# Thread: Sufficient Estimator for a Geometric Distribution

1. ## Sufficient Estimator for a Geometric Distribution

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

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

I don't believe that the factorization theroem can be applied here. Is there some trick to this that I'm not seeing?

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

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

$
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 $\theta$ is ONLY in conjunction with $T(X)$.

The pmf above is an exponential family distribution with

$h(x) = 1$, $\eta(\theta) = \ln(1-\theta)
$
, and $T(\hat{x}) = \sum x_i.$.

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

-Andy

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