1
$\begingroup$

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 $\mu$

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ 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

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.