Originally Posted by

**akbar** Let $\displaystyle \xi_1$, $\displaystyle \xi_2$... be independent identically distributed random variables with the Cauchy distribution. How do you prove that:

$\displaystyle \liminf_{n\rightarrow\infty}\mathbb{P}(\max(\xi_1, ...,\xi_n)> xn) \geq \exp(-\pi x)$

for any $\displaystyle x\geq 0$ ?

This is not only a liminf. I just did the usual procedure with this kind of problem, and it gives the limit. We have:

$\displaystyle P(\xi>T)=\int_T^\infty \frac{dt}{\pi(1+t^2)}=$ $\displaystyle \frac{1}{\pi}\arctan\frac{1}{T}\sim_{T\to\infty}\f rac{1}{\pi T}$

Hence $\displaystyle P(\max \xi_1,\ldots,\xi_n >xn)=1-P(\xi<xn)^n = 1-(1-P(\xi>xn))^n=$ $\displaystyle 1-\exp(n\log\left(1-\frac{1}{\pi x n}+o\left(\frac{1}{n}\right)\right))=1-\exp(-\frac{1}{\pi x}+o(1))$, which gives:

$\displaystyle P(\max \xi_1,\ldots,\xi_n> xn)\to_n 1-e^{-\frac{1}{\pi x}}$.

As a matter of fact, it seems that indeed $\displaystyle 1-e^{-\frac{1}{\pi x}}\geq e^{-\pi x}$ (study the function $\displaystyle x\mapsto e^{-x}+e^{-1/x}$ to prove it is less than 1? Maybe not easy, anyway the plots convinces me. ) I guess the question suggests a more direct way, but since the above limit is stronger... it satisfies me