# Complete sufficient statistic and unbiased estimator

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!

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

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.
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.