Accept-reject and subsets of iid samples I have some confusion about subsets of iid samples being distributed as the original sample. 
As an illustration, consider the accept-reject algorithm to produce iid samples from a pdf $f(x)$. We draw, say $M=100$, samples from a proposal $g(x)$, perform a test for each of these individual samples and say we end up with $N=30$ samples that passed the test, and hence they represent an iid sample from $f(x)$.
Now, my question is: these $N=30$ samples are iid from $f(x)$ since they resulted from the accept-reject algorithm, but at the same time, are they iid from $g(x)$ since they were all drawn independently from it? 
More generally, is any subset of an iid sample (from $g(x)$), also an iid sample (from $g(x)$)?
 A: The accepted and rejected samples are no longer distributed from $g$ because an event depending on the realisations $Y_i$ occurred (acceptance with probability $f(y_i)/Mg(y_i)$) or did not occur (rejection with probability $1-f(y_i)/Mg(y_i)$) which, marginally, changed their distribution. While the accepted $X_i$'s are distributed from $f$ rather than $g$ and iid, since
$$g(x_i)\times\frac{f(x_i)}{Mg(x_i)} \propto f(x_i)$$
the rejected $Z_i$'s are distributed from 
$$g(z_i)\times\left\{1-\frac{f(z_i)}{Mg(z_i)}\right\} \propto
\frac{g(z_i)-Mf(z_i)}{1-M}$$ 
and independent conditional on $N$.
Here is an excerpt from Monte Carlo Statistical Methods taking advantage of the distinction between accepted and rejected subsamples. Itself borrowed from our 1996 Biometrika Rao-Blackwellisation paper.

Consider an Accept-Reject method based on the instrumental
  distribution with density $g$. If the original sample produced by the
  algorithm is $(X_1,\ldots,X_m)$, it can be associated with two iid
  samples,  $$(U_1,\ldots,U_N)\quad\text{ and }\quad(Y_1,\ldots,Y_N)$$ 
  with corresponding distributions ${\cal U}_{[0,1]}$ and $g$; $N$ is
  then the stopping time associated with the acceptance of $m$ variables
  $Y_j$. An estimator of $\mathbb E_f[h]$ based on $(X_1,\ldots,X_m)$
  can therefore be written $$ \delta_1 = {1\over m} \; \sum_{i=1}^m \;
  h(X_i)  =  {1\over m}\; \sum_{j=1}^N\; h(Y_j)\; \mathbb I_{U_j\leq
  w_j}\,, $$  with $$w_j = f(Y_j)/Mg(Y_j).$$
A reduction of the variance of $\delta_1$ can be obtained by
  integrating out the $U_i$'s, which leads to the estimator $$ \delta_2
  = {1\over m} \; \sum_{j=1}^N \; \mathbb E[\mathbb I_{U_j \leq w_j} | N,Y_1,\ldots,Y_N] \; h(Y_j) = {1\over m} \sum_{i=1}^N \rho_i h(Y_i),
  $$ where, for $i =1, \ldots, n-1$, $\rho_i$ satisfies 
  \begin{align*}
  \rho_i &= \mathbb{P}(U_i\le w_i|N=n,Y_1,\ldots,Y_n) \\ &= w_i
  \frac{\sum_{(i_1,\ldots,i_{m-2})} \prod_{j=1}^{m-2} w_{i_j}
  \prod_{j=m-1}^{n-2} (1-w_{i_j})}{\sum_{(i_1,\ldots,i_{m-1})}
  \prod_{j=1}^{m-1} w_{i_j} \prod_{j=m}^{n-1} (1-w_{i_j})}, \tag{1}
  \end{align*} while $\rho_n = 1$. The numerator sum is over all subsets
  of $\{1,\ldots,i-1, i+1, \ldots, n-1\}$ of size $m-2$, and the
  denominator sum is over all subsets of size $m-1$. The resulting
  estimator $\delta_2$ is an average over all the possible permutations
  of the realized sample, the permutations being weighted by their
  probabilities. The Rao-Blackwellized estimator is then a function only
  of $(N,Y_{(1)},\ldots,Y_{(N-1)}, Y_N)$, where
  $Y_{(1)},\ldots,Y_{(N-1)}$ are the order statistics.
Although the computation of the $\rho_i$'s may appear formidable, a
  recurrence relation of order $n^2$ can be used to calculate the
  estimator. Define, for  $k\le m < n$, $$ S_k(m) =
  \sum_{(i_1,\ldots,i_k)} \prod_{j=1}^{k} w_{i_j} \prod_{j=k+1}^{m}
  (1-w_{i_j}), $$ with $\{i_1,\ldots,i_m\} = \{1,\ldots,m \}$,
  $S_k(m)=0$ for $k>m$, and $S^i_k(i)=S_k(i-1)$. Then we can recursively
  calculate \begin{align*} S_k(m) &= w_mS_{k-1}(m-1)+(1-w_m)S_k(m-1),
  \\ S^i_k(m) &= w_m S^i_{k-1}(m-1)+(1-w_m)S^i_k(m-1)  \end{align*} and
  note that weight $\rho_i$ of (1) is given by $$ \rho_i =w_i\;
  S^i_{t-2}(n-1)\big/S_{t-1}(n-1) \qquad (i<n). $$

A: Your final 30 points are not an i.i.d. sample from $g$ (since they are an i.i.d. sample from $f$), the accept-reject step does modify the distribution. So the answer to your question is: no, a subsample from an i.i.d. sample with distribution $g$ does not have the same distribution $g$. 
You can see that very easily on a simpler example. Suppose $B_1, ..., B_n$ are iid draw from a $\mathcal{B}ernouilli(p)$ (with $p$ not $0$ nor $1$), and select only the points which are equal to $1$. Your subsample is obviously not following a $\mathcal{B}ernouilli(p)$ since it is all equal to 1.
I guess what can be puzzling here is that if the subsample was draw uniformly and independently from the sample (in a bootstrap spirit for instance), then you would have that the distribution of your subsample is following the same distribution than the sample (without conditionning on the realization of the sample). 
