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Let $X_1,X_2... X_n$ be iid $\sim F$ where $F$ is any symmetric continuous distribution and let $\mid E(X)\mid<\infty$.

Is $\bar{x}$ complete sufficient for $E(x)=\int{xf(x)dx}$?

Assume that all parameters of $F$ except $E(x)=\mu$ are known and that the support of $X$ does not depend on $\mu$

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    $\begingroup$ I only ever see sufficiency defined for parameters that fully characterise the distribution. Here E(x) does not characterise F, and one could see the shape of F as another "parameter"; certainly it is additional information required to characterise F. In such a situation I only see joint sufficient statistics defined, for all characterising parameters together. I haven't come across a definition of sufficiency that covers one parameter only, if that parameter doesn't characterise the distribution. No such definition may exist, in which case the concept of sufficiency doesn't apply. $\endgroup$ – Lewian Dec 3 '19 at 16:16
  • $\begingroup$ I made some changes. Suppose E(x) fully characterizes the distribution (it can have other parameters but all of these are assumed known). Also, let the support of $X$ not depend on $\mu$ to remove uniform $\endgroup$ – Marj Dec 3 '19 at 16:41
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    $\begingroup$ @jbowman: Do you mean Uniform(0,a) or something? Uniform(0,1) is a single distribution and doesn't have a free parameter. $\endgroup$ – Lewian Dec 3 '19 at 16:49
  • $\begingroup$ @Lewian - oh, right, of course. Thanks! $\endgroup$ – jbowman Dec 3 '19 at 16:50
  • $\begingroup$ @Marj: Chances are one can still construct a counterexample based on the uniform. Take a mixture of 0.9*Uniform(0,2a)+0.1*something else symmetric about a with support on the full real line. Still the mean doesn't hold all information for estimating a. $\endgroup$ – Lewian Dec 3 '19 at 16:51
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Consider the Laplace distribution with known scale. It is symmetric, continuous, and has support equal to the real line. The unknown location parameter $\mu$ is equal to the expected value. Yet the sample mean is not the sufficient statistic for $\mu$; the sufficient statistic in this case is the entire sample.

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    $\begingroup$ Although I believe that you are right, it may be helpful to show that the mean is not sufficient. (I'm too lazy trying it... just saying... ;-) $\endgroup$ – Lewian Dec 4 '19 at 15:48
  • $\begingroup$ @Lewian - me too, that's why I didn't. Oh well, with a little prodding from you, maybe I'll get it done ;-) $\endgroup$ – jbowman Dec 4 '19 at 16:21

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