Question

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 \(a_i \in \mathbb{R}\).

Answer 1

I tried using induction (the usual, assume it's true for n=k, and prove it for n=k+1), so
I get \[E\left(\sum_{i=1}^{k+1} a_i X_i|Y=y\right)\\=E\left(\sum_{i=1}^{k} a_i X_i+a_{k+1}X_{k+1}|Y=y\right)\\=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_1x_1+...+a_kx_k+a_{k+1}x_{k+1})~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\=\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_1x_1+...+a_kx_k)~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1}\\~~~+\underbrace{\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}}_{k+1~ \text{integrals}}(a_{k+1}x_{k+1})~f_{X_1,...,X_k,X_{k+1}|Y}(x_1,...,x_{k+1}|y)~dx_1...dx_{k+1} \]
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 as 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).