The Gauss-Markov Theorem proof - matrix form - part 2

[@zzack4012]

Man, you are doing a life saving job here.

[@gordan99]

Thank you, you have saved my tenure!

[@ryancausey3804]

The condition that DXbeta= 0 does not imply that DX=0, even if beta is non-zero. Matrices can have divisors of zero.

[@kottelkannim4919]

Of course you are right and "beta" should merely belong to the null-space of DX in order for the equality DXbeta=0 to hold, but let me suggest an explanation for the validity of DX=0. I hope I am not misleading you. Suppose "beta" is a vector of "p" parameters. Key point : Since DXbeta=0 should hold for an ANY ARBITRARY "beta", the dimension of the space of the arbitrary "beta-s" is "p". DX is a p x p matrix. By the Rank-Nullity theorem: Rank(DX)+Nullity(DX)=p. but since Nullity(DX)=p we get Rank(DX)=0 meaning the column space of DX comprises the zero vector, and DX is the zero matrix of the linear space of p x p matrices.

[@WalterWhite-ov3sw]

Hi, thanks for the videos, The graduate econometrics course series has been an absolute lifesaver. One question I had on this video is that once we showed that Var(Btild) = sigma^2 CC', isn't that a sufficient condition to show that Var(Btild) > Var(Bhat) since by definition CC' is positive semidefinite?

[@KrishnaHariBaral3]

y = Xb + e y = y^ + e As X'e =0 Var(y) = var(y^) + var(e) Var(y) = var(Xb) + var(e) Var(y) = X vay(b) X' + var(e) Var(y) = X . d^2 . Inv(X'X) . X' + d^2 Variance of y should be the sum of variance of deterministic and residual parts. Please correct me if I am wrong.

[@Mc44499]

out here doing God's work mate. please keep it up (although this is 11 years ago haha)

[@poppyblop484]

why is var(y) = sigma^2 Identity matrix? Isnt it E(u|x) = sigma2ID matrix?