A comparison of Rao’s score test and likelihood ratio tests for three separable covariance matrix structures Katarzyna Filipiak 1 , Daniel Klein 2 , Anuradha Roy 3 1 Institute of Mathematics, Pozna´ n University of Technology, Poland 2 Institute of Mathematics, P. J. ˇ Saf´ arik University, Koˇ sice, Slovakia 3 Department of Management Science and Statistics, University of Texas at San Antonio, USA XLII Conference on Mathematical Statistics K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 1 / 20
Data q – number of characteristics p – number of time points n – number of individuals X i , i = 1 ,..., n – i.i.d. observation matrices K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 2 / 20
Observation matrices x 1 , 1 , 1 x 1 , 1 , 2 ... x 1 , 1 , p x 1 , 2 , 1 x 1 , 2 , 2 ... x 1 , 2 , p X 1 = . . . . . . . . ... . x 1 , q , 1 x 1 , q , 2 ... x 1 , q , p . . . x n , 1 , 1 x n , 1 , 2 ... x n , 1 , p x n , 2 , 1 x n , 2 , 2 ... x n , 2 , p X n = . . . . . . . . ... . x n , q , 1 x n , q , 2 ... x n , q , p K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 3 / 20
Two-level multivariate model Model X i ∼ N q , p ( M , Ω ) X i - matrix of observations M - matrix of means Ω - variance-covariance matrix (p.d.) K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 4 / 20
Two-level multivariate model Model vec X i ∼ N pq ( vec M , Ω ) vec X i ∼ N pq ( µ , Ω ) Vector of unknown parameters: � � � � θ 1 µ pq θ = = θ 2 pq ( pq + 1 ) / 2 vech Ω Estimability of Ω : n > pq K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 5 / 20
Covariance structure pq × pq = Ψ p × p ⊗ Σ Ω q × q Ψ – variance-covariance matrix of the repeated measurements on a given characteristic (the same for all characteristics) Σ – variance-covariance matrix of the q response variables at any given time point (the same for all time points) p ( p + 1 ) + q ( q + 1 ) Number of parameters: − 1 2 2 Roy & Khattree (2003), Lu & Zimmerman (2005), Roy (2007) K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 6 / 20
Covariance structure pq × pq = Ψ p × p ⊗ Σ Ω q × q Ψ – AR(1) structure ρ p − 1 ρ 2 ρ ... 1 ρ p − 2 ρ ρ ... 1 ρ p − 3 ρ 2 ρ ... Ψ = 1 . . . . ... . . . . . . . . ρ p − 1 ρ p − 2 ρ p − 3 ... 1 1 + q ( q + 1 ) Number of parameters: 2 Roy and Khatree, 2005, Roy and Leiva, 2008 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 7 / 20
Covariance structure pq × pq = Ψ p × p ⊗ Σ Ω q × q Ψ – CS structure 1 ρ ρ ... ρ ρ 1 ρ ... ρ ρ ρ 1 ... ρ Ψ = . . . . ... . . . . . . . . ρ ρ ρ ... 1 1 + q ( q + 1 ) Number of parameters: 2 Roy and Khatree, 2005, Roy and Leiva, 2008, Filipiak et al., 2016 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 8 / 20
Hypothesis Hypothesis H 0 : Ω = Ψ ⊗ Σ , Ψ UN H A : Ω UN Degrees of freedom: ν = pq ( pq + 1 ) − q ( q + 1 ) − p ( p + 1 ) − 1 2 2 2 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 9 / 20
