Conditional Monte Carlo simulation

Question

Suppose random variables $Z_i$ ~ $p(z)$ and $X_i$ ~ $p(x|z)$. $\frac{\sum_{i=1}^n X_i}{n}$ converges to $ E(X)$?

Answer 1

If you generate $Z$ from $p_Z(z)$, then $X$ from $p(x\, \vert Z=z)$ you're simply generating $(X, Z)$ samples from $p_{XZ}(x, z)$ since $p_{XZ}(x, z) = p(x\, \vert Z=z) p_Z(z)$. So the expected value of each $X_i$ is $\text{E}X$. Take the sample average of $n$ of them and the answer stays the same.