# 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 (\(\Omega)\) equal to the covariance between the measurements? Is \(\sigma^2\) 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 \(\sigma^2 = 0\) and the noisy case \(\sigma^2 = 1\) 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 \(\sigma^2=0\), then the off-diagonal elements of \(\Omega\) reproduce the correlations among the \(y\)’s. The estimate of \(\sigma^2\) is too high (about 0.13 instead of 0).
2. If \(\sigma^2 > 0\), then \(\hat C = \Omega + \hat\sigma^2 I(n_i)\) 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