Prove $\mathbb{E}(X_{1}+X_{2}|Y)=\mathbb{E}(X_{1}|Y)+\mathbb{E}(X_{2}|Y)$ I would like to prove
$$\mathbb{E}(X_{1}+X_{2}|Y)=\mathbb{E}(X_{1}|Y)+\mathbb{E}(X_{2}|Y)$$
This question is given in the text book as an exercise. It looks trivial but took me hours to think of a proof. I don't know what to do after expand it using definition of expectation...
 A: Use the definition of conditional expectation. in others word, you have to prove that $E(X_1|Y)+E(X_2|Y)$ is $\sigma(Y)$-mesurable and for any $Z$ $\sigma(Y)$-mesurable, $E(Z(X_1+X_2))=E(Z(E(X_1|Y)+E(X_2|Y)))$.
First, $E(X_1|Y)+E(X_2|Y)$ is $\sigma(Y)$-mesurable (by definition of conditional expectation). And for any $Z$ $\sigma(Y)$-mesurable, we have :
$E(Z(X_1+X_2))=E(ZX_1)+E(ZX_2)= E(E(ZX_1|Y))+E(E(ZX_2|Y))$.
Since Z is $\sigma(Y)$-mesurable, we have:
$E(E(ZX_1|Y))+E(E(ZX_2|Y))=E(ZE(X_1|Y))+E(ZE(X_2|Y))$
so $ E(Z(X_1+X_2))=E(Z(E(X_1|Y)+E(X_2|Y)))$.
A: Let $X,Y$ be two real random variables defined on a probability space $(\Omega,\mathcal{A},P)$. One possible definition of $\mathbb{E}[X|Y]$ is the following. 
The conditional expectation of $X$ given a $\sigma$-field $\mathcal{F}$ is the function of $X$, denoted $\mathbb{E}[X\vert \mathcal{F}]$ such that, for any $F\in\mathcal{F}$,
$$\int_{F}\mathbb{E}[X\vert\mathcal{F}]\text{d}P=\int_{F}X\text{d}P$$
If $\mathcal{F}=Y^{-1}(\mathcal{B}(\mathbb{R}))$, we say the conditional expectation of $X$ given $Y$ and we denote $\mathbb{E}[X\vert Y]$ (where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-field on $\mathbb{R}$)
Since, by definition, for all $B\in\mathcal{B}(\mathbb{R})$
$$\int_{Y^{-1}(B)}\mathbb{E}[X_{1}\vert Y]\text{d}P=\int_{Y^{-1}(B)}X_{1}\text{d}P$$
and
$$\int_{Y^{-1}(B)}\mathbb{E}[X_{2}\vert Y]\text{d}P=\int_{Y^{-1}(B)}X_{2}\text{d}P$$
and since we also have
$$\int_{Y^{-1}(B)}\mathbb{E}[X_{1}\vert Y]\text{d}P+\int_{Y^{-1}(B)}\mathbb{E}[X_{2}\vert Y]\text{d}P=\int_{Y^{-1}(B)}X_{1}\text{d}P+\int_{Y^{-1}(B)}X_{2}\text{d}P$$
the linearity of the integral allows us to conclude:
$$\begin{align}
\int_{Y^{-1}(B)}\mathbb{E}[X_{1}\vert Y]\text{d}P+\int_{Y^{-1}(B)}\mathbb{E}[X_{2}\vert Y]\text{d}P &= \int_{Y^{-1}(B)}X_{1}\text{d}P+\int_{Y^{-1}(B)}X_{2}\text{d}P\\
\int_{Y^{-1}(B)}\left(\mathbb{E}[X_{1}\vert Y]+\mathbb{E}[X_{2}\vert Y]\right)\text{d}P &=\int_{Y^{-1}(B)}\left(X_{1}+X_{2}\right)\text{d}P \tag{linearity}\\
&=\int_{Y^{-1}(B)}\mathbb{E}[X_{1}+X_{2}\vert Y]\text{d}P \tag{definition}
\end{align}$$
A: Hint #1: If $W = g(X_1,X_2)$, then
$$E[W \mid Y] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x_1,x_2)f_{X_1,X_2 \mid Y}(x_1, x_2 \mid y)dx_1dx_2$$  
Here you can find a proof for the non conditional case in page 6, proposition 2.
Hint #2: If you integrate the (conditional) joint PDF over one of the variables, you get the (conditional) PDF of the other one:
$$\int_{-\infty}^{\infty} f_{X_1,X_2 \mid Y}(x_1, x_2 \mid y)dx_1 = f_{X_2 \mid Y}(x_2 \mid y)$$ 
A: You might want to consider a realistic case for proving a such statement, because it'll help you visualize how the process works for similar question. 
Let $\:Y\:$: the number of different math courses offered next semester, $\:Y\longmapsto\mathbf{N\:\cup\{0\}}$
Let $X_i\:$: the number of courses chosen by student $i$, where $\:i\in\{1,2\}\:$
Ok, now let's let $\:x_1+x_2=\:\varphi(x_1,x_2)\:$
Thus, $\:\mathbb{E}[X_1+X_2\:|\:Y]\:=\large\sum_\limits{x_1\geq0}\:\sum_\limits{x_2\geq0}$$\varphi(x_1,x_2)\:\text{Pr}\{x_1,x_2\:|\:y\}$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\large\sum_\limits{x_1\geq0}\:\sum_\limits{x_2\geq0}$$(x_1+x_2)\:\text{Pr}\{x_1,x_2\:|\:y\}$
$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\large\sum_\limits{x_1\geq0}$$x_1\large\sum_\limits{x_2\geq0}$$\text{Pr}\{x_1,x_2\:|\:y\}+\large\sum_\limits{x_2\geq0}$$x_2\large\sum_\limits{x_1\geq0}$$\text{Pr}\{x_1,x_2\:|\:y\}\:\:\:(\color{teal}{linearity})$
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$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mathbb{E}[X_1\:|\:Y]+\mathbb{E}[X_2\:|\:Y]$
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