Score test for random changepoint in a mixed modelgdr-stat-sante.math.cnrs.fr/spip/IMG/pdf/corentin_se... · 2017-10-26 · Introduction Model Score test Simulations Application Conclusion

Introduction Model Score test Simulations Application Conclusion

Score test for random changepoint in a mixedmodel

Corentin Segalas and Helene Jacqmin-GaddaINSERM U1219, Biostatistics team, Bordeaux

GDR Statistiques et Sante

October 6, 2017

Biostatistics

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Introduction

Model

Score test

Simulations

Application

Conclusion

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Alzheimer’s Disease (AD)

• A major public health issue today and tomorrow

• A very long pre-diagnostic phase

• Heterogeneous and non-linear decline trajectories

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Different profiles?

Figure: Estimated mean BVRT score according to age for 2 subjectsdemented at 90 with low or high educational level (Jacqmin-Gadda et al.,2006)

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Objective

Propose a test for the existence of a random changepoint in amixed model for longitudinal data.

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The mixed model with random changepoint

Y (tij) = Yij = β0i + β1i tij + β2

√(tij − τi )2 + γ + εij (1)

with

βki = βTk Xki + αki for k = 0, 1,

αi = (α0i , α1i )T ∼ N (0,B),

τi = µτ + στ τi with αi independent from τi and τi ∼ N (0, 1),

γ = 0.1 a fixed smoothness parameter.

β1i is the mean slope and β2 half the difference of the slopes.

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The mixed model with random changepoint

0 5 10 15 20

1213

1415

1617

18

time (t)

mar

ker

Y(t

)

t=10

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Estimation

• Model estimated by MLE and integral computed by gaussianquadrature (15 nodes)

`n(Y ; θ) =n∑

i=1

log

∫∫ ni∏j=1

f (Yij |αi , τi )f (αi )f (τi )dαidτi .

• No known methods to test the existence of a random CP

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Classic score test

• (H0) : β2 = β02 vs. (H1) : β2 6= β0

2

• test statistic:

Sn =Un(β0

2 , θ0)2

Var(Un(β02 , θ0)2)

with Un(β02 , θ0) =

∂`n(Y ;β2, θ0)

∂β2

∣∣∣∣∣β2=β0

2

with θ0 the MLE of nuisance parameters under the null

• null distribution: χ2(1)

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Identifiability issue

Yij = β0i + β1i tij + β2

√(tij − µτ − στ τi )2 + γ + εij

Hypotheses:(H0) : β2 = 0 vs. (H1) : β2 6= 0

• nuisance parameters : β0, β1, σ, σ0, σ1, σ01, µτ , στ

• µτ and στ unidentifiable under the null: we can’t use theclassic score test statistic Sn which depends on them.

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The score under the null (β2 = 0)

Un(0; θ) =N∑i=1

[ ∫f (τi )

∫f (αi )

ni∏j=1

1√2πσ

exp

{− 1

2σ2(Yij − β0i − β1i tij)

2

}dαidτi

]−1

×∫∫

f (αi )f (τi )(√

2πσ)−nini∑j=1

[1

σ2exp

{− 1

2σ2(Yij − β0i − β1i tij)

2

}

(Yij − β0i − β1i tij)×√

(tij − µτ − στ τi )2 + γ∏k 6=j

exp

{− 1

2σ2(Yik − β0i − β1i tik)2

}]dαidτi

How to circumvent this problem?

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Score test with identifiability issue

Classic problem when testing homogeneity on mixture models.Two main approaches :

• replace µτ and στ by the MLE under the alternative (Conniffe,

2001)

• consider the supremum in (µτ , στ ) of the score test statistic(Hansen, 1996)

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The sup score test

• (H0) : β2 = 0 vs. (H1) : β2 6= 0

• test statistic:

Tn = sup(µτ ,στ )

Sn(0;µτ , στ , θ0)

with

Sn(0;µτ , στ , θ0) =Un(0;µτ , στ , θ0)2

Var(Un(0;µτ , στ , θ0))

with θ0 the MLE of identifiable nuisance parameters under thenull

• null distribution: approached by MC perturbation algorithm ormultiplier bootstrap (van der Vaart and Wellner, 1996).

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The complete procedure

1. estimation of the null model (linear mixed model) using nlme

2. computing the observed test statistic (optimization viaquasi-Newton and integral via pseudo-adaptive gaussianquadrature)

T obsn = sup

(µτ ,στ )

Un(0;µτ , στ , θ0)2

Var(Un(0;µτ , στ , θ0))

where Un(0;µτ , στ , θ0) =∑n

i=1 ui (0;µτ , στ , θ0) and thevariance is estimated by

n∑i=1

ui (0;µτ , στ , θ0)2.

