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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..

user135939
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    what if say $f$ is a constant function that always outputs $0$---something is wrong with this question... – Suvrit Oct 05 '14 at 20:54
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    the function $f(T)$ needs to be one-to-one, then the statement is correct – Carlo Beenakker Oct 05 '14 at 21:11
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    @CarloBeenakker: It doesn't have to be one-to-one (it needs to be measurable). The definition of "complete" (see the link you gave) has the form "for any function $g(T)$ [... something holds]", so completeness is trivially preserved under taking functions. – Christian Remling Oct 05 '14 at 22:00
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    I checked the definition, and the constant function is tautologically "complete", so please ignore my previous comment. – Suvrit Oct 06 '14 at 16:55

2 Answers2

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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 $\mathbb P_\theta(Z=0)\ne 1$.

Bjørn Kjos-Hanssen
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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

Carlo Beenakker
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