Page 241 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

(Continued)
a
Solutions at the end of iteration one Updated estimates after iteration one
Δh ˆ = 0.286869 h ˆ = 0.286869
2 2
Δh = 0.000000 h = 0.000000
1 1
Δh = −0.358323 h = −0.358323
2 2
Δu = −0.041528 u = −0.041528
1 1
Δu = 0.057853 u = 0.057853
2 2
Δu = 0.039850 u = 0.039850
3 3
Δu = −0.065178 u = −0.065178
4 4
a The updated estimates were obtained as the sum of the starting values and the solutions at the end
of the first iteration.
The following steps are involved in calculating P , which is required to calculate
jk
subsequent matrices in Eqn 13.4 for the example data. In each round of iteration and
for each subclass, i.e. for j = 1,... s:
1. Initially calculate (t − a ) in Eqn 13.5 for k = 1,... m − 1. Therefore:
k j
d = (t − a ) = t − x − z for k = 1,... m − 1
jk k j k j j
where x and z are the jth rows of X and Z.
j j
For the example data in the second iteration:
ˆ
d = t − h − h ˆ − uˆ
11 1 1 1 1
d = 0.441008 − 0 − 0 − (−0.041528) = 0.482536
11
ˆ
d = t − h − h ˆ − uˆ
12 2 1 1 1
d = 1.044792 − 0 − 0 − (−0.041528) = 1.086320
12
ˆ
d = t − h − h ˆ − uˆ
21 1 1 2 1
d = 0.441008 − 0 − (−0.358323) − (−0.041528) = 0.840859
21
ˆ
d = t − h − h ˆ − uˆ
22 2 1 2 1
d = 1.044792 − 0 − (−0.358323) − (−0.041528) = 1.444643
22

ˆ
d = t − h − h ˆ − u ˆ
201 1 2 1 4
d = 0.441008 − 0.286869 − 0 − (−0.065178) = 0.219317
201
ˆ
d = t − h − h ˆ − u ˆ
202 2 2 1 4
d = 1.044792 − 0.286869 − 0 − (−0.065178) = 0.823101
202
2. Using the values of d computed above, calculate f (see Eqn 13.6) and F , for
jk jk jk
k = 0,..., m. Note that in all cases, when k = 0, f = F = 0 and when k = m, f = 0
jk jk jk
and F = 1.
jk
In the second round of iteration for the example data:
f = f(0.482536) = 0.355099 and F = F(0.482536) = 0.685288
11 11
f = f(1.086320) = 0.221135 and F = F(1.086320) = 0.861331
12 12
Analysis of Ordered Categorical Traits 225