There are $n$ people who privately and independently draw a score from the uniform distribution on $[0,1]$. There are $q$ prices to be given to the people with the highest scores among those who apply. A person only applies if the probability of getting a price is above a fixed threshold $\overline{p}$. Hence, with some probability (approximately $\overline p$) there are less applications than prices.
I am interested in how the expected number of people who don't apply but could have won evolves for $n\to\infty$ (and $q$ fixed).
Intuitively, there are two countervailing effects. As n grows the confidence interval of the cutoff score which is just sufficient to win (so the q-th order statistic) shrinks. On the other hand there are more people in any interval.
Mathematically the question can be phrased as follows. An individual person does not apply if her score is in the lower $\overline p$ quantile of the q-th highest order statistic. So, let us call $F_n$ the distribution q-th highest order statistic of $n$ draws (with density $f_n$). The critical cut-off below which people don't apply be called $\omega_n=F_n^{-1}(\overline p)$ we get as the ex-ante probability that a person does not apply even if she could have won:
$ P=\int_0^{\omega_n} (w_n-x)f_n(x)dx=\overline p \Big( \omega_n-\mathbb E_{F_n}[X|X\leq \omega_n]\Big)$
Multiplying this with the number of $n$ students the question reduces to the following:
At which rate does the conditional expectation of the lower p-quantile of the q-th oder statistic approach the $p$-quantile as $n$ goes to infinity? Is it faster or slower than $n$?
Maybe there is a straightforward formal argument. My hands on approach so far:
If I have score $\omega$ I can win with probability
$F(\omega)=\sum_{k=0}^{q-1}{n-1 \choose k} (1- \omega)^k \omega^{n-1-k} =\frac{(n-1)!}{((n-1)-(q-1)-1)!(q-1)!}\int_0^\omega x^{(n-1)-(q-1)-1}(1-x)^{q-1} dx $,
where I use the Binomial distribution and its Beta distribution form. I call $\overline\omega=F^{-1}(\overline p)$ the crucial score at the $\overline p$ quantile such that I apply if and only if $\omega\geq\overline\omega$. I am a person who doesn't apply but could have won if the the necessary score $\omega$ to win is below $\overline\omega$ and my score is in $[\omega,\overline\omega]$. Integrating this up and multiplying by $n$, the expected number of people who don't apply but could have won should be $ \mathbb{E_n}=n \frac{(n-1)!}{((n-1)-(q-1)-1)!(q-1)!} \int_0^{\overline\omega} (\overline\omega-x) x^{(n-1)-(q-1)-1}(1-x)^{q-1} dx $
I now need to determine $\lim_{n\to\infty} \mathbb E_n$, but I am stuck here. Can you provide any help, either from here or in any other form?
Thanks in advance!
Jonas
PS: I am not happy with my title, feel free to edit!