suppose $T$ is a complete stats for a parameter $\theta$.
Is any function $f(T)$ again complete?
It sounds weird but the definition seems to confirm that $f(T)$ is indeed complete..



Is a function of complete statistics again complete?
Geometrically, completeness means something like this: if a vector $g(T)$ is orthogonal to the p.d.f. $f\_\theta$ of $T$ for each $\theta$,
$$\mathbb E\_\theta g(T) = \langle g(T),f\_\theta\rangle=0$$
then $g(T)=0$ i.e., the functions $f\_\theta$ for varying $\theta$ span the whole space of functions of $T$. So in a way it would be more natural to say that


*$\theta$ is complete for $T$*


than what we do say,


*$T$ is complete for $\theta$*.


This way it is not so strange that a constant function would be "complete"!




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Maybe an example helps.


Suppose $X$ and $Y$ are independent and identically distributed Bernoulli($\theta$) random variables taking values in $\{0,1\}$, and $Z=X-Y$. Then $Z$ is incomplete for $\theta$, because taking $g=\text{identity}$,
$$\mathbb E\_\theta(Z)=0$$ for all $0<\theta<1$, but nevertheless $\mathbb P\_\theta(Z=0)\ne 1$.


