Yi=Xi*beta1 + beta0 + error
Prove:
COV(average of Yi, estimated beta1) = 0
Thanks!
Last edited by hehe1223; 06-28-2010 at 08:03 AM.
Really? Cannot agree.
I actually figured out a troublesome way (don't think it is the expected way, thus asking here) of doing it and indeed get 0.
Some known and provable facts:
Var(estimated beta1) = sigma^2/Sxx
Cov(estimated beta1, estimated beta0) = -sigma^2*xbar/Sxx (something like what you claimed for Cov(Ybar, b1))
Cov(Ybar, b1') = E(Ybar*b1') - E(Ybar)*E(b1')
=E(xbar*b1' + b0')*b1' -(xbar*b1+b0)*b1
=xbar*E(b1'^2) +E(b1'*b0') -xbar*b1^2-b0*b1
=xbar(Var(b1')+b1^2) + b1*b0 + Cov(b1', b0') - xbar*b1^2 - b0*b1
= ...
=0
What do you think?
Have a look at this link. I sketched the proof in my last post.
http://www.talkstats.com/showthread.php?t=8666
Okay, I think I see what is going on now.
My original assertion (which is true) is for the special case of when the intercept term is zero. I missed that subtle point when I went back and looked at my previous post i.e. the model was specified without an intercept term.
So, yes, in general, I believe it is correct that the covariance should be zero.
I think a quick way to write this would be:
where the first term in the last part would not appear if the intercept term is zero.
Yi=Xi*beta1 + beta0 + error
Prove:
COV(estimated beta0, estimated beta1) = 0
Thanks!
Thanks by the answer. You are brilliant!
Last edited by natswim; 10-09-2010 at 03:48 PM.
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