Example 1: Bernoulli model

The Bernoulli model admits a complete statistic.^[2] Let X be a random sample of size n such that each X_i has the same Bernoulli distribution with parameter p. Let T be the number of 1s observed in the sample, i.e. ${\displaystyle \textstyle T=\sum _{i=1}^{n}X_{i}}$. T is a statistic of X which has a binomial distribution with parameters (n,p). If the parameter space for p is (0,1), then T is a complete statistic. To see this, note that

${\displaystyle \operatorname {E} _{p}(g(T))=\sum _{t=0}^{n}{g(t){n \choose t}p^{t}(1-p)^{n-t}}=(1-p)^{n}\sum _{t=0}^{n}{g(t){n \choose t}\left({\frac {p}{1-p}}\right)^{t}}.}$

Observe also that neither p nor 1 − p can be 0. Hence ${\displaystyle E_{p}(g(T))=0}$ if and only if:

${\displaystyle \sum _{t=0}^{n}g(t){n \choose t}\left({\frac {p}{1-p}}\right)^{t}=0.}$

On denoting p/(1 − p) by r, one gets:

${\displaystyle \sum _{t=0}^{n}g(t){n \choose t}r^{t}=0.}$

First, observe that the range of r is the positive reals. Also, E(g(T)) is a polynomial in r and, therefore, can only be identical to 0 if all coefficients are 0, that is, g(t) = 0 for all t.

It is important to notice that the result that all coefficients must be 0 was obtained because of the range of r. Had the parameter space been finite and with a number of elements less than or equal to n, it might be possible to solve the linear equations in g(t) obtained by substituting the values of r and get solutions different from 0. For example, if n = 1 and the parameter space is {0.5}, a single observation and a single parameter value, T is not complete. Observe that, with the definition:

${\displaystyle g(t)=2(t-0.5),\,}$

then, E(g(T)) = 0 although g(t) is not 0 for t = 0 nor for t = 1.

Lehmann–Scheffé theorem

Completeness occurs in the Lehmann–Scheffé theorem,^[4] which states that if a statistic that is unbiased, complete and sufficient for some parameter θ, then it is the best mean-unbiased estimator for θ. In other words, this statistic has a smaller expected loss for any convex loss function; in many practical applications with the squared loss-function, it has a smaller mean squared error among any estimators with the same expected value.

Examples exists that when the minimal sufficient statistic is not complete then several alternative statistics exist for unbiased estimation of θ, while some of them have lower variance than others.^[5]

See also minimum-variance unbiased estimator.

Basu's theorem

Bounded completeness occurs in Basu's theorem,^[6] which states that a statistic that is both boundedly complete and sufficient is independent of any ancillary statistic.

Bahadur's theorem

Bounded completeness also occurs in Bahadur's theorem. In the case where there exists at least one minimal sufficient statistic, a statistic which is sufficient and boundedly complete, is necessarily minimal sufficient. Another form of Bahadur's theorem states that any sufficient and boundedly complete statistic over a finite-dimensional coordinate space is also minimal sufficient.^[7]