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Normal Equation


Multivariate Linear 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 [math]\mathbf x_i \in \mathbb{R}^{n+1}[/math] 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

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
  • $J(\mathbf w) = \| \mathbf e \|^2 = \| \mathbf y - X \mathbf w \|^2$


Minimization

So our problem is

  • $\hat{\mathbf w} = \operatorname{arg \, max}\limits_{\mathbf w} J(\mathbf w) = \operatorname{arg \, max}\limits_{\mathbf w} \| \mathbf y - X \mathbf w \|^2$
  • let's expand $J(\mathbf w)$:
    • $J(\mathbf w) = \| \mathbf y - X \mathbf w \|^2 = ( \mathbf y - X \mathbf w )^T ( \mathbf y - X \mathbf w ) = \mathbf y^T \mathbf y - (X \mathbf w)^T \mathbf y - \mathbf y^T (X \mathbf w) + (X \mathbf w)^T (X \mathbf w) = \ ...$
    • $... \ = \mathbf y^T \mathbf y - 2 \mathbf w^T X^T \mathbf y + \mathbf w^T X^T X \mathbf w$
  • now minimize $J(\mathbf w)$ w.r.t. $\mathbf w$:
    • $\frac{\partial J(\mathbf w)}{\partial \mathbf w} = - 2 X^T \mathbf y + 2 X^T X \mathbf w \mathop{=}\limits^! \mathbf 0$
    • $X^T X \mathbf w = X^T \mathbf y$ or
  • 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 $X$


Linear Algebra Point of View

In Linear algebra we typically use different notation

  • Instead of $X$ we use $A$ - it's a System of Linear Equations that is very tall and thin
  • so we have an $m \times n$ matrix $A$ s.t. $m > n$ -
  • 618d1cdc2f5c4d2fb19a34eb118d5f5f.png
  • we need to solve the system $A \mathbf x = \mathbf b$
  • if $\mathbf b \not \in C(A)$ (Column Space) then there's no solution
  • how to find an approximate solution? Project onto $C(A)$!
  • it also gives the Normal Equation


Projection onto $C(A)$

Suppose we have a matrix $A$ with out observations

  • the system $A \mathbf x = \mathbf b$ has no solution
  • We project $\mathbf b$ on the Column Space $C(A)$
  • how do we do it? $C(A)$ is all the combinations of columns in $A$, so they form a hyperplane in $\mathbb R^m$
  • $\mathbf b$ is not on this hyperplane - otherwise we would not need to project on it


Normal Equation:

  • so we have $A \mathbf x = \mathbf b$
  • let's multiply both sides by $A^T$ - to find the best $\mathbf{\hat x}$ that approximates the solution $\mathbf x$ that doesn't exist
  • $A^T A \mathbf{\hat x} = A^T \mathbf b$ - this one usually has the solution, and it's called the Normal Equation
  • it projects $\mathbf b$ onto $C(A)$ and gives the solution $\mathbf{\hat x}$
  • it also happens to be the best solution in terms of Least Squares error: the projection error $\| \mathbf e \|^2 = \| \mathbf b - A \mathbf{\hat x} \|^2$ is minimal


Invertability of $A^T A$

When does $A^T A$ have no inverse?

Consider this example:

$A^T A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 1 & 3 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 9 & 27 \end{bmatrix}$

In this case $\text{rank}(A) = 1$ and $\text{rank}(A^T A) = 1$ so $A^T A$ is not invertible

  • $\text{rank}(A) = \text{rank}(A^T A)$


When it is invertible?

  • $N(A^T A) = N(A)$ (see the theorem in Projection onto Subspaces)
  • so when $N(A) = \{ \; \mathbf 0 \; \}$ then it's invertible
  • or, in other words, the columns of $A$ are linearly independent


$(A^T A)$ may be not invertible if

  • some columns are linearly dependent (i.e. we have redundant features)
    • solution: remove the linear dependency
  • too many features ($m < n$)
    • solution: delete some features, there are too many features for the amount of data we have


$\mathbb R^2$ Case

  • suppose that $\text{dim } C(A) = 2$, i.e. the basis made of columns of $A$: $\mathbf a_1$ and $\mathbf a_2$, $A = \Bigg[ \ \mathop{\mathbf a_1}\limits_|^| \ \mathop{\mathbf a_2}\limits_|^| \ \Bigg]$
  • 245834296b494b6a8f42522ff1feb119.png
  • $\mathbf b$ is not on the plane $C(A)$, but we project on it to get $\mathbf p$
  • $\mathbf e$ is our projection error



Example

$\mathbb R^2$ Case

Suppose we have the following dataset:

