Gram-Schmidt Procedure

Solving Regression using Gram-Schmidt Procedure | January 07, 2018


  • python
  • machine-learning
  • statistics

An interesting way to understand Linear Regression is Gram-Schmidt Method of successive projections to calculate the coefficients of regression. Gram-Schmidt procedure transforms the variables into a new set of orthogonal or uncorrelated variables. On applying the procedure, we should get exactly the same regression coefficients as with projection of predicted variable on the feature space.

Linear Model with number of inputs (p) such $p > 1$ is called multiple regression. We can represent least squares estimates of multiple regression in terms of estimates of univariate linear model. To understand this, let us assume a multivariate $(p > 1)$ linear model - $\mathbf{Y} = \mathbf{X}\beta + \epsilon$.

The least square estimate and residuals are:

\begin{equation} \hat \beta = \dfrac{\sum_{n=1}^{N} x_i y_i}{\sum_{n=1}^{N} x_i^2 } \end{equation}

and

\begin{equation} r_i = y_i - x_i \hat \beta \end{equation}

In convenient vector notation, we let $\mathbf{y} = (y_1, …, y_N)^\intercal$, $\mathbf{x} = (x_1, …, x_N)^\intercal$ and define:

\begin{equation} \langle \mathbf{x}, \mathbf{y} \rangle = \sum_{n=1}^{N} x_i y_i = \mathbf{x}^\intercal \mathbf{y} \end{equation}

Hence, we can write the parameters in terms of inner product of x and y.

\begin{equation} \hat \beta = \dfrac{\langle \mathbf{x}, \mathbf{y} \rangle}{\langle \mathbf{x}, \mathbf{x} \rangle}; \mathbf{r} = \mathbf{y} - \mathbf{x} \hat \beta \end{equation}

The inner product notation generalizes the linear regression to different metric spaces, as well as to probability spaces.

If the inputs $\mathbf{x_1}, \mathbf{x_2}, …, \mathbf{x_p}$ are orthogonal, i.e. $\langle \mathbf{x_j}, \mathbf{x_k} \rangle = 0$ for all $j \neq k$, then it is easy to check that the multiple least squares estimates $\beta_j$ are equal to $\langle \mathbf{x_j}, \mathbf{y} \rangle / \langle \mathbf{x_j}, \mathbf{x_j} \rangle$.

Now if we have an intercept and a single input $\mathbf{x}$, we can find that

\begin{equation} \hat \beta_1 = \dfrac{\langle \mathbf{x} - \bar x \mathbb{1}, \mathbf{y} \rangle}{\langle \mathbf{x} - \bar x \mathbb{1}, \mathbf{x} - \bar x \mathbb{1} \rangle} \end{equation}

where $\bar x = \sum_{n=1}^{N} x_i / N$ and $\mathbb{1} = \mathbf{x}_0$, the vector of N ones.

The steps to generate the regression using this procedure -

  1. Regress $\mathbf{x}$ on $\mathbb{1}$ to produce the residual $\mathbf{z} = \mathbf{x} - \bar x \mathbb{1}$
  2. Regress $\mathbf{y}$ on the residual $\mathbf{z}$ to give the coefficient $\hat \beta_1$

where, “regress $\mathbf{b}$ on $\mathbf{a}$” means a single univariate regression of $\mathbf{b}$ on $\mathbf{a}$ with no intercept.

This process also generalizes to $p$ points and is called Gram - Schmidt Process. It can be understood as a process of Successive orthogonalization of the inputs, starting from $\mathbb{1}$.

Gram-Schmidt Process

ALGORITHM

  • Initialize $\mathbf{z_0} = \mathbf{x_0} = \mathbb{1}$.
  • For all $\mathbf{x_j}$ s.t. $j$ in ${1, 2, 3, …, p}$ for $p$ inputs, regress $\mathbf{x_j}$ on the residuals after $j_th$ step, where the coefficients $\hat \gamma_{lj}$ are:

\begin{equation} \hat \gamma_{lj} = \dfrac{\langle \mathbf{z_l}, \mathbf{x_j} \rangle}{\langle \mathbf{z_l}, \mathbf{z_l} \rangle} \end{equation}

and residuals at each step are:

\begin{equation} \mathbf{z_j} = \mathbf{x_j} - \sum_{k=0}^{j-1} \hat \gamma_{kj}\mathbf{z_k} \end{equation}

  • Finally, we can calculate $\hat \beta_p$ as:

\begin{equation} \hat \beta_p = \dfrac{\langle \mathbf{z_p}, \mathbf{y} \rangle}{\langle \mathbf{z_p}, \mathbf{z_p} \rangle} \end{equation}

Let us test this procedure with $p = 2$.

