I've just started reading up on Monte Carlo methods (sparse background in probability and stats back in undergraduate).
Let $\phi(X)$ be a function under a complex distribution $P(X)$ where $X \in \mathbb{R}^{N}$.
The expectation, $\Phi$, to this function $\phi(X)$ is:
$\Phi = E[\phi(X)] = \int d^{N} X P(X) \phi(X)$
Since $P(X)$ is sufficiently complex, an exact method to solving for $\Phi$ is difficult so an approach is to sample from the sufficiently complex distribution $P(X)$ first.
If the sampling is done $R$ times, we have $X^{R}$ so that an estimate to this estimator may be described by
$\hat{\Phi} = \frac{1}{R} \sum_{r=1}^{R} \phi(X^{r})$.
So, the expectation of $\hat{\Phi}$ is $\Phi$
How should I convince myself of the claim in bold? I could only imagine that this is true in the case where $R$ is extremely large so that the variance of $\hat{\Phi}$ decreases. There appears to be more to this claim.
This is a well-studied and well-known concept in statistical inference.
Suppose you want to estimate a quantity $\phi$, let it be a population parameter of some statistical model with distribution $\mathcal{P}$ (say, the mean) and suppose we know the true value, the $\Phi$.
Say you come up with an estimator $\hat\phi$ based on samples from distribution $\mathcal{P}$.
Then, $\hat\phi$ is an unbiased estimator of $\phi \hspace{1pt}$ if $ \hspace{1pt} \mathbb{E}_{{\mathcal{P}}}\big[\hat\phi\big] = \Phi$.
Relevant links: Estimator, Bias of an estimator, Unbiased Estimator
The sample mean is an unbiased estimator of the population mean: Proof. It is also the minimum variance estimator, i.e. the "best" estimator, but that doesn't pertain to this question.