# 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 $\displaystyle P(x; \theta) = (1-\theta)^x\theta$. Show that $\displaystyle Y = \prod X_i$ is a sufficient estimator of theta.

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

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

Thank you in advance.

2. Originally Posted by 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

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