L0 (Linear Regression)

Predicting y using x and 2 points given on same line

Deriving Formulas

To start predicting the value of y with x lets just see the 2 points given.

Let’s assume the points are A (1,5) and B (2,9)
now put the points on graph we get:

So how much we traveled on x-axis from the initial point ‘A’ till the final point ‘B’
we traveled(change in x): x2-x1 on the x-axis.
change in x = 2–1 change in x = 1

Similarly,

How much we traveled on y-axis from the initial point ‘A’ till the final point ‘B’
we traveled(change in y): y2-y1 on the y-axis.
change in y = 9–5 chnage in y = 4

now let’s find the ratio of change in y with respect to change in x(m). (Because we wanna find y using x).

    cahnge in y   y2-y1
m = ----------- = -----
    change in x   x2-x1

    cahnge in y   4
m = ----------- = - (putting the values)
    change in x   1

Thus m = 4
This means y = m(x) and m = (y2-y1)/(x2-x1)

Let’s try this:
for A (1,5) x = 1 and y = 5
using formula: y = m(x)
5 = 4(1)
5 = 4 ,here it’s not equal we must add 1 but why so

Let’s try for B (2,9) x = 2 and y = 9
using formula: y = m(x)
9 = 4(2)
9 = 8 ,here it’s not equal we must add 1 but why so

why we are adding 1 to make it equal, let’s again move towards graph:

In the Graph if we extend the line AB, we can see it cuts the y-axis on 1
which tells that yes y = m(x) but we have to add the starting of y(c)to get the value which is 1 in this case.

so the new formula to get y with respect to x is: y = m(x) +c (proof for Slope Intercept form of straight line)

But how we can get c for other cases: By now we know y = m(x) + c
we can write:
c = y-m(x)

Let’s try this:
for A (1,5) c = 5–4(1), using c = y-m(x)
c = 1, it is satisfying.

Let’s try this for B (2,9):
c = 9–4(2), using c = y-m(x)
c = 1, it is satisfying.

The logic is we are drawing a line from c where the line cuts y-axis (x=0) till x given with the slope m and then finding the y at the end.

Formulas are:

• m = (y2-y1)/(x2-x1) :slope of the line
• c = y-m(x) :starting point of the line
• x = given :total length of line (x-0, as the value of x at starting point is 0)
• y = m(x) +c

Let’s code this to find m and c

In [1]:

def learn(x1,y1,x2,y2):
    m = (y2-y1)/(x2-x1)
    c = y1-(m*x1)
    return m,c

In [2]:

m_c=learn(1,5,2,9)
print(m_c)

(4.0, 1.0)

Let’s code to predict the y using x (m and c)

In [3]:

def predict(x,m_c):
    return (m_c[0]*x)+m_c[1]

In [4]:

predict (3,m_c)

Out[4]:

13.0

Here we are trying to draw the below line:

Let’s do it another way

we know:

• m = (y2-y1)/(x2-x1) :slope of the line
• c = y-m(x) :starting point of the line
• x = given :total length of line (x-0, as the value of x at starting point is 0)
• y = m(x) +c

lets say:

• m = (y2-y1)/(x2-x1) : slope of the line
• c = y1 : let starting point be y1
• total length of line = x-x1 (as value of x at the starting of point is x1)
• y = y1 + m(x-x1)
• So the formula will be: y = y1 + m(x-x1) (proof for Point slope form of straight line)

Here we are trying to draw the below line:

Let’s code to predict y using x (without c)

In [5]:

def Predict(x1,y1,x2,y2,x):
    return y1 + ((y2-y1)/(x2-x1)*(x-x1))

In [6]:

Predict(1,5,2,9,3)

Out[6]:

13.0