Multivariate Regression Model

y = x1 + x2 + x3 +… +

The OLS estimates b0,b1 ,b2 , b3 .. …. are sample statistics used to estimate respectively

y is the DEPENDENT variableEach of the xj is an INDEPENDENT variable

Conditions:

Each explanatory variable Xj is assumed

(1A) to be deterministic or non-random

(1B) : to come from a ‘fixed’ population

(1C) : to have a variance V(xj) which is not ‘too large’The above assumptions are best suited to a situation of a controlledexperiment

Assumptions concerning the random term

(IIA) E(i ) = 0 for all i

(IIB) Var(i) = constant for all i

(IIC) Covariance (k) = for any iand k (IID) Each of the i has a normal

distribution

Properties of b0 , b1 , b2 , b3

1. Each of these statistics is a linear functions of the Y values.

2. Therefore, they all have normal distributions

3. Each is an unbiased estimator.

That is, E(bk) =

4. Each bk is the most efficient estimator of all unbiased estimators.

Best

Linear

Unbiased

Estimator of the respective parameter

Thus, each of b0 , b1 , b2 ….is

Conclusion

Each estimator bi has a normal distribution with mean = and variance = bi

2 where bi2 is

unknown.

Income (£ per week) of an individual is regressed on a constant, education (in years), age (in years) and wealth inheritance (in £), using EViews.

Number of observations is 20 and the regression output is given below:

Variable Coefficient Std.Error t-Stats Prob. C -1001.87 520.71 -1.92 0.0654

AGE 8.85 5.45 1.62 0.1168

EDUCATION 95.17 38.54 2.46 0.0252

WEALTH 1.51 0.46 3.26 0.0031

SignificanceLevel (

The Maximum Type 1 Error= SignificanceLevel

The smaller the p-value the more significant is the test

p-value

The proposed regression model is: 

Income = ß0 + ß1(Age) +ß2(Education)

+ ß3(Wealth Inheritance)

… . . (A) 

We are proposing that Income is the variable dependent on three independent variables: Age, Education and Wealth.

0 is a constant.  

It measures the effect of other deterministic factors on Income not included in the model.

1 , 2, 3 measure the effect of a marginal

change in Age, Education and Wealth,

respectively.

However, we recognise that there may be other random factors affecting the dependent variable Income.

So we add a random variable to the model which now becomes:

Income = ß0 + ß1(Age) +ß2(Education)

+ ß3(Wealth Inherited) + … . . (B)

 

We use the least squares technique to estimate the model B.

Therefore, our estimation of the proposedmodel B is Ye = -1001.87 + 8.85*AGE + 95.17*EDUCATION + 1.51*WEALTH INHERITANCE

Here Ye is the estimated value of income

-1001.87 is the estimate of ß0, 8.85 is

the estimate of ß1,; 95.17 is the estimate

of ß2 and 1.51 is the estimate of ß3

The least-squares estimates of the ß-values are denoted by b-values. Thus, b1 is the estimate of ß1 and b2 is the

estimate of ß2 . In our case, b1 = 8.85

and b2 = 95.17.

We next make the following assumptions on the specification of model B so that the least-squares method produces ‘good’ estimators.

i.      is normally distributed with mean 0 and an unknown variance 2

. In the context of the model B, can be thought

of as a luck factor which can be good (positive values) or bad (negative values),

If the positive and negative values cancel out on average, we can say that mean value is 0.

The values are uncorrelated across the population

(Whether or not you are lucky does not influence my being lucky/unlucky)

i.   The values have the same variance (2)

across it. (Every individual is exposed to the same extent/chance of good or bad luck)

The values are uncorrelated with the independent variables Age, Education and Wealth Inheritance.

(For example, an old person is as likely to be lucky as a young one;

or a university graduate is as likely to be unlucky as someone with no A-levels).

We now test (at 10% significance) the following hypothesis:

Education has a positive effect on income  Step 1: Set up the hypotheses

H0 : ß2 = 0 (Education has no effect)

H1 : ß2 > 0(Education has a positive effect)

one-tailed test

Step 2: Select statistic

The estimator b2 is the test-statistic

Step3 : Identify the distributionof b2

Best Linear in the dependent variable incomeUnbiased Estimator of 2 

Assumptions i-iii above imply that b2 is

Since b2 is unbiased, E(b2) = 2

b2 has a normal distribution because it

is linear in Income

Thus, b2~ N(2, 22) where

is unknown.

Therefore, the test statistic ist (b2- 2) / (standard error of b2)

has a Student’s t-distribution with 20-4 = 16 d.o.f.

Step 4: Construct test statistic We use the standard error of b2

because we do not know what is

EViews therefore gives us a t-statistic regarding education of 2.46907

As 2 = 0 under the null hypothesis (H0)

t = b2 / (standard error of b2)

The corresponding probability value is 0.0252.

Select fx /TDIST. For X, enter 2.469607,

the t-Statistic value. The degree of freedomis 16. EViews calculates two-tail probabilitySo number of tails is 2. You now get the 2-tail probability of 0.025165 from Excel.

Since we are performing a one-tail test, take half the probability value, or 0.0126 .

Step 5: Compare with critical value tC

tC = 1.336757for a one-tailed test with significance level () = 0.1 and d.o.f. = 16

tC = 1.336757 < 2.469607

Step 6 : Draw conclusion

The test is significant. Reject H0 at 10% and at 5% (1.745884 < 2.469607) butnot at 1% (2.583492 > 2.469607)

Step 7: Interpret result

The data supports (with at least 98% accuracy) the hypothesis that EDUCATION is an important explanatory variable affecting income.

The probability of a type 1 error is nothing but the area to the right of t-statistic, or 0.0126.

In rejecting H0, we are prone to make a

Type 1 Error.

Example 2: Use output 2 to test the hypothesis (at 5% significance) that weightgain is proportional to foodvalue.

H0 : a = 0 (proportionality) H1 : a 0 (non-proportionality)

The estimator a is the test-statistic

Step 1:

Step 2:

The Model :: y = x + and add the assumptions (Lec17)

Conditions:

The explanatory variable X is assumed

(1A) to be deterministic or non-random

(1B) : to come from a ‘fixed’ population

(1C) : to have a variance V(x) which is not ‘too large’The above assumptions are best suited to a situation of a controlledexperiment

Assumptions concerning the random term

(IIA) E(i ) = 0 for all i

(IIB) Var(i) = constant for all i

(IIC) Covariance (j) = for any iand j (IID) Each of the i has a normal

distribution

Step 3:Thus, a~ N(, ) where is unknown.

Step 4:Therefore, the test statistict (a- ) / (standard error of a) has a Student’s t-distribution with 10-2 = 8 d.o.f.

The p-value is 0.0169 < 0.05

Foodvalue is not the only variable that affects weightgain

Step 6: Draw conclusion

Step 5: Compare with critical value tC

tC = -2.31 > -3.005262

tC = -2.31 for a two-tailed test with significance level () = 0.05 and d.o.f.= 8

The test is significant. Reject H0 at 5%

Step 7: Interpret

Use output 3 to test (at 5% significance) the following hypothesis: Exercise has a negative effect on weight

gain  The proposed regression model is:

 Weightgain

= ß0 + ß1(Foodvalue) +ß2(Exercise) +

Example 3:

Step 1: Set up the hypotheses

H0 : ß2 = 0 (Exercise has no effect)

H1 : ß2 < 0(Exercise has a negative effect)