Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Basically, I have that $\ Y_i = \alpha +\beta(x_i-x_{bar}) + \epsilon_i $ where $\epsilon_i$ are i.i.d normally distributed with mean 0 variance $\sigma^2$ $\ Y_i ~~has ~a~normal~distribution~as ~follows~; N(\alpha + \beta(x_i-x_{bar}),\sigma^2 )$ Where the $\ x_i $ are just given numbers $\ \alpha^{hat}$ has distribution $\ N(\alpha,\sigma^2/n)$ AND $\beta^{hat}$ has a distribution $\ N(\beta,\sigma^2/\Sigma x_i^2)$

Let Z=AY where A an orthogonal matrix.

Show what $\Sigma (Y_i -\alpha_{hat} -\beta_{hat}(x_i))^2 ~~~{i=1...n} ~~~~= ~~\Sigma Z_i^2 ~~~~{i=3...n} $ and show then that $\Sigma Z_i^2 ~~~~{i=3...n} $ has a chi sqared distribution with n-2 degrees of freedom.

The thing is, I can do this up until the last part, what I dont understand is the degrees of freedom, I can see by showing this equation that $\Sigma Z_i^2 ~~~~{i=3...n} $ is a sum of squared standard normals, namely each of the terms $\ (Y_i -\alpha_{hat} -\beta_{hat}(x_i))^2$ are standard normals, but if the left = the right,. then isnt it the sum of n squared standard normals not n-2?!?!

Really confused by this.

Thanks

share|improve this question
add comment

1 Answer

That would be the case if you had n independent normals with mean 0 and variance 1 squared. But these normals are not independent because of the common αhat and βhat being in the part subtracted for each Yi. The general rule is that the degrees of freedom - number of observations-number of parameters estimated in your case n observations minus 2 parameters estimated. That is where the n-2 comes from. In the case of a sample mean from n independent normals it is (Xi-Xbar)$^2$ that are summed S$^2$ = ∑(Xi-Xbar)$^2$/(n-1) satisfies (n-1)S$^2$/σ$^2$ is chi-square with n-1 degrees of freedom. Here there is just one parameter estimated, the mean.

share|improve this answer
    
thanks for your reply, the thing Im not sure of though is we were told that it has chi squared distributionwith n-2 degrees of freedom because it was the sum of n-2 squares of standard normals, but where are these sums of squares of random normals?! –  Rosie Jun 6 '12 at 20:14
    
When you take the sum of squares of n independent standard normals you get chi sqaure n degrees of freedom. But in this case you are taking sum of squares for n DEPENDENT standard normals with the dependence coming from the two estimated parameters that appear in the formula for each standard normal. –  Michael Chernick Jun 6 '12 at 20:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.