In a longitudinal study, outcomes $Y_{it}$ of units $i$ are repeatedly measuret at time points $t$ with a total of $m$ fixed measurement occasions (fixed = measurements on units are taken at the same time).

Units are randomly assigned either to a treatment, $G=1$, or to a control group, $G=0$. I want to estimate and test the average effect of treatment, i.e. $$ATE=E(Y | G=1) – E(Y | G=0),$$ where expectations are taken across time and individuals. I consider using a fixed-occasion multilevel (mixed-effects) model for this purpose:

$$Y_{it} = \alpha + \beta G_i + u_{0i} + e_{it}$$

with $\alpha$ the intercept, $\beta$ the $ATE$, $u$ a random intercept across units, and $e$ the residual.

Now I am considering alternative model

$$Y_{it} = \tilde{\beta} G_i + \sum_{j=1}^m \kappa_j d_{ij} + \sum_{j=1}^m \gamma_j d_{ij} G_i + \tilde{u}_{0i} + \tilde{e}_{it}$$

which contains the fixed effects $\kappa_j$ for each occasion $t$ where dummy $d_t=1$ if $j=t$ and $0$ else. In addition this model contains an interaction between treatment and time with parameters $\gamma$. So this model takes into account that the effect of $G$ may differ across time. This is informative in itself, but I believe that it should also increase precision of estimation of the parameters, because the heterogeneity in $Y$ is taken into account.

However, in this model the $\tilde{\beta}$ coefficient does not seem to equal the $ATE$ anymore. Instead it represents the ATE at the first occasion ($t=1$). So the estimate of $\tilde{\beta}$ may be more efficient than $\beta$ but it does not represent the $ATE$ anymore.

My questions are:

- What is the best way to estimate the treatment effect in this longitudinal study design?
- Do I have to use model 1 or is there a way to use (perhaps more efficient) model 2?
- Is there a way to have $\tilde{\beta}$ have the interpretation of the $ATE$ and $\gamma$ the occasion specific deviation (e.g. using effect coding)?

**Answer**

Addressing your question “I wonder how to get the ATE out of model 2” in the comments:

First of all, in your model 2, not all $\gamma_j$ is identifiable which leads to the problem of rank deficiency in design matrix. It is necessary to drop one level, for instance assuming $\gamma_j =0$ for $j=1$. That is, using the contrast coding and assume the treatment effect at period 1 is 0. In R, it will code the interaction term with treatment effect at period 1 as the reference level, and that is also the reason why $\tilde{\beta}$ has the interpretation of treatment effect at period 1. In SAS, it will code the treatment effect at period $m$ as the reference level, then $\tilde{\beta}$ has the interpretation of treatment effect at period $m$, not period 1 anymore.

Assuming the contrast is created in the R way, then the coefficients estimated for each interaction term (I will still denote this by $\gamma_j$, though it is not precisely what you defined in your model) has the interpretation of treatment effect difference between time period $j$ and time period 1. Denote ATE at each period $\mathrm{ATE}_j$, then $\gamma_j= \mathrm{ATE}_j – \mathrm{ATE}_1$ for $j=2,\dots, m$. Therefore an estimator for $\mathrm{ATE}_j$ is $\tilde{\beta} + \gamma_j$. (ignoring the notation difference between true parameter and estimator itself because laziness) And naturally your $\mathrm{ATE}=\beta=\frac{1}{m} \sum_{j=1}^m \mathrm{ATE}_j=\frac{\tilde{\beta}+(\tilde{\beta}+\gamma_2)+\cdots+(\tilde{\beta}+\gamma_m)}{m} = \tilde{\beta}+\frac{1}{m}(\gamma_2 + \cdots + \gamma_m)$.

