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 .