In statistics, completeness is a property of a statistic computed on a sample dataset in relation to a parametric model of the dataset. It is opposed to the concept of an ancillary statistic. While an ancillary statistic contains no information about the model parameters, a complete statistic contains only information about the parameters, and no ancillary information. It is closely related to the concept of a sufficient statistic which contains all of the information that the dataset provides about the parameters.[1]
Consider a random variable X whose probability distribution belongs to a parametric model Pθ parametrized by θ.
Say T is a statistic; that is, the composition of a measurable function with a random sample X1,...,Xn.
The statistic T is said to be complete for the distribution of X if, for every measurable function g,[2]
if\operatorname{E}\theta(g(T))=0forall\thetathenP\theta(g(T)=0)=1forall\theta.
The statistic T is said to be boundedly complete for the distribution of X if this implication holds for every measurable function g that is also bounded.
The Bernoulli model admits a complete statistic.[3] Let X be a random sample of size n such that each Xi has the same Bernoulli distribution with parameter p. Let T be the number of 1s observed in the sample, i.e.
styleT=
| n | |
| \sum | |
| i=1 |
Xi
\operatorname{E}p(g(T))=
| n | |
| \sum | |
| t=0 |
{g(t){n\chooset}pt(1-p)n-t
Observe also that neither p nor 1 - p can be 0. Hence
Ep(g(T))=0
| n | |
| \sum | |
| t=0 |
g(t){n\chooset}\left(
| p | |
| 1-p |
\right)t=0.
On denoting p/(1 - p) by r, one gets:
| n | |
| \sum | |
| t=0 |
g(t){n\chooset}rt=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, a single observation and a single parameter value, T is not complete. Observe that, with the definition:
g(t)=2(t-0.5),
then, E(g(T)) = 0 although g(t) is not 0 for t = 0 nor for t = 1.
This example will show that, in a sample X1, X2 of size 2 from a normal distribution with known variance, the statistic X1 + X2 is complete and sufficient. Suppose (X1, X2) are independent, identically distributed random variables, normally distributed with expectation θ and variance 1.The sum
s((X1,X2))=X1+X2
is a complete statistic for θ.
To show this, it is sufficient to demonstrate that there is no non-zero function
g
g(s(X1,X2))=g(X1+X2)
remains zero regardless of the value of θ.
That fact may be seen as follows. The probability distribution of X1 + X2 is normal with expectation 2θ and variance 2. Its probability density function in
x
\exp\left(-(x-2\theta)2/4\right).
The expectation of g above would therefore be a constant times
| infty | |
| \int | |
| -infty |
g(x)\exp\left(-(x-2\theta)2/4\right)dx.
A bit of algebra reduces this to
k(\theta)
| infty | |
| \int | |
| -infty |
h(x)ex\thetadx
where k(θ) is nowhere zero and
| -x2/4 | |
| h(x)=g(x)e |
.
As a function of θ this is a two-sided Laplace transform of h(X), and cannot be identically zero unless h(x) is zero almost everywhere.[4] The exponential is not zero, so this can only happen if g(x) is zero almost everywhere.
By contrast, the statistic is sufficient but not complete. It admits a non-zero unbiased estimator of zero, namely
Suppose Then regardless of the value of Thus is not complete.
For some parametric families, a complete sufficient statistic does not exist (for example, see Galili and Meilijson 2016 [5]).
For example, if you take a sample sized n > 2 from a N(θ,θ2) distribution, then
| n | |
| \left(\sum | |
| i=1 |
Xi,
| n | |
| \sum | |
| i=1 |
| 2\right) | |
| X | |
| i |
| n | |
| 2\left(\sum | |
| i=1 |
| 2 | |
| X | |
| i |
If there is a minimal sufficient statistic then any complete sufficient statistic is also minimal sufficient. But there are pathological cases where a minimal sufficient statistic does not exist even if a complete statistic does.
The notion of completeness has many applications in statistics, particularly in the following two theorems of mathematical statistics.
Completeness occurs in the Lehmann–Scheffé theorem,[6] 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.[7]
See also minimum-variance unbiased estimator.
Bounded completeness occurs in Basu's theorem,[8] which states that a statistic that is both boundedly complete and sufficient is independent of any ancillary statistic.
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.[9]