Page 282 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition
P. 282

−1
                                        2
                                     −1
            ˆ
            u  = 0  and  (z′ z  + A −1  a) s  = (3.667) 40 = 10.908
             1            1 1    1,1    e
                             −1
                                 −1
         The value of (z′ z  + A a)  is taken from the diagonal element of the coefficient
                      1 1    1,1
         matrix of the MME for Example 3.1. Assuming the RN from N(0,1) is 0.2067:
              1 []
            u =+     .       .    0 683
                 0 0 2067 10 98 =
                                    .
             1
                                                                −1
                                                                         −1
                                                                   2
                                                             −1
            For animal 2, u  from Eqn 3.8 = −0.171, and (z′ z  + A a) s  = (4) 40 = 10.
                         ˆ
                          2                           1 1    1,1   e
         Then from Eqn 16.12, assuming RN from N(0,1) is −1.8025:
              1 []
            u =-  0 171+ - 1 8025 10 = - 5 871
                   .
                                        .
                           .
             2
                                                                     −1
                                                                2
                                                         −1
                                                             −1
         Similarly, given that u  from Eqn 3.8 = 1.468, (z′ z  + A a) s  = (4) 40 = 10 and
                           ˆ
                            3                      1 1   1,1    e
         RN = −0.5558, then:
              1 []
            u =  1 468 0 5558 10 = - 0 290
                      -
                                      .
                         .
                  .
             3
                                                                   −1
                                                       −1
                                                   −1
                     ˆ
                                                          2
         For animal 4, u  = 0.0976 from Eqn 3.8, (z′ z  + A a) s  = (4.667) 40 = 8.571 and
                      4                      1 1   1,1    e
         RN = −1.8654, then:
              1 []
                       -
            u =  0 0976 1 8654 8 571 = - 5 364
                  .
                                          .
                          .
                                 .
             4
                                                           [1]
                                                               [1]
         Similar calculations using Eqn 16.12 gave estimates of u , u , u  and u  to be
                                                                           [1]
                                                                   [1]
                                                           5   6   7       8
         −3.097, −2.577, −1.621 and 0.697, respectively.
            The vector of residuals, eˆ = y −Xb − Zu, is:
             ˆ e æ  4 ö  æ  . 45ö æ 4 955 ö æ -5 364 ö  æ 4 908ö
                           .
                                     .
                                              .
            ç  ÷  ç   ÷ ç      ÷ ç       ÷  ç     ÷
                           .
                                              .
            ç  ˆ e 5 ÷  ç  . 29 ÷ ÷ ç 2 821 ÷ ç -3.0097 ÷  ç 3 176 ÷
                                  ç
             ˆ e ç  ÷ = ç  . 39 ÷ - ç 2 821 ÷ - - 2 577 ÷ =.  ç 3 656 ÷
                           .
                                              .
            ç  6  ÷  ç  ÷ ç    ÷ ç       ÷  ç     ÷
                                    1 621
             ˆ e ç  7 ÷  ç  . 35 ÷ ç 4 955 ÷ ç - .  ÷  ç 01655 ÷
                                              .
                           .
            ç  ÷  ç   ÷ ç      ÷ ç       ÷  ç     ÷
                                              .
                                    0 697
                           .
            è  ˆ e 8 ø  è  . 50 ø è 4 955 ø è - .  ø  è 0 742 ø
                                                   2
              ˆ
             ˆ
         and e′e = 48.118. Sampling from the inverted χ  distribution with three degrees of
         freedom (Eqn 16.13) gave an estimate of 39.870 for the residual variance.
                                                                 2
                                        2
            Using Eqn 16.14, sampling for s  is again from the inverted χ  distribution, with
                                        u
            −1
         u′A u = 93.11 and degrees of freedom being 6. An estimate of 23.913 was obtained
                                                   −1
                                                                             −1
              2
         for s . Note that it is easier to compute  u′A u using Eqn 2.3. Thus  u′A u =
              u
                −1
            −1
                                        −1
                   −1
         u′(T )′D T u = m′Dm where m = T u, with m being a vector of Mendelian sampling
         for animals calculated using Eqn 2.2.
            The next round of iteration is then commenced using the updated values com-
         puted for the parameters.
         16.3   Multivariate Animal Model
         In this section, the Gibbs sampling algorithm developed by Jensen et al. (1994) for models
         with maternal genetic effects is generalized for a multivariate situation. Given that ani-
         mals are ordered within traits, the multivariate model for two traits could be written as:
            æ y ö  æ X 1  0 öæ b ö æ Z 1  0 öæ  u ö æ e ö
              1
                                                 +
                                               1
                               1
                                 +
                                                     1 1
            ç  ÷  =  ç      ÷ç  ÷ ç         ÷ç  ÷ ç   ÷
            è y 2 ø  è 0  X 2 øè b 2 ø è 0  Z 2 øè  u 2 ø è e 2 ø
          266                                                            Chapter 16
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