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).
use the complement, let M be the min
$\displaystyle F_M(y)=P(M\le y)=1-P(M>y)$
now if the min exceeds the value, y, they all do...
$\displaystyle =1-P(Y_1>y,\ldots ,Y_n>y)$
using independence
$\displaystyle =1-P(Y_1>y)\cdots P(Y_n>y)$
$\displaystyle =1-\bigl[P(Y_1>y)\bigr]^n$
$\displaystyle =1-\bigl[1-F_Y(y)\bigr]^n$
now differentiate and you have
$\displaystyle f_M(y)=P(M\le y)=n\bigl[1-F_Y(y)\bigr]^{n-1}f_Y(y)$
finally plug in your density/distribution.
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?