Prove that $P(X(n) ≤ k\ln(n)) = 0$ for $k < 1$ and that $P(X(n) ≤ k\ln(n)) = 1$ for $k > 1$

Question

Let $X_1,X_2,\ldots,X_n$ be a collection of independent exponentially distributed random variables each with rate parameter equal to $1$ ($ = 1$).

Also for each integer $n ≥ 1$, let $X_{(n)}$ = $\max\{X_1,X_2,\ldots,X_n\}$.

Show that for $k < 1$, $$\lim_{n\to\infty} P(X_{(n)}≤k\ln(n)) = 0$$

Also show that for $k > 1$, $$\lim_{n\to\infty} P(X_{(n)}≤k\ln(n)) = 1$$

I know how to figure out the CDF of the $X_{(n)}$ , which is (with $ = 1$) $$F_{X_{(n)}} = P(X_{(n)}≤x) = [1-e^{-x}]^n$$

But I got stuck here. I am not sure how to proceed to show the two separate limit values.

Help is much appreciated. Please and thank you.

Answer 1

So taking $x=k\ln(n)$, you have, for any $k>0$, $$P\Bigl(X_{(n)}\le k\ln(n)\Bigr)=\left(1-\mathrm e^{-k\ln(n)}\right)^{\!n}= \exp\left(n\ln\Bigl(1-n^{-k}\Bigr)\right).$$ Since $\ln(1-n^{-k})\sim-n^{-k}$ as $n\to\infty$, the last expression is $\exp\bigl(-n^{1-k}+o(n^{1-k})\bigr)$. Can you conclude?