Linearity of conditional expectation (proof for n joint random variables) Linearity of conditional expectation:
I want to prove $$E\left(\sum_{i=1}^n a_i X_i|Y=y\right)=\sum_{i=1}^n a_i~ E(X_i|Y=y)$$ where $X_i, Y$ are random variables and $a_i \in \mathbb{R}$.
I tried using induction (the usual, assume it's true for n=k, and prove it for n=k+1), so
I get, in the continuous case, $$E\left(\sum_{i=1}^{k+1} a_i X_i|Y=y\right)\\=E\left(\sum_{i=1}^{k} a_i X_i+a_{k+1}X_{k+1}|Y=y\right)\\=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_1x_1+...+a_kx_k+a_{k+1}x_{k+1})~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_1x_1+...+a_kx_k)~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\+\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_{k+1}x_{k+1})~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1} $$
On the last step I separated the $(k+1)^{\text{th}}$ "term" since I'm trying to find a way to use the induction hypothesis... but I need to do something to get rid of the $(k+1)^{\text{th}}$ integral, as well the $(k+1)^{\text{th}}$ random variable in the underlying conditional distribution
I know this is very long to write, I'm just hoping that I can get some hints on how to proceed further (or if there's perhaps a simpler method). 
 A: It might help to work with just two joint random variables before you generalize. Here we go:
$$ \begin{align*}
&E\left(\sum_{i=1}^{k+1} a_i X_i \middle| Y=y\right)\\
&= E\left(\sum_{i=1}^{k} a_iX_i+a_{k+1}X_{k+1} \middle| Y=y\right)\\
&=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_1x_1+...+a_kx_k+a_{k+1}x_{k+1})~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\
&=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_1x_1+...+a_kx_k)~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\
&+\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_{k+1}x_{k+1})~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\
&=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k~ \text{integrals}}
(a_1x_1+...+a_kx_k)~dx_1...dx_{k}\int_{-\infty}^{\infty}~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_{k+1}\\
&+a_{k+1}\int_{-\infty}^{\infty}x_{k+1}~dx_{k+1}\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k~ \text{integrals}}f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k}\\
&=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k~ \text{integrals}}
(a_1x_1+...+a_kx_k)f_{X_1,...,X_k|Y}(x_1,...,x_{k}|y)~dx_1...dx_{k}\\
&+a_{k+1}\int_{-\infty}^{\infty}x_{k+1}f_{X_{k+1}|Y}(x_{k+1}|y)~dx_{k+1}\\
&= \left( \sum_{i=1}^k a_i~E(X_i \mid Y=y) \right) + a_{k+1}~E(X_{k+1} \mid Y=y) \\
&= \sum_{i=1}^{k+1} a_i~E(X_i \mid Y=y) \\
\end{align*} $$
as desired.
A: Hint: Condition on $X_n$ and use Baye's rule then apply the induction hypothesis.
