Order Statistics/Change of Variable

**statmajor** · November 5th 2009, 12:29 PM

Let Y1<Y2<...<Yn be the order statistics of a random sample of size n from the pdf $f(x) = e^{-x}$ x ranging from 0 to infinity.

a) Show that Z1=nY1, Z2 = (n-1)(Y2 - y1) Z3= (n-2)(Y3-Y2)... Zn = Yn - Y_(n-1) are independent and that each Z has the exp distribution.

b) Demonstrate that all linear functions of Y1, Y2,...,Yn such as $\Sigma a_i Y_i$ can be expressed as a linear function of independent random variables.

a)

so $y_1 = z_1/n$ , $y_2 = z_2/(n-1) +z_1/n$ , $y_3 = z_3/(n-2) + z_2/(n-1) +z_1/n$ , etc...

So how would I find the jacobian?

**matheagle** · November 5th 2009, 03:10 PM

I showed and used this in a paper of mine.
The joint distribution of the order stats is n! the orginal density
WITH the restriction of y(1)<y(2)<....<y(n).

**statmajor** · November 5th 2009, 04:13 PM

so:

$h(y_1,y_2,...,y_n) = n! e^{-y_1 - y_2 - ... - y_n}$

The determinant of the Jacobian Matrix is 1/n!.

$g(z_1,z_2,...,z_n) = n! e^{-\frac{z_1}{n} - \frac{z_2}{n-1} - \frac{z_1}{n} - \frac{z_3}{n-2} - \frac{z_2}{n-1} - \frac{z_1}{n}...}\frac{1}{n!}$

but according to my textbook, the pdf of Z1, Z2...Zn is supposed to be (after the change of variable):

$g(z_1,z_2,...,z_n) = e^{-z_1 - z_2 - ... - z_n}$

If I add up $-\frac{z_1}{n} - \frac{z_2}{n-1} - \frac{z_1}{n} - \frac{z_3}{n-2} - \frac{z_2}{n-1} - \frac{z_1}{n}...$ would I just get z1, z2,...,zn?

Okay, I think I got part A. But how would I do part B?

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