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So basically the title is my question. lin-reg model: $$y_i = x^{T}_i\beta + \epsilon_i, i = 1,...,n$$

  1. Initalize $\hat{\beta^{[0]}}$ and the number of iterations $m_{stop}$.
  2. Compute: $$u = y - X\hat{\beta}^{[m-1]}$$ $$ \hat{b} = (X^TX)^{-1}X^Tu$$ Update $$\hat{\beta}^{[m]} = \hat{\beta}^{[m-1]} + v\hat{b}$$
  3. Repeat 2. for $m_{stop}$

Why are we estimating the beta cofficients based on the residuals and then using the estimatet $\hat{\beta}^{[m_{stop}]}$ for the normal linear regression model - where we originally try to estimate $y$?

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The core idea behind using residuals to update the $\beta$ coefficients is to move the coefficient estimates in a direction that reduces the difference between the actual and predicted values of $y$. This is based on the principle that the best-fitting model is the one that minimizes the sum of squared residuals (RSS).

By iteratively adjusting $\beta$ based on the residuals, the algorithm seeks to find the minimum RSS in a step-wise fashion. The adjustment vector $\hat{b}$ points in the direction that most rapidly decreases the RSS, given the current estimate $\hat{\beta}^{[m-1]}$. By moving in this direction, the algorithm iteratively reduces the error between the observed and predicted values.

Under suitable conditions (e.g., $X^TX$ is non-singular), this iterative process converges to the $\hat{\beta}$ that minimizes the RSS, which is the solution to the normal equations of linear regression: $$X^TX\hat{\beta} = X^Ty$$

This iterative method can be more numerically stable and efficient than directly solving the normal equations, especially when $X^TX$ is close to singular or when $n$ is very large. By adjusting $\hat{\beta}$ iteratively, the method can avoid the numerical difficulties associated with directly inverting $X^TX$.

In the context of linear regression, the gradient of the RSS with respect to $\beta$ is given by $-2X^T(y-X\beta)$, and the Hessian matrix (the second derivative of RSS with respect to $\beta$) is $2X^TX$. The adjustment $\hat{b}$ is derived by applying a Newton-like step, assuming the Hessian is constant (which is true for linear regression), hence moving $\beta$ towards the minimum of RSS.

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