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Pawel
2026-02-18 20:36:45 -05:00
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@@ -140,13 +140,21 @@ 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 > **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
> $$ X = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } y = \begin{bmatrix} 1 \\ 2\\ 2 \\ 4 \end{bmatrix}. $$ > $$ X = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \text{ and } y = \begin{bmatrix} 1 \\ 2\\ 2 \\ 4 \end{bmatrix}. $$
> We want to fit a line $y = \beta_0 + \beta_1x$ to these data points. We will have our augmented matrix be > 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 > 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 > 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 > 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 $y = \frac{9}{10}x$. > 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. 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.