As a first step, we will run a Multiple Regression over a set of inputs and get the regression coefficients.

import numpy as np
from sklearn.linear_model import LinearRegression
x1 = np.array([2, 2.2, 3.2, 4.5, 5.0])
x2 = np.array([45.0, 20.0, 30.0, 10.0, 6.5])
x0 = np.ones(len(x1))
z0 = x0
y = np.array([2.3, 4.5, 6.7, 8.9, 10.11])

X = np.matrix([x0, x1, x2]).T
Y = np.matrix(y).T

lin_reg = LinearRegression(copy_X=True, fit_intercept=False, n_jobs=1, normalize=False)
lin_reg.fit(X, Y)
coeffs = lin_reg.coef_
print("Coefficients: {}".format(coeffs))
Coefficients: [[ 1.39782326  1.83576285 -0.04935882]]

Now, let us calculate the $\beta_2$ using the iterative procedure.

Step 1

gamma_01 = z0.dot(x1) / (z0.dot(z0))
z1 = x1 - gamma_01 * z0

Step 2:

gamma_02 = z0.dot(x2) / (z0.dot(z0))
gamma_12 = z1.dot(x2) / (z1.dot(z1))
z2 = x2 - gamma_02 * z0 - gamma_12 * z1

Step 3:

beta_p = z2.dot(y) / (z2.dot(z2))
print(beta_p)
-0.0493588203647

Similarly, we can calculate the $\beta_1$ using the iterative procedure.

gamma_01 = z0.dot(x2) / (z0.dot(z0))
z1 = x2 - gamma_01 * z0

gamma_02 = z0.dot(x1) / (z0.dot(z0))
gamma_12 = z1.dot(x1) / (z1.dot(z1))
z2 = x1 - gamma_02 * z0 - gamma_12 * z1

beta_p = z2.dot(y) / (z2.dot(z2))
print(beta_p)
1.83576285118

As we can see that both $\beta_1$ and $\beta_2$ match the regression coefficients obtained via L2 - norm minimization.

We can represent the transformations on $\mathbf{Z}$ more generally as:

\begin{equation} \mathbf{X} = \mathbf{Z} \mathbf{\Gamma} \end{equation}

where $\mathbf{z_j}$ are the columns of $\mathbf{Z}$ and $\mathbf{Gamma}$ is an upper trangular matrix with the coefficients $\gamma_{lj}$.

For $p=2$ case, we have:

\begin{equation} \mathbf{\Gamma} = \begin{bmatrix} 1 & \hat \gamma_{01} & \hat \gamma_{02}
0 & 1 & \hat \gamma_{12}
0 & 0 & 1 \end{bmatrix} \end{equation}

This is similar to QR Decomposition. We can do a scaled QR-decomposition as:

\begin{equation} \mathbf{X} = \mathbf{Z} \mathbf{D^{-1}} \mathbf{D} \mathbf{\Gamma} = \mathbf{Q} \mathbf{R} \end{equation}

where,

$D_{jj} = ||\mathbf{z_j}||$, and $\mathbf{Q^\intercal} = \mathbf{Q^{-1}}$

From here, we can calculate $\hat \beta$:

\begin{equation} \hat \beta = (\mathbf{X^\intercal} \mathbf{X})^{-1} \mathbf{X^\intercal} y \\ = (\mathbf{(QR)^\intercal} \mathbf{QR})^{-1} \mathbf{(QR)^\intercal} y \\ = (\mathbf{R^\intercal} \mathbf{Q^\intercal} \mathbf{Q} \mathbf{R})^{-1} \mathbf{R^\intercal} \mathbf{Q^\intercal} y \\ = (\mathbf{R^\intercal} \mathbf{I} \mathbf{R})^{-1} \mathbf{R^\intercal} \mathbf{Q^\intercal} y \\ = (\mathbf{R^\intercal} \mathbf{R})^{-1} \mathbf{R^\intercal} \mathbf{Q^\intercal} y \\ = \mathbf{R^{-1}} (\mathbf{R^\intercal})^{-1} \mathbf{R^\intercal} \mathbf{Q^\intercal} y \\ = \mathbf{R^{-1}} \mathbf{Q^\intercal} y \\ \end{equation}

and:

\begin{equation} \hat {\mathbf{y}} = \mathbf{X} \hat \beta \\ = \mathbf{Q} \mathbf{R} \mathbf{R^{-1}} \mathbf{Q^\intercal} y \\ = \mathbf{Q} \mathbf{Q^\intercal} y \\ \end{equation}

Sources

  1. Elements of Statistical Learning

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