Minimum-variance unbiased estimator

Estimator selection

An efficient estimator need not exist, but if it does and if it is unbiased, it is the MVUE. Since the mean squared error (MSE) of an estimator δ is

$${\displaystyle \mathrm{MSE}(\delta )=\mathrm{var}(\delta )+[\mathrm{bias}(\delta ){]}^2}$$

the MVUE minimizes MSE among unbiased estimators. In some cases biased estimators have lower MSE because they have a smaller variance than does any unbiased estimator; see estimator bias.

Example

Consider the data to be a single observation from an absolutely continuous distribution on ${\displaystyle \mathbb{R}}$ with density

$${\displaystyle p_{\theta }(x)=\frac{\theta e^{-x}}{{\left(1+e^{-x}\right)}^{\theta +1}},}$$

where θ > 0, and we wish to find the UMVU estimator of

$${\displaystyle g(\theta )=\frac{1}{{\theta }^2}}$$

First we recognize that the density can be written as

$${\displaystyle \frac{e^{-x}}{1+e^{-x}}\exp \left(-\theta \log (1+e^{-x})+\log (\theta )\right)}$$

which is an exponential family with sufficient statistic ${\displaystyle T=\log (1+e^{-X})}$. In fact this is a full rank exponential family, and therefore ${\displaystyle T}$ is complete sufficient. See exponential family for a derivation which shows

$${\displaystyle \mathrm{E}(T)=\frac{1}{\theta },\mathrm{var}(T)=\frac{1}{{\theta }^2}}$$

Therefore,

$${\displaystyle \mathrm{E}(T^2)=\frac{2}{{\theta }^2}}$$

Here we use Lehmann–Scheffé theorem to get the MVUE. Clearly, ${\displaystyle \delta (X)=T^2/2}$ is unbiased and ${\displaystyle T=\log (1+e^{-X})}$ is complete sufficient, thus the UMVU estimator is

$${\displaystyle \eta (X)=\mathrm{E}(\delta (X)\mid T)=\mathrm{E}\left(\frac{T^2}{2}|T\right)=\frac{T^2}{2}=\frac{\log {\left(1+e^{-X}\right)}^2}{2}}$$

This example illustrates that an unbiased function of the complete sufficient statistic will be UMVU, as Lehmann–Scheffé theorem states.

Other examples

• For a normal distribution with unknown mean and variance, the sample mean and (unbiased) sample variance are the MVUEs for the population mean and population variance.

  However, the sample standard deviation is not unbiased for the population standard deviation – see unbiased estimation of standard deviation.

  Further, for other distributions the sample mean and sample variance are not in general MVUEs – for a uniform distribution with unknown upper and lower bounds, the mid-range is the MVUE for the population mean.

• If k exemplars are chosen (without replacement) from a discrete uniform distribution over the set {1, 2, ..., N} with unknown upper bound N, the MVUE for N is: $${\displaystyle \frac{k+1}{k}m-1,}$$ where m is the sample maximum. This is a scaled and shifted (so unbiased) transform of the sample maximum, which is a sufficient and complete statistic. See German tank problem for details.