# efficiency help

• Mar 1st 2010, 07:52 PM
Danneedshelp
efficiency help
Q) $Y_{1},Y_{2},...,Y_{n}$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^{2}$. Two unbiased estimators of $\sigma^{2}$ are

$\hat{\sigma_{1}}^{2}=S^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(Y_{i}-\bar{Y})^{2}$ and $\hat{\sigma_{2}}^{2}=\frac{1}{2}(Y_{1}-Y_{2})^{2}$.

Find the efficiency of $\hat{\sigma_{1}}^{2}$ relative to $\hat{\sigma_{2}}^{2}$.

I am stuck on finding the proper variance for $\hat{\sigma_{2}}^{2}$, $V(\hat{\sigma_{2}}^{2})$. My work indicates $V(\hat{\sigma_{2}}^{2})=\frac{\sigma^{4}}{2}$, because $\hat{\sigma_{2}}^{2}$ is simply $S^{2}$
with $n=2$. Moreover, $V(\hat{\sigma_{1}}^{2})=\frac{2\sigma^{4}(n-1)}{n^{2}}$.

So, the efficiency of $\hat{\sigma_{1}}^{2}$ relative to $\hat{\sigma_{2}}^{2}$ is $eff(\hat{\sigma_{1}}^{2},\hat{\sigma_{2}}^{2})=\fr ac{V(\hat{\sigma_{2}}^{2})}{V(\hat{\sigma_{1}}^{2} )}=\frac{n^{2}}{4(n-1)}$.

The book's answer is $n-1$.

I am stuck, so any help would be great.

Thanks
• Mar 1st 2010, 10:39 PM
matheagle
nope, use...

${(n-1)S^2\over \sigma^2}={\sum_{i=1}^n(X_i-\bar X)^2\over n-1}\sim \chi^2_{n-1}$

Thus

$V(S^2)=V\left[\left({(n-1)S^2\over \sigma^2}\right)\left({\sigma^2\over n-1}\right)\right]$

$=\left({\sigma^4\over (n-1)^2}\right)V\left(\chi^2_{n-1}\right)$

$=\left({\sigma^4\over (n-1)^2}\right)2(n-1)$

$= {2\sigma^4\over n-1}$

And, yes the other point estimator is a sample variance with n=2.

So $V(\hat\sigma_2^2)= {2\sigma^4\over 2-1} =2\sigma^4$

So, the ratio is n-1 or 1/(n-1) depending on which way you divide.