Page 212 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition
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ˆ e æ  1 ö  æ  9.0 ö æ 9.888ö æ 0.45ö  æ -1.388 ö
            ç  ÷  ç     ÷ ç     ÷ ç     ÷  ç       ÷
            ç  ˆ e 2 ÷  ç 13.4 ÷ ç 9.888 ÷ ç 0.30 ÷  ç  3.213 ÷
                               8
              ˆ e ç  ÷  ç 1 12.7 ÷ ç 9.888÷ ç 0.55÷  ç 2.263÷
            ç  3 ÷  ç   ÷ ç     ÷ ç     ÷  ç       ÷
              ˆ e ç  4 ÷  ç 15.4 ÷ ç 9.888 ÷ ç 0.45 ÷  ç  5.063 ÷
                                 - -
                         -
            ç  ÷  = ç   ÷ ç     ÷ ç     ÷ = ç      ÷
            ç  ˆ e 5 ÷  ç  5.9 ÷ ç 9.888 ÷ ç 0.45 ÷  ç  -4.438 ÷
              ˆ e ç  ÷  ç  7.7 ÷ ç 9.888÷ ç 0.50 ÷  ç ç  -2.688 ÷
            ç  6 ÷  ç   ÷ ç     ÷ ç     ÷  ç       ÷
            ç  ˆ e 7 ÷  ç 10.2 ÷ ç 9.888 ÷ ç 0.40 ÷ ÷  ç  -0.088 ÷
            ç  ÷  ç     ÷ ç     ÷ ç     ÷  ç       ÷
            è  ˆ e 8 ø  è  4.8  ø è 9.888 ø è 0.35 ø  è  -5.437 ø
                        ˆ
         From the above, e′eˆ = 99.345, and thus given the value of 8.131 sampled from the
                 2                                       2[1]
                                                         e
         inverted χ  distribution with n – 2 degrees of freedom, s   = 99.345/8.131 = 12.218,
         using Eqn 11.21. The superscript in brackets denotes the iteration number.
                                                ˆ
                                                                  −1
                         [1]
            Then sample b  using Eqn 11.20, with b  calculated as (x ′x) 1′eˆ = 9.456 after
                                                               j
         initially updating eˆ, the vector of residuals to include information on b as:
                     ˆ
            e = eˆ  + Xb with i = 1, n
            ˆ
             i  i
            Assuming the random number generated from a normal distribution is 0.873 and
                2
             −1
         (x ′x) s  = 12.218/8 = 1.527, then b  [1]  = (9.456 + 0.873 (.527 ) = 10.535.
                                                            1
          j     e                        1
            After sampling for b, the ê is updated to exclude the information on b as:
                     ˆ
            e = eˆ  − Xb with i = 1, n
            ˆ
             i  i
                          2
                                                                         2
         Using Eqn 11.22, s  for the ith SNP effect is sampled from the inverted χ  distribu-
                          gi
         tion with degrees of freedom 5.012 and S = 0.352 computed earlier. For the first SNP,
          2
         g  = 0.003, thus given the value of 11.422 sampled from the inverted χ  distribution
                                                                      2
         ˆ
          1
                    ˆ
                     2
          2 [1]  = (S + g )/11.422 = 0.031. The variance estimates for other SNPs in the first
         s g1
         iteration are shown in Table 11.5.
            Finally, estimates of gˆ are sampled from the normal distribution using Eqn 11.23.
         First update the vector of residuals to include information on the jth SNP. Thus for
         the jth SNP effect:
         Table 11.5. SNP solutions and variances from BayesA and BayesB.
                            BayesA                             BayesB
                  First iteration  Posterior means  First iteration  Posterior means
         SNP    Effects   Var    Effects   Var     Effects   Var    Effects   Var
          1      0.289   0.031    0.018   0.170     2.187   1.105    0.038   0.316
          2      0.279   0.049   −0.064   0.179    −1.565   0.516   −0.107   0.319
          3     −0.010   0.070    0.058   0.179    −0.156   0.124    0.067   0.293
          4      0.023   0.097   −0.023   0.176    −0.309   0.118   −0.034   0.300
          5      0.045   0.052    0.022   0.167     0.413   0.363    0.047   0.328
          6     −0.321   0.050    0.025   0.171    −0.521   0.161    0.031   0.283
          7      0.411   0.256   −0.006   0.186     0.000   0.000    0.009   0.335
          8      0.408   0.056   −0.008   0.168    −0.010   0.431    0.008   0.261
          9      0.115   0.034   −0.003   0.162     0.000   0.000   −0.006   0.294
         10     −0.578   0.152   −0.008   0.165     0.000   0.000   −0.017   0.286
         Var, SNP variances.
          196                                                            Chapter 11
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