ECON60622: Further Econometrics Semester 2, 2020-21
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ECON60622:
Further Econometrics
Semester 2, 2020-21
Testing linear restrictions on β
In econometric models, it is often desired to tests hypotheses about whether the regression pa- rameters satisfy linear restrictions. It is common to express such restrictions using linear algebra. The purpose of this note is to illustrate this linear algebraic approach via a few examples.
As in lectures, we write the regression model of interest as
y = xβ + u,
where x is the 1 × (k +1) row vector x = (1,x1 ,x2 , . . . ,xk) and β is the (k +1) × 1 column vector
l β0 」
β = β2 .
「 βk
Suppose we wish to test the null hypothesis that the regression parameters satisfy the following set of m linear restrictions:
H0 : Ri,1β0 + Ri,2β1 + ... + Ri,k+1βk = ri , for i = 1, 2, . . . ,m,
where {Ri,j,ri; i = 1, 2, . . . ,m; j = 1, 2, . . . ,k + 1} are specified constants. Notice that these restrictions involve liuear combinations of {βi} and we can write H0 equivalently as:
k
H0 : 工 Ri,j+1βj = ri , for i = 1, 2, . . . ,m.
j=0
The alternative hypothesis is that H0 is not true that is,
k
H1 : 工 Ri,j+1βj ri for at least one i out of i = 1, 2, . . . ,m.
j=0
We now illustrate the types of hypotheses that fit within this framework, as will be evident these include hypotheses that are of interest in any linear regression model.
Example 1: H0 : β1 = 0 versus H1 : β1 0 . in a model with k = 3
(Notice that under the null hypothesis x1 does not affect y; this hypothesis is tested using the t-statistic and is routinely reported as part of the output from any regression computer package.) In this case, m = 1, R1,2 = 1, R1,j = 0 for j = 1, 3, 4, and r1 = 0.
Example 2: H0 : β1 = 10 versus H1 : β1 10 in a model with k = 4
In this case, m = 1, R1,2 = 1, R1,j = 0 for all j = 1, 3, 4, 5, r1 = 10.
Example 3: H0 : βj = 0 for j = 1, 2, . . . ,k versus H1 : βj 0 for at least one j out of j = 1, 2, . . . ,k in a model with k = 5.
(Notice that under this null hypothesis, y is not related to any of the explanatory variables (except trivially the intercept). This hypothesis is tested using the F-statistic that is routinely reported in the ANOVA part of the output from any regression computer packages.)
In this case, m = k , Rj,j+1 = 1 for j = 1, 2, 3, 4, 5, and Rj,l = 0 for all ! j + 1, and rj = 0 for j = 1, 2, . . . ,k .
Example 4: H0 : β1 + β2 = 1 versus H1 : β1 + β2 1 in a model with k = 2 .
In this case, m = 1, R1,2 = 1, R1,3 = 1, R1,1 = 0, r1 = 1.
Example 5: H0 : β1 + β2 = β3 and 1 + β4 /3 = β5 in a model with k = 5
In this case, m = 2, R1,2 = 1, R1,3 = 1, R1,4 = −1, R1,j = 0 for j = 1, 5, 6, r1 = 0, R2,5 = 1/3, R2,6 = −1, R2,j = 0 for j = 1, 2, 3, 4, r2 = −1.
The generic null hypothesis above can be written very compactly using linear algebra. To this end, define R to be the m × (k +1) matrix with i − jth element Ri,j and r to be the m × 1 vector with ith element ri . We can then write the null and alternative hypotheses as follows:
H0 : Rβ = r
H1 : Rβ r
As an exercise, try to write out R and r for the five examples above . The answers are on the next page.
We now give the relevant choices of R and r for the five examples.
Example 1:
R = [0, 1, 0, 0] , and r = 0.
Example 2:
R = [ 0, 1, 0, 0, 0] , and r = 10.
Example 3:
l 0 1 0 0 0 0 」
0 0 0 0 1 0
0 0 0 0 0 1
and
0
Example 4:
R = [0, 1, 1] , and r = 1.
Example 5: R = [ 0(0) |
1 0 |
1 0 |
−1 0 |
0 1/3 |
1 ] , |
and
r = [ 1 ] .
2022-08-29
Testing linear restrictions on β