Ordinary Least Squares Regression


Regression Problem

Suppose we have

  • $m$ training examples $(\mathbf x_i, y_i)$
  • $n$ features, $\mathbf x_i = \big[x_{i1}, \ ... \ , x_{in} \big]^T \in \mathbb{R}^n$
  • We can put all such $\mathbf x_i$ as rows of a matrix $X$ (sometimes called a design matrix)
  • [math]X = \begin{bmatrix} - \ \mathbf x_1^T - \\ \vdots \\ - \ \mathbf x_m^T - \\ \end{bmatrix} = \begin{bmatrix} x_{11} & \cdots & x_{1n} \\ & \ddots & \\ x_{m1} & \cdots & x_{mn} \\ \end{bmatrix}[/math]
  • the observed values: [math]\mathbf y = \begin{bmatrix} y_1 \\ \vdots \\ y_m \end{bmatrix} \in \mathbb{R}^{m}[/math]
  • Thus, we expressed our problem in the matrix form: $X \mathbf w = \mathbf y$
  • Note that there's usually additional feature $x_{i0} = 1$ - the slope,
    • so $\mathbf x_i \in \mathbb{R}^{n+1}$ and [math]X = \begin{bmatrix} - \ \mathbf x_1^T - \\ - \ \mathbf x_2^T - \\ \vdots \\ - \ \mathbf x_m^T - \\ \end{bmatrix} = \begin{bmatrix} x_{10} & x_{11} & \cdots & x_{1n} \\ x_{20} & x_{21} & \cdots & x_{2n} \\ & & \ddots & \\ x_{m0} & x_{m1} & \cdots & x_{mn} \\ \end{bmatrix} \in \mathbb R^{m \times n + 1}[/math]


Thus we have a system

  • $X \mathbf w = \mathbf y$
  • how do we solve it, and if there's no solution, how do we find the best possible $\mathbf w$?


Least Squares

Normal Equation

There's no solution to the system, so we try to fit the data as good as possible

  • Let $\mathbf w$ be the best fit solution to $X \mathbf w \approx \mathbf y$
  • we'll try to minimize the error $\mathbf e = \mathbf y - X \mathbf w$ (also called residuals)
  • we take the square of this error, so the objective is
  • $J(\mathbf w) = \| \mathbf e \|^2 = \| \mathbf y - X \mathbf w \|^2$


The solution:

  • $\mathbf w = (X^T X)^{-1} X^T \mathbf y = X^+ \mathbf y$
  • where $X^+ = (X^T X)^{-1} X^T$ is the Pseudoinverse of $X$


From the Linear Algebra point of view:

  • we need to solve $X \mathbf w = \mathbf y$
  • if $\mathbf y \not \in C(X)$ (Column Space) then there's no solution
  • How to solve it approximately? Project on $C(A)$!
  • again, it gives us the Normal Equation: $X^T X \mathbf w = X^T \mathbf y$


Gradient Descent

Alternatively, we can use Gradient Descent:

  • objective is $J(\mathbf w) = \| \mathbf y - X \mathbf w \|^2$
  • the derivative w.r.t. $\mathbf w$ is $\cfrac{\partial J(\mathbf w)}{\partial \mathbf w} = 2 X^T X \mathbf w - 2 X^T \mathbf y$
  • so the update rule is $\mathbf w \leftarrow \mathbf w - \alpha 2 (X^T X \mathbf w - X^T \mathbf y)$
  • where $\alpha$ is the learning rate


Example

Suppose we have the following dataset:

  • ${\cal D} = \{ (1,1), (2,2), (3,2) \}$
  • the matrix form is [math]\begin{bmatrix} 1 & 1\\ 1 & 2\\ 1 & 3\\ \end{bmatrix} \begin{bmatrix} w_0 \\ w_1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}[/math]
  • no line goes through these points at once
  • so we solve $X^T X \mathbf{\hat w} = X^T \mathbf y$
  • [math]\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \end{bmatrix} \begin{bmatrix} 1 & 1\\ 1 & 2\\ 1 & 3\\ \end{bmatrix} = \begin{bmatrix} 3 & 6\\ 6 & 14\\ \end{bmatrix}[/math]
  • this system is invertible, so we solve it and get $\hat w_0 = 2/3, \hat w_1 = 1/2$
  • thus the best line is $h(t) = w_0 + w_1 t = 2/3 + 1/2 t$


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Normal Equation vs Gradient Descent

Gradient Descent:

  • need to choose learning rate $\alpha$
  • need to do many iterations
  • works well with large $n$


Normal Equation:

  • don't need to choose $\alpha$
  • don't need to iterate - computed in one step
  • slow if $n$ is large $(n \geqslant 10^4)$
  • need to compute $(X^T X)^{-1}$ - very slow
  • if $(X^T X)$ is not-invertible - we have problems


See Also


Sources

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