# Risk/Loss functions

• January 5th 2010, 06:33 PM
statmajor
Risk/Loss functions
Let X1,...,Xn denote a random sample from a $N(\mu , \sigma)$ distribution. Let $Y = \Sigma \frac{(X_i - \overline{X})^2}{n}$and let $L(\theta , \delta (y)) = [\theta - \delta(y)]^2$ and $\delta (y) = by$ where b does not depend on y.

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

So:

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

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

Any help would be appreciated.
• January 10th 2010, 08:52 PM
matheagle
first of all ....

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

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

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

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

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

So, if you want $E(Y)={\sigma^2\over n} (n-1)$
• January 11th 2010, 06:41 AM
statmajor
Thanks for the help, but I already solved the problem by doing the following:

$\frac{n-1}{n} \frac{1}{n-1} \Sigma (X_i - \overline{X})^2 = \frac{n-1}{n}s^2$ 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
• January 11th 2010, 03:00 PM
matheagle
Quote:

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

$\frac{n-1}{n} \frac{1}{n-1} \Sigma (X_i - \overline{X})^2 = \frac{n-1}{n}s^2$ 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