VARIANCE OF MARKER BREEDING VALUES
Xiaochen Sun
December, 2013
1. Variance of a random vector and variance of vector
elements
Assume that
a ∼ (µ, Σσ 2 )
(1)
is an n × 1 random vector, Σ being positive definite matrix. Let V(a) denote
the variance among the n elements in a. It follows that
1 0 0 1 0 1
a − 11 a a − 11 a
V(a) =
n
n
n
1 0
1 0 1 0
= a I − 11 I − 11 a
n
n
n
1 0
1 0
= a I − 11 a.
n
n
1
(2)
(3)
(4)
The expected value of V(a) is
1 E V(a) = µ0 I −
n
1 = µ0 I −
n
1 = µ0 I −
n
1 0
= µ I−
n
1 0
11 µ +
n
1 0
11 µ +
n
1 0
11 µ +
n
1 0
11 µ +
n
σ2 1 0
tr I − 11 Σ
n
n
2
1 0 σ
tr Σ − tr 11 Σ
n
n
2
1 0 σ
tr Σ − tr 1 Σ1
n
n
2
1 0
σ
tr Σ − 1 Σ1 .
n
n
(5)
(6)
(7)
(8)
When µ = 0,
σ2
1 0
E V(a) =
tr Σ − 1 Σ1
n
n
1
σ2
n · diag −
n · diag + n(n − 1) · offdiag
=
n
n
n − 1
=
diag − offdiag σ 2 ,
n
(9)
(10)
(11)
in which diag is the average diagonal elements of Σ, and offdiag is the average off-diagonal elements of Σ.
For example, when a includes breeding values of n individuals and Σ equals
the numerator relationship matrix A, σ 2 is the additive genetic variance in
pedigree founders (independent individuals), denoted as σa2 . If the n individ
uals are independent from each other, E V(a) = σa2 given large n. If the n
individuals are paternal half-sibs, the diagonal elements in A equals 1.0 and
off-diagonal elements equals 0.25, therefore the expected variance of breeding
2
values among half-sibs equals to 0.75σa2 .
2. Variance of marker breeding values
Consider the statistical model for phenotypes
y = a + e,
(12)
a = Mα,
(13)
e ∼ 0, Iσe2 ,
(14)
in which y is an n × 1 vector of phenotypic values of n individuals adjusted
for fixed non-genetic effects, a is an n × 1 vector of marker breeding values,
M is an n × m matrix with row i including the genotypes of m markers for
individual i, α is an m × 1 vector of m marker effects, and e is an n × 1
vector of random residuals. Assuming each single locus in HWE, column j
of M, denoted as mj , includes n independent binomial variables with mij ∼
Bin(2, pj ), in which pj is allele frequency for marker j. Further we have
E mj = 2qj 1,
Var mj = 2pj qj I,
(15)
(16)
in which qj = 1 − pj . We will discuss about the variance of a, Var(a), and
variance of breeding values, V(a), under different assumptions on M and α.
3
2.1. Random α and fixed M (BayesC)
Assume α ∼ (η, Iσα2 ). The variance of marker breeding value Var(a) can be
written as
Var(a) = Var(Mα) = MM0 σα2 .
(17)
n
n − 1 X (i) 0 (i)
1 X (i) 0 (j) 2
V(a) =
m m − 2
m m σα ,
n2 i=1
n i6=j
(18)
According to (11),
in which m(i) is the vector of ith row of M, i.e. marker genotypes for individual i.
4
2.2. Fixed α and random M
In this case, the variance of marker breeding value Var(a) can be written as
Var(a) = Var(Mα)
m
X
= Var
m j αj
(19)
(20)
j=1
=
=
m
X
j=1
m
X
Var(mj αj ) + 2
X
Cov(mi αi , mj αj )
(21)
αi αj Cov(mi , mj )
(22)
i<j
αj2 Var(mj ) + 2
j=1
m
X
=I
X
i<j
2pj qj αj2 + 2
j=1
m
X
= 2I
X
αi αj rij
p
p
2pi qi I · 2pj qj I
(23)
i<j
pj qj αj2 + 4I
j=1
X√
pi qi pj qj αi αj rij ,
(24)
i<j
in which rij is linkage disequilibrium measured as correlation coefficient between genotypes of marker i and marker j.
According to equation (1),
Σ = I,
2
σ =2
(25)
m
X
pj qj αj2 + 4
X√
j=1
i<j
5
pi qi pj qj αi αj rij .
(26)
2.3. Random α and random M
Assume α ∼ (η, Iσα2 ), and α and M are independent. The variance of marker
breeding value Var(a) can be written as
Var(a) = EM Var(a|M) + VarM E(a|M)
= EM MM0 σα2 + VarM Mη
= σα2 EM MM0 + VarM Mη .
(27)
(28)
(29)
The first term in equation (25) can be written as
σα2 E(MM0 )
=
m
X
σα2 E
(30)
(31)
mj m0j
j=1
=
σα2
m
X
E mj m0j
j=1
m n
X
0 o
2
= σα
Var mj + E(mj ) E(mj )
(32)
j=1
=
=
σα2
m
X
2pj qj I +
j=1
m
X
2
2σα I
pj qj
j=1
σα2
m
X
(2qj )1 · (2qj )10
(33)
j=1
+
4σα2 110
m
X
qj2 .
(34)
j=1
The second term in equation (25) has similar form as equation (22) and can
be written as
Var Mη = 2I
m
X
pj qj ηj2 + 4I
j=1
X√
i<j
6
pi qi pj qj ηi ηj rij .
(35)
Assuming η = 0, we have Var Mη = 0. The diagonal elements of Var(a)
equal to
2σα2
m
X
pj qj +
4σα2
j=1
m
X
qj2
=
2σα2
j=1
m
X
qj + qj2 .
(36)
j=1
The off-diagonal elements of Var(a) equal to
4σα2
m
X
qj2 .
(37)
j=1
According to (11), when n is large, the variance of breeding values is
V(a) =
2σα2
m
X
qj +
qj2
−
j=1
4σα2
m
X
j=1
7
qj2
=
2σα2
m
X
j=1
pj q j .
(38)