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APPENDIX E
SOLUTIONS TO PROBLEMS
E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write
X = 12n ⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭x x x , X ' = (1'x 2'x n 'x ), and y = 12n ⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭
y y y
Therefore, X 'X = 1
n
t t t ='∑x x and X 'y = 1
n
t t t ='∑x y . An equivalent expression for ˆβ is
ˆβ = 1
11n
t t t n --=⎛⎫' ⎪⎝⎭∑x x 11n t t t n y -=⎛⎫' ⎪⎝⎭
∑x which, when we plug in y t = x t β + u t for each t and do some algebra, can be written as
ˆβ= β + 1
11n t t t n --=⎛⎫' ⎪⎝⎭∑x x 11n
t t t n u -=⎛⎫' ⎪⎝⎭
∑x . As shown in Section E.4, this expression is the basis for the asymptotic analysis of OLS using matrices.
E.2 (i) Following the hint, we have SSR(b ) = (y – Xb )'(y – Xb ) = [ˆu
+ X (ˆβ – b )]'[ ˆu + X (ˆβ – b )] = ˆu
'ˆu + ˆu 'X (ˆβ – b ) + (ˆβ – b )'X 'ˆu + (ˆβ – b )'X 'X (ˆβ – b ). But by the first order conditions for OLS, X 'ˆu
= 0, and so (X 'ˆu )' = ˆu 'X = 0. But then SSR(b ) = ˆu 'ˆu + (ˆβ – b )'X 'X (ˆβ – b ), which is what we wanted to show.
(ii) If X has a rank k then X 'X is positive definite, which implies that (ˆβ
– b ) 'X 'X (ˆβ – b ) > 0 for all b ≠ ˆβ
. The term ˆu 'ˆu does not depend on b , and so SSR(b ) – SSR(ˆβ) = (ˆβ– b ) 'X 'X (ˆβ
– b ) > 0 for b ≠ˆβ.
E.3 (i) We use the placeholder feature of the OLS formulas. By definition, β = (Z 'Z )-1Z 'y =
[(XA )' (XA )]-1(XA )'y = [A '(X 'X )A ]-1A 'X 'y = A -1(X 'X )-1(A ')-1A 'X 'y = A -1(X 'X )-1X 'y = A -1ˆβ
.
(ii) By definition of the fitted values, ˆt y = ˆt x β and t y = t
z β. Plugging z t and β into the second equation gives t
y = (x t A )(A -1ˆβ
) = ˆt x β = ˆt
y .
(iii) The estimated variance matrix from the regression of y and Z is 2σ(Z 'Z )-1 where 2σ is the error variance estimate from this regression. From part (ii), the fitted values from the two
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regressions are the same, which means the residuals must be the same for all t . (The dependent
variable is the same in both regressions.) Therefore, 2σ = 2ˆσ
. Further, as we showed in part (i), (Z 'Z )-1 = A -1(X 'X )-1(A ')-1, and so 2σ(Z 'Z )-1 = 2ˆσ
A -1(X 'X )-1(A -1)', which is what we wanted to show.
(iv) The j β are obtained from a regression of y on XA , where A is the k ⨯ k diagonal matrix
with 1, a 2, , a k down the diagonal. From part (i), β = A -1ˆβ
. But A -1 is easily seen to be the k ⨯ k diagonal matrix with 1, 1
2a -, , 1k a - down its diagonal. Straightforward multiplication
shows that the first element of A -1ˆβ
is 1ˆβ and the j th element is ˆj
β/a j , j = 2, , k .
(v) From part (iii), the estimated variance matrix of β is 2ˆσ
A -1(X 'X )-1(A -1)'. But A -1 is a symmetric, diagonal matrix, as described above. The estimated variance of j β is the j th
diagonal element of 2ˆσA -1(X 'X )-1A -1, which is easily seen to be = 2ˆσc jj /2
j
a -, where c jj is the j th diagonal element of (X 'X )-1
. The square root of this, a j |, is se(j β), which is simply se(j β)/|a j |.
(vi) The t statistic for j β is, as usual,
j β/se(j β) = (ˆj β/a j )/[se(ˆj
β)/|a j |],
and so the absolute value is (|ˆj β|/|a j |)/[se(ˆj β)/|a j |] = |ˆj β|/se(ˆj
β), which is just the absolute value of the t statistic for ˆj
β
. If a j > 0, the t statistics themselves are identical; if a j < 0, the t statistics are simply opposite in sign.
E.4 (i) 垐?E(|)E(|)E(|).====δ
X GβX G βX Gβδ
(ii) 2121垐?Var(|)Var(|)[Var(|)][()][()].σσ--'''''====δ
X GβX G βX G G X X G G X X G
(iii) The vector of regression coefficients from the regression y on XG -1 is
111111111111[()]()[()]() ()[()]()ˆ ()()().------------''''''='''''=''''''''===XG XG XG y G X XG G X y G X X G G X y
G X X G G X y G X X X y δ
Further, as shown in Problem E.3, the residuals are the same as from the regression y on X , and
so the error variance estimate, 2ˆ,σ
is the same. Therefore, the estimated variance matrix is
