GLM notes of JKKs lecture I am using I for the incidence matrix, - PDF document

GLM notes of JKK’s lecture I am using I for the incidence matrix, while John used X in lecture. C − 1 2 ln | V | − 1 2( Y − Iη ) T V − 1 ( Y − Iη ) = (1) ℓ   y 11 y 12     y 21   = (2) Y   y 22     y 31   y 32   µ 1 = (3) η µ 2   µ 3  1 0 0  1 0 0     0 1 0   = (4) I   0 1 0     0 0 1   0 0 1  µ 1  µ 1     µ 2   E [ Y ] = Iη = (5)   µ 2     µ 3   µ 3 (6) Diet C T Fam 1 y C 11 y T 11 y C 12 y T 12 2 y C 21 y T 21 y C 22 y T 22 = µ + τ i + F j + C ij + ǫ ijk (7) Y ijk 1

 y C 11  y C 12     y C 21     y C 22   = (8) Y   y T 11     y T 12     y T 21   y T 22   µ η = τ C (9)   τ T σ 2 F + σ 2 C + σ 2 Var[ Y C 11 ] = ν = (10) E Cov[ Y C 11 , Y C 12 ] = c 1 = = Cov[ F 1 + C C 1 + ǫ C 11 , F 1 + C C 1 + ǫ C 12 ] (11) σ 2 F + σ 2 = (12) C Cov[ Y T 11 , Y T 12 ] = Cov[ F 1 + C T 1 + ǫ T 11 , F 1 + C T 1 + ǫ T 12 ] (13) σ 2 F + σ 2 = C = c 1 (14) Cov[ Y C 11 , Y C 21 ] = = Cov[ F 1 + C C 1 + ǫ C 11 , F 2 + C T 2 + ǫ C 22 ] (15) = 0 (16) Cov[ Y C 11 , Y T 11 ] = = Cov[ F 1 + C C 1 + ǫ C 11 , F 1 + C T 1 + ǫ T 12 ] (17) σ 2 = (18) F σ 2 σ 2 0 0 0 0  ν c 1  F F σ 2 σ 2 0 0 0 0 c 1 ν  F F  σ 2 σ 2   0 0 ν c 1 0 0  F F   σ 2 σ 2  0 0 0 0 c 1 ν   F F V = (19)   σ 2 σ 2 0 0 ν c 1 0 0   F F   σ 2 σ 2 0 0 0 0 c 1 ν   F F   σ 2 σ 2 0 0 0 0 ν c 1   F F σ 2 σ 2 0 0 0 0 c 1 ν F F Reordering the order of data could give us a block diagonal. Which is nice because matrix inversion is O ( N 3 ):  y C 11  y C 12     y T 11     y T 12   = (20) Y   y C 21     y C 22     y T 21   y T 22 2

σ 2 σ 2  ν c 1 0 0 0 0  F F σ 2 σ 2 0 0 0 0 c 1 ν F F    σ 2 σ 2  ν c 1 0 0 0 0  F F   σ 2 σ 2  0 0 0 0 c 1 ν  F F  = (21) V   σ 2 σ 2 0 0 0 0 ν c 1   F F   σ 2 σ 2 0 0 0 0 c 1 ν   F F   σ 2 σ 2 0 0 0 0 ν c 1   F F σ 2 σ 2 0 0 0 0 c 1 ν F F 3