General Gauss-Markov theorem [closed]

$$Y=XB+u$$ where $$X$$ is a non random $$n\times k$$ Matrix, $$\textrm{rank}(X)=k, E(u)=0, E(uu')=\sigma^2\Omega$$, How to form $$(1)$$ How to proof $$(2)$$ the general Gauss-Markov theorem?

closed as off-topic by StubbornAtom, Xander Henderson, Adrian Keister, dantopa, Lee David Chung LinMay 16 at 5:02

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The Gauss-Markov Theorem states that the OLS estimator:

$$\hat{\boldsymbol{\beta}}_{OLS} = (X'X)^{-1}X'Y$$

is Best Linear Unbiased. For the proof, I will focus on conditional expectations and variance: the results extend easily to non conditional. Also, for the proof, I consider $$I_{n}$$ $$=$$ $$\Omega$$, but the result extends easily to the non equal case as well.

Proof:

Is it unbiased?

$$E(\hat{\boldsymbol{\beta}}_{OLS} \mid X) = E[(X'X)^{-1}X'Y \mid X] = E[(X'X)^{-1}X'(X\boldsymbol{\beta} + u) \mid X] = \\ \boldsymbol{\beta} + (X'X)^{-1}X'E(u \mid X) = \boldsymbol{\beta}$$

Yes! Is it, then, among the unbiased, that with the smallest variance? Consider, for this purpose, a general linear unbiased estimator $$\boldsymbol{b}$$:

$$\boldsymbol{b} = C\boldsymbol{y}$$

where $$C$$ is a generic $$k$$ $$\times$$ $$n$$ matrix that depends only on the sample information in $$X$$ and, given unbiasedness, such that $$CX$$ $$=$$ $$I_{k}$$ to guarantee unbiasedness. Note that for $$\hat{\boldsymbol{\beta}}_{OLS}$$, $$C_{Ols}$$ $$=$$ $$(X'X)^{-1}X'$$. It can be proved that:

$$Var(\hat{\boldsymbol{\beta}}_{OLS} \mid X) = \sigma^{2}(X'X)^{-1}$$

and, for the generic linear estimator:

$$Var(\boldsymbol{b} \mid X) = \sigma^{2}(C'C)^{-1}$$.

We can additionally define $$D$$ $$=$$ $$C$$ $$-$$ $$C_{ols}$$. It is immediate that $$DX$$ $$=$$ $$0$$. From it, we can finally conclude that:

\begin{align} Var(\boldsymbol{b} \mid X) &= \sigma^{2}[D - (X'X)^{-1}X'][D - X(X'X)^{-1}] \\ &= \sigma^{2}(X'X)^{-1} + \sigma^{2}DD' \\ &= Var(\hat{\boldsymbol{\beta}}_{OLS} \mid X) + \sigma^{2}DD' \\ &> Var(\hat{\boldsymbol{\beta}}_{OLS} \mid X) \end{align}

Since DD' is non negative defined.

• Hello. We discourage giving full answers to questions with zero input from the asker. – StubbornAtom May 15 at 13:24
• Thanks for the note, StubbornAtom: a form of netiquette I wasn't aware of. – Nicg May 15 at 13:27