Check perfect model

Objective

In general, the broken stick model smoothes the observed growth trajectory. What happens of all observations are already aligned to the break ages? Does the model perfectly represent the data? Is the covariance matrix of the random effects (Ω) equal to the covariance between the measurements? Is σ2 equal to zero?

Data generation

We adapt code from http://www.davekleinschmidt.com/sst-mixed-effects-simulation/simulations_slides.pdf to generate test data:

library("plyr")
## ------------------------------------------------------------------------------
## You have loaded plyr after dplyr - this is likely to cause problems.
## If you need functions from both plyr and dplyr, please load plyr first, then dplyr:
## library(plyr); library(dplyr)
## ------------------------------------------------------------------------------
## 
## Attaching package: 'plyr'
## The following objects are masked from 'package:dplyr':
## 
##     arrange, count, desc, failwith, id, mutate, rename, summarise,
##     summarize
library("mvtnorm")
make_data_generator <- function(resid_var = 1,
                                ranef_covar = diag(c(1, 1)), n = 100
                                ) {
  ni <- nrow(ranef_covar)
  generate_data <- function() {
    # sample data set under mixed effects model with random slope/intercepts 
    simulated_data <- rdply(n, {
      b <- t(rmvnorm(n = 1, sigma = ranef_covar))
      epsilon <- rnorm(n = length(b), mean = 0, sd = sqrt(resid_var))
      b + epsilon
    })
  data.frame(
    subject = rep(1:n, each = ni),
    age = rep(1:ni, n),
    simulated_data)
  }
}

We choose between the perfect situation where σ2 = 0 and the noisy case σ2 = 1 and where the ages align perfectly.

resid_var <- 0
resid_var <- 1
set.seed(77711)
covar <- matrix(c(1, 0.7, 0.5, 0.3,
                  0.7, 1, 0.8, 0.5,
                  0.5, 0.8, 1, 0.6,
                  0.3, 0.5, 0.6, 1), nrow = 4)
gen_dat <- make_data_generator(n = 10000, 
                               ranef_covar = covar,
                               resid_var = resid_var)
data <- gen_dat()
head(data)
##   subject age .n      X1
## 1       1   1  1 -0.9478
## 2       1   2  1 -2.0837
## 3       1   3  1 -2.6512
## 4       1   4  1 -2.5526
## 5       2   1  2 -0.0825
## 6       2   2  2 -1.2707

We wish to reproduce the correlation matrix among the y’s from the mixed model estimates. The target correlation matrix is:

library("tidyr")
library("dplyr")
d <- as_tibble(data[,-3])
broad <- t(spread(d, subject, X1))[-1,]
cor(broad)
##       [,1]  [,2]  [,3]  [,4]
## [1,] 1.000 0.350 0.255 0.161
## [2,] 0.350 1.000 0.406 0.246
## [3,] 0.255 0.406 1.000 0.313
## [4,] 0.161 0.246 0.313 1.000

Fit model

Fit broken stick model, with knots specified at ages 1:4.

library("brokenstick")
knots <- 1:3
boundary <- c(1, 4)
fit <- brokenstick(X1 ~ age | subject, data, 
                   knots = knots, boundary = boundary,
                   method = "lmer")
## Warning: number of observations (=40000) <= number of random effects (=40000)
## for term (0 + age_1 + age_2 + age_3 + age_4 | subject); the random-effects
## parameters and the residual variance (or scale parameter) are probably
## unidentifiable
## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl = control$checkConv, :
## unable to evaluate scaled gradient
## Warning in checkConv(attr(opt, "derivs"), opt$par, ctrl = control$checkConv, :
## Model failed to converge: degenerate Hessian with 1 negative eigenvalues
omega <- get_omega(fit, hide = "no")
beta <- coef(fit, hide = "no")
sigma2 <- fit$sigma2
round(beta, 2)
## age_1 age_2 age_3 age_4 
## -0.02 -0.02  0.01  0.01
round(sigma2, 4)
## [1] 0.818
# correlation random effects
round(covar, 3)
##      [,1] [,2] [,3] [,4]
## [1,]  1.0  0.7  0.5  0.3
## [2,]  0.7  1.0  0.8  0.5
## [3,]  0.5  0.8  1.0  0.6
## [4,]  0.3  0.5  0.6  1.0
round(omega, 2)
##       age_1 age_2 age_3 age_4
## age_1  1.23  0.71  0.52  0.33
## age_2  0.71  1.16  0.82  0.49
## age_3  0.52  0.82  1.22  0.63
## age_4  0.33  0.49  0.63  1.17
# covariances measured data
round(omega + diag(sigma2, 4), 3)
##       age_1 age_2 age_3 age_4
## age_1 2.052 0.706 0.521 0.326
## age_2 0.706 1.982 0.816 0.489
## age_3 0.521 0.816 2.034 0.630
## age_4 0.326 0.489 0.630 1.992
round(cov(broad), 3)
##       [,1]  [,2]  [,3]  [,4]
## [1,] 2.052 0.706 0.521 0.326
## [2,] 0.706 1.982 0.816 0.489
## [3,] 0.521 0.816 2.034 0.630
## [4,] 0.326 0.489 0.630 1.992
# convert to time-to-time correlation matrix
round(cov2cor(omega + diag(sigma2, 4)), 3)
##       age_1 age_2 age_3 age_4
## age_1 1.000 0.350 0.255 0.161
## age_2 0.350 1.000 0.406 0.246
## age_3 0.255 0.406 1.000 0.313
## age_4 0.161 0.246 0.313 1.000
round(cor(broad), 3)
##       [,1]  [,2]  [,3]  [,4]
## [1,] 1.000 0.350 0.255 0.161
## [2,] 0.350 1.000 0.406 0.246
## [3,] 0.255 0.406 1.000 0.313
## [4,] 0.161 0.246 0.313 1.000
z <- predict(fit, x = "knots", include_data = FALSE, shape = "wide")[, -1]
# off-diagonal elements of covariance of broken stick estimates approach correlation
# not enough variance in the diagonal because of smoothing
cov(z)
##       1     2     3
## 1 0.795 0.595 0.475
## 2 0.595 0.791 0.681
## 3 0.475 0.681 0.821
# correlations of broken stick estimates are inflated because of smoothing
cor(z)
##       1     2     3
## 1 1.000 0.750 0.587
## 2 0.750 1.000 0.845
## 3 0.587 0.845 1.000

Conclusions

  1. If σ2 = 0, then the off-diagonal elements of Ω reproduce the correlations among the y’s. The estimate of σ2 is too high (about 0.13 instead of 0).
  2. If σ2 > 0, then Ĉ = Ω + σ̂2I(ni) reproduces the sample covariance matrix between y’s exactly.
  3. cov2cor(hatC) reproduces the sample time-to-time correlation matrix.
  4. Conclusion: The unbiased estimate of the time-to-time correlation matrix among the (unobserved) measurements at the knots:
cov <- get_omega(fit)
chat <- cov + diag(fit$sigma2, nrow(cov))
r <- cov2cor(chat)
r
##       age_1 age_2 age_3
## age_1 1.000 0.350 0.255
## age_2 0.350 1.000 0.406
## age_3 0.255 0.406 1.000