Show $\mathbb{E}(X \mid Y,Z) = \mathbb{E}(X \mid Y)$ if $Z$ is independent of $X$ and $Y$

Question

Let $X,Y,Z$ be random variables, $X$ integrable, $Z$ independent of $X$ and $Y$. Then we have $E[X\mid Y,Z]=E[X\mid Y]$. Why is only assuming $Z$ independent of $Y$ not enough.

I was able to verify this for random variables that have a joint density, but I have no idea how to verify this one. I tried using the tower property to no avail. I only want a hint to get started, no full solution.

Answer 1

Hints:

1. The $\sigma$-algebra $\sigma(Y,Z)$ is generated by sets of the form $$\{Y \in A\} \cap \{Z \in B\}$$ for Borel sets $A,B \in \mathcal{B}(\mathbb{R})$.
2. By step 1 and the definition of conditional expectation, it suffices to show that $$\int_{\{Y \in A\} \cap \{Z \in B\}} \mathbb{E}(X \mid Y) \, d\mathbb{P} = \int_{\{Y \in A\} \cap \{Z \in B\}} X \, d\mathbb{P} \tag{1} $$ for all $A,B \in \mathcal{B}(\mathbb{R})$.
3. Use the given assumption on the independence and the equality $$\int_{\{Y \in B\}} \mathbb{E}(X \mid Y) \, d\mathbb{P} = \int_{\{Y \in B\}} X \, d\mathbb{P}$$ to prove $(1)$.