APPENDIX E
SOLUTIONS TO PROBLEMS E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write
⎛ x1 ⎞
⎛ y1 ⎞
X
=
⎜ ⎜ ⎜
x2
⎟ ⎟ ⎟
,
X′
=
( x1′
网 of the t statistic for βˆ j . If aj > 0, the t statistics themselves are identical; if aj < 0, the t statistics 案 are simply opposite in sign. 答 E.4 (i) E(δˆ | X) = E(Gβˆ | X) = GE(βˆ | X) = Gβ = δ. 后 (ii) Var(δˆ | X) = Var(Gβˆ | X) = G[Var(βˆ | X)]G′ = G[σ 2 (X′X)−1]G′ = σ 2G[(X′X)−1]G′. 课 (iii) The vector of regression coefficients from the regression y on XG-1 is
k × k diagonal matrix with 1, a2−1 , … , ak−1 down its diagonal. Straightforward multiplication
shows that the first element of A-1 βˆ is βˆ1 and the jth element is βˆ j /aj, j = 2, … , k.
da se( β j )/|aj|.
kh (vi) The t statistic for β j is, as usual, www. β j /se( β j ) = ( βˆj /aj)/[se( βˆj )/|aj|],
and so the absolute value is (| βˆ j |/|aj|)/[se( βˆ j )/|aj|] = | βˆ j |/se( βˆ j ), which is just the absolute value
x′2
…
x′n
),
and
y
=
⎜ ⎜
y
2
⎜
⎟ ⎟ ⎟
⎜⎜⎝ xn ⎟⎟⎠
⎜⎜⎝ yn ⎟⎟⎠
m n
n
o ∑ ∑ Therefore, X′X = x′txt and X′y = x′tyt . An equivalent expression for βˆ is
c t=1
t =1
w. ∑ ∑ βˆ
=
⎛ ⎜⎝
245
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.k xtG−1 = [xt1 − (c1 / ck )xtk , xt2 − (c2 / ck )xtk ,..., xt,k−1 − (ck−1 / ck )xtk , xtk / ck ].
www E.5 (i) By plugging in for y, we can write
网 β = (Z′X)−1Z′y = (Z′X)−1Z′(Xβ + u) = β + (Z′X)−1Z′u. 案 Now we use the fact that Z is a function of X to pull Z outside of the conditional expectation:
0 for all b ≠ βˆ . The term uˆ ′ uˆ does not depend on b, and so SSR(b) – SSR( βˆ ) = ( βˆ – b) ′X′X
课后 (βˆ – b) > 0 for b ≠βˆ .
E.3 (i) We use the placeholder feature of the OLS formulas. By definition, β = (Z′Z)-1Z′y =
n−1
n t =1
x′t xt
⎞−1 ⎟⎠
⎛ ⎜⎝
n−1
n t =1
x ′t
yt
⎞ ⎟⎠
da which, when we plug in yt = xtβ + ut for each t and do some algebra, can be written as
kh ∑ ∑ βˆ⎝
E.2 (i) Following the hint, we have SSR(b) = (y – Xb)′(y – Xb) = [ uˆ + X( βˆ – b)]′[ uˆ + X( βˆ –
b)] = uˆ ′ uˆ + uˆ ′X( βˆ – b) + ( βˆ – b)′X′ uˆ + ( βˆ – b)′X′X( βˆ – b). But by the first order conditions
⎛ 1 0 0 ... 0 ⎞
⎜ ⎜
0
1
0 ...
0
⎟ ⎟
G
=
⎜ ⎜ ⎜
⋅
⋅
⋅
⋅
⋅
⋅ ... ⋅
⋅
⋅
⋅⎟
⋅
⋅
⎟ ⎟
m ⎜ 0 ... 0 1 0 ⎟
o ⎜⎝ c1 c2 c3 ... ck ⎟⎠
.c does the trick: if δ = Gβ then δj = βj, j = 1,…,k−1, and δk = c1β1 + c2β2 + ... + ck βk . daw (v) Straightforward matrix multiplication shows that, for the suggested choice of G-1, h G−1G = In. Also by multiplication, it is easy to see that, for each t,
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, σ 2 = σˆ 2 . Further, as we showed in part (i), (Z′Z)-1 = A-1(X′X)-1(A′)-1, and so σ 2 (Z′Z)-1 = σˆ 2 A-1(X′X)-1(A-1)′, which is what we wanted to show.
.c diagonal
element
of
σˆ
2
A-1(X′X)-1A-1,
which
is
easily
seen
to
be
=
σˆ
2
cjj/
a
−2 j
,
where
cjj
is
the
jth
w diagonal element of (X′X)-1. The square root of this, σˆ cjj /|aj|, is se( β j ), which is simply
n−1
n t =1
x′t xt
⎞−1 ⎟⎠
⎛ ⎜⎝
n−1
n t =1
x′tut
⎞ ⎟⎠
.
. As shown in Section E.4, this expression is the basis for the asymptotic analysis of OLS using
www matrices.
(iv) The β j are obtained from a regression of y on XA, where A is the k × k diagonal matrix
with 1, a2, … , ak down the diagonal. From part (i), β = A-1 βˆ . But A-1 is easily seen to be the
答 E(β | X) = β + E[(Z′X)−1Z′u | X] = β + (Z′X)−1Z′E(u | X) = β. 后 (ii) We start from the same representation in part (i): β = β + (Z′X)−1Z′u and so
σˆ 2[(XG−1)′XG−1]−1 = σˆ 2G(X′X)−1G′,
which is the proper estimate of the expression in part (ii). (iv) It is easily seen by matrix multiplication that choosing
244
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