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