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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?

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    $\begingroup$ 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. $\endgroup$
    – knrumsey
    Jul 6, 2018 at 22:24

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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 $\theta$.

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

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  • $\begingroup$ 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? $\endgroup$
    – son520804
    Jan 24, 2018 at 19:11
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    $\begingroup$ 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 $\bar{X}_n$. $\endgroup$
    – Xi'an
    Jan 25, 2018 at 14:06
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    $\begingroup$ For $N(\theta,\theta^2)$, a similar argument would work showing that $(\bar X,S^2)$ is minimal sufficient but not complete for $\theta$. $\endgroup$ Jul 20, 2018 at 12:48
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    $\begingroup$ 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. $\endgroup$ Jul 20, 2018 at 13:02
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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 $\mu$.

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  • $\begingroup$ 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 $\mu$? $\endgroup$
    – son520804
    Jan 25, 2018 at 4:18
  • $\begingroup$ Correct: I added a function to make the expectation to exist! $\endgroup$
    – Xi'an
    Jan 25, 2018 at 5:47
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    $\begingroup$ Addendum: Actually the order statistics have expectations except for the extreme ones. $\endgroup$
    – Xi'an
    Jan 25, 2018 at 7:08

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