2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $T(X)=V(X)$. $\endgroup$ – StubbornAtom Nov 14 '19 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 '19 at 11:33
1
$\begingroup$

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 $S$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.