 in statistics, completeness is a property of a statistic in relation to a model for a set of observed data. in essence, it is a condition which ensures that the parameters of the probability distribution representing the model can all be estimated on the basis of the statistic: it ensures that the distributions corresponding to different values of the parameters are distinct. it is closely related to the idea of identifiability, but in statistical theory it is often found as a condition imposed on a sufficient statistic from which certain optimality results are derived. ==definition== consider a random variable x whose probability distribution belongs to a parametric family of probability distributions p''Î¸ parametrized by&nbsp;Î¸. formally, a statistic s is a measurable function of x; thus, a statistic s is evaluated on a random variable x, taking the value s(x), which is itself a random variable. a given realization of the random variable x(Ï) is a data-point (datum), on which the statistic s takes the value&nbsp;s(x(Ï)). the statistic s is said to be complete for the distribution of x if for every measurable function g (which must be independent of Î¸) the following implication holds:young, g. a. and smith, r. l. (2005). essentials of statistical inference. (p. 94). cambridge university press. :e(g(s(x))) = 0 for all Î¸ implies that pÎ¸(g(s(x)) = 0) = 1 for all Î¸. the statistic s is said to be boundedly complete if the implication holds for all bounded functions g. === example 1: bernoulli model === the bernoulli model admits a complete statistic.casella, g. and berger, r. l. (2001). statistical inference. (pp. 285-286). duxbury press. 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. 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 : \operatorname{e}(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&nbsp;&minus;&nbsp;p can be 0. hence e(g(t)) = 0 if and only if: :\sum_{t=0}^n g(t){n \choose t}\left(\frac{p}{1-p}\right)^t = 0. on denoting p/(1&nbsp;&minus;&nbsp;p) by r, one gets: :\sum_{t=0}^n g(t){n \choose t}r^t = 0 . first, observe that the range of r is all 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)&nbsp;=&nbsp;0 for all&nbsp;t. it is imp
Statistical_terminology Statistical_theory