Instatistics,completeness is a property of astatistic computed on asample dataset in relation to a parametric model of the dataset. It is opposed to the concept of anancillary 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 asufficient statistic which contains all of the information that the dataset provides about the parameters.^[1]

Consider arandom variableX whose probability distribution belongs to aparametric modelP_θ parametrized by θ.

SayT is astatistic; that is, the composition of ameasurable function with a random sampleX₁,...,X_n.

The statisticT is said to becomplete for the distribution ofX if, for every measurable functiong,^[1]

${\displaystyle \text{if }{\mathrm{E}}_{\theta }(g(T))=0\text{ for all }\theta \text{ then }{\mathbf{P}}_{\theta }(g(T)=0)=1\text{ for all }\theta .}$

The statisticT is said to beboundedly complete for the distribution ofX if this implication holds for every measurable functiong that is also bounded.

The Bernoulli model admits a complete statistic.^[1] LetX be arandom sample of sizen such that eachX_i has the sameBernoulli distribution with parameterp. LetT be the number of 1s observed in the sample, i.e.${\displaystyle T=\sum _{i=1}^n{X}_i}$.T is a statistic ofX which has abinomial distribution with parameters (n,p). If the parameter space forp is (0,1), thenT is a complete statistic. To see this, note that

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

Observe also that neitherp nor 1 − p can be 0. Hence${\displaystyle E_p(g(T))=0}$ if and only if:

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

On denotingp/(1 − p) byr, one gets:

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

First, observe that the range ofr is thepositive reals. Also, E(g(T)) is apolynomial inr 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 ofr. Had the parameter space been finite and with a number of elements less than or equal ton, it might be possible to solve the linear equations ing(t) obtained by substituting the values ofr and get solutions different from 0. For example, ifn = 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 althoughg(t) is not 0 fort = 0 nor fort = 1.

This example will show that, in a sampleX₁, X₂ of size 2 from anormal distribution with known variance, the statisticX₁ + X₂ is complete and sufficient. SupposeX₁,X₂ areindependent, identically distributed random variables,normally distributed with expectationθ and variance 1.The sum

${\displaystyle s((X_1,X_2))=X_1+X_2}$

is acomplete statistic forθ.

To show this, it is sufficient to demonstrate that there is no non-zero function${\displaystyle g}$ such that the expectation of

${\displaystyle g(s(X_1,X_2))=g(X_1+X_2)}$

remains zero regardless of the value ofθ.

That fact may be seen as follows. The probability distribution ofX₁ + X₂ is normal with expectation 2θ and variance 2. Its probability density function in${\displaystyle x}$ is therefore proportional to

${\displaystyle \exp \left(-(x-2\theta {)}^2/4\right).}$

The expectation ofg above would therefore be a constant times

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}g(x)\exp \left(-(x-2\theta {)}^2/4\right)dx.}$

A bit of algebra reduces this to

${\displaystyle k(\theta ){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}h(x)e^{x\theta }dx,}$

wherek(θ) is nowhere zero and

${\displaystyle h(x)=g(x)e^{-x^2/4}.}$

As a function ofθ this is a two-sidedLaplace transform ofh, and cannot be identically zero unlessh is zero almost everywhere.^[2] The exponential is not zero, so this can only happen ifg is zero almost everywhere.

By contrast, the statistic$(X_1,X_2)$ is sufficient but not complete. It admits a non-zero unbiased estimator of zero, namely$X_1-X_2$.

Most parametric models have asufficient statistic which is not complete. This is important because theLehmann–Scheffé theorem cannot be applied to such models. Galili and Meilijson 2016^[3] propose the following didactic example.

Consider${\displaystyle n}$ independent samples from the uniform distribution:

${\displaystyle k}$ is a known design parameter. This model is ascale family (a specific case ofa location-scale family) model: scaling the samples by a multiplier${\displaystyle c}$ multiplies the parameter${\displaystyle \theta }$.

Galili and Meilijson show that the minimum and maximum of the samples are together a sufficient statistic:${\displaystyle X_{(1)},X_{(n)}}$ (using the usual notation fororder statistics). Indeed, conditional on these two values, the distribution of the rest of the sample is simply uniform on the range they define:${\displaystyle \left[X_{(1)},X_{(n)}\right]}$.

However, their ratio has a distribution which does not depend on${\displaystyle \theta }$. This follows from the fact that this is a scale family: any change of scale impacts both variables identically. Subtracting the mean${\displaystyle m}$ from that distribution, we obtain:

${\displaystyle \mathbb{E}\left[\frac{X_{(n)}}{X_{(1)}}\right]-m=0}$

We have thus shown that there exists a function${\displaystyle g\left(X_{(1)},X_{(n)}\right)}$ which is not${\displaystyle 0}$ everywhere but which has expectation${\displaystyle 0}$. The pair is thus not complete.

The notion of completeness has many applications in statistics, particularly in the following theorems of mathematical statistics.

Completeness occurs in theLehmann–Scheffé theorem,^[1]which states that if a statistic that is unbiased,complete andsufficient for some parameterθ, then it is the best mean-unbiased estimator for θ. In other words, this statistic has a smaller expected loss for anyconvex loss function; in many practical applications with the squared loss-function, it has a smaller mean squared error among any estimators with the sameexpected value.

Examples exists that when the minimal sufficient statistic isnot complete then several alternative statistics exist for unbiased estimation ofθ, while some of them have lower variance than others.^[3]

See alsominimum-variance unbiased estimator.

Bounded completeness occurs inBasu's theorem,^[1] which states that a statistic that is bothboundedly complete andsufficient isindependent of anyancillary statistic.

Bounded completeness also occurs in Bahadur's theorem. In the case where there exists at least oneminimal sufficient statistic, a statistic which issufficient and boundedly complete, is necessarily minimal sufficient.^[4]

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