Hypothesis Hypothesis H 0 : Ω = Ψ ⊗ Σ , Ψ AR(1) or CS H A : Ω UN Degrees of freedom: ν = pq ( pq + 1 ) − q ( q + 1 ) − 1 2 2 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 10 / 20
Likelihood ratio test X = [ vec X 1 , vec X 2 ,..., vec X n ] ∈ R pq , n – data matrix ln L ( µ , Ω ; X ) – log-likelihood function (partially differentiable with respect to each coordinate of θ for every X ). Likelihood ratio (LR) Λ = max H 0 L max H A L Likelihood ratio test statistics χ 2 − 2ln Λ ∼ ν app K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 11 / 20
Rao’s score test Rao’s score (RS) s ′ ( � θ ) F − 1 ( � θ ) s ( � θ ) � θ – MLE of θ � � ′ � � ′ = ∂ ln L ∂ ln L s ′ 1 ( θ ) , s ′ s ( θ ) = 2 ( θ ) ∂ vec ′ µ , – score vector ∂ vech ′ Ω � � ∂ s ( θ ) F ( θ ) = − E – Fisher information matrix (invertible) ∂θ ′ Rao’s score test statistics s ′ ( � θ ) F − 1 ( � θ ) s ( � χ 2 θ ) ∼ ν app K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 12 / 20
Hypothesis H 0 : Ω = Ψ ⊗ Σ , Ψ UN, AR(1) or CS H A : Ω UN RS test statistics �� � 2 � RS = n Ψ − 1 ⊗ � n XQ 1 n X ′ ( � 1 Σ − 1 ) − I pq 2tr n 1 n 1 ′ Q 1 n = I n − 1 n K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 13 / 20
Remark RST statistic depends only on the data matrix X , an estimate of Ψ and Σ . Thus, the minimum number of samples needed to calculate RST statistic is: max { p , q } + 1 if Ψ is UN; q + 1 if Ψ is AR(1) or CS, whereas the minimum number of samples needed to calculate LRT statis- tic is pq + 1 , since it depends on the MLE of a pq × pq variance-covariance matrix Ω . K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 14 / 20
Theorem Theorem The distributions of both LRT and RST statistics under the null hypothesis of separability of Ψ ⊗ Σ with first component as UN, AR(1) or CS, do not depend on the true values of µ or Σ . Moreover, if Ψ is UN or CS, the distributions of both test statistics do not depend on the true value of Ψ . K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 15 / 20
Simulation study - empirical Type I error α LRT � α RST � � α LRT α RST � H 0 p q n n UN ⊗ UN 5 5 6 — 0.536 30 1.000 0.028 AR(1) ⊗ UN 12 2 — 0.361 0.999 0.030 CS ⊗ UN 12 2 — 0.363 0.999 0.030 UN ⊗ UN 5 5 10 — 0.150 50 0.794 0.019 AR(1) ⊗ UN 12 2 — 0.131 0.691 0.020 CS ⊗ UN 12 2 — 0.132 0.691 0.020 UN ⊗ UN 5 5 15 — 0.067 100 0.187 0.014 AR(1) ⊗ UN 12 2 — 0.069 0.151 0.014 CS ⊗ UN 12 2 — 0.069 0.150 0.014 UN ⊗ UN 5 5 20 — 0.045 150 0.083 0.013 AR(1) ⊗ UN 12 2 — 0.046 0.069 0.014 CS ⊗ UN 12 2 — 0.046 0.069 0.014 UN ⊗ UN 5 5 25 — 0.033 200 0.052 0.012 AR(1) ⊗ UN 12 2 1.000 0.034 0.044 0.013 CS ⊗ UN 12 2 1.000 0.034 0.045 0.013 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 16 / 20
