$E(2\xi_1\mid\xi_1, \xi_1+\xi_2)$. If $\xi_1$ and $\xi_2$ are independent identically distributed random variables with $P(\xi=1)=P(\xi=-1)=1/2$ Find $E(2\xi_1\mid\xi_1, \xi_1+\xi_2)$. 
I don't understand what the notation $\xi_1, \xi_1+\xi_2$ means in this context. I read something about it being a decomposition induced by random variables, but it wasn't really helpful. 
 A: What is the meaning of $\operatorname{E}(Y\mid X)$ when $X$ and $Y$ are random variables?  That seems to be the main question here.
Suppose each of two random variables $X$ and $Y$ is uniformly distributed on the set $\{1,2,3\}$ and the probability that they are equal is $0$, so we have this joint distribution:
$$
\begin{array}{c|ccccc}
_X\backslash^Y & 1 & 2 & 3 \\
\hline
1 & 0\ & 1/6\ & 1/6 \\[4pt]
2 & 1/6 & 0 & 1/6 \\[4pt]
3 & 1/6 & 1/6 & 0
\end{array}
$$
Thus for example, $\Pr(X = 1) = 1/3$ and $\Pr(Y=2\mid X=1) = 1/2 = \Pr(Y=3\mid X=1)$.
Then
\begin{align}
\operatorname{E}(Y\mid X=1) & = 2.5, \\
\operatorname{E}(Y\mid X=2) & = 2, \\
\operatorname{E}(Y\mid X=3) & = 1.5.
\end{align}
So we have $\operatorname{E}(Y\mid X=x) = 3 - \frac 1 2 x$.
Then we have the random variable $\operatorname{E}(Y\mid X) = 3 - \frac 1 2 X$.
That's what it means.
One basic fact about that random variable $\operatorname{E}(Y\mid X)$ is that its expected value must be the same as that of $Y$, provided $\operatorname{E}(|Y|)<\infty$, i.e. provided $\operatorname{E}(Y)$ exists.
