Is a function of complete statistics again complete? 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..
 A: 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"!

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 $\mathbb P_\theta(Z=0)\ne 1$.
A: Statistics $T$ and $T'$ are called equivalent if there exists a one-to-one function $f$ such that $T'=f(T)$. Equivalent statistics give equivalent information, in particular, if $T$ and $T'$ are equivalent statistics and $T$ is complete for $\theta$ then $T'$ is complete for $\theta$.
See, for example, http://www.randomservices.org/random/point/Sufficient.html