I did a simple simulation in R to verify this:

```
set.seed(1234)
time <- 4
n <-2000
trt.period <- c(2,3,4,5) #ATE=3.5
kj <- c(1,2,3,4)
intercept <- rep(rnorm(n, 1, 1), each=time)
eij <- rnorm(n*time, 0, 1.5)
trt <- rep(c(rep(0,n/2),rep(1,n/2)), each=time)
y <- intercept + trt*(rep(trt.period, n))+rep(kj,n)+eij
sim.data <- data.frame(id=rep(1:n, each=time), period=factor(rep(1:time, n)), y=y, trt=factor(trt))
library(lme4)
fit.model1 <- lmer(y~trt+(1|id), data=sim.data)
beta <- getME(fit.model1, "fixef")["trt1"]
fit.model2 <- lmer(y~trt*period + (1|id), data=sim.data)
beta_t <- getME(fit.model2, "fixef")["trt1"]
gamma_j <- getME(fit.model2, "fixef")[c("trt1:period2","trt1:period3","trt1:period4")]
results <-c(beta, beta_t+sum(gamma_j)/time)
names(results)<-c("ATE.m1", "ATE.m2")
print(results)
```

And the results verifies this:

```
ATE.m1 ATE.m2
3.549213 3.549213
```

I don’t know how to directly change contrast coding in model 2 above, so to illustrate how one can directly use a linear function of the interaction terms, as well as how to obtain the standard error, I used the multcomp package:

```
sim.data$tp <- interaction(sim.data$trt, sim.data$period)
fit.model3 <- lmer(y~tp+ (1|id), data=sim.data)
library(multcomp)
# w= tp.1.1 + (tp.2.1-tp.2.0)+(tp.3.1-tp.3.0)+(tp.4.1-tp.4.0)
# tp.x.y=interaction effect of period x and treatment y
w <- matrix(c(0, 1,-1,1,-1,1,-1,1)/time,nrow=1)
names(w)<- names(getME(fit.model3,"fixef"))
xx <- glht(fit.model3, linfct=w)
summary(xx)
```

And here is the output:

```
Simultaneous Tests for General Linear Hypotheses
Fit: lmer(formula = y ~ tp + (1 | id), data = sim.data)
Linear Hypotheses:
Estimate Std. Error z value Pr(>|z|)
1 == 0 3.54921 0.05589 63.51 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Adjusted p values reported -- single-step method)
```

I think the standard error is obtained by $\sqrt{w \hat{V} w^T}$ with $w$ being the above linear combination form and $V$ the estimated variance-covariance matrix of the coefficients from model 3.

**Deviation coding**

Another way to make $\tilde{\beta}$ having directly the interpretation of $\mathrm{ATE}$ is to use deviation coding, so that later covariates represent $\mathrm{ATE}_j – \mathrm{ATE}$ comparison:

```
sim.data$p2vsmean <- 0
sim.data$p3vsmean <- 0
sim.data$p4vsmean <- 0
sim.data$p2vsmean[sim.data$period==2 & sim.data$trt==1] <- 1
sim.data$p3vsmean[sim.data$period==3 & sim.data$trt==1] <- 1
sim.data$p4vsmean[sim.data$period==4 & sim.data$trt==1] <- 1
sim.data$p2vsmean[sim.data$period==1 & sim.data$trt==1] <- -1
sim.data$p3vsmean[sim.data$period==1 & sim.data$trt==1] <- -1
sim.data$p4vsmean[sim.data$period==1 & sim.data$trt==1] <- -1
fit.model4 <- lmer(y~trt+p2vsmean+p3vsmean+p4vsmean+ (1|id), data=sim.data)
```

Output:

```
Fixed effects:
Estimate Std. Error t value
(Intercept) 3.48308 0.03952 88.14
trt1 3.54921 0.05589 63.51
p2vsmean -1.14774 0.04720 -24.32
p3vsmean 1.11729 0.04720 23.67
p4vsmean 3.01025 0.04720 63.77
```

**Attribution***Source : Link , Question Author : tomka , Answer Author : jujae*