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which is equivalent to the selection index. Thus BLUP is the selection index with the
         GLS solution of b substituted for b.

                           ˆ
         C.2   Proof that b and a ˆ from MME are the GLS
         of b and BLUP of a, Respectively

         In computation terms, the use of Eqn c.3 to obtain the BLUP of k′b + a is not feasible
         because the inverse of V is required. Henderson (1950) formulated the MME that are
                                                                     ˆ
         suitable for calculating solutions for b and a, and showed later that k′b and a, where
                                                                          ˆ
         ˆ
         b and aˆ are solutions from the MME, are the best linear unbiased estimator (BLUE)
         of k′b and BLUP of a, respectively.
            The usual MME for Eqn c.1 are:
            ⎡ X R X       X R Z′⎤ ⎡ b⎤ ⎡ X R y ⎤                              (c.5)
                           ′
               ′
                                         ′
                                   ˆ
                              −1
                 −1
                                           −1
                                      =
            ⎢    −1      −1     −1⎥ ⎢ ⎥ ⎢  −1 ⎥
                       ′
               ′
                                   a ˆ
                            +
                                          R
            ⎣ Z R X Z R Z G ⎦ ⎣ ⎦ ⎣     Z′R y ⎦
                                                                      ˆ
                          ˆ
            The proof that b from the MME is the GLS of b and therefore k′b is the BLUE
         of k′b was given by Henderson et al. (1959). From the second row of Eqn c.5:
                                       ˆ
                −1
                               −1
                       −1
            (Z′R Z + G )aˆ = Z′R (y − Xb)
                                        ˆ
                    −1
                          −1 −1
                                 −1
            â = (Z′R Z + G ) Z′R (y − Xb)                                     (c.6)
         From the first row of Eqn c.5:
                         −1
                −1
            X′R Xb + Z′R Zâ = X′R y
                                   −1
         Substituting the solution for aˆ into the above equation gives:
                                                  −1
                          −1
                −1
                                  −1
            X′R Xb + X′R Z(WZ′R )(y − Xb) = X′R y
                              −1 −1
                       −1
         where W = (Z′R Z + G ) :
                          −1
                                    −1
                                                       −1
                                                               −1
                                              −1
                −1
            X′R Xb − (X′R Z)(WZ′R )Xb = X′R y − X′R ZWZ′R y
                              −1
                −1
                                                      −1
                                              −1
                                         −1
                      −1
            X′(R  − R ZWZ′R )Xb = X′(R  − R ZWZ′R )y
                −1
                          −1
            X′V Xb = X′V y
               −1
                                 −1
                    −1
                         −1
         with V  = R  – R ZWZ′R :
            ˆ
                            −1
                    −1
                       −1
            b = (X′V X) X′V y                                                 (c.7)
         It can be shown that:
                        −1
              −1
                   −1
            V  = R  − R ZWZ′R   −1
         by pre-multiplying the right-hand side by  V and obtaining an identity matrix
         (Henderson et al., 1959):
                     −1
                                              −1
                                                           −1
                                                   −1
                −1
                             −1
            V[R  − R ZWZ′R ] = (R + ZGZ′)(R  − R ZWZ′R )
                                           −1
                                                                −1
                                                      −1
                                = I + ZGZ′R  − ZWZ′R  − ZGZ′R ZWZ′R     −1
                                = I + ZGZ′R  − Z(I + GZ′RZ)WZ′R −1
                                           −1
                                           −1
                                                    −1
                                = I + ZGZ′R  − ZG(G  + Z′RZ)WZ′R   −1
                                = I + ZGZ′R  − ZG(W )WZ′R   −1
                                           −1
                                                    −1
                                = I + ZGZ′R  − ZGZ′R −1
                                           −1
                                = I
          312                                                            Appendix C
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