# Thread: Unbiased estimator involving chi square distribution

1. ## 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.

2. 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....

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

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

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

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

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

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

5. 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

6. 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.