Hypothesis testing in variance components with constraints
M. Przystalski1 M. Fonseca2
1The Research Center for Cultivar Testing, S lupia Wielka, Poland
2Centro de Mathem´atica e Aplica¸c˜oes, Faculdade de Ciˆencias e Tecnologia, Universidade Nova de Lisboa, Lisboba, Portugal
LinStat 2014
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 1 / 32
Introduction
For testing hypotheses related to variance components one uses e.g.
permutation test (e.g. Fitzmaurice at al., 2007), or generalized p-values (see e.g. Volaufova, 2009).
For inference on variance component the usual chi-square distribution of likelihood ratio statistic under the null is used.
E.g. Self and Liang (1987) had shown that this not necessarily hold. For LMM with one variance component, Crainiceanu and Ruppert (2004) had shown that the distribution of the LR statistic is mixture of two chi-squares.
This is due to the fact that the values of the variance component under the null lie on the boundry of the parameter space.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 2 / 32
Introduction
For testing hypotheses related to variance components one uses e.g.
permutation test (e.g. Fitzmaurice at al., 2007), or generalized p-values (see e.g. Volaufova, 2009).
For inference on variance component the usual chi-square distribution of likelihood ratio statistic under the null is used.
E.g. Self and Liang (1987) had shown that this not necessarily hold. For LMM with one variance component, Crainiceanu and Ruppert (2004) had shown that the distribution of the LR statistic is mixture of two chi-squares.
This is due to the fact that the values of the variance component under the null lie on the boundry of the parameter space.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 2 / 32
Introduction
For testing hypotheses related to variance components one uses e.g.
permutation test (e.g. Fitzmaurice at al., 2007), or generalized p-values (see e.g. Volaufova, 2009).
For inference on variance component the usual chi-square distribution of likelihood ratio statistic under the null is used.
E.g. Self and Liang (1987) had shown that this not necessarily hold.
For LMM with one variance component, Crainiceanu and Ruppert (2004) had shown that the distribution of the LR statistic is mixture of two chi-squares.
This is due to the fact that the values of the variance component under the null lie on the boundry of the parameter space.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 2 / 32
Introduction
For testing hypotheses related to variance components one uses e.g.
permutation test (e.g. Fitzmaurice at al., 2007), or generalized p-values (see e.g. Volaufova, 2009).
For inference on variance component the usual chi-square distribution of likelihood ratio statistic under the null is used.
E.g. Self and Liang (1987) had shown that this not necessarily hold.
For LMM with one variance component, Crainiceanu and Ruppert (2004) had shown that the distribution of the LR statistic is mixture of two chi-squares.
This is due to the fact that the values of the variance component under the null lie on the boundry of the parameter space.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 2 / 32
Introduction
For testing hypotheses related to variance components one uses e.g.
permutation test (e.g. Fitzmaurice at al., 2007), or generalized p-values (see e.g. Volaufova, 2009).
For inference on variance component the usual chi-square distribution of likelihood ratio statistic under the null is used.
E.g. Self and Liang (1987) had shown that this not necessarily hold.
For LMM with one variance component, Crainiceanu and Ruppert (2004) had shown that the distribution of the LR statistic is mixture of two chi-squares.
This is due to the fact that the values of the variance component under the null lie on the boundry of the parameter space.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 2 / 32
Introduction
Aims
1 Construct statiscal tests for variance components with posistivity constraints.
2 Investigate the perfomance of the proposed procedures.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 3 / 32
Introduction
Aims
1 Construct statiscal tests for variance components with posistivity constraints.
2 Investigate the perfomance of the proposed procedures.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 3 / 32
One-way model
Statisitcal model
y =µ1+Zα+e, where
y - vector of observed values, µ- general mean
1N - vector of one’s
α - vector of random effects, Z=Ia⊗1n is design matrix, e - vector of errors
We assume that α ande are indendent and that α∼N 0, σa2Ia and e ∼N 0, σ2eIN
,respectively.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 4 / 32
One-way model
Statistical model
The problem of interest is:
H0:σ2a = 0 vs. H1:σ2a >0.
Alternatively, the problem of interest can be written as: H00 :γ1 =γ2 vs. H10 :γ1 > γ2,
whereγ1 =nσ2a+σe2 andγ2=σe2 (see Khuri et al., 1998).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 5 / 32
One-way model
Statistical model
The problem of interest is:
H0:σ2a = 0 vs. H1:σ2a >0.
