Re: two-parameter exponential

On Sun, 25 Mar 2007 22:58:46 +0200, André <aajd@xxxxxxxxx> wrote:

Can someone explain to me why the distribution of the sample mean from a
random sample, X1,...,Xn, from a two-parameter distribution EXP(1,n)
equals the GAM(1/n,n) ditribution?

Two questions for you:

1. if each X_i is distributed as EXP(1,n), what is the distribution of
T=X_1+X_2+...+X_n?
2. if W is distributed as GAM(a,b), what is the distribution of cW?
Then consider that you get the sample mean from the sample total by
dividing by n.

Cheers,
Ken.

--
Ken Butler, Lecturer (Statistics)
University of Toronto at Scarborough
butler (at) utsc.utoronto.ca
http://www.utsc.utoronto.ca/~butler
.

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