A GAM for time trends in a stepped-wedge trial with a binary outcome

Data generation

The data generation process I am using here follows along pretty closely with the earlier post, except, of course, the outcome has changed from continuous to binary. In this example, I’ve increased the correlation for between-period effects because it doesn’t seem like outcomes would change substantially from period to period, particularly if the time periods themselves are relatively short. The correlation still decays over time.

Here are the necessary libraries:

library(simstudy)
library(ggplot2)
library(data.table)
library(mgcv)
library(gratia)
library(patchwork)
library(mgcViz)
library(DHARMa)
library(itsadug)

The data generation, based on 24 sites, 25 time periods, and 100 individuals per site per time period, is formulated this way:

\[ y_{ijk} \sim Bin\left(p_{jk}\right) \\ \ \\ log\left( \frac{p_{ijk}}{1-p_{ijk}} \right) = -1.5 + a_j + b_{jk} + 0.65 A_{jk} \]

\(y_{ijk} \in \{0,1\}\) is the outcome, and \(p(y_{ijk} = 1) = p_{ijk}\). The log-odds ratio is a linear function of the site specific random intercept \(a_{j}\), the site-specific period \(k\) effect \(b_{jk}\), and treatment status of site \(j\) in period \(k\), \(A_{jk} \in \{ 0, 1\}\) depending the the stage of stepped wedge. The treatment effect in this case (an odds ratio) is \(exp(0.65) = 1.9\). The \(a_j \sim N(0, 0.6)\). The vector of site-period effects \(\mathbf{b_j} \sim N(0, \Sigma_b)\), where \(\Sigma_b = DRD\) is a \(25 \times 25\) covariance matrix based on a diagonal matrix \(D\) and an auto-regressive correlation structure \(R\):

\[ D = \sqrt{0.1} * I_{25 \times 25} \]

and

\[ R =\begin{bmatrix} 1 & \rho & \rho^2 & \dots & \rho^{24} \\ \rho & 1 & \rho & \dots & \rho^{23} \\ \rho^2 & \rho & 1 & \dots & \rho^{22} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \rho^{24} & \rho^{23} & \rho^{22} & \dots & 1 \\ \end{bmatrix}, \ \ \rho = 0.95 \]

Here is the implementation of this data generation process using simstudy:

def <- defData(varname = "a", formula = 0, variance = 0.6)
def <- defData(def, varname = "mu_b", formula = 0, dist = "nonrandom")
def <- defData(def, varname = "s2_b", formula = 0.1, dist = "nonrandom")
  
defOut <- defDataAdd(varname = "y", 
  formula = "-1.5 + a + b + 0.65 * A", 
  dist = "binary", link="logit"
)

set.seed(1913)

ds <- genData(24, def, id = "site")
ds <- addPeriods(ds, 25, "site", perName = "k")
ds <- addCorGen(
  dtOld = ds, idvar = "site", 
  rho = 0.95, corstr = "ar1",
  dist = "normal", param1 = "mu_b", param2 = "s2_b", cnames = "b"
)

ds <- trtStepWedge(ds, "site", nWaves = 24, 
  lenWaves = 1, startPer = 1, 
  grpName = "A", perName = "k"
)

ds$site <- as.factor(ds$site)
  
dd <- genCluster(ds, "timeID", numIndsVar = 100, level1ID = "id")
dd <- addColumns(defOut, dd)

dd
##        site  k     a mu_b s2_b timeID      b startTrt A    id y
##     1:    1  0 0.142    0  0.1      1 -0.867        1 0     1 0
##     2:    1  0 0.142    0  0.1      1 -0.867        1 0     2 0
##     3:    1  0 0.142    0  0.1      1 -0.867        1 0     3 0
##     4:    1  0 0.142    0  0.1      1 -0.867        1 0     4 0
##     5:    1  0 0.142    0  0.1      1 -0.867        1 0     5 0
##    ---                                                         
## 59996:   24 24 0.879    0  0.1    600 -0.546       24 1 59996 0
## 59997:   24 24 0.879    0  0.1    600 -0.546       24 1 59997 0
## 59998:   24 24 0.879    0  0.1    600 -0.546       24 1 59998 0
## 59999:   24 24 0.879    0  0.1    600 -0.546       24 1 59999 1
## 60000:   24 24 0.879    0  0.1    600 -0.546       24 1 60000 0

Here is visualization of the observed proportions of a good outcome (\(y = 1\)) by site and period:

