# Classical Linear Regression Model

Classical Linear Regression Model we described

## Assumption No.1

The regression model is linear in the parameters, as shown in sample/example.

Yi=β1 + β2xi+ui                      LRM                (Linear Regression Model)

Yi= β1 + β22xi+ui                   NLRM             (Non-Linear Regression Model)

Yi= β1 + β2xi2+ui                   LRM                (Linear Regression Model)

## Assumption No.2

The x value is fixed in repeated sampling

Values taken by the regression X are considered fixed in repeated sampling/samples. More technically, X is assumed to be non-stochastic.

For example, there are 10 person whose salary or income / X is 80 but their consumptions are different from each other just as

X                     Y

80                    65.

80                    70

80                    75

80                    80

So we say that the value of X is fixed in repeated sampling.

Classical Linear Regression Model

The regression model is linear in the parameters, as shown in sample/example.

Yi=β1 + β2xi+ui                      LRM                (Linear Regression Model)

## Assumption No.3

Zero mean value of disturbance ui

Given the value of X, the mean, or expected value of the random disturbance term ui is zero.

Technically the conditional mean value of ui is zero symbolically, we have

E(ui/xi) = 0

For example,

X         :           2          4          6          8          10        =          30/5    =          6 mean

x-X =U:           6-2      6-6      6-8      6-10

=          4          0          -2         -4                     =          0          =          0 mean

Assumption 3 stats that the mean value of ui, conditional upon the given xi is zero. Geometrically this assumption can be pictured as:

In passing, note that the assumption E(ui/xi) =0 implies that E(yi/xi) = B1+B2xi (why?) therefore, the two assumptions are equivalent.

If E(ui/xi) = 0

Then E(yi/xi) = β1 + β2xi

We prove that

Yi= β1 + β2xi + ui

E(yi/xi) =         yi = β1 + β2xi + ui

### Applying expectations both sides

E(yi/xi)            =          E(β1 + β2xi + ui) / xi

E(yi/xi)            =          E(β1/xi) + E(β2/xi) + E(ui/xi) + E(xi/xi)

E(yi/xi)            =          β1 + β2xi + E(ui/xi)

E(yi/xi)            =          β1 + β2xi + 0

E(yi/xi)            =          β1 + β2xi

Where :          –           E(β1/xi) = β1

E(β2/xi) = β2

E(xi/xi)   = E(ui/xi)     = 0

So we prove that E(yi/xi)        = β1 + β2xi