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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 $1le k_n le n$ for all n, and


${lim_{xto infty}frac{k_n}{n}} = pin(0,1)$


Show that as n$toinfty$

$qquadqquadqquadqquadmathit{{X_{(n, k_n)}to-frac{1}{lambda}log(1-p)}}$


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.