Simulation study - Empirical null distribution LRT 95 th RST 95 th n p q H 0 4 5 2 UN ⊗ UN — — 3 3 AR(1) ⊗ UN — ( 62.223 ) 3 3 CS ⊗ UN — (62.262) 10 5 2 UN ⊗ UN — 55.044 3 3 AR(1) ⊗ UN ( 145.064 ) ( 56.266 ) 3 3 CS ⊗ UN (145.131) (56.237) 25 5 2 UN ⊗ UN 69.799 54.298 3 3 AR(1) ⊗ UN ( 66.399 ) ( 54.563 ) 3 3 CS ⊗ UN (66.379) (54.496) 50 5 2 UN ⊗ UN 60.457 53.907 3 3 AR(1) ⊗ UN ( 59.065 ) ( 53.734 ) 3 3 CS ⊗ UN (59.034) (53.727) 100 5 2 UN ⊗ UN 56.523 53.603 3 3 AR(1) ⊗ UN ( 55.923 ) ( 53.558 ) 3 3 CS ⊗ UN (55.892) (53.527) 200 5 2 UN ⊗ UN 54.861 53.474 3 3 AR(1) ⊗ UN ( 54.533 ) ( 53.348 ) 3 3 CS ⊗ UN (54.536) (53.373) 5 (3) 2 (3) (Sep. struc) 53.384 53.384 ∞ K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 17 / 20
Monkey data set (McKiernan et al, 2009) n = 12 subjects (rhesus monkeys) q = 3 characteristics (body weight, % body fat, % maximum mass of upper leg) p = 3 time points 6th year 9th year 12th year Weight % BF % MM Weight % BF % MM Weight % BF % MM 1 15.97 28.31 92.9 15.80 30.58 64.3 14.93 32.64 54.6 2 12.25 16.52 94.6 13.98 25.12 93.2 13.96 26.74 82.6 3 18.52 31.55 96.9 19.09 33.54 88.6 19.09 34.64 85.4 4 11.85 27.25 93.6 14.69 39.39 85.8 14.79 44.14 71.5 5 14.48 19.36 99.0 15.06 23.42 88.7 14.56 24.62 90.3 6 14.70 26.38 94.9 13.90 26.93 88.4 13.06 27.11 78.2 7 13.26 22.54 96.3 14.00 30.82 81.8 13.46 30.54 74.6 8 13.67 21.42 88.4 16.70 26.62 82.0 15.47 27.32 77.8 9 15.33 24.85 98.9 14.68 30.58 69.1 11.58 24.11 60.5 10 10.96 16.79 95.3 10.69 17.55 90.7 8.32 9.46 82.0 11 12.92 22.54 94.9 12.02 20.43 87.9 11.61 24.15 84.5 12 11.10 15.65 94.6 14.58 29.52 83.5 13.30 30.93 77.1 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 18 / 20
Monkey data set (McKiernan et al, 2009) LRT RST UN ⊗ UN AR(1) ⊗ UN CS ⊗ UN UN ⊗ UN AR(1) ⊗ UN CS ⊗ UN 1,2,3 Value 86.706 106.979 118.609 49.894 49.862 61.388 p − val : χ 2 0 . 039 0 . 094 0 . 0095 0 0 0 ν p − val : END ( 0 . 05 , 0 . 1 ) ( 0 . 01 , 0 . 05 ) < 0 . 01 ( 0 . 05 , 0 . 1 ) ( 0 . 1 , 0 . 15 ) ( 0 . 01 , 0 . 05 ) 1,2 Value 11.493 21.753 30.251 11.295 17.532 26.254 p − val : χ 2 0 . 430 0 . 194 0 . 025 0 . 414 0 . 419 0 . 070 ν p − val : END ( 0 . 85 , 0 . 9 ) ( 0 . 6 , 0 . 65 ) ( 0 . 2 , 0 . 25 ) 0 . 75 ( 0 . 5 , 0 . 55 ) ( 0 . 05 , 0 . 1 ) 2,3 Value 19.241 45.215 54.675 16.458 30.364 40.096 p − val : χ 2 0 . 116 0 0 0 . 225 0 . 024 0 . 0012 ν p − val : END ( 0 . 45 , 0 . 50 ) ( 0 . 01 , 0 . 05 ) ( < 0 . 01 ) ( 0 . 30 , 0 . 35 ) ( 0 . 01 , 0 . 05 ) < 0 . 01 1,3 Value 28.619 47.578 57.264 25.842 29.381 36.715 p − val : χ 2 0 . 007 0 0 0 . 018 0 . 031 0 . 004 ν p − val : END ( 0 . 1 , 0 . 15 ) ( 0 . 01 , 0 . 05 ) < 0 . 01 ( 0 . 01 , 0 . 05 ) ( 0 . 01 , 0 . 05 ) < 0 . 01 K. Filipiak (Pozna´ n, Poland) Rao’s score test 01.12.2016 19 / 20
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