# Convergence in distribution sequence max

Let $(X_n)_{n≥1}$ be a sequence of i.i.d. random variables with standard Cauchy distribution and let $M = \max\{X_1 ,...,X_n\}$. Prove that $(n M^{-1}_n)_{n \ge 1}$ converges in distribution and identify the limit.

I am a little confused. is this supposed to be inverse Cauchy?

• $P(nM_n^{-1}\ge t)=P(M_n\le n/t)=(\frac 1 2 +\frac 1 \pi \arctan(n/t))^n\longrightarrow e^{-t/\pi}1_{t\ge 0}$ hence the limit distribution is $\exp(1/\pi)$. – A.S. Dec 4 '15 at 13:31
• @A.S. What about $t<0$? – Math1000 Dec 4 '15 at 13:32
• @Math $P(nM_n^{-1}< 0)=2^{-n}\to 0$ so negative $t$ don't happen in the limit distribution. – A.S. Dec 4 '15 at 13:39

For $x>0$ we have \begin{align} \mathbb P(nM_n^{-1} \leqslant x) &= 1- \mathbb P(nM_n^{-1}>x)\\ &= 1-\mathbb P\left(M_n\leqslant \frac nx\right)\\ &= 1-\left(\frac1\pi\arctan\left(\frac nx\right) + \frac12 \right)^n\\&\stackrel{n\to\infty}\longrightarrow 1 - e^{-\frac1\pi x}, \end{align} so that $nM_n^{-1}$ converges in distribution to $\operatorname{Exp}\left(\frac1\pi\right)$.