Econometric analysis Linear Least Squares参考.ppt

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Partitioned Solution Method of solution (Why did FW care? In 1933, matrix computation was not trivial!) Direct manipulation of normal equations produces b2 = (X2?X2)-1X2?(y - X1b1) What is this? Regression of (y - X1b1) on X2 Important result (perhaps not fundamental). Note the result if X2?X1 = 0. Useful in theory: Probably Likely in practice? Not at all. Partitioned Inverse Use of the partitioned inverse result produces a fundamental result: What is the southeast element in the inverse of the moment matrix? Partitioned Inverse The algebraic result is: [ ]-1(2,2) = {[X2’X2] - X2’X1(X1’X1)-1X1’X2}-1 = [X2’(I - X1(X1’X1)-1X1’)X2]-1 = [X2’M1X2]-1 Note the appearance of an “M” matrix. How do we interpret this result? Note the implication for the case in which X1 is a single variable. (Theorem, p. 28) Note the implication for the case in which X1 is the constant term. (p. 29) Frisch-Waugh Result Continuing the algebraic manipulation: b2 = [X2’M1X2]-1[X2’M1y]. This is Frisch and Waugh’s famous result - the “double residual regression.” How do we interpret this? A regression of residuals on residuals. “We get the same result whether we (1) detrend the other variables by using the residuals from a regression of them on a constant and a time trend and use the detrended data in the regression or (2) just include a constant and a time trend in the regression and not detrend the data” “Detrend the data” means compute the residuals from the regressions of the variables on a constant and a time trend. Important Implications Isolating a single coefficient in a regression. (Corollary 3.3.1, p. 28). The double residual regression. Regression of residuals on residuals – ‘partialling’ out the effect of the other variables. It is not necessary to ‘partial’ the other Xs out of y because M1 is idempotent. (Th