  • ${\cal D} = \big\{ (1,1), (2,2), (3,2) \big\}$

so we have this system:

  • [math]\left\{\begin{array}{l} x_0 + x_1 = 1\\ x_0 + 2 x_1 = 2\\ x_0 + 2 x_1 = 3\\ \end{array}\right.[/math]
  • first column is always 1 because it's our intercept term $x_0$, and $x_1$ is the slope
  • the matrix form is [math]\begin{bmatrix} 1 & 1\\ 1 & 2\\ 1 & 3\\ \end{bmatrix} \begin{bmatrix} x_0 \\ x_1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}[/math]
  • no line goes through these points at once
  • so we solve [math]A^T A \mathbf{\hat x} = A^T \mathbf b[/math]
  • [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 $\hat x_0 = 2/3, \hat x_1 = 1/2$
  • thus the best line is $y = x_0 + x_1 t = 2/3 + 1/2 t$

9af9633b58c04fc9b5ba4aa720e63a8f.png


Is this indeed the best straight line through these points?

  • we want to make the overall error as small as possible
  • recall that $\mathbf e$ is our projection error - so we want to minimize it
  • usually we minimize the square: $\min \| \mathbf e \|^2 = \min \big[ e_1^2 + e_2^2 + e_3^2 \big]$
  • so we minimize this: $\| \mathbf e \|^2 = \| A \mathbf x - \mathbf b \|^2$
  • we claim that the solution to $A^T A \mathbf{\hat x} = A^T \mathbf b$ minimizes $\| A \mathbf x - \mathbf b \|^2$


ae0b635a2e81493bb363d898b0e6369c.png


Let's check if $\mathbf p \; \bot \; \mathbf e$

  • [math]\mathbf{\hat x} = \begin{bmatrix} \hat x_0 \\ \hat x_1 \end{bmatrix} = \begin{bmatrix} 2/3 \\ 1/2 \end{bmatrix}[/math]
  • thus [math]\mathbf p = A \mathbf{ \hat x } = \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix} = \begin{bmatrix} 7/6 \\ 5/3 \\ 13/6 \end{bmatrix}[/math]
  • $\mathbf p + \mathbf e = \mathbf b$, so $\mathbf e = \mathbf b - \mathbf p = \begin{bmatrix} 1 - 7/6 \\ 2 - 5/3 \\ 2 - 13/6 \end{bmatrix} = \begin{bmatrix} - 1/6 \\ 2/3 \\ -1/6 \end{bmatrix} $
  • $\mathbf p \; \bot \; \mathbf e$ $\Rightarrow$ $\mathbf p^T \mathbf e = 0$.
  • Check: [math]\begin{bmatrix} 7/6 & 5/3 & 13/6 \end{bmatrix} \begin{bmatrix} - 1/6 \\ 2/3 \\ -1/6 \end{bmatrix} = - 7/6 \cdot 1/6 + 5/3 \cdot 2/3 - 13/6 \cdot 1/6 = 0[/math]


We can also verify that $\mathbf e \; \bot \; C(A)$

  • let's take one vector from $C(A)$, e.g. $\mathbf 1 = [1, 1, 1]^T \in C(A)$,
  • $\mathbf e^T \cdot \mathbf 1 = -1/6 + 2/6 - 1/6 = 0$


Python code  
import matplotlib.pylab as plt
import numpy as np

class Line:
    def __init__(self, slope, intercept):
        self.slope = slope
        self.intercept = intercept

    def calculate(self, x1):
        x2 = x1 * self.slope + self.intercept
        return x2

A = np.array([[1, 1], [1, 2], [1, 3]])
b = np.array([1, 2, 2])

x0, x1 = np.linalg.solve(A.T.dot(A), A.T.dot(b))
lsq = Line(x1, x0)

# figure
plt.scatter(A[:, 1], b, marker='x', color='black')

points = np.array([0.5, 3.5])
plt.plot(points, lsq.calculate(points))

plt.scatter(A[:, 1], lsq.calculate(A[:, 1]), marker='o', color='red')
plt.vlines(A[:, 1], b, lsq.calculate(A[:, 1]))

plt.show()

x = np.array([[x0], [x1]])
p = A.dot(x).reshape(-1)
e = p - b
print p.dot(e)



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


Additional

Orthogonalization

How to speed up computation of $(X^T X)^{-1}$?

