Explanation of the concept of regression analysis using python Part 2

Search for α and β values

Grab an image with a graph

In [Part 1] of this article (http://qiita.com/kenmatsu4/items/8b4e908d7c93d046110d), see that you can find the minimum value if you fix each of $ \ alpha $ and $ \ beta $ with some value. However, in order to actually find the parameters $ \ alpha, \ beta $ of the approximate straight line (regression line) for this data, do not look for the case where $ \ alpha $ and $ \ beta $ have the minimum values at the same time. Must not be.

The function $ S $ handled in [Part 1](http://qiita.com/kenmatsu4/items/8b4e908d7c93d046110d#Let's calculate) is organized as a two-variable function of $ \ alpha $ and $ \ beta $ as follows. It looks like.

S(\alpha, \beta) = 
\left( \sum_i^n x_i^2 \right) \alpha^2 + n\beta^2
+ 2 \left( \sum_i^n x_i \right)\alpha \beta
- 2 \left( \sum_i^n x_i y_i \right)\alpha 
- 2 \left( \sum_i^n y_i \right)\beta
+ \sum_i^n y_i^2

In addition, each coefficient can be obtained from the data (calculated in [Part 1](http://qiita.com/kenmatsu4/items/8b4e908d7c93d046110d#Let's calculate)).

n=50

\left( \sum_i^n x_i^2 \right) =34288

\left( \sum_i^n y_i^2 \right)=11604

\left( \sum_i^n x_iy_i \right)=18884

\left( \sum_i^n x_i \right)=1240

\left( \sum_i^n y_i \right)=655

And if you substitute this

S(\alpha, \beta) = 
34288 \alpha^2 + 50\beta^2
+ 2480\alpha \beta
- 37768\alpha 
- 1310 \beta
+ 11604

It will be. Let's draw a graph of this two-variable quadratic curve. The vertical axis $ S $ is the sum of squares of the distance (error) between the points of each data and the straight line. Find the place where this is the smallest.

from mpl_toolkits.mplot3d.axes3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
from matplotlib import cm

# set field
X = np.linspace(0.2, 1.3, 100)
Y = np.linspace(-25, 15, 100)

# set data
#sum(x**2)
sum_x_2 = 34288.2988
#sum(y**2)
sum_y_2 = 11603.8684051
#dot(x,y)
sum_xy = 18884.194896
#sum(x)
sum_x = 1239.7
#sum(y)
sum_y = 655.0152
X, Y = np.meshgrid(X, Y)
#S(α,β)=34288α^2 + 50β^2 + 2480αβ − 37768α − 1310β + 11604
S = (sum_x_2 * (X**2)) + (50 * (Y**2)) + (2 * sum_x * X * Y) + (-2 * sum_xy  * X) + (-2 * sum_y * Y) + sum_y_2

# prepare plot
fig = plt.figure(figsize=(18,6))
ax = fig.add_subplot(121, projection='3d', azim=60)
ax.set_xlabel("alpha")
ax.set_ylabel("beta")
ax.set_zlabel("S")

# draw 3D graph
surf = ax.plot_surface(X, Y, S, rstride=1, cstride=1, cmap=cm.coolwarm,
        linewidth=0, antialiased=False)

# draw contour
ax = fig.add_subplot(122)
plt.contour(X,Y,S,50)
ax.set_xlabel("alpha")
ax.set_ylabel("beta")

plt.show()

The 3D graph on the left is a bit confusing, but looking at the contour graph on the right, the minimum value in the center of the ellipse is somehow close to the previously visually predicted $ \ alpha = 0.74, \ beta = -5 $. I will do it!

Try to solve by calculation

To solve it by calculation, partial derivative of $ S $ with $ \ alpha $ and $ \ beta $, respectively, to obtain a value that becomes 0. If you write $ S $ as a function of $ \ alpha $ and $ \ beta $, respectively, it will be as follows.

S(\alpha) = \left( \sum_i^n x_i^2 \right) \alpha^2
 + 2\left( \sum_i^n (x_i\beta - x_i y_i ) \right) \alpha 
 + n\beta^2 - 2\beta\sum_i^n y_i + \sum_i^n y_i^2

S(\beta) = n\beta^2
+ 2 \left( \sum_i^n (x_i\alpha - y_i) \right) \beta
+ \alpha^2\sum_i^n x_i^2 - 2\alpha \sum_i^n x_iy_i + \sum_i^n y_i^2

Since this is partially differentiated and set as 0,

\frac{\partial S}{\partial \alpha} = 0,

\frac{\partial S}{\partial \beta } = 0,

Will solve the simultaneous equations of. In other words

\frac{\partial S}{\partial \alpha} =  2\left(\sum_i^n x_i^2 \right) \alpha +  2\left( \sum_i^n x_i \right) \beta - 2\sum_i^n x_i y_i = 0

\frac{\partial S}{\partial \beta} = 2n\beta + 2\left( \sum_i^n x_i\right) \alpha - 2\sum_i^ny_i　= 0

Will be solved. Substituting the value obtained from the data

\frac{1}{2}\frac{\partial S}{\partial \alpha} =  34288 \alpha +  1240 \beta - 18884 = 0

\frac{1}{2}\frac{\partial S}{\partial \beta} = 50 \beta + 1240 \alpha - 655 = 0

If you solve the simultaneous equations using Python,

from sympy import *
a, b = symbols('a b')
init_printing()

# 34288α + 1240β − 18884 = 0
#    50β + 1240α −   655 = 0
solve([34288 * a + 1240 * b - 18884, 50* b + 1240 * a - 655], [a, b])

a --> 0.746606334842
b --> -5.41583710407

\alpha = 0.746606334842
\beta  = -5.41583710407

You can get the solution like

Substituting the values of $ \ alpha and \ beta $ again and plotting a straight line on the scatter plot, it will be as follows.

import numpy as np
import matplotlib.pyplot as plt

data= np.loadtxt('cars.csv',delimiter=',',skiprows=1)
data[:,1] = map(lambda x: x * 1.61, data[:,1])    #km from mph/Convert to h
data[:,2] = map(lambda y: y * 0.3048, data[:,2])  #Convert from ft to m

fig = plt.figure(figsize=(10,6))
ax = fig.add_subplot(111)
ax.set_xlim(0,50)
ax.set_title("Stopping Distances of Cars with estimated regression line")
ax.set_xlabel("speed(km/h)")
ax.set_ylabel("distance(m)")
plt.scatter(data[:,1],data[:,2])

x = np.linspace(0,50,50)
y = 0.746606334842 * x -5.41583710407
plt.plot(x,y)

This is the regression line: smile:

Continued to explain with animation.