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The d.f/p.d.f. of the mean values in a random sample

 
 
Reply Thu 9 Dec, 2004 10:40 pm
I'm stuck.

Suppose that the variables X1, X2, ..., Xn form a random sample of size n from a distribution for which the p.d.f. is f and the d.f. is F.
I was successful in finding the d.f. and p.d.f for the largest value Yn where Yn=max{X1,...,Xn} and the smallest value Y1 where Y1=min{x1,...,Xn}.
I found them to be the following:
Gn(y) = [F(y)]^n and gn(y) = n{F(y)}^(n-1)f(y) for largest value.
G1(y) = 1- [1-F(y)}^n and g1(y) = n[1-F(y)]^(n-1)f(y) for the smallest value.

But the problem I'm facing is finding the pf or pdf of the mean value. Is it really simple and I'm just making it a lor harder than it is? Please give some advice so I can figure it out!

Thanks.
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Type: Discussion • Score: 1 • Views: 521 • Replies: 1
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spendius
 
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Reply Fri 10 Dec, 2004 06:04 am
Its nice to meet like minded people.

I can't figure it out either.

spendius.
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