@L|
∂λ lHe
- T tr[(w '"b + BbfW )(B'B)-1 ]
+ 212 u[IT
(W0Bb + Bb0W)]u^ = 0;
because
-1
u
∂λ lHe
⅛v 1T
. ^ ^
B'B) (b°IT (B'B)-1(W0B + B'W)(b'B)-1)
=IT (W0Bb+Bb0W)(Bb0Bb)
-1
-1 ∖ -1Le = ¾2IT (W'B + B'w):
Therefore, the score vector under H0e is given by
ʌ
D^ =
0
0
ʌ
D^ μ
The elements of the information matrix for this model using (A.5) are given by
@2L 1 2 -1 2 TN
J11 El @(σ°)2JlHe 2 [((σ°) IT IN)] 2b° ;
J22 = E h - @2LLi| = 1 tr [(It (W'B + B'W)(B'B)-1)2i
L @\ -llHe 2 L J
= T tr[((W 'b + B'w )(B'B)-1)2i ;
J33 = Eh- @2L] IH5 =1 tr[((b2)-1 JT bBi = 2Tb°tr[(B'B)2]'
J12
@2L l 1
E[- @ст°@Л] lHe = 2trh(b°)^ (IT IN)(IT
1
-tr ^(b° )-1IT (W 'B + B,W )(B'B)-1)J
(W'Bb + Bb' W)(Bb'Bb)-1
(A.30)
T
—2 tr (W'B + BfW)(B'B)-1∖ ;
2¾° l j
@2L 1
J13 = E[ - â^]|He = 2tr[(b°)-1(IT IN)((b2) JT B B
° pt 0
1T
2¾4 tr [JT B b] = 2¾4tr[B B];
25
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