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Pawel Sarkowicz
2026-02-19 01:38:25 +00:00
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@@ -140,21 +140,13 @@ Let us first see why we get a line of best fit.
> **Example**. Let us show why this describes a line of best fit when we are working with one feature and one target. Suppose that we observe four data points
> $$ X = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } y = \begin{bmatrix} 1 \\ 2\\ 2 \\ 4 \end{bmatrix}. $$
> We want to fit a line $y = \beta_0 + \beta_1x$ to these data points. We will have our augmented matrix be
> $$
> \tilde{X} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4 \end{bmatrix},
> $$
> $$ \tilde{X} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4 \end{bmatrix}, $$
> and our parameter be
> $$
> \tilde{\beta} = \begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix}.
> $$
> $$ \tilde{\beta} = \begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix}. $$
> We have that
> $$
> \tilde{X}^T\tilde{X} = \begin{bmatrix} 4 & 10 \\ 10 & 30 \end{bmatrix} \text{ and } \tilde{X}^Ty = \begin{bmatrix} 9 \\ 27 \end{bmatrix}.
> $$
> $$ \tilde{X}^T\tilde{X} = \begin{bmatrix} 4 & 10 \\ 10 & 30 \end{bmatrix} \text{ and } \tilde{X}^Ty = \begin{bmatrix} 9 \\ 27 \end{bmatrix}. $$
> The 2x2 matrix $\tilde{X}^T\tilde{X}$ is easy to invert, and so we get that
> $$
> \tilde{\beta} = (\tilde{X}^T\tilde{X})^{-1}\tilde{X}^Ty = \frac{1}{10}\begin{bmatrix} 15 & -5 \\ -5 & 2 \end{bmatrix}\begin{bmatrix} 9 \\ 27 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{9}{10} \end{bmatrix}.
> $$
> $$ \tilde{\beta} = (\tilde{X}^T\tilde{X})^{-1}\tilde{X}^Ty = \frac{1}{10}\begin{bmatrix} 15 & -5 \\ -5 & 2 \end{bmatrix}\begin{bmatrix} 9 \\ 27 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{9}{10} \end{bmatrix}. $$
> So our line of best fit is of them form $y = \frac{9}{10}x$.
Although the above system was small and we could solve the system of equations explicitly, this isn't always feasible. We will generally use python in order to solve large systems.