  • let's make the columns of $X$ orthonormal: orthogonal to each other and of length 1
  • we can do the QR Factorization and obtain matrix $X = QR$
  • $Q^T Q = I$, and it simplifies the calculation a lot!
  • usual case: $\mathbf w = (X^T X)^{-1} X^T \mathbf y$
  • with $X = QR$: $X^T X = R^T Q^T Q R = R^T R$
  • so,
    • $X^T X \mathbf w = X^T \mathbf y$
    • $\cancel{R^T} R \mathbf w = \cancel{R^T} Q^T \mathbf y$
    • $\mathbf w = R^{-1} Q^T \mathbf y$
  • so it becomes much simpler: no need to invert $X^T X$ directly


Singular Value Decomposition

Let's apply SVD to $X$:

  • $X = U \Sigma V^T$, with $\text{dim } X = \text{dim } \Sigma$
  • [math]\begin{align} X \mathbf w - \mathbf y & = U \Sigma V^T \mathbf w - \mathbf y \\ & = U \Sigma V^T \mathbf w - U U^T \mathbf y \\ & = U (\Sigma V^T \mathbf w - U^T \mathbf y) \\ \end{align}[/math]
  • let $\mathbf v = V^T \mathbf w$ and $\mathbf z = U^T \mathbf y$
  • then we have $U (\Sigma \mathbf v - \mathbf z)$


Orthogonal Matrices preserve the $L_2$-norm

  • i.e. $\| U \mathbf x \| = \| \mathbf x \|$
  • thus, $\| \mathbf e \| = \| X \mathbf w - \mathbf y \| = \| U (\Sigma \mathbf v - \mathbf z) \| = \| \Sigma \mathbf v - \mathbf z\|$.
  • $\| X \mathbf w - \mathbf y \| = \| \Sigma \mathbf v - \mathbf z\|$
  • $\| \Sigma \mathbf v - \mathbf z\|$ is easier to minimize than $\| X \mathbf w - \mathbf y \|$

So we reduced OLS Regression problem to a diagonal form


Minimization $\| \Sigma \mathbf v - \mathbf z\|$:

  • $\text{diag}(\Sigma) = (\sigma_1, \ ... \ , \sigma_r, 0, \ ... \ , 0)$
  • $\Sigma \mathbf v = \begin{bmatrix} \sigma_1 \mathbf v_1 \\ \vdots \\ \sigma_r \mathbf v_r \\ 0 \\ \vdots \\ 0 \end{bmatrix}$ and therefore $\Sigma \mathbf v - \mathbf z = \begin{bmatrix} \sigma_1 v_1 - z_1 \\ \vdots \\ \sigma_r v_r - z_r \\ -z_{r+1} \\ \vdots \\ -z_{m} \end{bmatrix}$
  • since we minimizing it w.r.t. $\mathbf v$, only first $r$ components of $\Sigma \mathbf v - \mathbf z$ matter
    • we can make these $\sigma_i v_i - z_i$ as small as possible by using $v_i = z_i / \sigma_i$
    • so first $r$ components become 0, and the rest are $-c_i$, thus, $\| \Sigma \mathbf v - \mathbf z \|^2 = \sum\limits_{i = r+1}^m c_i^2$
    • when $r = m$, $\| \Sigma \mathbf v - \mathbf z \| = 0$, but in this case there's no need to Normal Equation


Summary:

  • calculate $X = U \Sigma V^T$ and $\mathbf z = U^T \mathbf y$
  • use $\mathbf v = \left( \cfrac{z_1}{\sigma_1}, \ ... \ , \cfrac{z_r}{\sigma_r}, 0, \ ... \ , 0 \right)$ to minimize $\| \Sigma \mathbf v - \mathbf z\|$
  • since $\mathbf v = V^T \mathbf w$, we recover $\mathbf w$ as $\mathbf w = V \mathbf v$
  • this gives solution $\mathbf w$ and residual error $\| \Sigma \mathbf v - \mathbf z\|$


Compact solution:

  • if $X = U \Sigma V^T$, then $\mathbf w = V \Sigma^+ U^T \mathbf y$
  • where $\Sigma^+ = \Big[ \text{diag}(\Sigma) \Big]^{-1}$ (invert only non-zero elements on the diagonal of $\Sigma$)


Regularization

We find $\mathbf w$ by calculating $\mathbf w = (X^T X + \lambda E^*)^{-1} \cdot X^T \cdot y$

  • where $E^* \in \mathbb{R}^{(n + 1) \times (n + 1)}$
    • and $E$ is almost identity matrix (1s on the main diagonal, the rest is 0s), except that the very first element is 0
    • i.e. for $n = 2$ : $\left[\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$
    • because we don't regularize for the bias input $x_{i0} = 1$
  • $(X^T X + \lambda E^*)$ is always invertible


This is called Ridge Regression


Implementation

Implementation in Octave

pinv(X' * X) * X' * y


See Also

Sources

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