Re: Expected value
- From: PARASHAR <parashar.dave@xxxxxxxxxxxxxxx>
- Date: Tue, 31 Jan 2006 03:22:35 EST
(1)[F_n](u) = (n-1) integral(x=(n-1) u...1-u, [F_(n-1)](u/x)
> x^(n-2)) for 0 <= u <= 1/n. F1(u) = 1 for
> 0 <= u < 1.
(2)By inspection then induction, [F_n](u) = (1-n u)^(n-1) for 0 <= u
> <= 1/n. Thus, EM_n = 1/n^2.
Stephen, Sorry but I couldn't get the above two. Canj u plz explain?
.
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