Introduction to Machine Learning (Part-7)

 Linear Regression : In Linear regression, we are bound with the straight line which is our decision boundary. The value of y can be same for different values of x. Can be represented as : 

According to Gaussian :
 
P(y|x=x1) = (1/(√2π)*σ)* e-(y-mean)2/2σ2

Through linear regression equation we define yi' = β1x1 + β0

We are predicting value of y as   yi' = β1x1 + β0+ e  where e is the error or noise.

Let us assume  σ for every value of x is same and  β1 = 0 then,

                              β0*argmin C = Σni=1(yi -β0)2

                                By differentiating on both sides

                               dC/dβ0 = Σni=1(0 + 2β0 - 2yi) = 0   

                                     2Σni=1β0 - 2Σni=1yi = 0

                                         nβ0 - Σyi = 0

                                         β0 = Σyi/n

Here it comes out to be the best value for   β0 which is the mean average of y values observed in the data.

Now comes the question , How can we determine the Goodfit of our linear regression Model??

We can check the good fit of the linear regression model by the R2 matrix which generalize Linear regression.

                       R2  = 1 -Σni=1(yi - (Σβjxj + β0))2 / Σni=1(yi - mean)2

Where βjxj  is independent variable. 

The good fit can be done by matrices. The more the value of R close to 1 , it means independent variable has sizable impact on predictions of output. We don't want  βj  to be dependent on the range or domain of the variable.For the solution of the same we do normalization of variables beforehand. Normalization can be done for this using Min-Max Sealing . For that we define variable Z.

                                   Z = x - xmin/xmax - xmin

Now for the actual value of yi, 

                                        yi = β0 + Σmi=1βjxj + e

                                  Therefore  yi = yi' + e

                                Cost function (C) = Σni=1(yi - (Σβjxj + β0))2

                                                          =   yi - (Σβjxj + β0) - e



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