Conditional expectation $\mathbb{E}[\mathbb{E}[X\vert Y,Z]\vert Z]$

Question

Problem

When proving one result in the statistical learning theory course, the instructor uses $$ \mathbb{E}[\mathbb{E}[X\vert Y,Z]\vert Z]=\mathbb{E}[X\vert Z] $$ but I am not sure why this is true.

What I Have Done

I know I could do the following $$ \mathbb{E}[X\vert Y]=\int xf_{X\vert Y}(x\vert y)dx $$ But when $X$ becomes complicated like $\mathbb{E}[X\vert Y,Z]$ (sorry for the abuse of variable name), I do not know how to proceed.

Could someone help me, thank you in advance.

Answer 1

This is just a special case of the usual $$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X \mid Y]]$$ except all expectations are taken under the conditional distribution given the event $Z=z$. If you are still unsure, take your favorite proof of the above equality and replace all PDFs/PMFs with the conditional distribution given $Z=z$.

Answer 2

by tower property

if $F_1 \subset F_2$ so $E(E(X|F_2)|F_1)=E(X|F_1)$

now

$\mathbb{E}[\mathbb{E}[X\vert Y,Z]\vert Z]=\mathbb{E}[\mathbb{E}[X\vert \sigma (Y,Z)]\vert \sigma(Z)]$

$=\mathbb{E}[\mathbb{E}[X\vert F_2]\vert F_1]=E(X|F_1)$

$=\mathbb{E}[X\vert \sigma(Z)]=\mathbb{E}[X\vert Z]$

since $F_1=\sigma(Z) \subset F_2=\sigma (Y,Z)$

this proof valid for all type of random variable(continues,discrete and mixture). so you do not need any assumption about type of random variable.