Simple linear regression

In case of one feature (\(d=1\)) linear regression is written as

(7)\[ y = a x + b.\]

The parameters of this model \(\boldsymbol \theta = (a, b)\), where \(b\) is intercept (or bias), \(a\) is slope.

The feature matrix here has only one column, denote it \(\boldsymbol x\) and let \(\boldsymbol y\) be the vector of corresponding labels. Also denote

• \(\overline {\boldsymbol x} = \frac 1n \sum\limits_{i=1}^n x_i\) — sample mean of predictors;

• \(\overline {\boldsymbol y} = \frac 1n \sum\limits_{i=1}^n y_i\) — sample mean of targets.

MSE fit

Use MSE to fit parameters \(\boldsymbol \theta = (a, b)\):

(8)\[\mathcal L(a, b) = \frac 1n\sum\limits_{i=1}^n (y_i - ax_i - b)^2 \to \min\limits_{a, b}.\]

What about some calculus?

We have

\[ \frac{\partial \mathcal L}{\partial a} = -\frac 2n\sum\limits_{i=1}^n x_i(y_i - ax_i - b) = 0, \]

\[ \frac{\partial \mathcal L}{\partial b} = -\frac 2n\sum\limits_{i=1}^n (y_i - ax_i - b) = 0. \]

From the last equality it follows that

\[ b = \overline {\boldsymbol y} - a \overline {\boldsymbol x} \]

TODO: Finish the proof

The optimal parameters are

(9)\[ \widehat a = \frac{\sum\limits_{i=1}^n (x_i - \overline {\boldsymbol x})(y_i - \overline {\boldsymbol y})}{\sum\limits_{i=1}^n (x_i - \overline {\boldsymbol x})^2}, \quad \widehat b = \overline {\boldsymbol y} - \widehat a \overline {\boldsymbol x}.\]

Note that the slope is equal to the ratio of sample correlation between \(\boldsymbol x\) and \(\boldsymbol y\) to the sample variance of \(\boldsymbol x\).

Question

Does (9) work for all possible values of \(\boldsymbol x\) and \(\boldsymbol y\)?

Dummy model

The model (7) is simple but we can do even simpler! Let \(a=0\) and predict labels by a constant (or dummy) model \(\widehat y = b\). In fact, for constant predictions you don’t need any features, only labels do the job.

Question

Which value of \(b\) minimizes the MSE (8)?

Linear regression can be used with different loss functions. For example, we can choose mean absolute error (MAE) instead of MSE:

(10)\[\mathcal L(a, b) = \frac 1n\sum\limits_{i=1}^n \vert y_i - ax_i - b\vert \to \min\limits_{a, b}.\]

This time it is unlikely that we can find the analytical solution. But maybe it can be done for the dummy model?

Question

For which value of \(b\) the value of MAE

\[ \frac 1n\sum\limits_{i=1}^n \vert y_i - b\vert \]

is minimal?

Answer

\(\widehat b = \mathrm{med}(\boldsymbol y)\) (see this discussion for details)

RSS and \(R^2\)-score

Putting the optimal weights \(\widehat a\) and \(\widehat b\) into the loss function (8), we obtain residual square error (RSE). Multiplying by \(n\) we get residual sum of squares

\[ RSS = \sum\limits_{i=1}^n(y_i - \widehat a x_i - \widehat b)^2. \]

Also, total sum of squares is defined as

\[ TSS = \sum\limits_{i=1}^n(y_i - \overline {\boldsymbol y})^2. \]

A popular metric called coefficient of determination (or \(R^2\)-score) is defined as

(11)\[R^2 = 1 - \frac{RSS}{TSS} = 1 - \frac{\sum\limits_{i=1}^n(y_i - \widehat y_i)^2}{\sum\limits_{i=1}^n(y_i - \overline {\boldsymbol y})^2}.\]

The coefficient of determination shows proportion of variance explained. \(R^2\)-score does not exceed \(1\) (the greater — the better).

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

\(R^2\)-score measures the amount of variability that is left unexplained after performing the regression. It shows how better the model works in comparison with dummy prediction.

