[R-sig-ME] Understanding variance components

Chen, Gang (NIH/NIMH) [C] gangchen at mail.nih.gov
Thu Jan 19 16:14:26 CET 2017

Nice, Henrik!

One thing we need to resolve, though, is this.

The variances for model m1:

(1) intercept

[1] 2.565887

(2) residuals

attr(summary(m1)$varcor, "sc")^2
[1] 2.196212

And the variances for model m2:

(1) female

[1] 0.9944589

(2) male

[1] 4.137316

(3) residuals

attr(summary(m2)$varcor, "sc")^2
[1] 2.196212

It’s great to see that the residual variance matches between the two models m1 and m2. However, the intercept variance in m1, 2.565887, is not equal to the sum of female and male variances, 0.9944589 + 4.137316 = 5.131775. However, if we divide the total variance of female and male by 2, we have (0.9944589 + 4.137316)/2 = 2.565887. Why is that?

If we code the two groups as

obk.long$gender_F <- sqrt(2)*as.numeric(obk.long$gender == "F")
obk.long$gender_M <- sqrt(2)*as.numeric(obk.long$gender == "M”)

then we have the desired result,

m2 <- lmer(value ~ gender*phase+(0+gender_F|id)+(0+gender_M|id), data=obk.long)
[1] 2.565888

Even though the variance part is reconciled, I cannot come up with a good explanation as to why this coding strategy is required. Any thought?


On Jan 19, 2017, at 6:13 AM, Henrik Singmann <singmann at psychologie.uzh.ch<mailto:singmann at psychologie.uzh.ch>> wrote:

Hi Gang,

I have an idea which is based on the last example given on ?lmer:
## Fit sex-specific variances by constructing numeric dummy variables

I am not sure if this is entirely correct, but it looks good to me. If not, hopefully someone more knowledgeable will jump in.

## Original model without gender specific variance:
data(obk.long, package = "afex")

m1 <- lmer(value ~ gender*phase+(1|id), data=obk.long)
## Groups   Name        Std.Dev.
## id       (Intercept) 1.6018
## Residual             1.4820
## [1] 911.1599

## to get gender specific vari8ances, we construct two dummy variables:
obk.long$gender_F <- as.numeric(obk.long$gender == "F")
obk.long$gender_M <- as.numeric(obk.long$gender == "M")
m2 <- lmer(value ~ gender*phase+(0+gender_F|id)+(0+gender_M|id), data=obk.long)
## Groups   Name     Std.Dev.
## id       gender_F 0.99723
## id.1     gender_M 2.03404
## Residual          1.48196
## [1] 908.297

So far, looks reasonably close. Same for the conditional modes (thx Phillip). Left two columns is separate, right is joint variance.
cbind(ranef(m2)$id, rep(NA, 16), ranef(m1)$id)
##      gender_F   gender_M rep(NA, 16) (Intercept)
## 1   0.0000000 -3.0986753          NA  -3.0351426
## 2   0.0000000 -2.1328544          NA  -2.0891242
## 3   0.0000000  0.1207276          NA   0.1182523
## 4  -0.9806229  0.0000000          NA  -1.0642708
## 5  -0.3995130  0.0000000          NA  -0.4335918
## 6   0.0000000  2.6962499          NA   2.6409683
## 7   0.0000000  1.4084888          NA   1.3796103
## 8  -0.3995130  0.0000000          NA  -0.4335918
## 9  -0.6900680  0.0000000          NA  -0.7489313
## 10  0.0000000  0.4426679          NA   0.4335918
## 11  0.0000000 -1.1670336          NA  -1.1431057
## 12  0.0000000  1.7304291          NA   1.6949498
## 13  1.3438166  0.0000000          NA   1.4584452
## 14 -0.6900680  0.0000000          NA  -0.7489313
## 15  0.4721518  0.0000000          NA   0.5124267
## 16  1.3438166  0.0000000          NA   1.4584452

And, finally, the same for the fixed effects:
##       (Intercept)           genderM         phasepost
##      6.000000e+00      7.500000e-01     -6.250000e-01
##          phasepre genderM:phasepost  genderM:phasepre
##     -2.000000e+00     -1.243450e-15     -1.687539e-15
##      (Intercept)           genderM         phasepost
##     6.000000e+00      7.500000e-01     -6.250000e-01
##         phasepre genderM:phasepost  genderM:phasepre
##    -2.000000e+00      1.829648e-14      1.820766e-14

Hope that helps,

On Jan 18, 2017, at 5:18 PM, Chen, Gang (NIH/NIMH) [C] <gangchen at mail.nih.gov<mailto:gangchen at mail.nih.gov>> wrote:

Happy New Year, Henrik! Thanks for explaining the details. A couple of days after I posted the question, I realized that my question was silly! Once I laid out the LME model equation, my original confusion was resolved.

Actually I meant to ask a slightly different question. Let me use the dataset embedded in your ‘afex’ package as an example:

data(obk.long, package = "afex”)

Suppose that my base model is

lmer(value ~ gender*phase+(1|id), data=obk.long)

Is there a way to specify a different variance for each gender in one model?


On Jan 13, 2017, at 12:06 PM, Henrik Singmann <singmann at psychologie.uzh.ch<mailto:singmann at psychologie.uzh.ch>> wrote:

Hi Gang,

Sorry that I so am late to the party, but in case you are still interested I will reply (and, of course, for the archive).

The answer is basically given in the old faq:

(1|site/block) = (1|site)+(1|site:block)

Which is exactly what is given in your output. A random intercept for Worker and a random intercept for each worker:Machine interaction.

To answer your questions. The random intercepts do not have base or reference levels. They are increments or decrements to the overall intercept for each level of Worker or the Machine:Worker combination. The reported variance is the estimated variance of these increments, which is most likely unequal to the actual variance you would obtain by calculating it from the estimated increments, which are sometimes called BLUPs (I wonder if a better term for those exist).

Hope that helps,

PS: Belated Happy New Year to everyone.

Am 05.01.2017 um 17:28 schrieb Chen, Gang (NIH/NIMH) [C]:
Suppose that I have the following dataset in R:

mydata <- Machines[Machines$Machine!='C’,]

With the following model:

print(lmer(score ~ 1 + (1|Worker/Machine), data=mydata), ranef.comp="Var")

I have the variance components as shown below:

Random effects:
Groups         Name        Variance
Machine:Worker (Intercept) 46.00
Worker         (Intercept) 13.84
Residual                    1.16

I have trouble understanding exactly what the first two components are: Machine:Worker and Worker? Specifically,

1) What is the variance for Worker: corresponding to the base (or reference) level of the factor Machine? If so, what is the base level: the first level in the dataset or alphabetically the first level (it happens to be the same in this particular dataset)?

2) What is the variance for Machine:Worker? Is it the variance for the second level of the factor Machine, or the extra variance relative to the variance for Worker?

Furthermore, for the model:

print(lmer(score ~ 1 + (1|Worker/Machine), data=Machines), ranef.comp="Var")

what is the variance for Machine:Worker in the following result since there are 3 levels involved in the factor Machine?

Random effects:
Groups         Name        Variance
Machine:Worker (Intercept) 60.2972
Worker         (Intercept)  7.3959
Residual                    0.9246

R-sig-mixed-models at r-project.org mailing list

R-sig-mixed-models at r-project.org mailing list

R-sig-mixed-models at r-project.org<mailto:R-sig-mixed-models at r-project.org> mailing list

	[[alternative HTML version deleted]]

More information about the R-sig-mixed-models mailing list