CDF of max of $n$ cauchy variates Suppose $X_1,X_2,\cdots,X_n$ are iid copies of a standard cauchy variate with pdf
$$ f(x)=\frac{1}{\pi(1+x^2)},0<x< \infty. $$
Define:
$$ Y=1+ \underset{1 \leq i \leq n}\max X_i.$$ I want to find the cdf of $Y.$
My attempt:- First we try to find the cdf of $Z=\underset{1 \leq i \leq n}{\max} X_i$ as follows;
if the cdf of $X_i$ is $F_X(x)$, then
$$F_Z(z)= (F_X(x))^n$$ Or,
$$F_Z(z)=  \int_0^z \left(\frac{1}{\pi(1+x^2)} \right )^n$$
Now I am not sure whether I am right and how to proceed further .For any help I will be  greatly obliged
 A: You ask about the distribution of the maximum order statistic. There are many similar questions, for other distribution, so follow the outlines of for instance Distribution of extreme values, case of uniform  or Distribution of sample maximum from exponential distribution.
Doing so, simply differentiate the $n$th power of the Cauchy cumulative distribution function, and you will find
$$
   \frac{\left(\frac{1}{2}+\frac{\arctan \left(t \right)}{\pi}\right)^{n} n}{\pi  \left(t^{2}+1\right) \left(\frac{1}{2}+\frac{\arctan \left(t \right)}{\pi}\right)}
$$
You can then take care of adding 1 yourself.
As for the cdf, the above density must be integrated. Using maple, I do not find a very useful expression with a general $n$, but for specific values of $n$, I can get complicated expressions. As an example, for $n=10$ I get
$$
    \frac{\pi^{10}+20 \arctan \! \left(t \right) \pi^{9}+180 \arctan \! \left(t \right)^{2} \pi^{8}+960 \arctan \! \left(t \right)^{3} \pi^{7}+3360 \arctan \! \left(t \right)^{4} \pi^{6}+8064 \arctan \! \left(t \right)^{5} \pi^{5}+13440 \arctan \! \left(t \right)^{6} \pi^{4}+15360 \arctan \! \left(t \right)^{7} \pi^{3}+11520 \arctan \! \left(t \right)^{8} \pi^{2}+5120 \arctan \! \left(t \right)^{9} \pi +1024 \arctan \! \left(t \right)^{10}}{1024 \pi^{10}}
$$
Even for $n=100$ I get an expression, But I doubt it is very useful, as it will fit a complete page in a printed book.
