lecture 12 blup prediction - university of...
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Lecture 12 1
BLUP Best Linear Unbiased Prediction-Estimation
References
Searle, S.R. 1971 Linear Models, Wiley
Schaefer, L.R., Linear Models and Computer Strategies in Animal Breeding
Lynch and Walsh Chapter 26
Lecture 12 2
OLS Independently and Identically Distributed Errors with Mean 0 and variance σ2
[ ]
2
2
2
2
2
)(
000
000000
)(
)()(
e
e
e
e
V
V
EEV
σε
σ
σσ
ε
εεε
I=
=
−=
MOMM
The residual distribution from which each observation is sampled is the same
Residuals independent
Inbreeding Changes These
Related Individuals Cause these to be nonzero
Lecture 12 3
Solutions
• GLS– Fixes problem with changing variances and
correlations in the data• What about fixed effects?
– How does one correct for • Environmental trend without a control• Herd effects• Year effects• Hatch effects
Lecture 12 4
Confounding of data• Herd effects
– Balanced design no problem– Require sample of every family in every herd– Old solution was within herd deviations– What if better herds have better genetics
• Fixed effects must be adjusted for genetic differences
• Random effects must be adjusted for fixed effects
Lecture 12 5
Simultaneous Adjustment of Fixed and Random effects
• Separate Independent variable into those that are – Fixed Xb– Random Zu
eZuXbY ++=
Lecture 12 6
Fixed and Random Effects
• Fixed Effect– Inference Space only to those levels– Herd, Year, Season, Parity, and Sex effect
• Random Effect– Effect Sampled From A Distribution Of Effects– Inference Space To The Population From
Which The Random Effect Was Sampled
Lecture 12 7
Random Effect
Gametes
GoodBadSample
Inference is to the genetic worth of the bull
Lecture 12 8
Variances In Mixed Models
eZuXbY ++=
Ree'eGuu'u
0b
====
=
)()()()(
)(
EVEV
V
RZGZ'eZuXbY +=++= )()( VV
Lecture 12 9
Example 11 2 3
4 5
(7) (9) (10)
(6) (9)
=
961097
Y
=
11111
X
=
1000001000001000001000001
Z
=
5
4
3
2
1
aaaaa
U
[ ]µ=b
=
5
4
3
2
1
eeeee
e
Lecture 12 10
Example 11 2 3
4 5
(7) (9) (10)
(6) (9)
[ ]
+
+
=
5
4
3
2
1
5
4
3
2
1
1000001000001000001000001
11111
961097
eeeee
aaaaa
µ
eZuXbY ++=
Lecture 12 11
ML Derivation of BLUP)()()( uy/uuy, hgf =Joint density of y and u
)()( ey/u gg =
( )
e(e)e'e1
21
212
1
)(2
1)(−−= V
eVeg
Nπ ( )
u(u)u'
uu
121
212
1
)(2
1)(−−= V
Veh
Nπ
( ) ( )
uGu'eRe'
uGu'eRe'
uGu'
G
eRe'
R
uy,
uy,
uy,
1211
21
1211
21
121
212
1
121
212
1
)(
)(
)(
21
2
1
2
1
−−
−−
−−
−−
−−
−−
=
=
=
ecef
ececf
eefNN
ππ
Lecture 12 12
Maximize w.r.t b and u
uGu'eRe'uy,1
211
21
)(−− −−== eceLf
uGu'eRe' 1211
21)ln()ln( −− −−= cL
ZuXbYe −−=
( ) ( )ZuXbYRZuXbY −−−−−= −1'21)ln()ln( cL
uGu' 121 −−
Lecture 12 13
( ) ( )( ) ( )[ ] ( ) uGu'ZuXbYRZuXbY
uGu'ZuXbYRZuXbY11'''
11'
−−
−−
+−−−−=
+−−−−
Take Derivative w.r.t b
( ) 0ln=
∂∂bL ( )
( ) 01'1'1'
1'1'1'
=++
++−−−−−
−−−
XRZuZuRXXbRX
XRXbYRXXRY
0222 1'1'1' =++− −−− ZuRXXbRXYRX
