Order statistics for exponential [duplicate]

If I have {X_1, X_2 ... X_n} i.i.d exponential random variable with parameter $\lambda=1$. And define $Y = \max\{X_1, X_2 ,... X_n\}$. I am interested in $\lim\limits_{n\to\infty}P(Y-\ln n \leq x) $

$$ f(x) = \lambda e^{-\lambda x} , \quad \lambda =1$$

$ f(x) = e^{-x} $, $ F(x) = 1-e^{-x} $.

It follows that, $g(y_n) = nf(y_n)F(y_n)^{n-1}, \quad y_n > 0$.

And, $ g(y) = n e^{-y} ( 1-e^{-y})^{n-1}, \quad y > 0$

Thus, $P(Y-\ln n \leq x) = P(Y \leq x +\ln n)$

Hence:

\begin{align*} P(Y-\ln n \leq x) &= P(Y \leq x +\ln n)\\ &= \int_0^{x+\ln n} n e^{-y} ( 1-e^{-y})^{n-1} dy\\ &= n\int_0^{x+\ln n} e^{-y} ( 1-e^{-y})^{n-1} dy\\ &= p^n \bigg\vert_0^{-\ln{(1-x-\ln n)}}\\ &= p^{-\ln{(1-x-\ln n)}} - 1 \end{align*}

\begin{align*} \lim\limits_{n\to \infty }p^{-\ln{(1-x-\ln n)}} - 1 &= \lim\limits_{n\to \infty } p^{-\ln{(1-x-\ln n)}} - 1 \\ &=\lim\limits_{n\to \infty} e^{-(1-Inn -x)\ln p} \\ &= \infty \end{align*}

Approach 2

$P(Y-\ln n \leq x) = P(Y-x \leq \ln n)$

Let $r=y-x$, $t=y$. Then it follows that: $x=t-r,\, y=t$. Then the Jacobaian, $\mid{J} \mid=1$

Since $X,Y$ are i.i.d, $$f(x,y) = e^{-x}e^{-y}(1-e^{-y})$$

It follows that: $$f(r,t) = e^{-(t-r)}e^{-t}(1-e^{-t})$$ Then $$f_r = \int_0^\infty e^{-(t-r)}e^{-t}(1-e^{-t}) dt = e^r\int_0^\infty e^{-2t}(1-e^{-t}) dt = \frac{1}{6}e^r $$

$$P(r\leq In n) = \frac{1}{6}\int_0^{\ln n} e^r dr = \frac{ n-1}{6} $$

Thus $$ \lim\limits_{n\to \infty} P(Y-\ln n \leq x) = \lim\limits_{n\to \infty}P(r \leq In n ) = \lim\limits_{n\to \infty} \frac{ n-1}{6} = \infty $$

Are both approaches equivalent. Which is better? Thank you in advance.

Solution 1:

Let $X_{(n)}$ be the $n$-th order statistic. We find its distribution: $$P(X_{(n)}\leq y)=P\bigg(\bigcap_{k\leq n}\{X_k\leq y\}\bigg)=(1-e^{-\lambda y})^n$$ So $$P(X_{(n)}\leq y+\ln(n))=(1-e^{-\lambda (y+\ln(n))})^n=\bigg(1-\frac{e^{-\lambda y}}{n^{\lambda}}\bigg)^n$$ If $\lambda = 1$ we have $P(X_{(n)}\leq y+\ln(n))\to e^{-e^{-y}}$