I have a question regarding variance components I obtained after fitting a mixed linear model with random slope and random intercept using R software. My model is the following:

$ y_{ij} = β_0 + β_1 time_{ij}+ u_{0j} + u_{1j} time_i + ε_{ij}$

Results from fitting this linear mixed model is:

rs <- lmer(y ~ time + (time| id), data=dataset, REML=FALSE)

Linear mixed model fit by maximum likelihood  ['lmerMod']
Formula: y ~ time + (time | id)
   Data: dataset
      AIC       BIC    logLik  deviance  df.resid 
  66.2413   70.0756   -27.1206  54.2413         8 
Random effects:
 Groups   Name       Variance  Std.Dev.   Corr 
 id      (Intercept) 8.846024. 2.97423        
          time       0.006426  0.08016   -1.00
 Residual            1.473614. 1.21393     
Number of obs: 14, groups:  id, 4

Fixed Effects:
              Estimate Std.Error t value
(Intercept)   99.04580  1.57192   63.01      
time          -0.16086  0.04457  -3.61

My questions are:

  1. What is the slope variance $σ^2(β_1)$? It is 0.006426 or the result from vcov(rs)[2,2]?
  2. Is it correct to say that the variance of residuals $σ^2(ε)$ is 1.473614?

1 Answer 1


You have answered your own questions. Indeed, the estimated variance of the random slope $u_{1j}$ (and not $\beta_1$ as you write) is 0.006426 and the estimated residual variance, i.e. the estimated variance of $\varepsilon_{ij}$ is 1.473614. Obviously, the estimated variance of the random intercept $u_{0j}$ is 8.846024. Notice that the variance of the estimate of the fixed effect $\beta_1$ is $0.04457^2$, which you can also get by the command vcov(rs)[2,2].

  • $\begingroup$ Thank you for your reply. But σ2(β1) can be obtained as vcov(rs)[2,2] in R? $\endgroup$
    – user140911
    Dec 2, 2016 at 21:31
  • $\begingroup$ Yes! See the last sentence of my answer. $\endgroup$
    – utobi
    Dec 2, 2016 at 22:00

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