Econometric take home APPS_Part_3

Econometric take home APPS_Part_3 - 3. The OLS estimator...

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3. The OLS estimator fit without a constant term is b = x y / x x . Assuming that the constant term is, in fact, zero, the variance of this estimator is Var[ b ] = σ 2 / x x . If a constant term is included in the regression, then, b = ( ) ( ) 1 n ii i x xy y = Σ− / () 2 1 n x x = The appropriate variance is σ 2 / ( 2 1 n ) x x = as always. The ratio of these two is V a r [ b ]/Var[ b ] = [ σ 2 / x x ] / [ σ 2 / 2 1 n x x = ] But, ( 2 1 n ) x x = = x x + n x 2 so the ratio is Var[ b ]/Var[ b ] = [ x x + n x 2 ]/ x x = 1 - n x 2 / x x = 1 - { n x 2 /[ S xx + n x 2 ]} < 1 It follows that fitting the constant term when it is unnecessary inflates the variance of the least squares estimator if the mean of the regressor is not zero. 4. We could write the regression as y i = ( α + λ ) + β x i + ( ε i - λ ) = α * + β x i + ε i * . Then, we know that E [ ε i * ] = 0, and that it is independent of x i . Therefore, the second form of the model satisfies all of our assumptions for the classical regression. Ordinary least squares will give unbiased estimators of α * and β . As long as λ is not zero, the constant term will differ from α . 5. Let the constant term be written as a = Σ i d i y i = Σ i d i ( α + β x i + ε i ) = αΣ i d i + βΣ i d i x i + Σ i d i ε i . In order for a to be unbiased for all samples of x i , we must have Σ i d i = 1 and Σ i d i x i = 0. Consider, then, minimizing the variance of a subject to these two constraints. The Lagrangean is L * = Var[ a ] + λ 1 ( Σ i d i - 1) + λ 2 Σ i d i x i where Var[ a ] = Σ i σ 2 d i 2 . Now, we minimize this with respect to d i , λ 1 , and λ 2 . The ( n +2) necessary conditions are L * / d i = 2 σ 2 d i + λ 1 + λ 2 x i , L * / ∂λ 1 = Σ i d i - 1, L * / ∂λ 2 = Σ i d i x i The first equation implies that d i = [-1/(2 σ 2 )]( λ 1 + λ 2 x i ). Therefore, Σ i d i = 1 = [-1/(2 σ 2 )][ n λ 1 + ( Σ i x i ) λ 2 ] and Σ i d i x i = 0 = [-1/(2 σ 2 )][( Σ i x i ) λ 1 + ( Σ i x i 2 ) λ 2 ]. We can solve these two equations for λ 1 and λ 2 by first multiplying both equations by -2 σ 2 then writing the resulting equations as The solution is nx xx Σ ΣΣ 2 1 2 2 2 0 = λ λ σ . -1 λ σ 1 2 2 2 2 0 = Σ . Note, first, that Σ i x i = n x . Thus, the determinant of the matrix is n Σ i x i 2 - ( n x ) 2 = n ( Σ i x i 2 - n x 2 ) = nS xx where S xx ( 2 1 n ) x x = . The solution is, therefore, 22 1 2 2 1 00 xx xn x nS nx λ ⎤⎡ Σ −− σ ⎛⎞ = ⎜⎟ λ ⎝⎠ ⎦⎣ or λ 1 = (-2 σ 2 )( Σ i x i 2 / n )/ S xx λ 2 = (2 σ 2 x )/ S xx Then, d i = [ Σ i x i 2 / n - x x i ]/ S xx This simplifies if we write Σ x i 2 = S xx + n x 2 , so Σ i x i 2 / n = S xx / n + x 2 . Then, d i = 1/ n + x ( x - x i )/ S xx , or, in a more familiar form, d i = 1/ n - x ( x i - x )/ S xx . This makes the intercept term Σ i d i y i = (1/ n ) Σ i y i - x ( ) 1 n i x xy = / S xx = y - b x which was to be shown.
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This note was uploaded on 11/13/2011 for the course ECE 4105 taught by Professor Dr.fang during the Spring '10 term at University of Florida.

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Econometric take home APPS_Part_3 - 3. The OLS estimator...

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