Unbiased estimator involving chi square distribution

**statmajor** · January 6th 2010, 08:50 AM

Let X1,...,Xn denote a random sample from a normal distribution with mean zero and variance $\theta$ . Show that $\Sigma \frac{X^2_i}{n}$ is an unbaised estimator of $\theta$ and it's variance is $\frac{2\theta^2}{n}$

I know that $\frac{X^2_i}{\theta}$ is a chi square distribution with n degrees of freedom, and hence, it's expectation is just n.

I'm just not sure how to use this fact.

Any help would be appreciated.

**matheagle** · January 6th 2010, 03:07 PM

correction.....

$\frac{X^2_i}{\theta}$ is a chi square distribution with 1 degrees of freedom

$\frac{\sum_{i=1}^nX^2_i}{\theta}$ is a chi square distribution with n degrees of freedom

so $E\left[\frac{\sum_{i=1}^nX^2_i}{\theta}\right]=n$

now switch the constants....

$E\left[\frac{\sum_{i=1}^nX^2_i}{n}\right]=\theta$

**matheagle** · January 6th 2010, 03:26 PM

Likewise $V\left(\frac{\sum_{i=1}^nX^2_i}{\theta}\right)=2n$

So $V\left(\frac{\sum_{i=1}^nX^2_i}{n}\right)=V\left(\ frac{\theta\sum_{i=1}^nX^2_i}{n\theta}\right)$

$={\theta^2\over n^2}V\left(\frac{\sum_{i=1}^nX^2_i}{\theta}\right)$

$={\theta^2\over n^2}\left(2n\right)={2\theta^2\over n}$

**statmajor** · January 6th 2010, 03:41 PM

and this is because the variance is theta and not one?

**matheagle** · January 6th 2010, 03:42 PM

Originally Posted by statmajor

and this is because the variance is theta and not one?

I'm not sure what you're asking.
Theta is our unknown variance that we are estimating.

$\frac{\sum_{i=1}^nX^2_i}{n}$ is the MLE of theta

It looks like it's the UMVUE also

**statmajor** · January 6th 2010, 03:45 PM

Sorry about that. I'm working on a different question on an assigment and mixed the two of them up. You can just safely ignore my last post.

Thanks for your help once more.

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