Does conditional independence lead to equality of expectations?

Question

In my probability class I was tackled with this question of the type to prove or disprove the following:

Given random variables with PDF $ X,Y,Z $ and we know that given Y the variables X and Z are independent. We are asked to prove or disprove the following:

$ \forall y,z \in R $ : $ E[X | Y=y , Z=z] = E[X | Y=y ] $

I have no real idea how to do this as I do not even know if it is true or false so I need the help on this and thanks to all.

Answer 1

Yes, the statement is true. Let me abuse notation slightly and denote $p(x, y, z):=p_{X,Y,Z}(x, y, z)$ and analogously for other marginal distributions.

To prove this, we can see that even the PDFs coincide:

$$ p(x|y, z)=\frac{p(x, y, z)}{p(y, z)}=\frac{p(x,z|y)p(y)}{p(y,z)}=/\text{Conditional independence}/=\frac{p(x|y)\underbrace{p(z|y)p(y)}_{p(y,z)}}{p(y,z)}=p(x|y). $$ Now, taking expectations of the left- and right-hand-side, we get the claim.

Answer 2

By definition, X and Z are independent if the probability distribution of X is equal to the probability distribution of X given Z=z for any z. This would mean that $E[X]=E[X|Z=z]$.

Since here X and Z are independent given Y, we have $E[X]$ given (Y=y) equals $E[X|Z=z]$ given (Y=y) or in other words $E[X|Y=y]=E[X|Z=z,Y=y]$