# Unbiased estimator involving chi square distribution

• Jan 6th 2010, 08:50 AM
statmajor
Unbiased estimator involving chi square distribution
Let X1,...,Xn denote a random sample from a normal distribution with mean zero and variance $\displaystyle \theta$. Show that $\displaystyle \Sigma \frac{X^2_i}{n}$ is an unbaised estimator of $\displaystyle \theta$ and it's variance is $\displaystyle \frac{2\theta^2}{n}$

I know that $\displaystyle \frac{X^2_i}{\theta}$ 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, 03:07 PM
matheagle
correction.....

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

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

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

now switch the constants....

$\displaystyle E\left[\frac{\sum_{i=1}^nX^2_i}{n}\right]=\theta$
• Jan 6th 2010, 03:26 PM
matheagle
Likewise $\displaystyle V\left(\frac{\sum_{i=1}^nX^2_i}{\theta}\right)=2n$

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

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

$\displaystyle ={\theta^2\over n^2}\left(2n\right)={2\theta^2\over n}$
• Jan 6th 2010, 03:41 PM
statmajor
and this is because the variance is theta and not one?
• Jan 6th 2010, 03:42 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.

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

It looks like it's the UMVUE also
• Jan 6th 2010, 03:45 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.