# Interpretation of $E[X \mid Y, Z]$

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). – Michael Hardy Oct 8 '12 at 17:16
Thanks, I hope it's correct now. – somebody Oct 8 '12 at 17:21
Why should $E[X|Y=y,Z]$ be a constant? It is a function of the r.v. $Z$. – emrea Oct 8 '12 at 17:25
$E[X|Y,Z]$ will be a constant only when both $Y$ and $Z$ are instantiated/observed. – emrea Oct 8 '12 at 18:21
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. $Y$. – emrea Oct 9 '12 at 16:20

A simple example might help: assume that $X=Y+Z$ where the random variables $Y$ and $Z$ are integrable and independent.
Then $\mathbb E(X\mid Y,Z)=X$ because $X=Y+Z$ is $\sigma(Y,Z)$-measurable.
On the other hand, $Y$ is $\sigma(Y)$-measurable hence $\mathbb E(Y\mid Y)=Y$, and $Z$ is independent of $\sigma(Y)$ hence $\mathbb E(Z\mid Y)=\mathbb E(Z)$. Thus, $\mathbb E(X\mid Y)=\mathbb E(Y\mid Y)+\mathbb E(Z\mid Y)=Y+\mathbb E(Z)$.
Likewise, $\mathbb E(X\mid Z)=\mathbb E(Y)+Z$.