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$(X_n)_{n\geq 1}$ is a sequence of random variables independent and identically distributed with Cauchy distribution of parameter $\alpha$ , $\alpha > 0$.

Let $$ Y_n= \dfrac{\max_{1 \leq k \leq n} X_k}{n}. $$
How to prove that $(Y_n)_{ n \geq 1 }$ converge in law to random variable $Z$ such that $\dfrac{1}{Z}$ has an exponential distribution?

So \begin{align*} F_{Y_n}(y)=& \mathbb{P}(\max_{1 \leq k \leq n} X_k \leq n y) \\ =& \mathbb{P}(X_k \leq n y, \, \forall k, \, 1 \leq k \leq n) \\ =& (F_{X_1}(ny))^n \\ =& \left(\dfrac{1}{\pi}\arctan(\dfrac{ny}{\alpha}) + \dfrac{1}{2}\right)^n \\ \sim & \left(1+ \dfrac{\alpha}{y}\dfrac{1}{n} \right)^n \\ =& \exp \dfrac{\ln \left(1+ \dfrac{\alpha}{ny}\right)}{\frac{1}{n}} \rightarrow & e^{\dfrac{\alpha}{y}} \end{align*} But the last expression is not a cumulative fonction. I'm very grateful if a kindly sool should be answer this question.

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    $\begingroup$ $\arctan(\frac{ny}{\alpha})$ does not behave like $\frac{ny}{\alpha}$ for large $n$. $\endgroup$ – André Nicolas Feb 17 '14 at 7:33
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    $\begingroup$ and in fact for large $x$, $\arctan(x) \approx \dfrac{\pi}{2}-\dfrac1x$ $\endgroup$ – Henry Feb 17 '14 at 8:47
  • $\begingroup$ @Henry Thank's, So I have improve it $\endgroup$ – Zbigniew Feb 17 '14 at 16:10

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