Degrees of Freedom in t-Distribution for Simple Regression without a Constant

  • Thread starter Thread starter MaxManus
  • Start date Start date
  • Tags Tags
    Square Statistics
MaxManus
Messages
268
Reaction score
1

Homework Statement



Simple regression without a constant
Yi = Bxi + epsi for i = 1,2,...n
epsi are independent and N(0, sigma^2) distributed, B and sigma^2 are unknown.

All my sums are from i = 1 to n
The question is: Explain why:
\frac{\hat{B} - B}{S} \sqrt{\sum{x_i^2}}
is t-distirbuted with n-1 degrees of freedom.

\hat{B} is the least square estimator for B, and S^2 is the least square estiamtor for sigma^2


I'm not sure how to start solving the problem. My first idea was that this looket like a standard t-distribution for \hat{B}, but \sqrt{n} \neq \sqrt{\sum{x_i^2}}

Homework Statement





Homework Equations




The Attempt at a Solution

 
Physics news on Phys.org
Can you say;
If
\frac{\hat{B} - B}{S}\sqrt{n} is t-distributed then:
\frac{\hat{B} - B}{S} \sqrt{\sum{x_i^2}}
is t-distributed since n and the x are just numbers?
And can you go further and say that if the first have (n-1) degrees of freedom then the second equation also has to?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top