# Relationship between completeness and sufficiency

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?

• 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 $T(X)=V(X)$. Nov 14, 2019 at 21:15
• @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?
– asdf
Nov 15, 2019 at 11:33

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 $$S$$.

• 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? Jan 20, 2022 at 15:54
• That's right. It just highlights that $S$ is a function of $T$. Jan 20, 2022 at 16:21