Showing a distribution is not complete for a parameter $\theta$ I am trying to show that a normal distribution with parameters $\mu = 0$ and variance $\theta$ is not complete. I am looking for a function $u(x)$ that is not equal to 0 such that $\mathbb E(u(x)) = 0$. 
I have done some research on this problem and I have found that $\bar{X}$ and $S$ (sample standard deviation) are independent and can help yield me a function that will give me $\mathbb E(u(x)) = 0$. I know that $\bar{X}$ is normally distributed with mean 0 and variance $\theta/n$. I know that $S^2$ has a chi-squared distribution. I was trying to take an expectation $\mathbb E(\bar{X} S^2)$ to yield a value of 0, but I am not sure how exactly to do that.
Also, would this approach be correct in showing that it is not complete for $\theta$?
EDIT:
An iid sample is taken and theta > 0.
 A: If you mean a single observation and not an i.i.d. sample, so that you've got $X\sim N(0,\theta)$, then $X$ itself is an unbiased estimator of $0$, so this family of distributions is not complete.
You could use $X^2$ as an unbiased estimator of $\theta$.  You could also use $X^2+6X$.  Having two different unbiased estimators based on this statistic shows that its family of distributions is not complete.
If, on the other hand, you've got an i.i.d. sample $X_1,\ldots,X_n$, then $X_1-X_2$ is an unbiased estimator of $0$, so that family of distributions is not complete.  The sufficient statistic in that case would be $X_1^2+\cdots+X_n^2$, and I don't think any function of that is an unbiased estimator of $0$.
A: I think your question is not accurate. The question is, show that the family of distributions $\{f(x,\theta); 0<\theta<\infty\}$ where $X$ is distributed as $\operatorname{Normal}(0, \theta)$ is NOT complete.
A test for completeness can be done by finding a function $u(x)$ not equal to the zero-function such that $E[u(x)]=0$ $ \forall \theta$.
For this normal family this is rather easy, since you know the expected value of X is $0$ and it does not matter who $\theta$ is. Just use $u(x)=x$.
And you do not need a sample at all. But if you want to, use it and use the joint distribution. The same will hold.
