# Expectation of k-th order statistic of Negative Binomial Distribution

Let $$X_1, X_2,...$$ be i.i.d $$NB(k,q)$$. I am interested in calculating the expectation of their k-th order statistic $$X_{k:n}$$. From my understanding of order statistics, the CDF of $$X_{k:n}$$ is given as \ $$P[X_{k:n} \leq x] = \sum_{j=k}^n {n \choose j} (F(x))^j(1-F(x))^{n-j}$$

From CDF, the PMF can be derived as: $$P[X_{k:n} = x] = P[X_{k:n} \leq x] - P[X_{k:n} \leq x-1]$$

and the expectation : $$E[X_{k:n}] = \sum_{x = k}^\infty x P[X_{k:n} = x]$$

I fear the direct calculation in such manner would involve complex algebra. Is there any simple way to derive such expectation. I also read the "The Order Statistics of the Negative Binomial Distribution" by D. H. Young (link to the paper - The Order Statistics of the Negative Binomial Distribution). Although, the author uses the recurrence relation to calculate m-th moments.

Any help in this matter, even a simple algebra trick would be much appreciated.