# Is a minimal sufficient statistic also a complete statistic

I know that if a statistic is both sufficient and complete then it must also be minimal sufficient. But on the other hand, could I say a minimal sufficient statistic must also be a complete statistic?

• I agree with the answers below, however it is interesting to note that the converse is true: If a minimal sufficient statistic exists, then any complete statistic is also minimal sufficient. Jul 6, 2018 at 22:24

Examples of minimal sufficient statistic which are not complete are aplenty.

A simple instance is $$X\sim U (\theta,\theta+1)$$ where $$\theta\in \mathbb R$$.

It is not difficult to show $$X$$ is a minimal sufficient statistic for $$\theta$$. However, $$E_{\theta}(\sin 2\pi X)=\int_{\theta}^{\theta+1} \sin (2\pi x)\,\mathrm{d}x=0\quad,\forall\,\theta$$

And $$\sin 2\pi X$$ is not identically zero almost everywhere, so that $$X$$ is not a complete statistic.

Another example for discrete distribution can be found in textbooks as an exercise or otherwise:

Let $$X$$ have the mass function

$$f_{\theta}(x)=\begin{cases}\theta&,\text{ if }x=-1\\\theta^x(1-\theta)^2&,\text{ if }x=0,1,2,\ldots\end{cases}\quad,\,\theta\in (0,1)$$

It can be verified that $$X$$ is minimal sufficient for $$\theta$$.

Suppose $$\psi$$ is any measurable function of $$X$$. Then

\begin{align} &\qquad\quad E_{\theta}(\psi(X))=0\quad,\forall\,\theta \\&\implies \theta\psi(-1)+\sum_{x=0}^\infty \psi(x)\theta^x(1-\theta)^2=0\quad,\forall\,\theta \\&\implies \sum_{x=0}^\infty \psi(x)\theta^x=\frac{-\theta\psi(-1)}{(1-\theta)^2}=-\sum_{x=0}^\infty\psi(-1)x\theta^x\quad,\forall\,\theta \end{align}

Comparing coefficient of $$\theta^x$$ for $$x=0,1,2,\ldots$$ we have $$\psi(x)=-x\psi(-1)\quad,\, x=0,1,2,\ldots$$

If $$\psi(-1)=c\ne 0$$, then $$\psi(x)=-cx\quad,\, x=0,1,2,\ldots$$

That is, $$\psi$$ is non-zero with positive probability. Hence $$X$$ is not complete for $$\theta$$.

• This sine example is the best of all answers: very short and easy. Sep 11, 2020 at 16:55

Consider $N(\theta,\theta)$ where $\theta>0$.Of course $\dfrac{1}{n}\sum_{i=1}^n X_i$ is minimal sufficient but not complete. To see why it is not complete, find $a$ and $b$ such that:
$$E\Big(a\sum_{i=1}^n (X_i-\overline{X})^2 \Big)=E\Big(b\sum_{i=1}^nX_i^2\Big)=\theta^2$$ and therefore $E\Big(a\sum_{i=1}^n (X_i-\overline{X})^2-b\sum_{i=1}^nX_i^2\Big)=0$ for all $\theta$.

• Well-formulated, although it is a little weird to have a normal distribution with same mean and variance. Also, could you give an example of the choice of a and b to complete the solution? Jan 24, 2018 at 19:11
• I do not understand the argument since your function is a function of the pair $(\sum_{i=1}^n (X_i-\overline{X})^2,\sum_{i=1}^nX_i^2)$ not of $\bar{X}_n$. Jan 25, 2018 at 14:06
• For $N(\theta,\theta^2)$, a similar argument would work showing that $(\bar X,S^2)$ is minimal sufficient but not complete for $\theta$. Jul 20, 2018 at 12:48
• If a minimal sufficient statistic is not complete, then a complete statistic simply does not exist. But for $N(\theta,\theta)$, a minimal complete sufficient statistic is $\sum X_i^2$, as can be seen from the one-parameter exponential family setup. Jul 20, 2018 at 13:02

In the Cauchy distribution with unknown location, $$f(x;\mu) = \frac{1}{\pi} \, \frac{1}{1+(x-\mu)^2}$$ for a sample $(X_1,\ldots,X_n)$ the order statistic $(X_{(1)},\ldots,X_{(n)})$ is minimal sufficient, but it is incomplete since $$\mathbb{E}_\mu[\phi(X_{(i)} - X_{(j)})]\qquad i\ne j$$is constant in $\mu$ for bounded functions $\phi$. Or since $$\mathbb{E}_\mu[\phi(X_{(i)} - X_{(j)})]\qquad 1< i\ne j <n$$is (well-defined and) constant in $\mu$.

• Could you elaborate since I am a little bit lost? In my understanding, the expectation of Cauchy distribution should be infinity and how could you subtract the expectation of two order statistics, given $\mu$? Jan 25, 2018 at 4:18
• Correct: I added a function to make the expectation to exist! Jan 25, 2018 at 5:47
• Addendum: Actually the order statistics have expectations except for the extreme ones. Jan 25, 2018 at 7:08