Page 158 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

9.3.2 Partitioning animal solutions from random regression model

Equations for calculating the contribution of information from various sources to the
solutions (random regression coefficients) of an animal from an RRM were presented
by Mrode and Swanson (2002). These equations are the same as those presented in
Section 5.2.3 for the multivariate model. Test day records of cows contribute to ran-
dom regressions for the animal effect through the yield deviations. The calculation of
the vector of yield deviations (YD) is first examined. Using the same argument for
deriving Eqn 5.7, the equation for YD for an RRM is:

ˆ
−1
−1
−1
YD = (Q′R Q) ( Q′R ( y − Xb − Zpeˆ )) (9.3)
While this equation is similar to Eqn 5.6 for yield deviation under a multivariate
model, here YD is a vector of weighted regressions of the animal’s TD yields adjusted
for all effects other than additive genetic effect, on orthogonal polynomials for DIM.
Since YD is a vector of regressions, it can be used to generate actual yield deviations
for any DIM using Eqn 9.2. Thus actual yield deviation (yd*) for day m, for instance,
equals v′YD, where v is a vector of order nr with v = f and j = 1,nr. The actual
m mj
yield deviation for 305-day yield can be calculated using Eqn 9.2 but with û replaced
with YD.
The calculation of YD for cow 6 in Example 9.2 is illustrated below. First, the
vector of TD records for cow 6 corrected for all effects (y ) other than the additive
c
genetic effect is:
ˆ
ˆ
y = y − X b − X b − p ˆ e
c 6 1 1 2 2
⎡ 10.4⎤ ⎡ 3.0101⎤ ⎡ 10.7725⎤ ⎡ − 2.7251⎤ ⎡ −0.6576⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ ⎥
⎢ 12.3 ⎥ ⎢ 3.1085 ⎥ ⎢ 12.5295 ⎥ ⎢ 2.3920 ⎥ ⎢ −0.9460 ⎥
⎢
y = 13.2⎥ − ⎢ 3.1718⎥ − ⎢ 12.7890⎥ − − ⎢ 2.0752⎥ = ⎢ −0.6856⎥
8
c ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ 11.6 ⎥ ⎢ 0.5044 ⎥ ⎢ 12.34454 ⎥ ⎢ − 1.7746 ⎥ ⎢ 0.5249 ⎥
⎢ ⎣ 8.4 ⎥ ⎦ ⎢ ⎣ 0.0000 ⎥ ⎦ ⎢ ⎣ 11.7641 ⎥ ⎦ ⎢ ⎣ − −1.4904 ⎥ ⎢ −1.87738 ⎥ ⎦
⎦ ⎣
ˆ
ˆ
where b and b are vectors of solutions for HTD and fixed regression coefficients.
1 2
−1
−1
The matrices Q′R Q and Q′R y are:
c
⎡ 0.6738 − 0.6484 0.1674⎤ ⎡− 0.6934⎤
⎢
Q R Q′ −1 = 0.6484 0.8235 − 0.5906 ⎥ ⎥ and Q R y c = ⎢ ⎢ 0.5967 ⎥ ⎥
−
′
−1
⎢
⎢ ⎣ 0.1674 − 0.59906 1.0177⎥ ⎦ ⎣ − ⎢ ⎢ 0.1237⎥ ⎦
Using Eqn 9.3, yield deviation for cow 6 (YD ) is:
6
é -5.0 004ù
YD = ( ′ - 1 -1 Q R y = ′ - 1 ê - . 4 6419 ú
Q R Q)
6 c ê ú
ê ë - . 1 9931ú û
142 Chapter 9