Alternatively, the problem of interest can be written as:
H00 :γ1=γ2 vs. H10 :γ1 > γ2,
whereγ1=nσ2a+σ2e andγ2=σe2 (see Khuri et al., 1998).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 5 / 32
One-way model
Statistical model
The problem of interest is:
H0:σ2a = 0 vs. H1:σ2a >0.
Alternatively, the problem of interest can be written as:
H00 :γ1=γ2 vs. H10 :γ1 > γ2, whereγ1=nσ2a+σ2e andγ2=σe2 (see Khuri et al., 1998).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 5 / 32
One-way model
Statistical model
The maximum likelihood estimators (MLE)
bµ= 1
N10Ny, γb1 = 1
a−1y0P1y bγ2= 1
N−ay0P2y, where
P1 = Ia⊗Jn n − JN
N P2=Ia⊗In−Ia⊗Jn n .
E(bγ1) =γ1 andV(bγ1)≈ a−12γ21;
E(bγ2) =γ2 and; V(γb2)≈ N−a2γ22
. By the independence of bγ1 andbγ2 (see Theorem 2.4.1, Khuri et al., 1998) we have
Vγb =diag2γ2 1
a−1,N−a2γ22
V−1/2
γb (γb−γ)→d N(0,I2). (1)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 6 / 32
One-way model
Statistical model
The maximum likelihood estimators (MLE)
bµ= 1
N10Ny, γb1 = 1
a−1y0P1y bγ2= 1
N−ay0P2y, where
P1 = Ia⊗Jn n − JN
N P2=Ia⊗In−Ia⊗Jn n . E(bγ1) =γ1 andV(bγ1)≈ a−12γ21;
E(bγ2) =γ2 and; V(γb2)≈ N−a2γ22
.
By the independence of bγ1 andbγ2 (see Theorem 2.4.1, Khuri et al., 1998) we have
Vγb =diag2γ2 1
a−1,N−a2γ22
V−1/2
γb (γb−γ)→d N(0,I2). (1)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 6 / 32
One-way model
Statistical model
The maximum likelihood estimators (MLE)
bµ= 1
N10Ny, γb1 = 1
a−1y0P1y bγ2= 1
N−ay0P2y, where
P1 = Ia⊗Jn n − JN
N P2=Ia⊗In−Ia⊗Jn n . E(bγ1) =γ1 andV(bγ1)≈ a−12γ21;E(bγ2) =γ2 and; V(γb2)≈ N−a2γ22 .
By the independence of bγ1 andbγ2 (see Theorem 2.4.1, Khuri et al., 1998) we have
Vγb =diag2γ2 1
a−1,N−a2γ22
V−1/2
γb (γb−γ)→d N(0,I2). (1)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 6 / 32
One-way model
Statistical model
The maximum likelihood estimators (MLE)
bµ= 1
N10Ny, γb1 = 1
a−1y0P1y bγ2= 1
N−ay0P2y, where
P1 = Ia⊗Jn n − JN
N P2=Ia⊗In−Ia⊗Jn n . E(bγ1) =γ1 andV(bγ1)≈ a−12γ21;E(bγ2) =γ2 and; V(γb2)≈ N−a2γ22 . By the independence of bγ1 andbγ2 (see Theorem 2.4.1, Khuri et al., 1998) we have
Vγb =diag2γ2 1
a−1,N−a2γ22
V−1/2
γb (γb−γ)→d N(0,I2). (1)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 6 / 32
One-way model
Statistical model
The maximum likelihood estimators (MLE)
bµ= 1
N10Ny, γb1 = 1
a−1y0P1y bγ2= 1
N−ay0P2y, where
P1 = Ia⊗Jn n − JN
N P2=Ia⊗In−Ia⊗Jn n . E(bγ1) =γ1 andV(bγ1)≈ a−12γ21;E(bγ2) =γ2 and; V(γb2)≈ N−a2γ22 . By the independence of bγ1 andbγ2 (see Theorem 2.4.1, Khuri et al., 1998) we have V
γb =diag2γ2 1
a−1,N−a2γ22
V−1/2
γb (γb−γ)→d N(0,I2). (1)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 6 / 32
One-way model
Statistical model
The maximum likelihood estimators (MLE)
bµ= 1
N10Ny, γb1 = 1
a−1y0P1y bγ2= 1
N−ay0P2y, where
P1 = Ia⊗Jn n − JN
N P2=Ia⊗In−Ia⊗Jn n . E(bγ1) =γ1 andV(bγ1)≈ a−12γ21;E(bγ2) =γ2 and; V(γb2)≈ N−a2γ22 . By the independence of bγ1 andbγ2 (see Theorem 2.4.1, Khuri et al., 1998) we have V
γb =diag2γ2 1
a−1,N−a2γ22
V−1/2
γb (γb−γ)→d N(0,I2). (1)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 6 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2
⇒ eγ1 =bγ1 andeγ2=bγ2 ⇒`11= 0.
