Suppose there are two distributions with cdf $F_1$ and $F_2$ and I draw $n_i$ times from distribution $F_i$.
Distribution 1 has support $[a,b]$ and distribution 2 has support $[c,d]$ with $a \leq c <b\leq d$. For example, let the random variables described by the cdf $F_i$ be the age of a randomly drawn person of group $i$.
Let us look at the $k$-th highest value of $n_1+n_2$ draws with $n_i$ draws from distribution $F_i$. The cdf of this random variable should be: \begin{equation} H^k (x) := \sum\limits_{i=0}^{k-1} {n_1 \choose i} F_1(x)^{n_1-i} (1-F_1(x))^i \left( \sum\limits_{j=0}^{k-i-1} {n_2 \choose j} F_2(x)^{n_2-j} (1-F_2(x))^j \right) \end{equation} I can now easily form an expected value of this order statistic.
Suppose one of the two scenarios [where (2.) is the more general form of (1.)]:
- Some interval $X= [x, d]$ with $x \in (c,d)$ is announed.
- Some interval $X' = [\underline x, \overline x]$ is announced.
Then either no realization (person's age) is in the corresponding interval and no draw (person) is removed, or at least one draw is within the interval. In the latter case, one person with age within the interval is removed uniformly at random. In case no person was removed, the expected value of the $k$-th oldest person in the room is easy to calculate (just update $F_1$ and $F_2$ that no draw is within the interval).
How does the cdf of the $k$-th highest value of the remaining $(n_1+n_2-1)$ values look like if a draw was removed?
Clearly, if $X \subset (b,d] \cup [a,c)$, this is an easy task (assuming $H^k$ as above is correct), because I know with certainty from which distribution the removed draw came from. So I just replace the corresponding $n_i$ with $(n_i-1)$. Call this updated cdf $H^k_{-i}$.
However, what if the interval $X$ does not make it obvious from which distribution the draw was removed? How does the distribution look like that I, as a Bayesian, use to calculate expectations of the $k$-th order statistic.