I'm trying to figure out whether:
$$ \mathbb{E}\left[X\mid Y, Z\right] = \frac{\mathbb{E}\left[X\mid Y\right] * \mathbb{E}\left[X\mid Z\right]}{\mathbb{E}\left[X\right]} $$
if $Y$ and $Z$ are independent. This is true for probabilities with Bayes' Rule, and seems intuitively plausible, but I'm having trouble proving it here for expectations. Any thoughts?
No, it generally doesn't hold. Let $Y$ and $Z$ be two i.i.d. Bernoulli(1/2) random variables and let $X = Y+Z$. Then,
$$E[X|Y,Z] = Y+Z \neq (Y+1/2)(Z+1/2) = \frac{(Y+1/2)(Z+1/2)}{1} = \frac{E[X|Y]E[X|Z]}{E[X]}.$$
It does hold when $X,Y$ and $Z$ are all mutually independent. Or more generally when $X$ is independent of $Z$ and also conditionally independent of $Z$ given $Y$.
Edit: fixed a typo in the last sentence