Page 160 - Applied Probability
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145
                                                                     8. The Polygenic Model
                                        ∂
                              Note that
                                          µ and
                                       ∂µ i
                                Since E(Y )= Aµ, the expected information matrix has entries
                                          2
                                                             t
                                                        ∂
                                         ∂
                                                                      ∂



                                                                 −1
                                                               t
                                                                        µ
                                                                   A
                                                  =
                                                          µ A Ω
                                 E −
                                                                     ∂µ i
                                                       ∂µ j
                                       ∂µ i ∂µ j  L   ∂σ ∂ i 2 Ω=Γ i are treated as constants in these derivations.
                                         ∂              ∂      t  −1  −1
                                          2                  t
                                 E −     2   L    =       µ A Ω    Γ i Ω  [E(Y ) − Aµ]
                                      ∂σ ∂µ j          ∂µ j
                                         i
                                                  = 0                                      (8.3)
                                         ∂             1    −1    −1         −1    −1   −1
                                          2
                                 E −     2  2  L  = − tr(Ω     Γ i Ω  Γ j ) + tr(Ω  Γ i Ω  Γ j Ω  Ω)
                                      ∂σ ∂σ            2
                                         i  j
                                                      1    −1   −1
                                                  =    tr(Ω  Γ i Ω  Γ j ).
                                                      2
                                Some simplification in the above formulas can be achieved by defining
                              the partial score d µ L =(  ∂  L, ... ,  ∂  L) and the corresponding partial
                                                      ∂µ 1     ∂µ p
                                                                     ∂
                                                           2
                                                                              ∂
                              observed information matrix −d L. Since (  ∂µ 1  µ,...,  ∂µ p  µ) is the identity
                                                           µ
                              matrix I,
                                                               t
                                                    d µ L t  = A Ω −1 (y − Aµ)
                                                     2
                                                               t
                                                   −d L = A Ω    −1 A.
                                                     µ
                              The expected information matrix J evidently has the block diagonal form
                                                             2
                                                         E[−d L]      0
                                                J   =        µ         2     ,
                                                            0     E[−d 2 L]
                                                                       σ
                                                                                   2
                                       2
                                                                              2
                                                                                          2 t
                              where −d 2 L is the observed information matrix on σ =(σ ,...,σ ) .
                                       σ                                           1      r
                                Since the block form of J is retained under matrix inversion, the current
                              µ is updated in the scoring algorithm by
                                           t
                                                     t
                                                                                   t
                                                                         t
                                     µ +(A Ω −1 A) −1 A Ω −1 (y − Aµ)  = (A Ω −1 A) −1 A Ω −1 y.  (8.4)
                                                                          t
                                                                                     t
                              If there is more than one pedigree, the quantities A Ω −1 A and A Ω −1 y must
                              be summed over all pedigrees before matrix inversion and multiplication.
                                                     2
                                                           2
                              The scoring increment ∆σ to σ is similarly expressed as
                                                                          t
                                                                2
                                                  ∆σ 2  =E(−d 2L)   −1 d σ L .
                                                                        2
                                                                σ
                                                          2
                                                                        t
                                It is convenient to initialize σ at (0,... , 0, 1) . Since Ω = I in this case,
                              the first iteration of the scoring algorithm (8.4) produces the standard lin-
                              ear regression estimate of µ. Of course, this estimate does not take into
                              account the correlational architecture of the pedigree. In those unlikely cir-
                                                                                2
                              cumstances when µ is known exactly, the initial value σ =(0,... , 0, 1) t
                                                                2
                              leads to a least-squares estimate of σ after a single iteration of scoring.
                              (See Problem 4.) Computation of the score dL and expected information
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