# How to interpret variance and correlation of random effects in a mixed-effects model?

I hope you all don't mind this question, but I need help interpreting output for a linear mixed effects model output I've been trying to learn to do in R. I am new to longitudinal data analysis and linear mixed effects regression. I have a model I fitted with weeks as the time predictor, and score on an employment course as my outcome. I modeled score with weeks (time) and several fixed effects, sex and race. My model includes random effects. I need help understanding what the variance and correlation means. The output is the following:

Random effects
Group   Name    Variance
EmpId intercept 680.236
weeks           13.562
Residual 774.256

The correlaton is .231.

I can interpret the correlation as there is a a positive relationship between weeks and score but I want to be able to say it in terms of "23% of ...".

I really appreciate the help.

Thanks "guest" and Macro for replying. Sorry, for not replying, I was out at a conference and I’m now catching up. Here is the output and the context.

Here is the summary for the LMER model I ran.

>summary(LMER.EduA)
Linear mixed model fit by maximum likelihood
Formula: Score ~ Weeks + (1 + Weeks | EmpID)
Data: emp.LMER4

AIC     BIC   logLik   deviance   REMLdev
1815     1834  -732.6     1693    1685

Random effects:
Groups   Name       Variance Std.Dev. Corr
EmpID   (Intercept)  680.236  26.08133
Weeks         13.562 3.682662  0.231
Residual             774.256  27.82546
Number of obs: 174, groups: EmpID, 18

Fixed effects:
Estimate Std. Error  t value
(Intercept)  261.171      6.23     37.25
Weeks          11.151      1.780    6.93

Correlation of Fixed Effects:
(Intr)
Days -0.101

I don’t understand how to interpret the variance and residual for the random effects and explain it to someone else. I also don’t know how to interpret the correlation, other than it is positive which indicates that those with higher intercepts have higher slopes and those with those with lower intercepts have lower slopes but I don’t know how to explain the correlation in terms of 23% of . . . . (I don’t know how to finish the sentence or even if it makes sense to do so). This is a different type analysis for us as we (me) are trying to move into longitudinal analyses.

I hope this helps.

Thanks for your help so far.

Zeda

-
Zeda, it would be helpful to see more of the R output here, including the output's summary of the fixed effects – guest Mar 10 '12 at 21:35
One thing I can see is that the estimated intraclass correlation for EmpID is $\hat{\rho} = 680.236/(680.236+13.562+774.256)$. That is, the estimated correlation between two individuals in the same level of EmpID is $\hat{\rho}$. I agree with @guest that more output (and some context) would be helpful. – Macro Mar 10 '12 at 22:27
Zeda, I've converted your reply as an edit, and merged your two unregistered accounts. Please, register this one so that you can follow up and update your post yourself. – chl Mar 19 '12 at 15:30

## 1 Answer

Your fitted model with lme() can be expressed as

$y_{ij} = \alpha_0 + \alpha_1 x_j + \delta_{0i} + \delta_{1i} x_j + \epsilon_{ij}$

where $y_{ij}$ is the score of $i$th employee at $x_j$ weeks, $\alpha_0$ and $\alpha_1$ are the fixed intercept and slope respectively, $\delta_{0i}$ and $\delta_{1i}$ are the random intercept and slope, and $\epsilon_{ij}$ is the residual. The assumptions for the random effects $\delta_{0i}$, $\delta_{1i}$ and residual $\epsilon_{ij}$ are

$(\delta_{0i}, \delta_{1i})^T \stackrel{d}{\sim} N((0, 0)^T, G)$ and $\epsilon_{ij} \stackrel{d}{\sim} N(0, \sigma^2)$,

where the variance-covariance structure $G$ is a 2 x 2 symmetric matrix of form

$$\begin{pmatrix} g_1^2&g_{12}^2\\ g_{12}^2&g_2^2 \end{pmatrix}$$

You can get this from VarCorr(LMER.EduA)$EmpID. Your result basically says that$\alpha_0$= 261.171,$\alpha_1$= 11.151,$g_1^2$= 680.236,$g_2^2$= 13.562, and$\sigma^2$= 774.256.$g_{12}^2$can be found in VarCorr(LMER.EduA) or calculated as$0.23\times \sqrt{g_1^2 g_2^2}$. Specifically$g_1^2$= 680.236 shows the variability of the intercept across employees,$g_2^2$= 13.562 is the amount of variability in the slope across employees, and 0.231 indicates the positive correlation between intercept and slope (when an employee's intercept increases by one unit of standard deviation, that employee's slope would increase by 0.231 standard deviations). - Please check that recent edits didn't alter the meaning of your reply (Personally, I just fixed some$\LaTeX\$ expressions). – chl Mar 20 '12 at 20:51
@chl: I really appreciate you for structuring my response in such a nice format (I know nothing about LaTex). More importantly you corrected my sloppy response regarding the covariance part. Thanks again, chl! – bluepole Mar 20 '12 at 21:26
Credits should go to @GGeco who provided details about the VC matrix; as I said, I only texified part of your reply (and +1). – chl Mar 20 '12 at 21:42
How would this work if you many random effects? – user1987097 Jun 19 '14 at 13:26