Find the first order statistic

**jmyers12345** · March 21st 2011, 01:52 PM

Suppose Y_1,..., Y_n is a random sample where the density of each random variable Y_i is f(y) = 2*x^2*y^(-3), y >= 0 for some parameter x > 1. Find the first order statistic, Y_(1).

**matheagle** · March 21st 2011, 03:37 PM

use the complement, let M be the min

$F_M(y)=P(M\le y)=1-P(M>y)$
now if the min exceeds the value, y, they all do...

$=1-P(Y_1>y,\ldots ,Y_n>y)$

using independence

$=1-P(Y_1>y)\cdots P(Y_n>y)$

$=1-\bigl[P(Y_1>y)\bigr]^n$

$=1-\bigl[1-F_Y(y)\bigr]^n$

now differentiate and you have

$f_M(y)=P(M\le y)=n\bigl[1-F_Y(y)\bigr]^{n-1}f_Y(y)$

finally plug in your density/distribution.

**jmyers12345** · March 21st 2011, 03:51 PM

Thanks, that's what I had but I'm still getting the wrong answer. When computing the pdf and I'm integrating the cdf given what should my limits of integration be?

**matheagle** · March 21st 2011, 03:56 PM

show your work and we can better answer your question
y goes from zero to infinity, x is a fixed constant.
NEVERMIND
this is NOT a valid density
MY guess is that y is bounded below by x, clearly it is not bounded below by 0.

**jmyers12345** · March 21st 2011, 04:53 PM

Originally Posted by matheagle

show your work and we can better answer your question
y goes from zero to infinity, x is a fixed constant.
NEVERMIND
this is NOT a valid density
MY guess is that y is bounded below by x, clearly it is not bounded below by 0.

Ok, so when I integrate (2*x^2*y^(-3)) with respect to y I get -(x^2/y^2) and plugging in the bounds of x to infinity, I get 0 - (-1) = 1. But this can't be right because then the entire pdf for the order statistic ends up equaling 0. Would I just use the indefinite integral of the pdf -(x^2/y^2) as my cdf and plug that in for the cdf found in the equation for the first order statistic that you listed above?

**matheagle** · March 21st 2011, 05:02 PM

IT is right
ONE means it is a valid density
that's how I figured out what your density must be

**matheagle** · March 21st 2011, 05:03 PM

$F_Y(y)=2x^2\int_x^yt^{-3}dt$

x is not a variable, it's just a fixed constant.
BUT you should look up the problem again, having y bounded below by zero, seems off.

$2x^2\int_x^{\infty}y^{-3}dy$

must be one.

**jmyers12345** · March 21st 2011, 05:05 PM

Thanks for all the help, I got it

**matheagle** · March 21st 2011, 05:09 PM

but is that the correct density, where y>x?

**jmyers12345** · March 21st 2011, 05:13 PM

Originally Posted by matheagle

$F_Y(y)=2x^2\int_x^yt^{-3}dt$

x is not a variable, it's just a fixed constant.
BUT you should look up the problem again, having y bounded below by zero, seems off.

$2x^2\int_x^{\infty}y^{-3}dy$

must be one.

Yes, you're right. y >= x, not y >= 0 as I typed in the original problem statement. My mistake

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