Least squire method and mean squire error

1. Chapter 2
Linear Classifiers
DR. AMRAN HOSSAIN
ASSOCIATE PROFESSOR
CSE, DUET-GAZIPUR

2. LEAST SQUARES METHODS
The least square method is the process of finding the best-fitting
curve or line of best fit for a set of data points by reducing the sum of
the squares of the offsets (residual part) of the points from the curve
also called regression line.
During the process of finding the relation between two variables,
the trend of outcomes are estimated quantitatively. This process is
termed as regression analysis.
Least square method is the process of finding a regression line or
best-fitted line for any data set that is described by an equation.
Our main objective in this method is to reduce the sum of the
squares of errors as much as possible. This is the reason this method
is called the least-squares method.
This method is often used in data fitting where the best fit result is
assumed to reduce the sum of squared errors that is considered to
be the difference between the observed values and corresponding
fitted value.
Fig: Least squares method example [1]
[1] https://www.cuemath.com/data/least-squares/

3. LEAST SQUARES METHODS
(Least Square Method Formula)
The least-square method states that the curve that best fits a
given set of observations, is said to be a curve having a
minimum sum of the squared residuals (or deviations or errors)
from the given data points.
Let us assume that the given points of data are (x1, y1), (x2, y2),
(x3, y3), …, (xn, yn) in which all x’s are independent variables,
while all y’s are dependent ones.
Also, suppose that f(x) is the fitting curve and d represents
error or deviation from each given point.
Now, we can write:
d1 = y1 − f(x1)
d2 = y2 − f(x2)
d3 = y3 − f(x3)
…..
dn = yn – f(xn)
The least-squares explain that the curve that best fits
is represented by the property that the sum of squares
of all the deviations from given values must be
minimum, i.e:

4. LEAST SQUARES METHODS
(Least Square Method Formula)
Suppose when we have to determine the equation of line of best fit for the given data, then we
first use the following formula.
The equation of least square line is given by Y = a + bX
Normal equation for ‘a’:
∑Y = na + b∑X
Normal equation for ‘b’:
∑XY = a∑X + b∑X2
Solving these two normal equations we can get the required trend line equation.
Thus, we can get the line of best fit with formula y = ax + b

5. LEAST SQUARES METHODS
(Least Square Method Example)
Solution:
Mean of xi values = (8 + 3 + 2 + 10 + 11 + 3 + 6 + 5 + 6 + 8)/10 = 62/10
= 6.2
Mean of yi values = (4 + 12 + 1 + 12 + 9 + 4 + 9 + 6 + 1 + 14)/10 =
72/10 = 7.2
Straight line equation is y = a + bx.
The normal equations are
∑y = an + b∑x
∑xy = a∑x + b∑x2

6. LEAST SQUARES METHODS
(Least Square Method Example)
Substituting these values in the normal equations,
10a + 62b = 72….(1)
62a + 468b = 503….(2)
(1) × 62 – (2) × 10,
620a + 3844b – (620a + 4680b) = 4464 – 5030
-836b = -566
b = 566/836
b = 283/418
b = 0.677
Substituting b = 0.677 in equation (1),
10a + 62(0.677) = 72
10a + 41.974 = 72
10a = 72 – 41.974
10a = 30.026
a = 30.026/10
a = 3.0026
Therefore, the equation becomes,
y = a + bx
y = 3.0026 + 0.677x Fig. This is the required trend line
equation.
Now, we can find the sum of squares of deviations
from the obtained values as:
d1 = [4 – (3.0026 + 0.677*8)] = (-4.4186)
d2 = [12 – (3.0026 + 0.677*3)] = (6.9664)
d3 = [1 – (3.0026 + 0.677*2)] = (-3.3566)
d4 = [12 – (3.0026 + 0.677*10)] = (2.2274)
d5 = [9 – (3.0026 + 0.677*11)] =(-1.4496)
d6 = [4 – (3.0026 + 0.677*3)] = (-1.0336)
d7 = [9 – (3.0026 + 0.677*6)] = (1.9354)
d8 = [6 – (3.0026 + 0.677*5)] = (-0.3876)
d9 = [1 – (3.0026 + 0.677*6)] = (-6.0646)
d10 = [14 – (3.0026 + 0.677*8)] = (5.5814)
∑d2 = (-4.4186)2 + (6.9664)2 + (-3.3566)2 + (2.2274)2 + (-
1.4496)2 + (-1.0336)2 + (1.9354)2 + (-0.3876)2 + (-
6.0646)2 + (5.5814)2 = 159.27990

7. MEAN SQUARE ERROR (MSE) ESTIMATION
In statistics, the mean squared error (MSE) or
mean squared deviation (MSD) of an estimator
(of a procedure for estimating an unobserved
quantity) measures the average of the squares
of the errors—that is, the average squared
difference between the estimated values and
the actual value.
MSE is a risk function, corresponding to the
expected value of the squared error loss.
The MSE is a measure of the quality of an
estimator. As it is derived from the square of
Euclidean distance, it is always a positive value
that decreases as the error approaches zero.
If a vector of n predictions is generated from a sample of n
data points on all variables, and Y is the vector of observed
values of the variable being predicted, with being the
predicted values, then the within sample MSE of the
prediction is computed as:

