hopefully this isn't a duplicate of another question (at least I didn't find one).

Here is a question I have about completeness and sufficiency:

Problem: Suppose $T(x)$ is complete sufficient for $\theta$ given data $x$. Show that if a minimal sufficient statistic $S(x)$ for $\theta$ exists, then $T(x)$ is also minimal sufficient.

My solution: Since $T(x)$ is complete we have that $T(X)$ is the unique MVUE for $\mathbb{E}[T(X)]=m(\theta)$ for a specific function $m$.

Consider now $$V(X)=\mathbb{E}[T(X)|S(X)].$$

By Rao-Blackwell we know that $Var(V(X))\leq Var(T(X))$. Hence, by uniqueness of MVUEs we must have that $V(X)=T(X)$, i.e. that $T(X)=g(S(X))$ from the definition of $V(X)$ (for some function $g$). However, as $T$ is a function of minimal sufficient statistic, it is also minimal sufficient.

The problem with my solution is that I don't use the minimal sufficiency of $S$ until the very end, in comparison to the author's solution. Its idea is to say that $V(X)=h(S(X))$ by definition of the conditional expectation and then argue that $V(X)=f(T(X))$ as $S$ is minimal sufficient. The result then follows from the completeness of $T$.

I also seem to prove that every complete sufficient statistic for $\theta$ is a function of any other sufficient statistic for $\theta$. Is that true or have I made a mistake somewhere?

  • 1
    $\begingroup$ How does $Var(V(X))\le Var(T(X))$ guarantee that $V(X)$ (a function of $S(X)$) is UMVUE? You are missing some details. The idea is correct, but I think it is a slightly convoluted way of showing $T(X)=V(X)$. $\endgroup$ Nov 14, 2019 at 21:15
  • $\begingroup$ @StubbornAtom, hey! Well both of them are unbiased for $m(\theta) $ and V has lower variance than T which is an MVUE (and is in particular, unique). Doesn't that suffice? $\endgroup$
    – asdf
    Nov 15, 2019 at 11:33

1 Answer 1


A complete sufficient statistic is a minimal sufficient statistic whenever a minimal sufficient statistic exists.

Suppose for a family of distributions parameterized by $\theta$, there exists a minimal sufficient statistic $S(X)$ and a complete sufficient statistic $T(X)$ based on the data $X$. We show that $T$ is also minimal sufficient.

As $S$ is minimal sufficient and $T$ is sufficient, by definition of minimal sufficiency there exists a measurable function $h$ such that $S=h(T)$.

Consider the function $g(T)=T-E_{\theta}[T\mid S]=T-E[T\mid S]$, so that $E_{\theta}[g(T)]=0$ for every $\theta$.

As $T$ is complete, this implies $g(T)=0$ almost everywhere. That is, $$T=E[T\mid S]\quad,\text{a.e.}$$

So $T$ is a function of $S$. And as $S$ is a function of any other sufficient statistic, so is $T$.

Therefore $T$ is minimal sufficient and equivalent to $S$.

  • $\begingroup$ Where are we using $S=h(T)$ in the proof? Or is that line there just to highlight the fact that $S$ is a function of any other sufficient statistic? $\endgroup$
    – user587389
    Jan 20, 2022 at 15:54
  • $\begingroup$ That's right. It just highlights that $S$ is a function of $T$. $\endgroup$ Jan 20, 2022 at 16:21

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