Order statistics help

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January 19th 2010, 06:28 PM
Danneedshelp

Order statistics help

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

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

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

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

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

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

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

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

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

Thanks
January 19th 2010, 09:58 PM
CaptainBlack

Quote:

Originally Posted by Danneedshelp

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

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

Observe:

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

CB
January 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 $y_{k}$ or $y_{n}$ .....

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

is the derivative of the CDF that CB derived.

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

Quote:

Originally Posted by matheagle

If you are curious....
If we let $X=Y/\theta$ 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?
January 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?

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

CB