Page 169 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

ˆ
ˆ
ˆ
ˆ
plus 2. In estimating cov(W , W ) and cov (W , W ), df = (s − 1) + 2 and s(n − 1)
a,ij a,kl e,ij e,kl
+ 2, respectively. The estimated V therefore is:
⎡ 3450.0 2256.4 2184.6 2256.4 1480.3 1434.7 2184.6 1434.7 1390.9⎤
⎢ 2256.4 2959.6 2430.9 2959.6 2903.5 2530.2 2430.9 2530.2 2180.1 ⎥
6
⎢ ⎥
⎢ 2184.6 2430.9 3889.7 2430.9 2249.1 3181.6 38889.7 3181.6 4051.5 ⎥
⎢ 2256.4 2959.6 2430.9 29559.6 2903.5 2530.2 2430.9 2530.2 2180.1 ⎥
⎢
ˆ
V = 1480.3 2903.5 2249.1 2903.5 5711.4 4410.0 2249.1 4410..0 3417.5 ⎥ ⎥
⎢
⎢ 1434.7 2530.2 3181.6 2530.2 4410.0 5818.8 3181.6 5818.8 6354.3⎥
6
⎢ ⎥ ⎥
⎢ 2184.6 2430.9 3889.7 2430.9 2249.1 3181.6 3889.7 3181.6 4051.5 ⎥
⎢ ⎢ 1434.7 2530.2 31881.6 2530.2 4410.0 58818.8 3181.6 5818.8 6354.3⎥
⎢ ⎣ 1390.9 2180.1 4051.5 2180.1 3417.5 6354.3 4051.5 6354.3 11835.0 ⎥ ⎦
˘
However, the symmetry of G resulted in redundancies in the vector g˘ such that
V is singular. The vector g˘ can be redefined to be of the order s by 1, which contains
˘
only the elements in the lower half of G, where s = t(t + 1)/2. Therefore, delete from
˘
˘
g˘ the elements G for which i < j. Thus for the example G, the vector g˘ becomes:
ij
g˘ = [132.3 127.0 136.6 172.8 200.8 288.0]
˘
Then delete from V those columns and rows corresponding to elements G with i < j.
ij
This involves deleting rows and columns 4, 7 and 8 from the matrix V given above.
The V of reduced order (s by s) is:
é 3450.0 2256.4 2184.6 1480.3 1434.7 1390.9ù
ê ú
9
ê 2256.4 2959.6 2430.9 2903.5 2530.2 2180.1 ú
ê 2184.6 2430.9 3889.7 2249.1 3181.6 4051.5ú
ˆ
V = ê ú
ê 14880.3 2903.5 2249.1 5711.4 4410.0 3417.5 ú
ê ê 1434.7 2530.2 3181.6 4410.00 5818.8 6354.3 ú
ê ú
ë ê 1390.9 2180.1 4051.5 3417.5 6354.3 11835.0ú û
˘
Similarly, the rows corresponding to those elements of g for which G has i < j
˘
ij
are deleted from X . In the example X , rows 4, 7 and 8 are deleted. Thus X becomes:
s s s
é 0.5000 - 0.8660 - 0.8660 1.4999ù
ê 0.5000 - - ú
ê 0.0577 0.8660 0.0999 ú ú
ê 0..5000 0.8660 - 0.8660 - 1.4999ú
X s = ê ú
ê 0.5000 - 0.0577 - 0.0577 0.0067 ú
ê 0.5000 0 0.8660 - 0.0577 - 0.0999 ú
ê ú
ë ê 0.5000 0.8660 0.8660 1.4999ú û
ˇ
Also, for each element of cˇ for which C has i < j, add the corresponding column
ij
ˇ
of X to the column corresponding to C , then delete the former column. For the beef
s ji
ˇ
ˇ
ˇ
cattle example, the vector of coefficients, c′ = [C 00 C ˇ 10 C 01 C ]. Therefore, the third
ˇ
11
ˇ
column of X corresponding to C is added to the second column and the third col-
s 01
umn is deleted. The matrix X then becomes:
s
Analysis of Longitudinal Data 153