GuassMarkov Theorem and Multiple Regression
GaussMarkov Theorem
GaussMarkov
 “Least squares estimates the parameters $\beta$ have the smallest variance among all linear unbiased estimates.”
 Unbiased estimation is not always good.
 ridge regression
Proof of 1
 Model: $\theta = a^T \beta$
 Least square estimate of $\theta$: $\hat\theta = a^T \hat \beta = a^T ( \mathbf X^T \mathbf X )^{1} \mathbf X^T \mathbf y = \mathbf c_0^T \mathbf y$
 This is unbiased: $E(a^T\hat\beta) = a^T\beta$
 GaussMarkov theorem: If we have any other linear estimator $\tilde \theta = \mathbf c^T \mathbf y$ and $E(\mathbf c^T \mathbf y)=a^T \beta$, then $Var(a^T\hat \beta)\leq Var(\mathbf c^T \mathbf y)$.
 To prove it we first write down the general form of a linear estimator. Question: is the general form of a linear estimator $\alpha (X^T X)^{1} X^T + D$?
Useful functions for the proof:

Variance: $Var(X) = E[ (X  \mu)^2 ]$.

MSE: $MSE(\tilde\theta) = E( (\tilde\theta \theta)^2 ) = E( (\tilde \theta  E(\theta) + E(\theta)  \theta)^2 ) = E( (\tilde\theta  E(\theta))^2 ) + \cdots$

MSE is $MSE(\tilde \theta) = Var(\tilde theta) + (E(\theta) \theta)^2$ the second term is bias.

Least square is good but we can trade some bias to get a smaller variance sometimes.

Choices are variable subset selection, ridge regression.

Suppose new data is biased from the original data by a value $\epsilon_0$, the MSE using the original estimator is only the original MSE differed by a constant. Eq. 3.22

We always have a larger MSE????? I don’t get this.
Multiple Regression
 Model: $f(X) = \beta_0 + \sum_{j=1}^p X_j \beta$
 For multiple dimensional inputs, the estimator has no correlations for different features.
esl/gaussmarkovtheorem
Links to:
OctoMiao (2016). 'GuassMarkov Theorem and Multiple Regression', Connectome, 09 April. Available at: https://hugoconnectome.kausalflow.com/esl/gaussmarkovtheorem/.