YRXZuRXXbRX 1'1'1' −−− =+
( ) ( ) ( )( ) ( ) ( ) uGu'ZuRZuXbRZuYRZu
ZuRXbXbRXbYRXb
ZuRYXbRYYRY
11'1'1'
1'1'1'
1'1'1'
−−−−
−−−
−−−
+++−
++−
−−=
Lecture 12 14
( ) 0ln=
∂∂uL ( ) ( ) ( )
( ) ( ) 02 11'1'
1'1'1'1'
=+++
+−+−−−−
−−−−
uGZRZuZuRZ
XbRZYRZZRXbZRY
02222 11'1'1' =+++− −−−− uGZuRZXbRZZRY
ZRYuGZuRZXbRZ 1'11'1' −−−− =++
( ) ( ) uGu'ZuXbYRZuXbY 11' −− +−−−−
( ) ( ) ( )( ) ( ) ( ) uGu'ZuRZuXbRZuYRZu
ZuRXbXbRXbYRXb
ZuRYXbRYYRY
11'1'1'
1'1'1'
1'1'1'
−−−−
−−−
−−−
+++−
++−
−−=
Take Derivative w.r.t u
Lecture 12 15
Mixed Model EquationsYRXZuRXXbRX 1'1'1' −−− =+
=
+ −
−
−−−
−−
ZRYYRX
ub
GZRZXRZZRXXRX
1'
1'
11'1'
1'1'
ZRYuGZuRZXbRZ 1'11'1' −−−− =++
Simplifications If2eσIR =
=
+ − ZYYX
ub
GZZXZZXXX
'
'
12''
''
eσ
Lecture 12 16
Normal Distribution of Y Not Necessary
It is possible to show with alternative BLUE estimation techniques that the
Same Results would be obtained without assuming normality
See Schaffer Notes
Lecture 12 17
Simplifications
2aσAG =
=
+ − ZYYX
ub
GZZXZZXXX
'
'
12''
''
eσ
Assuming Additivity
=
+ − ZYYX
ub
AZZXZZXXX
'
'
1''
''
2
2
a
e
σσ
Only Estimate of Ratio is Needed Only inverse is needed
Lecture 12 18
Example 11 2 3
4 5
(7) (9) (10)
(6) (9)
=
961097
Y
=
11111
X
=
1000001000001000001000001
Z
=
5
4
3
2
1
aaaaa
u
[ ]µ=b
=
5
4
3
2
1
eeeee
e
Lecture 12 19
=
+ − ZYYX
ub
AZZXZZXXX
'
'
1''
''
2
2
a
e
σσ
[ ]
5'11111
11111'
=
=
XX
XX [ ]
[ ]111111000001000001000001000001
11111
=
=
ZX'
ZX'
Lecture 12 20
=
+ − ZYYX
ub
AZZXZZXXX
'
'
1''
''
2
2
a
e
σσ
=
=
11111
'
11111
1000001000001000001000001
'
XZ
XZ
=
1000001000001000001000001
'ZZ
Lecture 12 21
1 2 3
4 5
(7) (9) (10)
(6) (9)
=
10100100
0100001
41
21
21
41
21
21
2121
2121
A
Assume heritability=.5
12
2
=a
e
σσ
−−−−
−−−
−
=+ −
3011003011100113010
25
21
21
21
21
25
1'2
2
AZZa
e
σσ
Lecture 12 22
[ ]
[ ]41961097
11111
=
=
YX'
YX'
=
961097
ZY'
=
+ − ZYYX
ub
AZZXZZXXX
'
'
1''
''
2
2
a
e
σσ
Lecture 12 23
MME
=
−−−−
−−−
−
96109741
301101030111100111310101111115
5
4
3
2
1
25
21
21
21
21
25
aaaaaµ
Lecture 12 24
Estimation of Error Variance if the Ratio is Known
2
2
a
e
σσλ =
−−−
==)(ˆˆ
ˆ 2XRN
MSEeZuXbYY'σ
λσσ2
2 ˆˆ ea =
Lecture 12 25
proc iml;start main;
y={ 7,9,
10,6,9};
X={1,1, 1,1,1};
A={1 0 0 .5 0,0 1 0 .5 .5,0 0 1 0 .5,.5 .5 0 1 .25,0 .5 .5 .25 1};
lam=1;
Z={1 0 0 0 0,0 1 0 0 0,0 0 1 0 0,0 0 0 1 0,0 0 0 0 1};
LHS=((X`*X)||(X`*Z))//((Z`*X)||(Z`*Z+INV(A)#LAM));
RHS=(X`*Y)//(Z`*Y);C=INV(HS);
BU=C*RHS;print C BU;
finish main;run;quit;
Lecture 12 26
Estimates
BU
8.3018868-0.9608130.07547170.8853411
-1.0624090.5529753
=
5
4
3
2
1
ˆˆˆˆˆ
aaaaa
U
[ ]µ̂=b
Lecture 12 27
Variance of the Estimates1
1''
''
2221
1211
2
2
−
−
+=
AZZXZ
ZXXXCCCC
a
e
σσ
222
211
)ˆ(
)ˆ(
e
e
CVV
σ
σ
=−
=
uuCb
Prediction Error Variance
222
2)ˆ( eaV σσ CAu += Prediction Error Variance Including Drift Variance
Kennedy and Sorensen Quantitative Genetics
Lecture 12 28
