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Beginner to Advance level – Steps to make regression model part 1


You must have heard about Regression models many times but you might have not heard about the techniques of solving or making a regression model step-wise.

First we will talk about Simple Linear Regression : is a model with single regressor (x) has a linear relationship with a response variable (y).

We all(who have idea about regression) the linear regression equation:

For a given x, the corresponding observation Y consists of the value. Check above

We know make some assumptions on the model.

Assumptions:- 

  • ε(i) is a random variable with mean zero & variance σ^2(sigma square) [σ^2 is unknow]
    • i.e. E(ε(i)) = 0 & V(ε(i)) = σ^2 {E<- expectation}
  • ε(i) and ε(j) are uncorrelated i≠j . So cov(ε(i), ε(j)) = 0. Here uncorrelated means independent to each other
  • ε(i) is a normally distributed random variable, with mean zero and variance σ^2 
    • ε(i)=N(0,σ^2) {N – indicates random distributed, σ^2 – variance followed by mean = zero}

Assumptions in terms of Y[ (ε(Y))] :- Here i am not going in details to write the equation, I will tell you what to do just replace the ε(i) to ε(Y).

And Mean will become = β(0)+β(1)*x and variance will e same and equal to σ^2

Here is the Least Square Estimation of the Parameter we are going to discuss further.

Least Square Estimation(LSE):- 

  • The parameters β(0), β(1) are unknown and must be estimated using same data.

(x1,y1),(x2,y2), – – – – , (xn,yn)

  • The line fitted by (LSE) is the one that makes the sum of squares of all Vertical discrepancies as small as possible. 

  • We estimate β(0), β(1), so that the sum of square of all the difference between observation Y(i) and the fitted line is minimum = SS(Res) , explained in the below snapshot
  • The least square estimator of β(0)& β(1), (β_0 ̂,β_1 ̂ ) must satisfy the following two equation a snapshot is added:Equation 1 and 2 are called normal equations and they are uniquely independent.

So, the estimator β_0 ̂,β_1 ̂  are the solution of the equation

∑(Y(i)-β0ˆ-βiˆ*xi) = 0 —–(1),

∑(Y(i)-β0ˆ-βiˆ*xi)*xi = 0 —-(2)

Solution of the above equations are attached as image,  I am attaching image here because i can’t write mathematical eqaution here, So enjoy snapshots 🙂

Equation Solution:

Equation 2 solution:

Above we have calculated the parameters using least square estimator.

We have not discussed the benefits of using LSE , in one line the the most important benefits of using LSE is the solution will be most correct almost 95% accuracy.

Properties of Least Square Estimator:-

  • Sum of residuals in any regression model that contains an intercept β_0 is always 0.

∑y(i) = ∑(y(i)−y_iˆ) = 0   (perfect regression line)

  • ∑y(i) = ∑(y_iˆ) means the observation and estimated line of regression graph lie on each other(perfect regression line)
  • ∑x(i)*e(i) = 0
  • ∑y(i)*e(i) = 0

Statistical properties of LS estimation:- 

  • Both (β_0)ˆ and (β_1)ˆ are unbiased estimator of (β_0) and (β_1) respectively. Which means they should be equal in values.
  • (β_0)ˆ and (β_1)ˆ are linear combination of observation of y(i)

β_1ˆ= [∑(x(i)-mean(x))*(y(i)-mean(y))]/∑((x(i)-mean(x))^2

         =  [∑(x(i)-mean(x))*y(i)]/∑(x(i)-mean(x))^2]

β_1ˆ = ∑(e(i)*y(i)

Similarly we can also do for β_0ˆ

β_0ˆ = mean(y)−β_1ˆ*mean(x)

  = (1/n)∑y(i) – β_iˆ.(mean(x))

Take the value of beta 1 parameter from above equation.

Note:- I am not going to prove this, if you proof need please message me @ irrfankhann29@gmail.com i will personally send my documents.

Similarly we will calculate the variance of beta_0 and beta_1

v(β_1ˆ) = [σ^2/S(xx)]             :where S(xx) = ∑[(x_i – mean(x)]^2

v(β_0ˆ) = σ^2[(1/n)+{mean(x)^2}/S(xx)]

NOTE:-β_0 & β_1 are unbiased estimator of σ^2

Estimation of σ^2:- is obtained from the residual sum of square

SS(res) = SS(yy) – (β_1ˆ)^2.

We got the values of Coefficients, sum of squared errors (Regressor, Regresson and total) using these we can calculate the the null hypothesis which is based on t and z test.

The t and z test value can be calculated using this formula:

Usually varinace(σ²) is unknow, if variance(σ²) is unknow then we will follow t-test hypothesis

t = (β_1ˆ-β_1)/√(MS_res)/S(xx)

if |t|>t[(α/2), (n-2)] we reject null hypothesis. 

∴ |t| is calculated value and t[(α/2), (n-2)] is tabulated value.

And when (σ²) is know we will follow z-test hypothesis

z = (β_1ˆ-β_1)/√(σ²)/S(xx) which follow random normal distribution(0,1)

if |z|>z(α/2) we reject null hypothesis. 

∴ |z| is calculated value and z(α/2) is tabulated value.

Now we have all the values to calculate ANOVA table which  will describe in next article, so stay tuned.

Queries or Docs/ notes related please shoot me an email on khanirfan.khan21@gmail.com.

 

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