# Unbiased estimator involving chi square distribution

• Jan 6th 2010, 09:50 AM
statmajor
Unbiased estimator involving chi square distribution
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.
• Jan 6th 2010, 04:07 PM
matheagle
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....

$E\left[\frac{\sum_{i=1}^nX^2_i}{n}\right]=\theta$
• Jan 6th 2010, 04:26 PM
matheagle
Likewise $V\left(\frac{\sum_{i=1}^nX^2_i}{\theta}\right)=2n$

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

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

$={\theta^2\over n^2}\left(2n\right)={2\theta^2\over n}$
• Jan 6th 2010, 04:41 PM
statmajor
and this is because the variance is theta and not one?
• Jan 6th 2010, 04:42 PM
matheagle
Quote:

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
• Jan 6th 2010, 04:45 PM
statmajor
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.