Convergence in distribution of a sequence of random variables

Let $Y_1,Y_2,...,$ be independent $C(0,1)$ random variables, determine the limit distribution of :

$Z_n = \dfrac{1}{n} \cdot max\{Y_1, Y_2,..Y_n \}$ as $n \rightarrow \infty$,

Here is my approach:

$F_{Z_n} (x) = \mathbb{P}(Z_n \leq x) = \mathbb{P}(\dfrac{1}{n} max\{ Y_1,Y_2,...Y_n \} \leq x ) = \mathbb{P}( Y_1 \leq x\cdot n , Y_2 \leq x\cdot n, ....,Y_n \leq x\cdot n ) = \Big(F_{Y}(x\cdot n)\Big)^n$ (where $Y \in C(0,1)$ )

I have used that they are independent and identically distributed.

The next step is to find $F_Y(x\cdot n)$ which can be found by integration of the density of $C(0,1)$ , that is $F_Y(x\cdot n) = \dfrac{1}{\pi} \int_{-\infty}^{x\cdot n} \dfrac{1}{1+x^2} =\dfrac{1}{\pi} \big( arctan(x \cdot n) + \dfrac{\pi}{2} \big)$ .

Here is where Iam stuck , however, i know that $\dfrac{\pi}{2} = arctan(u) + arctan(1/u)$ and that $arctan(z) \approx z - \dfrac{z^3}{3} + \dfrac{z^5}{5} + ...$

I tried to use it somehow to get some "nice" expression for: $\dfrac{1}{\pi} \big( arctan(x \cdot n) + \dfrac{\pi}{2} \big)$, but failed

With "nice" i refer to the fact that $lim_{n \rightarrow \infty} \Big(F_{Y}(x\cdot n)\Big)^n$ could be recognized "easily"

Can someone give me a helping hand ?

• Perhaps using $F_{Z_n}$, you compute the characteristic function, and then let $n\to\infty$ to see if a pattern emerges. – Kim Jong Un Sep 16 '14 at 18:31

Your approach is correct. You obtain that $F_Y(nx) = 1 - \arctan (1/nx) / \pi = 1 - 1/(\pi n x) + \ldots$, so you deduce an approximation of $\log F_Y(nx)$ since $\log(1+x)$ is equivalent to $x$, and you get the result.
• what do mean with since $log(1+x)$ is equivalent to $x$ , (surly Iam missing something) @Paul – Danny Sep 16 '14 at 21:46
• It means that $\log(1+x)/x$ tends to $1$ when $x$ goes to $0$. So here $\log F_Y(nx)$ is equivalent to $F_Y(nx) - 1$ which itself is equivalent to $-1/(\pi nx)$ when $n$ goes to $\infty$. – polmath Sep 16 '14 at 21:55
• My goal is to get the limit of $\Big(F_{Y}(x\cdot n)\Big)^n$ as $n \rightarrow \infty$ , i don't see how maybe you could elaborate on this @Paul – Danny Sep 16 '14 at 22:23
• I could approximate $F_Y(nx) \approx 1 - \dfrac{1}{\pi n x}$ and then find the limit of $\Big(1-\dfrac{1}{\pi n x} \Big)^n$ as $n\rightarrow \infty$ , however this approximation is quiet rough isn't it – Danny Sep 16 '14 at 22:51
• To conclude, you have $F_Y(n x)^n = \exp(n \log F_Y(n x))$, and the fact that $\log F_Y(n x)$ is equivalent to $-1/(\pi n x)$ shows that $n \log F_Y(n x)$ tends to $-1/(\pi x)$; hence that $F_Y(n x)^n$ tends to $\exp(-1/(\pi x))$. – polmath Sep 17 '14 at 7:41