Kesetaraan spesifikasi efek acak (0 + faktor | kelompok) dan (1 | kelompok) + (1 | kelompok: faktor) jika simetri senyawa

Douglas Bates menyatakan bahwa model-model berikut ini setara "jika matriks varians-kovarians untuk efek acak bernilai vektor memiliki bentuk khusus, yang disebut simetri gabungan" ( slide 91 dalam presentasi ini ):

m1 <- lmer(y ~ factor + (0 + factor|group), data)
m2 <- lmer(y ~ factor + (1|group) + (1|group:factor), data)

Khususnya Bates menggunakan contoh ini:

library(lme4)
data("Machines", package = "MEMSS")

m1a <- lmer(score ~ Machine + (0 + Machine|Worker), Machines)
m2a <- lmer(score ~ Machine + (1|Worker) + (1|Worker:Machine), Machines)

dengan output yang sesuai:

print(m1a, corr = FALSE)

Linear mixed model fit by REML ['lmerMod']
Formula: score ~ Machine + (0 + Machine | Worker)
   Data: Machines
REML criterion at convergence: 208.3112
Random effects:
 Groups   Name     Std.Dev. Corr     
 Worker   MachineA 4.0793            
          MachineB 8.6253   0.80     
          MachineC 4.3895   0.62 0.77
 Residual          0.9616            
Number of obs: 54, groups:  Worker, 6
Fixed Effects:
(Intercept)     MachineB     MachineC  
     52.356        7.967       13.917  

print(m2a, corr = FALSE)

Linear mixed model fit by REML ['lmerMod']
Formula: score ~ Machine + (1 | Worker) + (1 | Worker:Machine)
   Data: Machines
REML criterion at convergence: 215.6876
Random effects:
 Groups         Name        Std.Dev.
 Worker:Machine (Intercept) 3.7295  
 Worker         (Intercept) 4.7811  
 Residual                   0.9616  
Number of obs: 54, groups:  Worker:Machine, 18; Worker, 6
Fixed Effects:
(Intercept)     MachineB     MachineC  
     52.356        7.967       13.917

Adakah yang bisa menjelaskan perbedaan antara model dan bagaimana m1mengurangi ke m2(diberikan simetri gabungan) dengan cara yang intuitif?

$I_n$$n$$1_n$$n$$y$$\mu$$\gamma$$\gamma$$y$

$$y \sim \mathcal{N}(\mu + \gamma, \sigma^2_y I_{54})$$

$\sigma^2_y$ adalah varian "residual".

$\gamma$

$$y \sim \mathcal{N}(\mu, \sigma^2_y I_{54} + \Sigma)$$

$\Sigma$$\gamma$$\Sigma$

m1

$$\gamma = Z \theta$$

$Z = I_{18} \otimes 1_3$$\theta^T = [\theta_{1,A}, \theta_{1,B}, \theta_{1,C} \dots \theta_{6,A}, \theta_{6,B}, \theta_{6,C}]$

$$\theta \sim \mathcal{N}(0, I_6 \otimes \Lambda)$$

$\Lambda$$\Lambda$$\sigma_\theta$$\tau$

$$\Lambda = \left[\begin{matrix} \sigma^2_\theta + \tau^2 & \tau^2 & \tau^2 \\ \tau^2 & \sigma^2_\theta + \tau^2 & \tau^2 \\ \tau^2 & \tau^2 & \sigma^2_\theta + \tau^2 \end{matrix}\right]$$

$\Lambda$ memiliki semua elemen pada set offdiagonal dengan nilai yang sama.)

$\Sigma = Z (I_6 \otimes \Lambda) Z^T$$\sigma^2_\theta + \tau^2 + \sigma^2_y$$i, j$$u, v$

$$\mathrm{cov}(y_{i,u}, y_{j,v}) = \begin{cases} 0 & \text{if } i\neq j \\ \tau^2 & \text{if } i=j, u\neq v \\ \sigma^2_\theta + \tau^2 & \text{if } i=j, u=v \end{cases}$$

For your m2, the random effects decompose into:

$$\gamma = Z \omega + X \eta$$

Where Z is as before, $X = I_6 \otimes 1_9$ is a design matrix that maps the random intercepts per worker onto observations, $\omega^T = [\omega_{1,A}, \omega_{1,B}, \omega_{1,C}, \dots, \omega_{6,A}, \omega_{6,B}, \omega_{6,C}]$ is the 18-dimensional vector of random intercepts for every combination of machine and worker; and $\eta^T = [\eta_{1}, \dots, \eta_{6}]$ is the 6-dimensional vector of random intercepts for worker. These are distributed as,

$$\eta \sim \mathcal{N}(0, \sigma^2_\eta I_6)$$ $$\omega \sim \mathcal{N}(0, \sigma^2_\omega I_{18})$$ Where $\sigma_\eta^2, \sigma_\omega^2$ are the variances of these random intercepts.

The marginal covariance structure of m2 is $\Sigma = \sigma^2_\omega Z Z^T + \sigma^2_\eta X X^T$, so that the variance of a given observation is $\sigma^2_\omega + \sigma^2_\eta + \sigma^2_y$, and the covariance between two observations from workers $i, j$ and machines $u, v$ is:

$$\mathrm{cov}(y_{i,u}, y_{j,v}) = \begin{cases} 0 & \text{if } i\neq j \\ \sigma_\eta^2 & \text{if } i=j,u\neq v \\ \sigma^2_\omega + \sigma^2_\eta & \text{if } i=j,u=v \end{cases}$$

So ... $\sigma^2_\theta \equiv \sigma^2_\omega$ and $\tau^2 \equiv \sigma^2_\eta$. If m1 assumed compound symmetry (which it doesn't with your call to lmer, because the random effects covariance is unstructured).

Brevity is not my strong point: this is all just a long, convoluted way of saying that each model has two variance parameters for the random effects, and are just two different ways of writing of the same "marginal" model.

In code ...

sigma_theta <- 1.8
tau         <- 0.5
sigma_eta   <- tau
sigma_omega <- sigma_theta
Z <- kronecker(diag(18), rep(1,3))
rownames(Z) <- paste(paste0("worker", rep(1:6, each=9)), 
                     rep(paste0("machine", rep(1:3, each=3)),6))
X <- kronecker(diag(6), rep(1,9))
rownames(X) <- rownames(Z)
Lambda <- diag(3)*sigma_theta^2 + tau^2

# marginal covariance for m1:
Z%*%kronecker(diag(6), Lambda)%*%t(Z)
# for m2:
X%*%t(X)*sigma_eta^2 + Z%*%t(Z)*sigma_omega^2