# Guass-Markov Theorem and Multiple Regression

## #Statistics #Gauss-Markov Theorem

Gauss-Markov Theorem

## Gauss-Markov

- “
**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: $E(a^T\hat\beta) = a^T\beta$
- Gauss-Markov 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.

Published:
by OctoMiao;

## Table of Contents

**Current Ref:**

- esl/gauss-markov-theorem.md