In case γ1 ≤γ2
⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2
⇒ eγ1 =bγ1 andeγ2=bγ2 ⇒`11= 0.
In case γ1 ≤γ2
⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2
⇒ γe1 =bγ1 andeγ2=bγ2 ⇒`11= 0. In case γ1 ≤γ2
⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2 ⇒ γe1 =bγ1 andeγ2=γb2
⇒`11= 0. In case γ1 ≤γ2
⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2 ⇒ γe1 =bγ1 andeγ2=γb2 ⇒`11= 0.
In case γ1 ≤γ2
⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2 ⇒ γe1 =bγ1 andeγ2=γb2 ⇒`11= 0.
In case γ1 ≤γ2
⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2 ⇒ γe1 =bγ1 andeγ2=γb2 ⇒`11= 0.
In case γ1 ≤γ2 ⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2 ⇒ γe1 =bγ1 andeγ2=γb2 ⇒`11= 0.
In case γ1 ≤γ2 ⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
Using the Self and Liang approach (Self and Liang, 1987) and (1)
`1=−a−1 2
bγ1
γ1
−1 2
−N−a 2
γb2
γ2
−1 2
In case γ1 > γ2 ⇒ γe1 =bγ1 andeγ2=γb2 ⇒`11= 0.
In case γ1 ≤γ2 ⇒ γ1 =γ2 =γ
`12=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
d`12
dγ−1 =−bγ1(a−1)
bγ1
γ −1
−bγ2(N−a)
bγ2
γ −1
= 0.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 7 / 32
One-way model
Likelihood ratio
γ1≤γ2 ⇒
eγ = (a−1)bγ12+ (N−a)bγ22 (a−1)bγ1+ (N−a)bγ2
, `12=−(a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)γb22
Under the null hypothesis
`0=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
eγ = (a−1)γb12+ (N−a)bγ22
(a−1)γb1+ (N−a)bγ2 `0 =− (a−1) (N−a) (bγ1−bγ2)2 2 (a−1)bγ12+ (N−a)bγ22
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 8 / 32
One-way model
Likelihood ratio
γ1≤γ2 ⇒
eγ = (a−1)bγ12+ (N−a)bγ22 (a−1)bγ1+ (N−a)bγ2
, `12=−(a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)γb22
Under the null hypothesis
`0=−a−1 2
bγ1
γ −1 2
−N−a 2
bγ2
γ −1 2
eγ = (a−1)γb12+ (N−a)bγ22
(a−1)γb1+ (N−a)bγ2 `0 =− (a−1) (N−a) (bγ1−bγ2)2 2 (a−1)bγ12+ (N−a)bγ22
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 8 / 32
One-way model
Likelihood ratio
γ1≤γ2 ⇒
eγ = (a−1)bγ12+ (N−a)bγ22 (a−1)bγ1+ (N−a)bγ2
, `12=−(a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)γb22
Under the null hypothesis
`0=−a−1 2
bγ1
γ −1 2
−N−a 2
γb2
γ −1 2
eγ = (a−1)γb12+ (N−a)bγ22
(a−1)bγ1+ (N−a)bγ2 `0 =− (a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)bγ22
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 8 / 32
One-way model
Likelihood ratio
γ1≤γ2 ⇒
eγ = (a−1)bγ12+ (N−a)bγ22 (a−1)bγ1+ (N−a)bγ2
, `12=−(a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)γb22
Under the null hypothesis
`0=−a−1 2
bγ1
γ −1 2
−N−a 2
γb2
γ −1 2
eγ = (a−1)γb12+ (N−a)bγ22
(a−1)bγ1+ (N−a)bγ2 `0 =− (a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)bγ22
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 8 / 32
One-way model
Likelihood ratio
γ1≤γ2 ⇒
eγ = (a−1)bγ12+ (N−a)bγ22 (a−1)bγ1+ (N−a)bγ2
, `12=−(a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)γb22
Under the null hypothesis
`0=−a−1 2
bγ1
γ −1 2
−N−a 2
γb2
γ −1 2
eγ = (a−1)γb12+ (N−a)bγ22
(a−1)bγ1+ (N−a)bγ2 `0 =−(a−1) (N−a) (bγ1−γb2)2 2 (a−1)bγ12+ (N−a)γb22
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 8 / 32
One-way model
Likelihood ratio
