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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.

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    I have tried to rewrite your post to use math notation. It would be useful for you to learn enough $\LaTeX$ to do it yourself. Looking at examples here should be enough.2012-01-26
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    Note that there is *still more work to be done*. In particular you use both $\theta$ and $\sigma^2$ to refer to the variance. So, you should settle on a single notation. Also, *completeness* is a property of a *family* of distributions, not a single one. Other notes: (1) $S^2$ is *not* distributed chi-squared, but a scaled version of it is, (2) a "function" like $\bar{X} \sigma^2$ will not help you since it is not a *statistic*. **Please** make edits to your question accordingly. Cheers. :)2012-01-26
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    @icobes: It might be best not to materially change the question (even when the edit is small) unless you have confirmation that is what the OP intended. :)2012-01-26
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    My apologies! Will refrain from doing it in the future :)2012-01-26
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    @icobes: Just meant as a word of caution. You have probably correctly interpreted the OP's intent. Still, my opinion (which others may not agree with) is that it is better to allow the OP to edit such things where appropriate.2012-01-26
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    Thank you all for making my post more readable. Any help on how to proceed with this expectation would be great. Also, would X-bar * S^2 now be a correct statistic to take an expectation of?2012-01-26
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    Speaking of $\bar{X}$ and $S^2$ makes it look as if you're talking about an i.i.d. sample. Is that what you have in mind? I assumed at first you were talking about a single obervation. If you've got $X_1,\ldots,X_n\sim\operatorname{i.i.d.}$ with any distribution at all, the statistic $X_1-X_2$ is already an unbiased estimator of $0$, so you've got lack of completeness.2012-01-26

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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 $0$.

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    Sorry, an iid sample is taken and theta > 0. Apologies for not stating this upfront.2012-01-26
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    Would E((X1 - X2)) for a normal (0, theta) be zero? Somehow I am skeptical that the integral of the normal pdf (0, theta) multiplied by (X1-X2) would be 0...?2012-01-26
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    ALso, shouldn't I be using X-bar and S^2 as the statistic...?2012-01-26
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    The expected value of the difference equals the difference in expected values. The variance of the difference is the sum of the variances if the things being subtracted are independent (or if they're merely uncorrelated). The density for the difference is therefore that of the $N(0,2\theta)$ distribution. And the expectation of that is indeed $0$.2012-01-26
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    If you "should" be using $\bar{X}$ and $S^2$, then your question wasn't all that clear. But in that case, you still have incompleteness, since $\bar{X}$ is an unbiased estimator of $0$.2012-01-26
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    The sufficient statistic $\sum_i X_i^2$ is, indeed, complete since it is the sufficient statistic of a one-parameter exponential family.2012-01-26
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    @cardinal : So you're agreeing with my last comment in the answer. But just what question was being asked remains unclear.2012-01-27
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    Yes, I agree on both counts. Given the lack of clarity in the question, the downvote here is mystifying. It seems it makes little sense to speak of the "completeness" of a single-parameter multivariate distribution associated with an iid sample, so it makes me wonder what the question should really be.2012-01-28
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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.