Suppose $\{P_n\}$ and P are probability measures on the real line with corresponding distribution functions $\{F_n\}$ and $F$, respectively. $P_n$ converges weakly to P if and only $$\lim_{n \rightarrow \infty} F_n(x) = F(x)$$ at every point x where F is continuous (this can be seen from Portmanteau theorem). How can we use this fact to prove the following.
Let $X_1,X_2,...$ denote i.i.d. Cauchy random variables with parameter $a>0$. More clearly, the density of $X_i$ is $$P(X_i \in dx) = \frac{adx}{\pi(a^2+x^2)}, x \in \mathbb{R}$$ Prove that $$\frac{1}{n}\max_{1\leq i \leq n}X_i$$ converges in law to $\frac{1}{T}$ where $T$ is an exponential random variable. What is the rate of $T$ as a function of $a$?
Let $F_n(x)$ denote the distribution function of $Y_n = \frac{1}{n}\max\limits_iX_i$. Then,
$$ \begin{eqnarray} F_n(x) &=& \mathbb{P}(Y_n \leq x)\\ &=& \mathbb{P}\left(\max\limits_iX_i \leq nx\right)\\ &=& \mathbb{P}\left(X_i \leq nx, \;\; \forall i\right)\\ &=& (\mathbb{P}(X_1 \leq nx))^n\\ &=& \left(\frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{nx}{a}\right)\right)^n \end{eqnarray}$$
Note that $\arctan(x)$ behaves asymptotically like $\frac{\pi}{2} - \frac{1}{x}$. Then, you can show that $F_n(x) \xrightarrow{n\rightarrow\infty} e^{-\frac{a}{\pi x}}$, which is the distribution function of an inverse exponential.