Model estimation using a GAM

The first model will include a treatment effect and an overall smooth function of time, and then a site-specific smooth “effect”. I am using the function bam in the mgcv package, though I could use the gamm function, the gam function, or even the gamm4 function of the gamm4 package. In this case, all provide quite similar estimates, but bam has the advantage of running faster with this large data set. Here is the model:

\[ \text{log-odds}\left[P(y_{ijk} = 1)\right] = \beta_0 + \beta_1 A_{jk} + s(k) + s_j(k) \]

fit.A <- bam(
  y ~ A + s(k) + s(k, site, bs = "fs"), 
  data = dd, 
  method = "fREML",
  family = "binomial"
)
summary(fit.A)
## 
## Family: binomial 
## Link function: logit 
## 
## Formula:
## y ~ A + s(k) + s(k, site, bs = "fs")
## 
## Parametric coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  -1.4804     0.1583   -9.35   <2e-16 ***
## A             0.6951     0.0529   13.15   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Approximate significance of smooth terms:
##             edf Ref.df  Chi.sq p-value    
## s(k)       2.77   3.34    1.29    0.64    
## s(k,site) 91.69 238.00 5640.97  <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## R-sq.(adj) =  0.122   Deviance explained = 10.6%
## fREML =  85129  Scale est. = 1         n = 60000

The model does well to recover the true values of the parameters used in the data generation process (not always guaranteed for a single data set). The plot on the right shows the “main” smooth effect of time (i.e., across all sites), and the plot on the left shows the site-specific effects over time. The between-site variability is quite apparent.

draw(fit.A) +
  plot_annotation("") &
  theme(panel.grid = element_blank())

Goodness of fit

I am particularly interested in understanding if the model is a good fit based on the observed data. One way to do this is to repeatedly simulate predicted data from the model, and the visually assess whether the observed (i.e., actual) data falls reasonably into range of simulated data. To do this, I am using the simulate.gam function from the mgcViz package. I’ve created 95% bands based on the simulated data, and in this case it looks like the observed data fits into the bands quite well.

sim <- simulate.gam(fit.A, nsim = 1000)

ls <- split(sim, rep(1:ncol(sim), each = nrow(sim)))

dq <- lapply(ls, 
  function(x) {
    d <- cbind(dd, sim = x)
    d[, .(obs = mean(y), sim = mean(sim)), keyby = .(site, k)]
  }
)

dl <- rbindlist(dq, idcol = ".id")
df <- dl[, .(obs = mean(obs), min = quantile(sim, p = 0.025), 
             max = quantile(sim, 0.975)), keyby = .(site, k)]

ggplot(data = df, aes(x= k, y = obs)) +
  geom_ribbon(aes(x = k, ymin = min, ymax = max),
              alpha = 0.2, fill = "forestgreen") +
  geom_point(color = "forestgreen", size = 1) +
  
  facet_wrap( ~ site, ncol = 6) +
  theme(panel.grid = element_blank())

An alternative way to assess the goodness of fit is to generate a QQ-type plot that will alert us to any deviations. I am using the DHARMa package, which “uses a simulation-based approach to create readily interpretable quantile residuals for fitted generalized linear mixed models.” This residual is defined as “the value of the empirical density function at the value of the observed data.” The empirical density function comes from the same simulated data I just used to generate the 95% bands. It turns out that these residuals should be uniformly distributed if the model is a good fit. (See here for more details.)

The QQ-plot indicates a good fit if all the residuals lie on the diagonal line, as they do here:

simResp <- matrix(dl$sim, nrow = 600)
obsResp <- dq[[1]]$obs

DHARMaRes = createDHARMa(
  simulatedResponse = simResp, 
  observedResponse = obsResp, 
  integerResponse = F
)

plotQQunif(DHARMaRes, testDispersion = F, testOutliers = F)

A plot of the residuals against the predicted values also indicates a uniform distribution:

plotResiduals(DHARMaRes, quantreg = T)

fit.1curve <- bam(
  y ~ A + s(k, k = 4)  , 
  data = dd, 
  method = "fREML",
  family = "binomial"
)

compareML(fit.A, fit.1curve)
## fit.A: y ~ A + s(k) + s(k, site, bs = "fs")
## 
## fit.1curve: y ~ A + s(k, k = 4)
## 
## Chi-square test of fREML scores
## -----
##        Model Score Edf Difference    Df   p.value Sig.
## 1 fit.1curve 85146   4                                
## 2      fit.A 85129   7     16.823 3.000 2.352e-07  ***
## 
## AIC difference: -6166.12, model fit.A has lower AIC.

fit.noA <- bam(
  y ~ s(k) + s(k, site, bs = "fs"), 
  data = dd, 
  method = "fREML",
  family = "binomial"
)

compareML(fit.A, fit.noA)
## fit.A: y ~ A + s(k) + s(k, site, bs = "fs")
## 
## fit.noA: y ~ s(k) + s(k, site, bs = "fs")
## 
## Model fit.noA preferred: lower fREML score (5.899), and lower df (1.000).
## -----
##     Model Score Edf Difference    Df
## 1   fit.A 85129   7                 
## 2 fit.noA 85123   6     -5.899 1.000
## 
## AIC difference: -81.59, model fit.A has lower AIC.