Let $X_1, X_2,\ldots $ be independent, Cauchy $C(0,1)-$distributed random variables. Determine the limit distribution of $$Y_n=\frac 1 {n} \cdot \max{(X_1,X_2,\dots,X_n)}$$ as $n \rightarrow \infty$.
Remark. It may be helpful to know that
- $\arctan x + \arctan \frac 1 {x} = \frac \pi 2$ and that
- $\arctan y = y - \frac {y^3} 3 + \frac {y^5} 5 - \frac {y^7} 7 + \dots$
I not sure how to attack this question, nor how I should handle the $\max{(X_1,X_2,\ldots,X_n)}$ in order to find the limit distribution. Any advice appreciated.
\begin{align}F_{Y_n}(y)&=P(Y_n\le y)=P(\max\{X_1,X_2,\dots,X_n\}\le yn)\\&=P(X_1\le ny, X_2\le ny, \dots, X_n\le ny)=^{\text{independent}}\\&=P(X_1\le ny)P(X_2\le ny)\cdots P(X_n\le ny)=^{\text{identically $X\sim C(0,1)$ distributed}}\\&=\left(F_X(yn)\right)^n\end{align} So $$\lim_{n\to \infty}F_{Y_n}(y)=\lim_{n\to \infty}\left(F_X(yn)\right)^n=\lim_{n\to \infty}\left(\frac12+\frac1{\pi}\arctan{(ny)}\right)^n=e^{-\frac1{yπ}}\mathbf1_{y> 0}$$
Since, by the hints that you are given, for $y>0$: $$\left(\frac12+\frac1{\pi}\arctan(ny)\right)^n\overset{1.}=\left(1+\frac{-\arctan{(1/yn)}}{\pi}\right)^n\overset{2.}\sim_{\infty}\left(1+\frac{-1}{\pi yn}\right)^n\to e^{-1/yπ}$$