Derivation of the estimating equations for the CBS-estimator

A. Derivation of the estimating equations for the CBS-estimator

Using ∂ ∂θ j log |Σ| = tr Σ −1 ∂Σ ∂θ j and ∂ ∂θ j Σ −1 = −Σ −1 ∂Σ ∂θ j Σ −1 with θ j any element of θ, the vector of the FA parameters, as well as ∂ ∂γ j Σ = ∂Γ ∂γ j Γ T +Γ ∂Γ T ∂γ j the partial derivatives of the Lagrangian 7 are then ∂ ∂µ L = −λ 1 n P n i=1 u d i ; cΣ −1 x i − µ = 0, 18 ∂ ∂ψ 2 j L = tr[Σ −1 Z j ] − λ 2n P n i=1 u d i ; cx i − µ T Σ −1 Z j Σ −1 x i − µ = 0, 19 with Z j = ∂ ∂ψ 2 j Σ = ∂ ∂ψ 2 j ΓΓ T + Ψ, and ∂ ∂γ j L =tr Σ −1 ∂ ∂γ j Γ Γ T + Σ −1 Γ ∂ ∂γ j Γ T − λ 2n n X i=1 u d i ; cx i − µ T Σ −1 ∂ ∂γ j Γ Γ T + Γ ∂ ∂γ j Γ T Σ −1 x i − µ = 0. 20 ¿From 18 we get the estimating equation for µ given in 6. Then, multiplying 19 by ψ 2 j and using the properties of the trace we get tr[Σ −1 Z j ψ 2 j ] − λ 2n n X i=1 u d i ; cx i − µ T Σ −1 Z j ψ 2 j Σ −1 x i − µ = 0, r X j=1 tr[Σ −1 Z j ψ 2 j ] − λ 2n r X j=1 n X i=1 u d i ; cx i − µ T Σ −1 Z j ψ 2 j Σ −1 x i − µ = 0, tr[Σ −1 r X j=1 Z j ψ 2 j ] − λ 2n n X i=1 u d i ; cx i − µ T Σ −1 r X j=1 Z j ψ 2 j Σ −1 x i − µ = 0, tr[Σ −1 Ψ] − λ 2n n X i=1 u d i ; cx i − µ T Σ −1 ΨΣ −1 x i − µ = 0. 21 13 Similarly for 20 we get tr Σ −1 ∂Γ ∂γ j γ j Γ T + Σ −1 Γ ∂Γ T ∂γ j γ j − λ 2n n X i=1 u d i ; cx i − µ T Σ −1 ∂Γ ∂γ j γ j Γ T + Γ ∂Γ T ∂γ j γ j Σ −1 x i − µ = 0, q X j=1 tr Σ −1 ∂Γ ∂γ j γ j Γ T + Σ −1 Γ ∂Γ T ∂γ j γ j − λ 2n q X j=1 n X i=1 u d i ; cx i − µ T Σ −1 ∂Γ ∂γ j γ j Γ T + Γ ∂Γ T ∂γ j γ j Σ −1 x i − µ = 0, tr         Σ −1         q X j=1 ∂Γ ∂γ j γ j         Γ T + Σ −1 Γ         q X j=1 ∂Γ T ∂γ j γ j                 − λ 2n n X i=1 u d i ; cx i − µ T Σ −1                 q X j=1 ∂Γ ∂γ j γ j         Γ T + Γ         q X j=1 ∂Γ T ∂γ j γ j                 Σ −1 x i − µ = 0, 2tr h Σ −1 ΓΓ T i − λ 2n n X i=1 u d i ; cx i − µ T Σ −1 2ΓΓ T Σ −1 x i − µ = 0. 22 Multiplying 21 by 2 and adding the result to 22 we obtain 2tr[Σ −1 Ψ] + 2tr[Σ −1 ΓΓ T ] − λ n n X i=1 u d i ; cx i − µ T Σ −1 ΨΣ −1 x i − µ− λ n n X i=1 u d i ; cx i − µ T Σ −1 ΓΓ T Σ −1 x i − µ = 0, 2tr[Σ −1 Ψ + Σ −1 ΓΓ T ] − λ n n X i=1 u d i ; cx i − µ T Σ −1 Ψ + ΓΓ T Σ −1 x i − µ = 0, 2tr h Σ −1 Ψ + ΓΓ T i − λ n n X i=1 u d i ; cx i − µ T Σ −1 ΣΣ −1 x i − µ = 0, 2p − λ n n X i=1 u d i ; cd 2 i = 0, so that we finally get λ = 2p h 1 n P n i=1 u d i ; cd 2 i i −1 . Using this λ in 19 and 20 respectively yields tr[Σ −1 Z j ] = p n        1 n n X i=1 u d i ; cd 2 i        −1        n X i=1 u d i ; cx i − µ T Σ −1 Z j Σ −1 x i − µ        , 23 14 and tr Σ −1 ∂Γ ∂γ j Γ T + Σ −1 Γ ∂Γ T ∂γ j = p n        1 n n X i=1 u d i ; cd 2 i        −1 ,        n X i=1 u d i ; cx i − µ T Σ −1 ∂Γ ∂γ j Γ T + Γ ∂Γ T ∂γ j Σ −1 x i − µ        . 24 23 and 24 are the estimating equations for the parameters γ j and ψ j . To define an iterative system, we need now to extract ψ j and γ j from 23 and 24. First, we use Ψ = P r j=1 Z j ψ 2 j to reformulate the left hand side of 23 as tr[Σ −1 Z j ] = tr[Σ −1 Z j Ψ −1 Ψ] = tr[Σ −1 Z j Ψ −1 r X k=1 Z k ψ 2 k ], = r X k=1 tr[Σ −1 Z j Ψ −1 Z k ]ψ 2 k , 25 which defines an iterative system for each ψ 2 k , k = 1, . . . , r. Let C = h tr[Σ −1 Z j Ψ −1 Z k ] i j,k=1,...,r , 26 and D =        n X i=1 u d i ; cx i − µ T Σ −1 Z j Σ −1 x i − µ        j=1,...,r , 27 an expression for the ψ 2 j is given by P = p n h 1 n n X i=1 u d i ; cd 2 i i −1 C −1

D. 28