Editing Minimum-variance unbiased estimator (section)

==Example==
Consider the data to be a single observation from an [[Absolutely continuous random variable|absolutely continuous distribution]] on <math>\mathbb{R} </math> 
with density

:<math> p_\theta(x) = \frac{ \theta e^{-x} }{(1 + e^{-x})^{\theta + 1} }, </math>
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where ''θ > 0'', and we wish to find the UMVU estimator of

:<math> g(\theta) = \frac 1 {\theta^2} </math>
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First we recognize that the density can be written as

:<math> \frac{ e^{-x} } { 1 + e^{-x} } \exp( -\theta \log(1 + e^{-x}) + \log(\theta)) </math>
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which is an exponential family with [[sufficient statistic]] <math>T = \log(1 + e^{-X})</math>. In fact this is a full rank exponential family, and therefore <math> T </math> is complete sufficient. See [[exponential family]] 
for a derivation which shows

: <math> \operatorname{E}(T) = \frac 1 \theta,\quad \operatorname{var}(T) = \frac 1 {\theta^2} </math>
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Therefore,

:<math> \operatorname{E}(T^2) = \frac 2 {\theta^2} </math>
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Here we use Lehmann–Scheffé theorem to get the MVUE. Clearly, <math> \delta(X) = T^2/2</math> is unbiased and <math>T = \log(1 + e^{-X})</math> is complete sufficient, thus the UMVU estimator is

:<math> \eta(X) = \operatorname{E}(\delta(X) \mid T) = \operatorname{E} \left( \left. \frac{T^2} 2 \,\right|\, T \right) = \frac{T^2} 2 = \frac{\log(1 + e^{-X})^2} 2 </math>
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This example illustrates that an unbiased function of the complete sufficient statistic will be UMVU, as [[Lehmann–Scheffé theorem]] states.