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**statmajor** Let X1,..,Xn be a random sample of size n from a geometric distribution with pmf $\displaystyle P(x; \theta) = (1-\theta)^x\theta$. Show that $\displaystyle Y = \prod X_i$ is a sufficient estimator of theta.

So $\displaystyle \prod P(x_i, \theta) = (1-\theta)^{\Sigma x_i} \theta^n$

$\displaystyle

P(\hat{x} | \theta) = \prod_{i=1}^n \big[(1-\theta)^{x_i}\theta\big] =

(1-\theta)^{\sum x_i} \theta^n = e^{\ln(1-\theta) \sum x_i + n \ln \theta}

$

EDIT: Note that the dependence on $\displaystyle \theta $ is ONLY in conjunction with $\displaystyle T(X)$.

The pmf above is an exponential family distribution with

$\displaystyle h(x) = 1 $, $\displaystyle \eta(\theta) = \ln(1-\theta)

$ , and $\displaystyle T(\hat{x}) = \sum x_i. $.

EDIT: To clarify, $\displaystyle \eta $ is simply the natural parameter of the exponential family.

-Andy