Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – cardinal Jan 26 '12 at 0:57
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. :) – cardinal Jan 26 '12 at 0:59
@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. :) – cardinal Jan 26 '12 at 1:10
My apologies! Will refrain from doing it in the future :) – icobes Jan 26 '12 at 1:20
@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. – cardinal Jan 26 '12 at 1:25

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 $0$.

share|cite|improve this answer
Sorry, an iid sample is taken and theta > 0. Apologies for not stating this upfront. – helpme123321 Jan 26 '12 at 1:45
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...? – helpme123321 Jan 26 '12 at 1:50
ALso, shouldn't I be using X-bar and S^2 as the statistic...? – helpme123321 Jan 26 '12 at 1:51
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 $0$. – Michael Hardy Jan 26 '12 at 2:34
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 $0$. – Michael Hardy Jan 26 '12 at 2:36

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.