Can the measurable mapping in the definition of complete statistics depend on sample size?

Question

1. The definition of complete statistics is from http://en.wikipedia.org/wiki/Completeness_(statistics)#Definition

   The statistic $s$ is said to be complete for the distribution of $X$ if for every measurable function $g$ (which must be independent of $θ$) the following implication holds:

   $E(g(s(X))) = 0$ for all $θ$ implies that $P_θ(g(s(X)) = 0) = 1$ for all $θ$.

   Let the codomain of statistic $s$ be $\mathbb R^m$, then is $g$ a measurable mapping from $\mathbb R^m$ to $\mathbb R$?

   Since $g$ acts on the codomain of $s$, does $g$ not know the sample size $n$ of $X = \{X_1, \dots, X_n\}$? Only $s$ acts on the sample $X$,so does only $s$ know the sample size of $X$?

   But in a solution to the problem 6.15 in Casella and Berger's Statistical Inference, when proving a statistic is not complete, $g$ is chosen to depend on $n$.

   what is $g$ actually? A measurable mapping from $\mathbb R^m$ to $\mathbb R$ which doesn't know the sample size, or something like a statistic which knows the sample size?

   Or does the statistic $s$ also output the sample size $n$ of its input $X$? I.e. the codomain of a statistic $s$ is $\mathbb R^m \times \mathbb N$? So that $g$ is defined on $\mathbb R^m \times \mathbb N$, i.e. $g$ get to know the sample size from the output of statistic $s$?

2. In general (going beyond the concept of complete statistics), when talking about a mapping $g$ on a statistic $s(X)$, i.e. $g(s(X))$, do we always assume $g$ knows the sample size of $X$, i.e. is the sample size of $X$ always an input to $g$?

   Even further do we assume $g$ know the entire input sample $X$ (not just its size $n$) to the statistic $s$? e.g. $s(X) = \sum_i X_i$, and does it make sense that $g(s(X)) = (\sum_i X_i) + (\sum_i X_i^2)$?

See also https://math.stackexchange.com/questions/918632/in-composition-of-two-mappings-can-the-outer-mapping-access-the-arguments-of-th

Thanks.

Answer 1

Things are clearer if one thinks of completeness as a property of a parametric family of distributions $$ \mathcal F = \{F_\theta: \theta \in \Theta\}, $$ where $F_\theta$ is a probability distribution on (say) $\mathbb R^m$. Then $\mathcal F$ is complete if $$ \int g(x) \ dF_\theta(x) = 0 \iff F_\theta(g(x) = 0) = 1 \mbox{ for all $\theta$}. $$ When one talks about a statistic $s(X)$ being sufficient, where $X$ is an $\mathbb R^n$-valued random vector and $X \sim F^n_\theta$ (i.e. $X_i \stackrel{iid}{\sim} F_\theta$) for some unknown $\theta$, what one really means is that the parametric family $\{G_\theta: \theta \in \Theta\}$ is complete where $G_\theta$ is the distribution of $s(X)$ when $X \sim F^n_\theta$.

As such, then, the notion of completeness has nothing to do with the sample size and the function $s(X)$ needed to get a complete statistic might change with the sample size. For example if $X_i \stackrel{iid}{\sim} N(\mu, 1)$ then $s(X) = \sum_{i=1}^n X_i$ depends on the sample size. What the solution to 6.15 shows is that, irrespective of the value of $n$, the parametric family of distributions of the statistic $(\bar X, S^2)$ does not correspond to a complete family. A different $g$ is required for each value of $n$, but for each value of $n$ we know such a function exists so the statistic is not complete for any sample size.

The function $g$ is just a mapping from the space $\mathbb R^m$ to the real numbers. $g$ doesn't "know" anything other than its input. But, again, the function $g$ needed to show incompleteness depends on the distribution of $s(X)$ and $s(X)$ itself may depend on the sample size. Indeed, $s: \mathbb R^n \to \mathbb R^m$ so $s$ itself obviously must depend on the sample size.