1.12236 0.29509 0.32030 0.65827 0.390930.29509 1.14758 0.29509 0.68854 0.688540.32030 0.29509 1.12236 0.39093 0.658270.65827 0.68854 0.39093 1.2686 0.600260.39093 0.68854 0.65827 0.60026 1.2686
2.86014 0.29509 0.32030 1.52716 0.390930.29509 2.88536 0.29509 1.5574 1.5574
0.32030 0.29509 2.86014 0.39093 1.527161.52716 1.5574 0.39093 3.00643 1.034710.39093 1.5574 1.52716 1.03471 3.00643
PEV
EV
Lecture 12 29
Example 11 2 3
4 5
(7) (9) (10)
(6) (9)
3.0067.52
2.8608.661
VarianceMeanGeneration
Generation
1
2
Lecture 12 30
Lab Problem 6.1• How does changing the heritability affect the
estimates and PEV and PE variance? Set to each of the following and compare results
1002
2
=a
e
σσ 1.2
2
=a
e
σσ
Interpret the results
Lecture 12 31
Lab Problem 6.2aA B C D
E F
G H
Find the best estimate of the genetic worth of each animal, additive and error variance, PEV, and PV. Assume a heritability of .5
J
1
2
3
4
9 13 4 12
11 11
13 9
10
Lecture 12 32
Lab Problem 6.2bA B C D
E F
G H
Environmental trend can be found by fitting generation number as a covariate. Genetic trend is found by taking the average of all EBV’s in that generation and fitting the means to a linear regression. What are the genetic and environmental trend for this data
J
1
2
3
4
9 13 4 12
11 11
13 9
10
generation
Lecture 12 33
Missing Values (Sex Limited Traits)
1 2 3
4 5
(7) M (10)
(6) M
Generation
1
2
=6107
Y
=111
X
=
100000010000001
Z
=
5
4
3
2
1
aaaaa
U
=
5
3
1
eee
e [ ]µ=b
=
10100100
0100001
41
21
21
41
21
21
2121
2121
A
Lecture 12 34
proc iml;start main;
y={ 7,10,6};
X={1,1,1};
A={1 0 0 .5 0,0 1 0 .5 .5,0 0 1 0 .5,.5 .5 0 1 .25,0 .5 .5 .25 1};
lam=1;
Z={1 0 0 0 0,0 0 1 0 0,0 0 0 0 1};
LHS=((X`*X)||(X`*Z))//((Z`*X)||(Z`*Z+INV(A)#LAM));
RHS=(X`*Y)//(Z`*Y);C=INV(HS);
BU=C*RHS;print C BU;
finish main;run;quit;
Lecture 12 35
7.6153846
-0.307692-0.5897440.8974359-0.448718-0.435897
=
5
4
3
2
1
ˆˆˆˆˆ
aaaaa
U
[ ]µ̂=b
Estimates
Lecture 12 36
Lab Problem 6.3: Sex Limited Trait
A B C D
E F
G H
Estimate breeding values for the males
J
1
2
3
4
912
11
13
10
12
2
=a
e
σσ
Lecture 12 37
Extensions of Model• Inclusion of Dominance and Epistasis
– Dominance relationship needed• Reflect the probability that animals have the same
pair of alleles in common• Epistatic genetic effects are the result of
interactions of among additive and dominance genetic effects
– Useful to determine in crossbreeding programs but generally not useful in pure breeding programs
• An individual does not pass on a dominance or epistatic effect, it is the results of both parents
Lecture 12 38
Limitations• Based on infinitesimal model
– Does not work for traits determined by small number of loci
– Genetic variance is assumed constant except Bulmer effect (reduction in variance due to disequilibrium)
• Model needs to be correct– Garbage In Garbage Out (GIGO)– Typical Animal Model Assumes Additivity and
Independence of residuals