8. MEAN SQUARE ERROR (MSE) ESTIMATION
General steps to calculate the MSE from a set
of X and Y values:
1.Find the regression line.
2.Insert your X values into the linear regression
equation to find the new Y values (Y’).
3.Subtract the new Y value from the original to
get the error.
4.Square the errors.
5.Add up the errors (the Σ in the formula is
summation notation).
6.Find the mean.
Example Problem: Find the MSE for the following set of values:
(43,41), (44,45), (45,49), (46,47), (47,44).
Step 1: Find the regression line. got the regression line y = 9.2 + 0.8x.
Step 2: Find the new Y’ values:
•9.2 + 0.8(43) = 43.6
•9.2 + 0.8(44) = 44.4
•9.2 + 0.8(45) = 45.2
•9.2 + 0.8(46) = 46
•9.2 + 0.8(47) = 46.8
Step 3: Find the error (Y – Y’):
•41 – 43.6 = -2.6
•45 – 44.4 = 0.6
•49 – 45.2 = 3.8
•47 – 46 = 1
•44 – 46.8 = -2.8

9. MEAN SQUARE ERROR (MSE) ESTIMATION
Step 4: Square the Errors:
•-2.62 = 6.76
•0.62 = 0.36
•3.82 = 14.44
•12 = 1
•-2.82 = 7.84
Step 5: Add all of the squared errors up:
6.76 + 0.36 + 14.44 + 1 + 7.84 = 30.4.
Step 6: Find the mean squared error:
30.4 / 5 = 6.08

10. Sum of Error Squares Estimation

11. Sum of Error Squares Estimation
invertible square matrix.

12. Assignment
Page No: 99 and 109
Please submit within next 15 days
Write in details explanation to solve the problem
If copy is found the score will be zero.

13. LOGISTIC DISCRIMINATION
Self Study: 3.6 (Important )
Page number: 117

14. SUPPORT VECTOR MACHINE Algorithm
Support Vector Machine or SVM is one of the most popular
Supervised Learning algorithms, which is used for Classification as
well as Regression problems.
However, primarily, it is used for Classification problems in
Machine Learning.
The goal of the SVM algorithm is to create the best line or decision
boundary that can segregate n-dimensional space into classes so
that we can easily put the new data point in the correct category in
the future.
This best decision boundary is called a hyperplane.
SVM chooses the extreme points/vectors that help in creating the
hyperplane.
These extreme cases are called as support vectors, and hence
algorithm is termed as Support Vector Machine.
Consider the below diagram in which there are two different
categories that are classified using a decision boundary or
hyperplane:
SVM algorithm can be used for Face detection, image classification, text
categorization, etc.

15. SUPPORT VECTOR MACHINE Algorithm
Types of SVM
SVM can be of two types:
Linear SVM: Linear SVM is used for linearly separable data, which means if a dataset can be classified into
two classes by using a single straight line, then such data is termed as linearly separable data, and
classifier is used called as Linear SVM classifier.
Non-linear SVM: Non-Linear SVM is used for non-linearly separated data, which means if a dataset
cannot be classified by using a straight line, then such data is termed as non-linear data and classifier
used is called as Non-linear SVM classifier.

16. SUPPORT VECTOR MACHINE Algorithm
Hyperplane and Support Vectors in the SVM algorithm:
Hyperplane:
 There can be multiple lines/decision boundaries to segregate the classes in n-dimensional space, but we need to find out
the best decision boundary that helps to classify the data points. This best boundary is known as the hyperplane of SVM.
 The dimensions of the hyperplane depend on the features present in the dataset, which means if there are 2 features (as
shown in image), then hyperplane will be a straight line. And if there are 3 features, then hyperplane will be a 2-
dimension plane.
 We always create a hyperplane that has a maximum margin, which means the maximum distance between the data
points.
Support Vectors:
The data points or vectors that are the closest to the hyperplane and which affect the position of the hyperplane are
termed as Support Vector. Since these vectors support the hyperplane, hence called a Support vector.

17. SUPPORT VECTOR MACHINE Algorithm
(Linear SVM)
The working of the SVM algorithm can be
understood by using an example.
Suppose we have a dataset that has two tags (green
and blue), and the dataset has two features x1 and
x2.
We want a classifier that can classify the pair(x1, x2)
of coordinates in either green or blue. Consider the
below image:
So as it is 2-d space so by just using a straight line, we can easily separate these two classes.
But there can be multiple lines that can separate these classes. Consider the below image:
 Hence, the SVM algorithm helps to find the best line or decision boundary;
this best boundary or region is called as a hyperplane.
 SVM algorithm finds the closest point of the lines from both the classes.
These points are called support vectors.
 The distance between the vectors and the hyperplane is called as margin.
 And the goal of SVM is to maximize this margin.
 The hyperplane with maximum margin is called the optimal hyperplane.

18. SUPPORT VECTOR MACHINE Algorithm
(Non-Linear SVM)
If data is linearly arranged, then we can separate it by using a
straight line, but for non-linear data, we cannot draw a single
straight line. Consider the below image:
So to separate these data points, we need to add one more
dimension. For linear data, we have used two dimensions x and
y, so for non-linear data, we will add a third dimension z. It can
be calculated as:
z=x2 +y2
By adding the third dimension, the sample space will become as below
image:
So now, SVM will divide the datasets into classes in the
following way. Consider the below image:
Since we are in 3-d Space, hence it
is looking like a plane parallel to
the x-axis. If we convert it in 2d
space with z=1, then it will become
as: