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When fitting the following simple model, using the 'lme4' R package and including a fixed and random slope term, I get:

Linear mixed model fit by REML ['lmerMod']
Formula: s ~ -1 + v8 + ((0 + v8 | Test))
   Data: d

Random effects:
 Groups   Name Variance Std.Dev.
 Test     v8   0.07676  0.2771  
 Residual      3.56656  1.8885  
Number of obs: 1647, groups:  Test, 41

Fixed effects:
  Estimate Std. Error t value
v8  2.83955    0.07845    36.2

Accordingly to my understanding of the model, the random effects of the above model are assumed to be a sample from a normal distribution with zero mean and variance equal to 0.07676. However when computing the variance directly from the estimated random effects I get a different result:

> var(ranef(m)$Test[,1])
[1] 0.02310659

> ranef(m)$Test[, 1]
[1] -0.160035611  0.091979744  0.024448306  0.103303471 -0.209127498 -0.115072081 -0.100169758 -0.439242134 -0.099724601  0.025481345
[11]  0.029968465 -0.099253951  0.166403989 -0.219299028  0.223841767  0.153699949  0.264563114 -0.143304606  0.177523761  0.054762082
[21] -0.056088689 -0.079896085 -0.013745153  0.122520213  0.214254150  0.252858418 -0.082402046 -0.095554209  0.000000000  0.045516274
[31] -0.017687631  0.003380337 -0.034645355 -0.184548770  0.143998225 -0.178323836  0.144361920 -0.106175741 -0.030701297  0.000000000
[41]  0.222132549

I was expecting to see something close to 0.07676 rather than 0.02310659?Could anybody clarify why I'm seeing this discrepancy?

If I plot the results I get:

plot(density(ranef(m)$Test[,1]), col="green")
hist(ranef(m)$Test[,1], freq=F, add=T)
curve(dnorm(x,mean=0,sd=0.2771), -1,1, col='red', add=T)
curve(dnorm(x,mean=0,sd=sd(ranef(m)$Test[,1])), -1,1, col='blue',add=T)

enter image description here

Where the green curve is the random effects non parametric density estimate, the blue curve is a gaussian with the variance estimated directly from the random effects and the red curve is a gaussian with a variance equal to the one returned by the lmer model.

Improve this question
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3
  • $\begingroup$ It'd be useful to have a picture. Say a histogram of your random effects coefficients with the proposed normal distribution over-layed? $\endgroup$
    – Matthew Drury
    Mar 22, 2016 at 15:32
  • 1
    $\begingroup$ (+1) Nicely done. $\endgroup$
    – Matthew Drury
    Mar 22, 2016 at 16:18
  • 1
    $\begingroup$ I think this question and this question may be of help. $\endgroup$
    – EdM
    Mar 22, 2016 at 19:55

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