Proposition
The LRT rejects the null hypothesis for large values of λgiven by:
λ=
0 bγ1≤bγ2
(a−1)(N−a)(bγ1−bγ2)2
((a−1)bγ12+(N−a)bγ22) bγ1>bγ2.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 9 / 32
One-way model
Random permutations
y = [yij] – lexicographical order
i – first factor j – replicates Permutation
yl =Ply, where Pl is a random permutation matrix.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 10 / 32
One-way model
Permutation procedure
1 Calculate ˆγ1 and ˆγ2
2 Calculateλ
3 Generate a pseudo-sample of size M of
`l =`(yl) =`(Ply)
4 Retrieve cα, the pseudo-sample’s empirical (1−α)-th quantile
5 Compare with the observed value λwith cα
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 11 / 32
One-way model
Application of delta method
Combining the delta method (see van der Vaart, 1998) with the Self and Liang approach (Self and Liang, 1987) we obtain the following proposition.
Proposition
The LRT rejects the null hypothesis H0 for a large values of κgiven by
κ=
(0 bγ1 ≤bγ2,
(a−1)(N−a)(logbγ1−logbγ2)2
2(N−1) bγ1 >bγ2.
Theorem
The distribution of the LRT κ is a mixture of χ20 andχ21 with coefficients p and 1-p, with
p =P(bγ1 ≤bγ2).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 12 / 32
One-way model
Application of delta method
Combining the delta method (see van der Vaart, 1998) with the Self and Liang approach (Self and Liang, 1987) we obtain the following proposition.
Proposition
The LRT rejects the null hypothesis H0 for a large values ofκ given by
κ=
(0 γb1 ≤bγ2,
(a−1)(N−a)(logbγ1−logbγ2)2
2(N−1) γb1 >bγ2.
Theorem
The distribution of the LRT κ is a mixture of χ20 andχ21 with coefficients p and 1-p, with
p =P(bγ1 ≤bγ2).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 12 / 32
One-way model
Application of delta method
Combining the delta method (see van der Vaart, 1998) with the Self and Liang approach (Self and Liang, 1987) we obtain the following proposition.
Proposition
The LRT rejects the null hypothesis H0 for a large values ofκ given by
κ=
(0 γb1 ≤bγ2,
(a−1)(N−a)(logbγ1−logbγ2)2
2(N−1) γb1 >bγ2.
Theorem
The distribution of the LRT κ is a mixture of χ20 andχ21 with coefficients p and 1-p, with
p =P(bγ1 ≤bγ2).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 12 / 32
Two-fold nested model
Statistical model
y =µ1N+Z1α+Z2β+e, where
y - vector of observed values, µ- general mean,
1N - vector of one’s,
α andβ- vectors of random effects,
Z1 =Ia⊗1b⊗1r andZ2 =Ia⊗Ib⊗1r - are design matrices, e - vector of errors
We assume that α,β ande are indendent and thatα∼N 0, σa2Ia , β∼N 0, σb2Ib
ande ∼N 0, σe2IN
,respectively.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 13 / 32
Two-fold neste model
Statistical model
The problem of interest is:
H0:σ2b= 0 vs. H1:σ2b>0.
Alternatively, the problem of interest can be written as: H00 :γ2 =γ3 vs. H10 :γ2 > γ3,
whereγ1 =brσa2+rσ2b+σe2,γ2 =rσ2b+σ2e andγ3=σ2e, respectively (see Khuri et al, 1998).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 14 / 32
Two-fold neste model
Statistical model
The problem of interest is:
H0:σ2b= 0 vs. H1:σ2b>0.
