# Determine the limit distribution of $\frac{\max\{X_1, X_2,…,X_n \}}{n}$ as $n \to\infty$ [duplicate]

Let $X_1, X_2, . . .$ be independent, $C(0, 1)$-distributed random variables (Cauchy).

Determine the limit distribution of $$Y_n=\frac{\max\{X_1, X_2,...,X_n \}}{n}\qquad\text{ as }\quad n\to\infty$$

This is how far I have come:

$$F_{Y_n}(y)=P(\max\{X_1, X_2,...,X_n \}<ny)=F_{\max(X_i)}(ny)=(F_x(ny))^n$$ Since the cumulative distribution function of $X$ is $F_X(x)=\frac{1}{\pi}\arctan(x)+\frac{1}{2}, x \in \mathbb{R}$, we get :

$$F_{Y_n}(y)=\left(\frac{1}{\pi}\arctan(ny)+\frac{1}{2}\right)^n$$

Correct so far?

Now, we seek $\lim_{n\to\infty}F_{Y_n}(y)$ and its here im stuck. I have tried to use the identity $$\arctan(x)+\arctan(1/x)=\frac{\pi}{2}$$ but I just get complicated sums inside sums leading me nowhere. Any help is appreciated!

## marked as duplicate by Mike Earnest, StubbornAtom, Lord Shark the Unknown, Did limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 21 '18 at 17:02

As you have obtained we can write $$F_{Y_n}(y)=\left(\frac{1}{\pi}\arctan(ny)+\frac{1}{2}\right)^n=F_{Y_n}(y)=\left(1-\frac{1}{\pi}\arctan(\dfrac{1}{ny})\right)^n$$when $y\ge 0$ using Taylor series $$\left(1-\frac{1}{\pi}\arctan(\dfrac{1}{ny})\right)^n=\left(1-\frac{1}{\pi}\left(\dfrac{1}{ny}+o(\dfrac{1}{ny})\right)\right)^n$$therefore $$\lim_{n\to\infty}\left(1-\frac{1}{\pi}\arctan(\dfrac{1}{ny})\right)^n=\lim_{n\to\infty}\left(1-\frac{1}{\pi}\left(\dfrac{1}{ny}+o(\dfrac{1}{ny})\right)\right)^n=e^{-\frac{1}{\pi y}}$$and if $y<0$ we have that $$0\ge \frac{1}{\pi}\arctan(ny)+\frac{1}{2}<\dfrac{1}{2}$$so we have $$F_Y(y)=\begin{cases}0&,\quad y<0\\e^{-\frac{1}{\pi y}}&, \quad y\ge 0\end{cases}$$and $$f_Y(y)=\begin{cases}0&,\quad y<0\\\dfrac{1}{\pi y^2}e^{-\frac{1}{\pi y}}&, \quad y\ge 0\end{cases}$$