show that the distribution function x max is F (x) to the power n Sol: distribution function F (x) =p (X <=x) and random sample x1, x2, x3...taken from population. F (x1)=P (X <=x1) F (x2)=P (X <=x2) ..... ..... F (xn)=P (X <=×n) Now how Summation of F (x) where x= 1 to n becomes F (x) to the power n as asked by he question. I know its easy but I am not getting it.
Firstly, you should do product not summation, Now, we know x will be maximum iff (x1,x2,.......,xn <x) Solution:- F(xmax)=P(x1,x2,.........,xn<x)=P(x1<x,x2<x,........,xn<x) Now on the basis of assumption: Random samples x_i's are independently taken from identical population with Random Variable X. F(xmax)=P(X<x)*P(X<x)....n times =[P(X<x)]^n =[F(x)]^n
As for CDF of X_MIN we can't find directly .... Hence, we first get P(x_1,x_2,........x_n>x) then assumed IID for x_i's P(x_1>x)P(x_2>x)..........P(x_n>x) then get [1- F(x)] n times Then P(x_1,x_2......x_n<x)=1-[1-F(x)]^n
