TAILIEUCHUNG - Class Notes in Statistics and Econometrics Part 15

CHAPTER 29 Constrained Least Squares. One of the assumptions for the linear model was that nothing is known about the true value of β. Any k-vector γ is a possible candidate for the value of β. We ˜ used this assumption . when we concluded that an unbiased estimator By of β ˜ must satisfy BX = I. | CHAPTER 29 Constrained Least Squares One of the assumptions for the linear model was that nothing is known about the true value of 3. Any k-vector 7 is a possible candidate for the value of 3. We used this assumption . when we concluded that an unbiased estimator B y of 3 must satisfy BX I. Now we will modify this assumption and assume we know that the true value 3 satisfies the linear constraint R 3 u. To fix notation assume y be a n x 1 vector u a i x 1 vector X a n x k matrix and R a i x k matrix. In addition to our usual assumption that all columns of X are linearly independent . X has full column rank we will also make the assumption that all rows of R are linearly independent which is called R has full row rank . In other words the matrix of constraints R does not include redundant constraints which are linear combinations of the other constraints. 737 738 29. CONSTRAINED LEAST SQUARES . Building the Constraint into the Model PROBLEM 337. Given a regression with a constant term and two explanatory variables which we will call x and z . yt a fixt YZt t a. 1 point How will you estimate fi and y if it is known that fi y Answer. Write yt a 0 xt zt et b. 1 point How will you estimate fi and y if it is known that fi y 1 Answer. Setting 7 1 0 gives the regression yt zt a 0 xt zt et c. 3 points Go back to a. If you add the original z as an additional regressor into the modified regression incorporating the constraint fi y then the coefficient of z is no longer an estimate of the original y but of a new parameter 6 which is a linear combination of a fi and y. Compute this linear combination . express 6 . BUILDING THE CONSTRAINT INTO THE MODEL 739 in terms of a fi and y. Remark no proof required this regression is equivalent to and it allows you to test the constraint. ANSWER. It you add z as additional regressor into you get yt a fi xt zt 5zt et. Now substitute the right hand side from for y to get a fixt .

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