Alternatively, the problem of interest can be written as:
H00 :γ2=γ3 vs. H10 :γ2 > γ3,
whereγ1=brσa2+rσ2b+σ2e,γ2 =rσ2b+σ2e andγ3=σ2e, respectively (see Khuri et al, 1998).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 14 / 32
Two-fold neste model
Statistical model
The problem of interest is:
H0:σ2b= 0 vs. H1:σ2b>0.
Alternatively, the problem of interest can be written as:
H00 :γ2=γ3 vs. H10 :γ2 > γ3, whereγ1=brσa2+rσ2b+σ2e,γ2 =rσ2b+σ2e andγ3=σ2e, respectively (see Khuri et al, 1998).
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 14 / 32
Two-fold nested model
Statistical model
The maximum likelihood estimators (MLE)
µb= 1
N10Ny, bγ1 = y0P1y
a−1 , bγ2= yP2y
a(b−1), bγ3 = y0P3y ab(r−1), whereP1= Ia−1aJa
⊗br1Jbr,P2=Ia⊗ Ib−1bJb
⊗1rJr, P3=Iab⊗ Ir −1rJr
.
E(bγ1) =γ1 andV(bγ1)≈ 2γa12;
E(bγ2) =γ2 and;V (bγ2)≈ 2γab22; E(bγ3) =γ3 and; V(bγ3)≈ 2γabr23.
Vγb =diag2γ2 1
a ,2γab22,2γabr23
V−1/2
γb (γb−γ)→d N(0,I3). (2)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 15 / 32
Two-fold nested model
Statistical model
The maximum likelihood estimators (MLE)
µb= 1
N10Ny, bγ1 = y0P1y
a−1 , bγ2= yP2y
a(b−1), bγ3 = y0P3y ab(r−1), whereP1= Ia−1aJa
⊗br1Jbr,P2=Ia⊗ Ib−1bJb
⊗1rJr, P3=Iab⊗ Ir −1rJr
.
E(bγ1) =γ1 andV(bγ1)≈ 2γa12;
E(bγ2) =γ2 and;V (bγ2)≈ 2γab22; E(bγ3) =γ3 and; V(bγ3)≈ 2γabr23.
Vγb =diag2γ2 1
a ,2γab22,2γabr23
V−1/2
γb (γb−γ)→d N(0,I3). (2)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 15 / 32
Two-fold nested model
Statistical model
The maximum likelihood estimators (MLE)
µb= 1
N10Ny, bγ1 = y0P1y
a−1 , bγ2= yP2y
a(b−1), bγ3 = y0P3y ab(r−1), whereP1= Ia−1aJa
⊗br1Jbr,P2=Ia⊗ Ib−1bJb
⊗1rJr, P3=Iab⊗ Ir −1rJr
.
E(bγ1) =γ1 andV(bγ1)≈ 2γa12;E(bγ2) =γ2 and;V (bγ2)≈ 2γab22;
E(bγ3) =γ3 and; V(bγ3)≈ 2γabr23. Vγb =diag2γ2
1
a ,2γab22,2γabr23
V−1/2
γb (γb−γ)→d N(0,I3). (2)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 15 / 32
Two-fold nested model
Statistical model
The maximum likelihood estimators (MLE)
µb= 1
N10Ny, bγ1 = y0P1y
a−1 , bγ2= yP2y
a(b−1), bγ3 = y0P3y ab(r−1), whereP1= Ia−1aJa
⊗br1Jbr,P2=Ia⊗ Ib−1bJb
⊗1rJr, P3=Iab⊗ Ir −1rJr
.
E(bγ1) =γ1 andV(bγ1)≈ 2γa12;E(bγ2) =γ2 and;V (bγ2)≈ 2γab22; E(bγ3) =γ3 and;V (bγ3)≈ 2γabr23.
Vγb =diag2γ2 1
a ,2γab22,2γabr23
V−1/2
γb (γb−γ)→d N(0,I3). (2)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 15 / 32
Two-fold nested model
Statistical model
The maximum likelihood estimators (MLE)
µb= 1
N10Ny, bγ1 = y0P1y
a−1 , bγ2= yP2y
a(b−1), bγ3 = y0P3y ab(r−1), whereP1= Ia−1aJa
⊗br1Jbr,P2=Ia⊗ Ib−1bJb
⊗1rJr, P3=Iab⊗ Ir −1rJr
.
