4
$\begingroup$

I am now studying complete sufficient statistic. My question is: Is there any relationship between the existence of complete sufficient statistic and the existence of unbiased estimator?

I know that by Lehmann-Scheffe Theorem, if an unbiased estimator of $\theta$(parameter) exists as a function of a complete sufficient statistic, then it should be unique up to a.s. sense. What I am wondering is: Can we always find such an unbiased estimator if we have complete sufficient statistic? Or conversely, if an unbiased estimator of $\theta$(parameter) exists as a function of a sufficient statistic, does that imply that the sufficient statistic is complete?

I tried to come up with some examples but failed to think of any. For example, is there an example s.t. a complete sufficient statistic exists whereas an unbiased estimator does not exists as a function of the complete sufficient statistic? I'd really appreciate your comment. Thanks!

$\endgroup$
  • 2
    $\begingroup$ There are situations, where an unbiased estimator does not exist even though a sufficient statistic exists. One example is for $\vartheta$ for $Y \sim \text{Bin}(n, 1/\vartheta)$. $\endgroup$ – Björn Oct 25 '17 at 9:41
  • 1
    $\begingroup$ And the Binomial admits a complete statistic. $\endgroup$ – Xi'an Oct 25 '17 at 10:14
2
$\begingroup$

Can we always find such an unbiased estimator if we have complete sufficient statistic?

A slight modification of the one given in the comments. Let $X_1,X_2,...X_n$ follow $B(m,\theta)$. Then the function $g(\theta)=\frac{1}{\theta}$ doesn't admit an unbiased estimator while $\sum_{i=1}^{n}{X_i}$ is a Complete Sufficient Statistic.

Or conversely, if an unbiased estimator of θ(parameter) exists as a function of a sufficient statistic, does that imply that the sufficient statistic is complete?

Let $X_1,X_2,...X_n$ follow $P(\theta)$ then $S^2 = \frac{1}{n-1}\sum_{i=1}^{n}{{(X_i-\bar{X}})}^2$ is an unbiased estimator and is a function of a statistic $T(X) = (\sum_{i=1}^{n}{X_i},\sum_{i=1}^{n}{X_i}^2)$ which is sufficient. But $S^2$ is not a Complete Sufficient Statistic.

$\endgroup$
0
$\begingroup$

No, there are examples where there is a complete sufficient statistic but there is some function of the parameter that does not admit an unbiased estimator. One binomial example, noted in comments, is discussed here at math SE. There are many other examples, this is one.

A paper discussing the topic is A Class of Parameter Functions for Which the Unbiased Estimator Does Not Έxist.

$\endgroup$

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.