Value drawn from a distribution $P(z\mid x)$ and conditional expectation How would you explain in layman terms the meaning of the statement " $\hat{x}$ drawn from a conditional distribution $p(\hat{x}|x)$ " ? Secondly does $\mathbb{E}_{p(\hat{x}|x)}[\hat{x}]$ mean the conditional expectation of $p(\hat{x}|x)$? In any case it would be helpful if someone could explain $\mathbb{E}_{p(\hat{x}|x)}[\hat{x}]$.
 A: $\hat{x}$ is drawn from $p(\hat{x} \mid x)$ if it's drawn from a distribution that depends on $x$. Maybe it's mean is a function of $x$, maybe some parameter depends on $x$, maybe both, etc.
You could also write $\mathbb{E}_{p(\hat{x}|x)}[\hat{x}] = \mathbb{E}[\hat{x} \mid x]$. That's probably clearer. If you're dealing with continuous random variables, it equals $\int \hat{x} p(\hat{x} \mid x) d\hat{x}$. So it's a function of $x$. If you're dealing with discrete random variables, it equals $\sum_{\hat{x}} \hat{x} p(\hat{x} \mid x)$, which is still a function of $x$. 
Expectations are just weighted averages, only this time we're using weights that depend on $x$. In the continuous case, the weights are $p(\hat{x} \mid x) d\hat{x}$ where $p(\hat{x} \mid x)$ is a density. And if we're looking at a discrete variable, the weights are $p(\hat{x} \mid x)$, where $p(\hat{x} \mid x)$ is a probability mass function.
A: The conditional expectation gives you what you would expect to be an estimate $\hat{x}$ for some quantity measured by a random variable X, given that the "true" value of X would be x. 
Let's say I'm flipping a coin and X can either be 0 or 1 (modeling tails or heads respectively), then the expectation conditioned on e.g. (x=1) would give me an average value I would estimate for x, given that coin flipped heads. Of course, this value could be different from 0 or 1, e.g. 0.5 if I would expect the estimator to randomly choose between the two predictions when the coin flipped heads.
