# Order statistics help

• Jan 19th 2010, 06:28 PM
Danneedshelp
Order statistics help
Q: Let $\displaystyle Y_{1},...,Y_{n}$ by independent, uniformly distributed random variables on the interval $\displaystyle [0,\theta]$. Find the

a) Probability distribution function of $\displaystyle Y_{(n)}=max(Y_{1},Y_{2},...,Y_{n})$.

b) density function of $\displaystyle Y_{(n)}$.

c) mean and variance of $\displaystyle Y_{(n)}$.

A: If I can figure out a) and identify the pdf, the cdf (part b) and mean / variance (part c) should follow. So, here is my attempt

Let $\displaystyle Y_{(n)}=max(Y_{1},Y_{2},...,Y_{n})$. Then, by Thereom blah, we have

$\displaystyle f_{Y_{(n)}}=g_{(n)}(y_{n})$$\displaystyle = \frac{n!}{(n-1)!(n-k)!}[F(y_{k})]^{k-1}[1-F(y_{k})]^{n-k}f(y_{k})$,$\displaystyle -\infty<y_{k}<\infty$ with $\displaystyle k=n$.

Thus, $\displaystyle f_{Y_{(n)}}=g_{(n)}(y_{n})= \frac{n!}{(n-1)!(n-n)!}[\frac{y_{n}}{\theta}]^{n-1}[1-\frac{y_{n}}{\theta}]^{n-n} \frac{1}{\theta}= \frac{n}{\theta}[\frac{y_{n}}{\theta}]^{n-1}, 0<y_{k}<\theta$.

I do not recognize the density function $\displaystyle \frac{n}{\theta}[\frac{y_{n}}{\theta}]^{n-1}, 0<y_{k}<\theta$ and its support.

Any help would be great. Not sure where I went wrong.

Thanks
• Jan 19th 2010, 09:58 PM
CaptainBlack
Quote:

Originally Posted by Danneedshelp
Q: Let $\displaystyle Y_{1},...,Y_{n}$ by independent, uniformly distributed random variables on the interval $\displaystyle [0,\theta]$. Find the

a) Probability distribution function of $\displaystyle Y_{(n)}=max(Y_{1},Y_{2},...,Y_{n})$.

Observe:

$\displaystyle F_{Y_{(n)}}(y)=P(Y_{(n)}<y)=\prod_{i=1}^n P(Y_i<y)=\begin{cases}0&,\ y<0 \\ \left(\frac{y}{\theta}\right)^n & ,\ 0 \le y \le \theta \\1 & , \ y>\theta \end{cases}$

CB
• Jan 19th 2010, 11:03 PM
matheagle
Just because you can't recognize this, doesn't mean it's wrong.
AND there is only one y, whether you call it $\displaystyle y_{k}$ or $\displaystyle y_{n}$.....

$\displaystyle \frac{n}{\theta}[\frac{y}{\theta}]^{n-1}, 0<y<\theta$

is the derivative of the CDF that CB derived.

If you are curious....
If we let $\displaystyle X=Y/\theta$ then X is a Beta and you can obtain the mean and variance that way if you wish.
• Jan 20th 2010, 12:39 PM
Danneedshelp
Quote:

Originally Posted by matheagle
If you are curious....
If we let $\displaystyle X=Y/\theta$ then X is a Beta and you can obtain the mean and variance that way if you wish.

I am not seeing how X is a beta random variable. What would my alpha and beta be?
• Jan 20th 2010, 01:41 PM
CaptainBlack
Quote:

Originally Posted by Danneedshelp
I am not seeing how X is a beta random variable. What would my alpha and beta be?

$\displaystyle \beta=1$ and $\displaystyle \alpha=N$

CB