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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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    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.)2011-09-13

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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.