y,Z]$ is a constant when conditioning on some event $\{Y=y\} \in \sigma(Y)$, which seems to contradict the above interpretation as a r.v. depending on $Y$ and $Z$.

What am I missing here?

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    If $g(y,z)=\mathbb E(X\mid Y=y\ \&\ Z=z)$, then $\mathbb E(X\mid Y,Z)$ is the random variable $g(Y,Z)$ (not $g(y,z)$, which is not a random variable).2012-10-08
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    Thanks, I hope it's correct now.2012-10-08
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    Why should $E[X|Y=y,Z]$ be a constant? It is a function of the r.v. $Z$.2012-10-08
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    Let me see if I understand it now: $\sigma(Y,Z)$ does contain events of the form $\{Y=y\}$ since it contains $\sigma(Y)$. So it's true that the expression would be a constant **if** I was only be conditioning over such events (in $\sigma(Y)$). But for general events in $\sigma(Y,Z)$, $E[X \mid Y, Z]$ is not a constant.2012-10-08
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    $E[X|Y,Z]$ will be a constant only when both $Y$ and $Z$ are instantiated/observed.2012-10-08
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    @somebody Indeed, just like a general random variable on $\Omega$ is not a constant.2012-10-08
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    @emrea: What confuses me is the fact that $E[X \mid \sigma(Y,Z)]$ conditions on events in $\sigma(Y,Z)$, which also includes events $B \in \sigma(Y)$ that don't say anything about $Z$ at all. This seems to be different from thinking of it as the r.v. $g(Y,Z)$, which **always** conditions on both, $Y$ and $Z$...2012-10-09
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    It might be misleading to say that $g(Y,Z)$ *always* conditions on both $Y$ and $Z$. No, it doesn't have to. The correct statement is this: if you condition it on both $Y$ and $Z$, it becomes a constant; if you condition it on $Y$ it becomes a function of r.v. $Z$; if you condition it on $Z$ it becomes a function of r.v. $Y$.2012-10-09

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