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If you have $n$ random variables that are iid with density $\frac{1}{p}e^{-x/p}$, how do you show that the sum of the $x_i$'s is a sufficient statistic?

Attempt: Take likelihood function and express in terms of $g(p)h(x)$ and use factorization theorem to show that it is a sufficient statistic. So likelihood = $\frac{1}{p^n} \exp(\sum \frac{-x_i}{p}) = \frac{1}{p^n} \exp(\frac{1}{p} \sum (-x_i))$.

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If I messed up your calculation while trying to format it, let me know. – Mike Spivey Sep 13 '11 at 3:34
lord12, You have not accepted any answers to your previous questions. It is generally considered rude in this site to keep posting questions, without accepting answers to previous ones. Please pick the answers that helped you the most and accept them. – Srivatsan Sep 13 '11 at 3:38
How do I accept an answer? – lord12 Sep 13 '11 at 3:49
lord12, there's a check mark by each answer. Click on the one by the answer that you found most helpful. (And it would also be considered polite to upvote - click on the little up arrow next to the answer - each answer that you found helpful.) – Mike Spivey Sep 13 '11 at 3:51
up vote 3 down vote accepted

These are exponential random variables. Consider the joint probability density. This would be $$ \prod_{i=1}^{n} \frac{1}{p}e^{-\frac{x_i}{p}}=p^{-n}e^{-\frac{1}{p}T(x)},\qquad T(x)= \sum_{i=1}^{n} x_i$$

Let $h(x) =1$ and $g(p,t) = p^{-n}e^{-t/p}$. Then the result follows from the factorization theorem.

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