Neeraj,
It is easy to verify that if Y is exponential with mean t then Y/t is is
exponential with mean 1.
Also, the sum of n exponentials with parameter 1 has the distribution
Gamma(n,1). Most texts on probability and statistics (Feller Vol II, Mood
and Graybill) are references. It is a consequence of the fact that Gamma
(1,b) is the exponential density with parameter b and the convolution of
Gamma (r,b) and Gamma (s,b) is Gamma (r+s,b).
Ellen Hertz
Arun Nagarkatti wrote:
> I am looking through notes for confidence interval for the exponential
> mean.
>
> I have been given that:
>
> Suppose Y_1,...,Y_n ~ Exp(t^(-1)) independently. Then for each of the
> Y_i:
>
> f(y_i|t) = t^(-1) exp(-y_i/t).
>
> We are then finding the maximum likelihood estimator ^t^, and with
> various calculations (including logarithmic differentiation) we arrive
> at the sample mean:
> _
> y (y-bar) = ^t^.
>
> Now it is this part I don't understand:
>
> The sampling distribution of ^t^ (t-hat) is obtained as follows:
>
> Since Y_i ~ Exp(t^(-1))
>
> Y_i/t ~ Exp(1) ==> Sum Y_i/t ~ Gamma(n,1), or:
>
> n ^t^
> ----- ~ Gamma(n,1).
> t
>
> I just can't seem to follow the logic here. Can someone explain to me
> what is going on?
>
> Thanks,
>
> Neeraj.
=================================================================
Instructions for joining and leaving this list and remarks about
the problem of INAPPROPRIATE MESSAGES are available at
http://jse.stat.ncsu.edu/
=================================================================
