Paired t-test as a special case of linear mixed-effect modeling

Question

We know that a paired t-test is just a special case of one-way repeated-measures (or within-subject) ANOVA as well as linear mixed-effect model, which can be demonstrated with lme() function the nlme package in R as shown below.

#response data from 10 subjects under two conditions
x1<-rnorm(10)
x2<-1+rnorm(10)

# Now create a dataframe for lme
myDat <- data.frame(c(x1,x2), c(rep("x1", 10), rep("x2", 10)), rep(paste("S", seq(1,10), sep=""), 2))
names(myDat) <- c("y", "x", "subj")

When I run the following paired t-test:

t.test(x1, x2, paired = TRUE)

I got this result (you will get a different result because of the random generator):

t = -2.3056, df = 9, p-value = 0.04657

With the ANOVA approach we can get the same result:

summary(aov(y ~ x + Error(subj/x), myDat))

# the F-value below is just the square of the t-value from paired t-test:
          Df  F value Pr(>F)
x          1  5.3158  0.04657

Now I can obtain the same result in lme with the following model, assuming a positive-definite symmetrical correlation matrix for the two conditions:

summary(fm1 <- lme(y ~ x, random=list(subj=pdSymm(form=~x-1)), data=myDat))

# the 2nd row in the following agrees with the paired t-test
# (Intercept) -0.2488202 0.3142115  9 -0.7918878  0.4488
# xx2          1.3325786 0.5779727  9  2.3056084  0.0466

Or another model, assuming a compound symmetry for the correlation matrix of the two conditions:

summary(fm2 <- lme(y ~ x, random=list(subj=pdCompSymm(form=~x-1)), data=myDat))

# the 2nd row in the following agrees with the paired t-test
# (Intercept) -0.2488202 0.4023431  9 -0.618428  0.5516
# xx2          1.3325786 0.5779727  9  2.305608  0.0466

With the paired t-test and one-way repeated-measures ANOVA, I can write down the traditional cell mean model as

Yij = μ + αi + βj + εij, i = 1, 2; j = 1, ..., 10

where i indexes condition, j indexes subject, Y_ij is the response variable, μ is constant for the fixed effect for overall mean, α_i is the fixed effect for condition, β_j is the random effect for subject following N(0, σ_p²) (σ_p² is population variance), and ε_ij is residual following N(0, σ²) (σ² is within-subject variance).

I thought that the cell mean model above would not be appropriate for the lme models, but the trouble is that I can't come up with a reasonable model for the two lme() approaches with the correlation structure assumption. The reason is that the lme model seems to have more parameters for the random components than the cell mean model above offers. At least the lme model provides exactly the same F-value, degrees of freedom, and p-value as well, which gls cannot. More specifically gls gives incorrect DFs due to the fact that it does not account for the fact that each subject has two observations, leading to much inflated DFs. The lme model most likely is overparameterized in specifying the random effects, but I don't know what the model is and what the parameters are. So the issue is still unresolved for me.

Answer 1

The equivalence of the models can be observed by calculating the correlation between two observations from the same individual, as follows:

As in your notation, let $Y_{ij} = \mu + \alpha_i + \beta_j + \epsilon_{ij}$, where $\beta_j \sim N(0, \sigma_p^2)$ and $\epsilon_{ij} \sim N(0, \sigma^2)$. Then $Cov(y_{ik}, y_{jk}) = Cov(\mu + \alpha_i + \beta_k + \epsilon_{ik}, \mu + \alpha_j + \beta_k + \epsilon_{jk}) = Cov(\beta_k, \beta_k) = \sigma_p^2$, because all other terms are independent or fixed, and $Var(y_{ik}) = Var(y_{jk}) = \sigma_p^2 + \sigma^2$, so the correlation is $\sigma_p^2/(\sigma_p^2 + \sigma^2)$.

Note that the models however are not quite equivalent as the random effect model forces the correlation to be positive. The CS model and the t-test/anova model do not.

EDIT: There are two other differences as well. First, the CS and random effect models assume normality for the random effect, but the t-test/anova model does not. Secondly, the CS and random effect models are fit using maximum likelihood, while the anova is fit using mean squares; when everything is balanced they will agree, but not necessarily in more complex situations. Finally, I'd be wary of using F/df/p values from the various fits as measures of how much the models agree; see Doug Bates's famous screed on df's for more details. (END EDIT)

The problem with your R code is that you're not specifying the correlation structure properly. You need to use gls with the corCompSymm correlation structure.

Generate data so that there is a subject effect:

set.seed(5)
x <- rnorm(10)
x1<-x+rnorm(10)
x2<-x+1 + rnorm(10)
myDat <- data.frame(c(x1,x2), c(rep("x1", 10), rep("x2", 10)), 
                    rep(paste("S", seq(1,10), sep=""), 2))
names(myDat) <- c("y", "x", "subj")

Then here's how you'd fit the random effects and the compound symmetry models.

library(nlme)
fm1 <- lme(y ~ x, random=~1|subj, data=myDat)
fm2 <- gls(y ~ x, correlation=corCompSymm(form=~1|subj), data=myDat)

The standard errors from the random effects model are:

m1.varp <- 0.5453527^2
m1.vare <- 1.084408^2

And the correlation and residual variance from the CS model is:

m2.rho <- 0.2018595
m2.var <- 1.213816^2

And they're equal to what is expected:

> m1.varp/(m1.varp+m1.vare)
[1] 0.2018594
> sqrt(m1.varp + m1.vare)
[1] 1.213816

Other correlation structures are usually not fit with random effects but simply by specifying the desired structure; one common exception is the AR(1) + random effect model, which has a random effect and AR(1) correlation between observations on the same random effect.

EDIT2: When I fit the three options, I get exactly the same results except that gls doesn't try to guess the df for the term of interest.

> summary(fm1)
...
Fixed effects: y ~ x 
                 Value Std.Error DF   t-value p-value
(Intercept) -0.5611156 0.3838423  9 -1.461839  0.1778
xx2          2.0772757 0.4849618  9  4.283380  0.0020

> summary(fm2)
...
                 Value Std.Error   t-value p-value
(Intercept) -0.5611156 0.3838423 -1.461839  0.1610
xx2          2.0772757 0.4849618  4.283380  0.0004

> m1 <- lm(y~ x + subj, data=myDat)
> summary(m1)
...
            Estimate Std. Error t value Pr(>|t|)   
(Intercept)  -0.3154     0.8042  -0.392  0.70403   
xx2           2.0773     0.4850   4.283  0.00204 **

(The intercept is different here because with the default coding, it's not the mean of all subjects but instead the mean of the first subject.)

It's also of interest to note that the newer lme4 package gives the same results but doesn't even try to compute a p-value.

> mm1 <- lmer(y ~ x + (1|subj), data=myDat)
> summary(mm1)
...
            Estimate Std. Error t value
(Intercept)  -0.5611     0.3838  -1.462
xx2           2.0773     0.4850   4.283

Answer 2

You might also consider using function mixed in package afex to return p values with Kenward-Roger df approximation, which returns identical p values as a paired t test:

library(afex)
mixed(y ~ x + (1|subj), type=3,method="KR",data=myDat) 

Or

library(lmerTest)
options(contrasts=c('contr.sum', 'contr.poly'))
anova(lmer(y ~ x + (1|subj),data=myDat),ddf="Kenward-Roger")