Thread: Order statistics - determine the cdf

1. Order statistics - determine the cdf

Let $\displaystyle u_1,...,u_N$ be N independent random variables uniformly distributed on the interval $\displaystyle [0,1]$. Determine the cdf of $\displaystyle y:= max_n u_n$ and $\displaystyle z := min_n u_n$

So we are after the cdf of $\displaystyle F_y(s)=P(max_u \leq s)$ which is the probability that $\displaystyle max_u$ is greater then $\displaystyle s_1, s_2...$? not sure how to formalise/what distribution this is?

Thanks for any help..

2. Originally Posted by Robb
Let $\displaystyle u_1,...,u_N$ be N independent random variables uniformly distributed on the interval $\displaystyle [0,1]$. Determine the cdf of $\displaystyle y:= max_n u_n$ and $\displaystyle z := min_n u_n$

So we are after the cdf of $\displaystyle F_y(s)=P(max_u \leq s)$ which is the probability that $\displaystyle max_u$ is greater then $\displaystyle s_1, s_2...$? not sure how to formalise/what distribution this is?

Thanks for any help..
Hint:

$\displaystyle max \{u_1, u_2, \dots , u_n\} \leq y$
if and only if
$\displaystyle u_i \leq y$ for all $\displaystyle i = 1, 2, \dots , n$.

3. Originally Posted by Robb
Let $\displaystyle u_1,...,u_N$ be N independent random variables uniformly distributed on the interval $\displaystyle [0,1]$. Determine the cdf of $\displaystyle y:= max_n u_n$ and $\displaystyle z := min_n u_n$

So we are after the cdf of $\displaystyle F_y(s)=P(max_u \leq s)$ which is the probability that $\displaystyle max_u$ is greater then $\displaystyle s_1, s_2...$? not sure how to formalise/what distribution this is?

Thanks for any help..
You will find detailed proofs in most mathematical statistics books, using Google and also by searching this subforum (use the Search tool).

4. hmm, so after doing a bit of research...
The distribution of independent order statistics would be;
For the Maximum
$\displaystyle F_{Y_{N}}(s)=P(Y_{n}\leq s)=P(Y_1\leq s)P(Y_2\leq s)...P(Y_n\leq s)=[F_U(s)]^N$
So since since it is the uniform distribution with $\displaystyle F_U(s)=s$ the cdf is $\displaystyle F_{Y_{N}}(s)=s^N$

And the minimum will be
$\displaystyle F_{Z_1}(s)=1-P(Z_1 > s)=1-[1-F_U(s)]^N$

5. Originally Posted by Robb
hmm, so after doing a bit of research...
The distribution of independent order statistics would be;
For the Maximum
$\displaystyle F_{Y_{N}}(s)=P(Y_{n}\leq s)=P(Y_1\leq s)P(Y_2\leq s)...P(Y_n\leq s)=[F_U(s)]^N$
So since since it is the uniform distribution with $\displaystyle F_U(s)=s$ the cdf is $\displaystyle F_{Y_{N}}(s)=s^N$

And the minimum will be
$\displaystyle F_{Z_1}(s)=1-P(Z_1 > s)=1-[1-F_U(s)]^N$
Order Statistics 10/30