1. ## Risk/Loss functions

Let X1,...,Xn denote a random sample from a $\displaystyle N(\mu , \sigma)$ distribution. Let $\displaystyle Y = \Sigma \frac{(X_i - \overline{X})^2}{n}$and let $\displaystyle L(\theta , \delta (y)) = [\theta - \delta(y)]^2$ and $\displaystyle \delta (y) = by$ where b does not depend on y.

Show that $\displaystyle R(\theta^2, \delta) = \frac{\theta^2}{n^2}[(n^2 - 1)b^2 -2n(n-1)b + n^2]$

So:

$\displaystyle R(\theta^2, \delta) = E(L(\theta , \delta (y))) = E([\theta - \delta(y)]^2) = E((\theta -by)^2) = E(\theta^2 -2 \theta by +b^2y^2)$

Kinda stuck here. Not sure what the expectation of $\displaystyle Y = \Sigma \frac{(X_i - \overline{X})^2}{n}$ is.

Any help would be appreciated.

2. first of all ....

$\displaystyle R(\theta^2, \delta) = E(L(\theta^2 , \delta (y))) = E([\theta^2 - \delta(y)]^2) = E((\theta^2 -by)^2)$

As for $\displaystyle Y = \frac{\Sigma (X_i - \overline{X})^2}{n}$ we have....

$\displaystyle (n-1)S^2\sim\sigma^2 \chi^2_{n-1}$ and $\displaystyle (n-1)S^2=\Sigma (X_i - \overline{X})^2$

So $\displaystyle Y = \frac{\Sigma(X_i - \overline{X})^2}{n}$

$\displaystyle = \left({\sigma^2\over n}\right) \frac{\Sigma(X_i - \overline{X})^2}{\sigma^2} \sim {\sigma^2\over n} \chi^2_{n-1}$

So, if you want $\displaystyle E(Y)={\sigma^2\over n} (n-1)$

3. Thanks for the help, but I already solved the problem by doing the following:

$\displaystyle \frac{n-1}{n} \frac{1}{n-1} \Sigma (X_i - \overline{X})^2 = \frac{n-1}{n}s^2$ and I know what the expecation and variance of the sample variance (sample from normal distribution) is.

But thanks for your help none the less

4. Originally Posted by statmajor
Thanks for the help, but I already solved the problem by doing the following:

$\displaystyle \frac{n-1}{n} \frac{1}{n-1} \Sigma (X_i - \overline{X})^2 = \frac{n-1}{n}s^2$ and I know what the expecation and variance of the sample variance (sample from normal distribution) is.

But thanks for your help none the less

that's the same thing