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Math Help - Sufficient Estimator for a Geometric Distribution

  1. #1
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    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?

    Thank you in advance.
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  2. #2
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    Quote Originally Posted by statmajor View Post
    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

    <br />
P(\hat{x} | \theta) = \prod_{i=1}^n \big[(1-\theta)^{x_i}\theta\big] = <br />
(1-\theta)^{\sum x_i} \theta^n = e^{\ln(1-\theta) \sum x_i + n \ln \theta}<br />

    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)<br />
 , and  T(\hat{x}) = \sum x_i. .

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

    -Andy
    Last edited by abender; January 12th 2010 at 01:28 PM. Reason: A few extra comments.
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