What is the difference between complete statistics and complete family of distributions?

Question

I fail to understand when we call a family of distribution is complete and when a statistic is complete. What is the difference between both?, Is there a relation between them? Please provide examples along with your answer.

Answer 1

Suppose $X_1,\ldots,X_n \sim \text{i.i.d. } N(\mu,\sigma^2).$

The family of distributions is $$\left\{ N_n\left(\begin{bmatrix} \mu \\ \vdots \\ \mu \end{bmatrix},\sigma^2 \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}\right) \quad : \quad \mu\in \mathbb R \right\}$$ (so $\sigma$ is fixed, so that the only difference between one distribution and another in this family is a different value of $\mu$). This is the family of $n$-dimensional normal distributions in which the expected value is that column of $\mu$s and the matrix of covariances is $\sigma^2$ times the $n\times n$ identity matrix. This family of distributions of $n$-tuples is not complete since it admits nontrivial unbiased estimators of $0$; for example $X_1-X_2$ is such an estimator.

Now suppose $T = T(X_1,\ldots,X_n) = X_1+\cdots+X_n.$ It follows that $T\sim N(n\mu,n\sigma^2).$ The family of distributions of $T$ is $\{ N(n\mu,n\sigma^2) : \mu \in\mathbb R\},$ so again $\sigma$ is fixed, so the difference between two members of this family is a different value of $\mu$. This family is complete since it admits no nontrivial unbiased estimators of $0$. The fact that this family of distributions is complete is also expressed by saying that the statistic $T$ is complete. Any time you define a statistic that is a function of $(X_1,\ldots,X_n),$ having already defined a family of distributions for that $n$-tuple, that definition induces another family of distributions for the statistic you have defined. To say that that statistic is complete merely means that that induced family of distributions is complete.