2
$\begingroup$

If I run a mixed model including a level 2 predictor as a random slope, it is sometimes possible to get an estimation of the variance. For example:

require(lmerTest)
test = lmer(Sweetness ~ Income + (Income|Consumer) + (Income|product), data = carrots)
summary(test)

Here, I am including income, which has no variance within each consumer, as a random slope. The output of the test suggests that consumer variance of the slope is not zero (SD=0.01087).

Here is the full output:

Linear mixed model fit by REML t-tests use Satterthwaite approximations to degrees
  of freedom [lmerMod]
Formula: Sweetness ~ Income + (Income | Consumer) + (Income | product)
   Data: carrots

REML criterion at convergence: 3828.9

Scaled residuals: 
     Min       1Q   Median       3Q      Max 
-2.58301 -0.68129 -0.07647  0.69873  2.59386 

Random effects:
 Groups   Name        Variance  Std.Dev. Corr 
 Consumer (Intercept) 0.4154951 0.64459       
          Income      0.0001182 0.01087  -1.00
 product  (Intercept) 0.2850742 0.53392       
          Income      0.0003368 0.01835  -1.00
 Residual             1.3572890 1.16503       
Number of obs: 1161, groups:  Consumer, 97; product, 12

Fixed effects:
            Estimate Std. Error       df t value Pr(>|t|)    
(Intercept)  3.84828    0.27317 38.83000  14.088   <2e-16 ***
Income      -0.18368    0.08892 93.46000  -2.066   0.0416 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
       (Intr)
Income -0.815

why is this term in the output and how could it be interpreted?

****EDIT

Another example where intercept and slope are not correlated at -1, and where the Income value within each Consumer is the same, thus random slope should be 0/non-estimable. This is a subset of a real dataset I'm working with. The full dataset results in a -.47 correlation between intercept and slope.

datTest = structure(list(Sweetness = c(6.26, 0.49, 1.65, 2.08, 7.37, 10, 
9.65, 5.95, 0.74, 7.86, 0.19, 3.66, 10, 7.37, 2.86, 0.43, 7.76, 
5.02, 7.08, 10, 8.21, 6.05, 0.39, 9.61, 2.08, 7.12, 10, 8.99, 
4.47, 0.39, 2.08, 0.88, 5.33, 7.96, 9.86, 3.7, 0.56, 9.73, 1.83, 
7.67, 10, 8.29, 5.95, 0.43, 5.86, 2.35, 7.65, 10, 5.04), Income  = structure(c(0.085, 
-27.915, 37.085, -15.915, -7.915, -22.915, 2.085, 0.085, -27.915, 
37.085, -15.915, -7.915, -22.915, 2.085, 0.085, -27.915, 37.085, 
-15.915, -7.915, -22.915, 2.085, 0.085, -27.915, 37.085, -15.915, 
-7.915, -22.915, 2.085, 0.085, -27.915, 37.085, -15.915, -7.915, 
-22.915, 2.085, 0.085, -27.915, 37.085, -15.915, -7.915, -22.915, 
2.085, 0.085, -27.915, 37.085, -15.915, -7.915, -22.915, 2.085
), .Dim = c(49L, 1L)), Consumer = c(2, 3, 7, 16, 18, 20, 23, 2, 3, 
7, 16, 18, 20, 23, 2, 3, 7, 16, 18, 20, 23, 2, 3, 7, 16, 18, 
20, 23, 2, 3, 7, 16, 18, 20, 23, 2, 3, 7, 16, 18, 20, 23, 2, 
3, 7, 16, 18, 20, 23), product = c(1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 
2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 
5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7)), row.names = c(2L, 
3L, 7L, 16L, 18L, 20L, 23L, 203L, 204L, 208L, 217L, 219L, 221L, 
224L, 404L, 405L, 409L, 418L, 420L, 422L, 425L, 605L, 606L, 610L, 
619L, 621L, 623L, 626L, 806L, 807L, 811L, 820L, 822L, 824L, 827L, 
1007L, 1008L, 1012L, 1021L, 1023L, 1025L, 1028L, 1208L, 1209L, 
1213L, 1222L, 1224L, 1226L, 1229L), .Names = c("Sweetness", "Income", 
"Consumer", "product"), class = "data.frame")

lmList(Sweetness ~  Income|Consumer, data = datTest)

Call: lmList(formula = Sweetness ~ Income | Consumer, data = datTest) 
Coefficients:
   (Intercept) Income
2     5.034286     NA
3     0.490000     NA
7     6.364286     NA
16    2.061429     NA
18    6.554286     NA
20    9.708571     NA
23    8.201429     NA