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The complete procedure

4. perturbation algorithm: for k = 1, . . . ,K = 500

• generate n r.v. ξ(k)i ∼ N (0, 1)

• compute

T (k)n = sup

(µτ ,στ )

(∑ni=1 ui (0;µτ , στ , θ0)ξ

(k)i

)2

∑ni=1 ui (0;µτ , στ , θ0)2

5. compute the empirical p−value

pK =1

K

K∑k=1

1T

(k)n >T

(obs)n

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Tests for the variability of β2

If we reject the null hypothesis, we can test if there is a

1. random effect for the difference of slope:

β2i = β2 + α2i with αi = (α0i , α1i , α2i ) ∼ N (0,B)

⇒ corrected LR test for variance component (Stram and Lee, 1994)

2. dependance on covariates:

β2i = β20 + β21X2i

⇒ Wald test

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Simulation scenarios

Yij = β0i + β1i tij + β2

√(tij − τi )2 + γ + εij

with

β0i = 20 + α0i and β1i = −0.3 + α1i

αi = (α0i , α1i )T ∼ N

((00

),

(1 0.1

0.1 0.2

)),

τi = 10 + 2τi with αi independant from τi and τi ∼ N (0, 1),

γ = 0.1, σε = 1, tij = 0, 3, 6, 9, 12, 15, 18, 21 for all i ,

β2 = 0,−0.05,−0.075,−0.1,−0.2,

Probability of drop-out at each visit: 0.1 ⇒ around 50% of thesample remaining at t = 21.

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Simulation Scenarios

0 5 10 15 20

1214

1618

20M1

time

mar

ker

0 5 10 15 20

1214

1618

20

M2

time

mar

ker

0 5 10 15 20

1214

1618

20

M3

time

mar

ker

0 5 10 15 20

1214

1618

20

M4

time

mar

ker

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Results

N 100 200

drop-out no yes no yes

size M0 0.029 0.042 0.034 0.039

power M1 0.397 0.054 - -M2 0.749 0.067 - -M3 0.949 0.093 - -M4 1 0.206 - 0.425

Table: Size and power of the test computed on 1000 replicates of eachscenarios with K = 500 perturbations.

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The PAQUID cohort

• 3777 subjects older than 65 from the french departments ofGironde and Dordogne, 25 years follow-up

• Marker : Isaac 15s score

• sample selection: incident case of dementia between year 1and 25

• High education sample• 522 subjects with at least 1 measure• 1 to 12 measures by subject (mean = 5.8)

• Low education sample• 358 subjects with at least 1 measure• 1 to 12 measures by subject (mean = 4.6)

• model (1) with βki = βk + αki for k = 0, 1 (no covariate)

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Score test results

obs. statistic test* p-value

High education 14.059 0.001Low education 1.388 0.443

Table: Score test results with K = 1000

For the high education subjects, we clearly reject the nullhypothesis of no random changepoint.

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Estimation (nq = 15)

PAQUID demented sampleHigh education Low education

N 522 358Log-lik -8845.889 -4685.284

Est sd Est sdβ0 23.087 0.219 20.477 0.342β1 -0.838 0.026 -0.531 0.039β2 -0.559 0.022 -0.354 0.033µτ -4.101 0.375 -5.512 0.694σ 3.476 0.045 3.358 0.074σ0 4.195 0.134 3.873 0.172σ1 0.213 0.017 0.209 0.024στ 2.925 0.016 1.776 0.026σ01 0.275 0.343 0.178 0.675

slope 1/2 -0.279 / -1.397 -0.177 / -0.885

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Estimation of the mixed model with random CP

−15 −10 −5 0 5

1520

2530

Delay

Isaa

c 15

s

High educationLow education

Figure: Mean estimation trajectory of the mixed model with randomchangepoint on the two educational level subsamples.

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Variability of β2: random effect ?

On high education subsample

(H0) : σ2 = 0 vs. (H1) : σ2 6= 0

where β2i = β2 + α2i with α2i ∼ N (0, σ22).

LRS = -137.2 ⇒ p < 0.001⇒ We need to add a random effect on β2

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Next steps

• simulations with varying στ• extension to :

• joint models• joint multi-state models for interval censored data• models for multiple markers• etc.

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References

1. Conniffe, D., ”Score Tests When a Nuisance Parameter Is Unidentified underthe Null Hypothesis.” Journal of Statistical Planning and Inference (2001)

2. Hansen, Bruce E., ”Inference When a Nuisance Parameter Is Not Identifiedunder the Null Hypothesi.” Econometrica (1996)

3. Jacqmin-Gadda, H., Commenges, D. and Dartigues, J.-F., ”Randomchangepoint model for joint modeling of cognitive decline and dementia.”Biometrics (2006)

4. Stram, D. O., and Lee J.W., ”Variance Components Testing in the LongitudinalMixed Effects Model.” Biometrics (1994)

5. van der Vaart, A. W. and Wellner, J.A., ”Weak Convergence and EmpiricalProcesses.” Chapter 2.9, Springer Series in Statistics (1996).

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Thank you foryour attention!

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Documents

Score test for random changepoint in a mixed modelgdr-stat-sante.math.cnrs.fr/spip/IMG/pdf/corentin_se... · 2017-10-26 · Introduction Model Score test Simulations Application Conclusion