---
title: "Check perfect model"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Check perfect model}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

## 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:

```{r}
library("plyr")
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)
  }
}
```

Let us first model the perfect situation where $\sigma^2 = 0$ (so we set `resid_var` to zero) and where the ages align perfectly.

```{r}
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 = 2)
data <- gen_dat()
head(data)
```

Check the correlation matrix of the $y$'s.

```{r}
library("tidyr")
library("dplyr")
d <- as_tibble(data[,-3])
broad <- t(spread(d, subject, X1))[-1,]
cor(broad)
```

## Fit model

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

```{r}
library("brokenstick")
knots <- 1:3
boundary <- c(1, 4)
fit <- brokenstick(X1 ~ age | subject, data, 
                   knots = knots, boundary = boundary,
                   method = "lmer")
omega <- fit$omega
beta <- fit$beta
sigma2 <- fit$sigma2
round(beta, 2)
round(sigma2, 4)

# correlation random effects
round(covar, 3)
round(omega, 2)

# covariances measured data
round(omega + diag(sigma2, 4), 3)
round(cov(broad), 3)

# convert to time-to-time correlation matrix
round(cov2cor(omega + diag(sigma2, 4)), 3)
round(cor(broad), 3)
```

## 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.