Degrees of freedom: 49 total; 35 residual
Residual standard error: 1.890709

summary(lmer(Sweetness ~  Income + (Income|Consumer) + (Income|product), data = datTest))

Linear mixed model fit by REML t-tests use Satterthwaite approximations to degrees of freedom [
lmerMod]
Formula: Sweetness ~ Income + (Income | Consumer) + (Income | product)
   Data: datTest

REML criterion at convergence: 208.8

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-1.9550 -0.3096  0.1659  0.4416  1.8956 

Random effects:
 Groups   Name        Variance Std.Dev. Corr 
 Consumer (Intercept) 3.336092 1.82650       
          Income      0.011874 0.10897  -0.81
 product  (Intercept) 0.376941 0.61396       
          Income      0.001311 0.03621  1.00 
 Residual             2.264987 1.50499       
Number of obs: 49, groups:  Consumer, 7; product, 7

Fixed effects:
            Estimate Std. Error      df t value Pr(>|t|)   
(Intercept)  6.19235    0.99829 3.32300   6.203  0.00621 **
Income       0.03679    0.06247 0.09200   0.589  0.88019   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
       (Intr)
Income 0.037 

summary(lmer(Sweetness ~  Income + (Income||Consumer) + (Income||product), data = datTest))

Linear mixed model fit by REML t-tests use Satterthwaite approximations to degrees of freedom [
lmerMod]
Formula: Sweetness ~ Income + ((1 | Consumer) + (0 + Income | Consumer)) +  
    ((1 | product) + (0 + Income | product))
   Data: datTest

REML criterion at convergence: 212.7

Scaled residuals: 
     Min       1Q   Median       3Q      Max 
-2.28024 -0.29821  0.08234  0.47653  1.69613 

Random effects:
 Groups     Name        Variance  Std.Dev.
 Consumer   (Intercept) 2.5399675 1.59373 
 Consumer.1 Income      0.0266043 0.16311 
 product    (Intercept) 0.1078963 0.32848 
 product.1  Income      0.0006577 0.02565 
 Residual               2.6104516 1.61569 
Number of obs: 49, groups:  Consumer, 7; product, 7

Fixed effects:
            Estimate Std. Error      df t value Pr(>|t|)   
(Intercept)  6.37522    1.01119 2.81400   6.305  0.00974 **
Income       0.07441    0.09030 4.20200   0.824  0.45409   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
       (Intr)
Income 0.335 
$\endgroup$
  • $\begingroup$ How do you know there is no variance within each consumer? Have you run the output of lmList and verified that each consumer specific regression model has an "income" coefficient equal to that of the fixed effect of the mixed model? $\endgroup$ – AdamO Jun 19 '17 at 16:33
  • $\begingroup$ I just ran lmList as such: lmList(Sweetness ~ Income | Consumer, data = carrots) and the income coefficients are all NA. Each consumer has the same income across observations, which is the reason why I am saying there is no income variance within each consumer - and thus shouldn't be able to estimate coefficients for income within consumer. Unless I am misunderstanding something about the meaning of the random slope. $\endgroup$ – user137451 Jun 19 '17 at 16:42
  • 1
    $\begingroup$ I don't know but note that you get correlation -1 between intercept and slope for both grouping factors (consumer and product), possibly indicating problems with convergence. What happens if you drop both correlation parameters by using || instead of |? $\endgroup$ – amoeba says Reinstate Monica Jun 19 '17 at 19:55
  • 1
    $\begingroup$ @user137451 (1) Variance of a random effect going to exactly zero even though there is some non-zero (but small) variation in the dataset is a well-known effect. It happens, even in simulations when we know that the true variance is non-zero. See stats.stackexchange.com/a/115180 and links therein. (2) As I said, obtaining correlation -1 is a red flag, and removing correlation parameters from the model is a sensible thing to do. Good that the slope variance goes to zero with carrots. (3) What you describe in a different dataset is a bit weird. Could you share it + code? $\endgroup$ – amoeba says Reinstate Monica Jun 21 '17 at 7:50
  • 1
    $\begingroup$ @amoeba thanks for pointing that out and for the help and edit. I have added the requested subset of my data with the code I'm running, and illustrating an example where the random slope is still estimated when removing the correlation parameter. Based on my limited knowledge, I wouldn't assume anything is wrong with the convergence from this output. $\endgroup$ – user137451 Jun 21 '17 at 12:38

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