What is the distribution of sampling top $p$ proportion of overlapping normal distributions? I have a tricky distribution question. Say I have two normal distributions (call them $n_1$ and $n_2$) described by $\mu_1, \sigma_1$ and $\mu_2, \sigma_2$ such that $\mu_2 >\mu_1$ and the distributions are overlapping. Now suppose I have a random variable $\rho$ with unknown distribution $\phi_{\rho}$ such that $0<\rho<1$ (i.e. $\phi_{\rho}(x)=0$ $\forall x<0$ and $\forall x>1$).
I now have 1000 observations. $\rho*1000$ of them are distributed according to $n_1$ and the other $(1-\rho)*1000$ of them are distributed according to $n_2$. If $q\in [0,1]$ and I grab the greatest $q * 1000$ observations from my $1000$ total observations, how many of them are likely to come from $n_2$ (as a function of $\rho$)?
I know I am interested in the expected value of this but I am not sure exactly how to think about the distribution I am describing here (or what its expected value is). Intuitively I am thinking for instance that if $\rho$ is $0$, then they all come from $n_2$. If $\rho$ is $\frac{1}{1000}$, then only $1$ observation comes from $n_1$ so by taking, say, the $100$ largest observations it is highly highly unlikely I will get the one observation from $n_1$. But this becomes much less clear for less extreme values of $\rho$.
I know this is very general and theoretical but I appreciate any thoughts or comments about similar problems that might be helpful.
 A: I don't have a full solution but I can tell you "how to think about this question". The way is to use priors and the Bayes theorem:
Suppose you knew  the conditional probability to come from each distribution, given $x$ :
$P({\bf 1}|x)$ and $P({\bf 2}|x)$, where ${\bf 1};{\bf 2}$ represent which distribution was used. This is a useful quantity and you can ask questions like what the probability than (say) $n_1$ observations were above $x_0$ and were all from $\bf 1$, while the other $n_2$ were below it and from $\bf 2$, e.g. $$\begin{equation}{n_1+n_2 \choose n_1}\left(\int_{x_0} dxP({\bf 1}|x)\right)^{n_1}\left(\int^{x_0} dxP({\bf 2}|x)\right)^{n_2}\end{equation}\tag{1}$$
This is very close to what you want, and the best I can do.
In order to compute this, we will use Bayes's theorem several times:
Suppose we knew the joint probability $P(x,{\bf 1} )$, that is the probability to both observe $x$ and be in distribution $\bf 1$. Then, Bayes's theorem states that $ P({\bf 1}|x)= {P(x,{\bf 1} )\over P(x)}$
To compute the joint probabilities $P(x,{\bf 1} );P(x,{\bf 2} )$ let us start with the conditional probability to pick the value $x$ given  the observation was from distribution $\bf 1$. Thus $$P(x|{\bf 1})=N(x;\mu_1,\sigma_1)$$ (where $N$ is the normal distribution.  Similarly $$P(x|{\bf 2})=N(x;\mu_2,\sigma_2)$$ is the probability to pick $x$ given that the observation was from distribution $\bf 2$. The joint probabilities can be deduced ( again via Bayes's theorem):
$$P(x,{\bf 1})=P(x|{\bf 1})P({\bf 1})=P(x|{\bf 1})\rho \tag{2}$$
$$P(x,{\bf 2})=P(x|{\bf 2})P({\bf 2})=P(x|{\bf 2})(1-\rho) \tag{3}$$
So, we have $P(x,{\bf \{1,2\}})$, we are lacking $P(x)$, but this can be deduced by normalization: $P(x)=P(x|{\bf 1})P({\bf 1})+P(x|{\bf 2})P({\bf 2})$ ( the probability to see $x$ at all).
Finally, we can recover $$P({\bf 1}|x)={P(x|{\bf 1})\rho \over P(x|{\bf 1})\rho+P(x|{\bf 2})(1-\rho)}\tag{4}$$
