Suppose that $X_1,X_2$,....are iid from exponential($\lambda$).For n $\geq$ 1, let $X_{(n,1)}\le X_{(n,2)}\le X_{(n,3)}\le.......\le X_{(n,n)}$ be the order statistics of $X_1,X_2....X_n$. Suppose that $k_n$ is a sequence of integers satisfying $1\le k_n \le n$ for all n, and
${\lim_{x\to \infty}\frac{k_n}{n}} = p\in(0,1)$
Show that as n$\to\infty$
$\qquad\qquad\qquad\qquad\mathit{{X_{(n, k_n)}\to-\frac{1}{\lambda}\log(1-p)}}$
I am thankful for any help.
Fix $\epsilon > 0$.
Let $Y_1 \sim \text{Binom}(n, (1-p) e^{\lambda \epsilon})$ and $Y_2 \sim \text{Binom}(n, (1-p) e^{-\lambda \epsilon})$. \begin{align} P(|X_{n,k_n} + \log(1-p)/\lambda| > \epsilon) &\le P(X_{n,k_n} < - \log(1-p)/\lambda - \epsilon) + P(X_{n,k_n} > - \log(1-p)/\lambda + \epsilon) \\ &= P(\text{less than $n-k_n$ of the $X_i$ are $\ge - \log(1-p)/\lambda - \epsilon$}) + P(\text{more than $n-k_n$ of the $X_i$ are $\ge - \log(1-p)/\lambda + \epsilon$}) \\ &= P(Y_1 < n-k_n) + P(Y_2 > n-k_n) \\ &= P(Y_1/n < 1 - k_n/n) + P(Y_2/n > 1 - k_n/n). \end{align} By the law of large numbers, $Y_1/n \to (1-p) e^{\lambda \epsilon} > 1 - p$ in probability, so the first term will tend to zero. Similarly, the second term will tend to zero.