Page 159 - Applied Probability

8. The Polygenic Model
144
−1 ∂
∂
ln det B = tr(B
(h) If B is a square matrix, then
∂θ
is validated by
∂
∂

ln det B
=
b ij
∂θ
∂θ
ij ∂b ij ln det B ∂θ ∂ B). This formula
B ij ∂
= b ij
det B ∂θ
ij
∂
=tr B −1 B
∂θ
using property (a).
Applying the above facts to the loglikelihood (8.2) leads to the score.
With respect to the mean component µ i , we have
∂ 1 ∂ t −1 1 t −1 ∂
L = A µ Ω (y − Aµ)+ (y − Aµ) Ω A µ
∂µ i 2 ∂µ i 2 ∂µ i
∂ t −1
t
= µ A Ω (y − Aµ).
∂µ i
2
With respect to the variance component σ , we have
i
∂ 1 ∂ 1 t ∂ −1
L = − ln det Ω − (y − Aµ) Ω (y − Aµ)
∂σ 2 2 ∂σ 2 2 ∂σ 2
i i i
1 −1 1 t −1 −1
= − tr(Ω Γ i )+ (y − Aµ) Ω Γ i Ω (y − Aµ).
2 2
In similar fashion, the elements of the observed information matrix are
∂ 2 ∂ t t −1 ∂
− L = µ A Ω A µ
∂µ i ∂µ j ∂µ j ∂µ i
∂ 2 ∂ t t −1 −1
− 2 L = µ A Ω Γ i Ω (y − Aµ)
∂σ ∂µ j ∂µ j
i
∂ 2
= − 2 L
∂µ j ∂σ
i
∂ 2 1 −1 −1
− 2 2 L = − tr(Ω Γ i Ω Γ j )
∂σ ∂σ 2
i j
1 t −1 −1 −1
+ (y − Aµ) Ω Γ i Ω Γ j Ω (y − Aµ)
2
1 t −1 −1 −1
+ (y − Aµ) Ω Γ j Ω Γ i Ω (y − Aµ)
2
1 −1 −1
= − tr(Ω Γ i Ω Γ j )
2
t
+(y − Aµ) Ω −1 Γ i Ω −1 Γ j Ω −1 (y − Aµ).