Example: Boston dataset

import pandas as pd
boston = pd.read_csv("../datasets/ISLP/Boston.csv").drop("Unnamed: 0", axis=1)
boston.head()

	crim	zn	indus	chas	nox	rm	age	dis	rad	tax	ptratio	lstat	medv
0	0.00632	18.0	2.31	0	0.538	6.575	65.2	4.0900	1	296	15.3	4.98	24.0
1	0.02731	0.0	7.07	0	0.469	6.421	78.9	4.9671	2	242	17.8	9.14	21.6
2	0.02729	0.0	7.07	0	0.469	7.185	61.1	4.9671	2	242	17.8	4.03	34.7
3	0.03237	0.0	2.18	0	0.458	6.998	45.8	6.0622	3	222	18.7	2.94	33.4
4	0.06905	0.0	2.18	0	0.458	7.147	54.2	6.0622	3	222	18.7	5.33	36.2

Let predictors be lstat, target — medv. Let’s calculate the regression coefficients using (9):

import numpy as np

x = boston['lstat']
y = boston['medv']
a = np.sum((x -x.mean()) * (y - y.mean())) /  np.sum((x -x.mean()) ** 2)
b = y.mean() - a*x.mean()
a, b

(-0.9500493537579907, 34.5538408793831)

Now plot the data and the regression line:

import matplotlib.pyplot as plt

%config InlineBackend.figure_format = 'svg'
plt.scatter(x, y, s=10, c='b', alpha=0.7)
xs = np.linspace(x.min(), x.max(), num=10)
plt.plot(xs, a*xs + b, c='r', lw=2, label=r"$y = \widehat a x + \widehat b$")
plt.plot([xs.min(), xs.max()], [y.mean(), y.mean()], c='orange', lw=2, label=r"$y = \overline y$")
plt.xlabel("lstat")
plt.ylabel("medv")
plt.title("Simple linear regression vs dummy model")
plt.legend()
plt.grid(ls=":");

Calculate MSE:

mse_lin_reg = np.mean((y - a*x - b)**2)
mse_dummy = np.mean((y - y.mean())**2)
print("Linear regression MSE:", mse_lin_reg)
print("Dummy model MSE:", mse_dummy)

Linear regression MSE: 38.48296722989415
Dummy model MSE: 84.41955615616556

Coefficient of determination:

print("R2-score:", 1 - mse_lin_reg / np.mean((y - y.mean())**2))
print("Dummy R2-score:", 1 - mse_dummy / np.mean((y - y.mean())**2))

R2-score: 0.5441462975864798
Dummy R2-score: 0.0

Of course, the linear regression line can be found automatically:

import seaborn as sb

sb.regplot(boston, x="lstat", y="medv",
           scatter_kws={"color": "blue", "s": 10}, line_kws={"color": "red"}, 
           ).set_title('Boston linear regression');

Linear regression from sklearn:

from sklearn.linear_model import LinearRegression

LR = LinearRegression()
x_reshaped = x.values.reshape(-1, 1)
LR.fit(x_reshaped, y)
print("intercept:", LR.intercept_)
print("slope:", LR.coef_[0])
print("r-score:", LR.score(x_reshaped, y))
print("MSE:", np.mean((LR.predict(x_reshaped) - y) ** 2))

intercept: 34.5538408793831
slope: -0.9500493537579906
r-score: 0.5441462975864797
MSE: 38.48296722989415

Compare this with dummy model:

dummy_mse = np.mean((y - y.mean())**2)
print(dummy_mse)

84.41955615616556

Exercises

1. Prove that \(\frac 1n \sum\limits_{i=1}^n (x_i - \overline {\boldsymbol x})^2 = \overline {\boldsymbol x^2} - (\overline {\boldsymbol x})^2\) where \(\overline {\boldsymbol x^2} = \frac 1n\sum\limits_{i=1}^n x_i^2\).

TODO

• Finish analytical derivation of \(a\) and \(b\)

• Add some quizzes

• Add more datasets (may me even simulated)

• Think about cases where the performance of linear model is poor