E(bγ1) =γ1 andV(bγ1)≈ 2γa12;E(bγ2) =γ2 and;V (bγ2)≈ 2γab22; E(bγ3) =γ3 and;V (bγ3)≈ 2γabr23.
Vγb =diag2γ2 1
a ,2γab22,2γabr23
V−1/2
γb (γb−γ)→d N(0,I3). (2)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 15 / 32
Two-fold nested model
Statistical model
The maximum likelihood estimators (MLE)
µb= 1
N10Ny, bγ1 = y0P1y
a−1 , bγ2= yP2y
a(b−1), bγ3 = y0P3y ab(r−1), whereP1= Ia−1aJa
⊗br1Jbr,P2=Ia⊗ Ib−1bJb
⊗1rJr, P3=Iab⊗ Ir −1rJr
.
E(bγ1) =γ1 andV(bγ1)≈ 2γa12;E(bγ2) =γ2 and;V (bγ2)≈ 2γab22; E(bγ3) =γ3 and;V (bγ3)≈ 2γabr23.
Vγb =diag2γ2 1
a ,2γab22,2γabr23
V−1/2
γb (γb−γ)→d N(0,I3). (2)
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 15 / 32
Two-fold nested model
Likelihood ratio. H1
bγ1 ≥γb2 ≥bγ3
eγ1 =bγ1 eγ2 =bγ2 eγ3 =bγ3
−2`1 = 0
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 16 / 32
Likelihood Ratio
H1
bγ1 <bγ2,bγ2 ≥bγ3
eγ12= bγ21+bγb22 bγ1+bγb2
eγ3 =bγ3
−2`1 = ab2(bγ1γb2−bγ22)2+ab(bγ1bγ2−bγ12)2 (bγ12+bγb22)2
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 17 / 32
Two-fold nested model
Likelihood ratio. H1
bγ1 ≥ bγ22+rbγ32
bγ2+rbγ3,bγ2 <bγ3
eγ1 =bγ1
eγ23= bγ22+rbγ32 bγ2+rbγ3
−2`1= abr2(bγ2γb3−bγ23)2+abr(bγ2bγ3−bγ22)2 (bγ22+rbγ32)2
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 18 / 32
Two-fold nested model
Likelihood ratio. H1
bγ1 < bγ22+rbγ32
bγ2+rbγ3,bγ2 <bγ3
γe123= bγ12+bbγ22+brbγ32 bγ1+bbγ2+brbγ3
−2`1 = a b(bγ1bγ2−bγ22) +br(γb1bγ3−bγ32)2
(bγ12+bγb22+brbγ32)2 +ab (bγ1bγ2−bγ12) +br(γb2bγ3−bγ32)2
(bγ12+bbγ22+brγb32)2
+abr (bγ1bγ2−bγ12) +b(γb1bγ2−bγ22)2
(bγ12+bbγ22+brγb32)2
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 19 / 32
Two-fold nested model
Likelihood ratio. H0 :γ2 =γ3
bγ1 ≥ bγ22+rbγ32
bγ2+rbγ3
eγ1 =bγ1
eγ23= bγ22+rbγ32 bγ2+rbγ3
−2`1= abr2(bγ2γb3−bγ23)2+abr(bγ2bγ3−bγ22)2 (bγ22+rbγ32)2
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 20 / 32
Two-fold nested model
Likelihood ratio. H0 :γ2 =γ3
bγ1 < bγ22+rbγ32
bγ2+rbγ3
γe123= bγ12+bbγ22+brbγ32 bγ1+bbγ2+brbγ3
−2`0 = a b(bγ1bγ2−bγ22) +br(γb1bγ3−bγ32)2
(bγ12+bγb22+brbγ32)2 +ab (bγ1bγ2−bγ12) +br(γb2bγ3−bγ32)2
(bγ12+bbγ22+brγb32)2
+abr (bγ1bγ2−bγ12) +b(γb1bγ2−bγ22)2
(bγ12+bbγ22+brγb32)2
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 21 / 32
Two-fold nested model
Random Permutations
y = [yijk] – lexicographical order
i – first factor j – second factor k – replicates Selective permutation
yl = (Ia⊗Pl)y, where Pl is a random permutation matrix.
M. Przystalski, M. Fonseca Hypothesis testing in variance components with constraints LinStat 2014 22 / 32