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I am using the glmer() function from the lme4 package to run a GLMM using the poisson distribution. In all the examples that I see, the random effects part of the output has a residual part that has been estimated from the data (surrounded by 2 asterisks on either side in the example below). This information can then be used in interpreting the amount of variation explained by the random effect. Here is an example:

> summary(M1)
Linear mixed model fit by REML 
Formula: Richness ~ NAP * fExp + (1 | fBeach) 
Data: RIKZ 
AIC   BIC    logLik   deviance REMLdev
236.5 247.3 -112.2    230.3    224.5
Random effects:
Groups   Name          Variance Std.Dev.
fBeach   (Intercept)   3.3072   1.8186  
**Residual             8.6605   2.9429**
Number of obs: 45, groups: fBeach, 9

Fixed effects:
              Estimate Std. Error t value
(Intercept)   8.8611     1.0208   8.681
NAP          -3.4637     0.6279  -5.517
fExp11       -5.2556     1.5451  -3.401
NAP:fExp11    2.0005     0.9461   2.114

Correlation of Fixed Effects:
           (Intr) NAP    fExp11
NAP        -0.181              
fExp11     -0.661  0.120       
NAP:fExp11  0.120 -0.664 -0.221

However, when I use my own data, I get output that does not include this information, and I am not sure why. I want to know how much variation is explained by my random effects, but can't figure out how to access the information necessary to answer the question. Any clues? Is this a data/statistics issue or is this a knowing how to access the information issue? I apologize if I'm asking in the wrong place. The output I get looks similar to the following output:

Generalized linear mixed model fit by the Laplace approximation 
Formula: y ~ z.score(x1) + z.score(x2) + z.score(x3) + z.score(x4) + z.score(x5) +      z.score(x6) + (1 | RE) 
Data: p 
AIC   BIC logLik deviance
419.5 454.7 -201.8    403.5
Random effects:
Groups Name        Variance Std.Dev.
RE     (Intercept) 0.021605 0.14699 
Number of obs: 600, groups: RE, 40

Fixed effects:
                  Estimate   Std. Error z value Pr(>|z|)    
(Intercept)       1.70591    0.02911    58.60   < 2e-16 ***
z.score(x1)       0.19087    0.03595    5.31    1.10e-07 ***
z.score(x2)      -0.14302    0.04083   -3.50    0.000460 ***
z.score(x3)      -0.16562    0.04020   -4.12    3.79e-05 ***
z.score(x4)       0.13229    0.03425    3.86    0.000112 ***
z.score(x5)      -0.10588    0.03985   -2.66    0.007885 ** 
z.score(x6)       0.17600    0.05798    3.04    0.002401 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Correlation of Fixed Effects:
            (Intr) z.(x1) z.(x2 z.s(x3) z.(x4 z.(x5
z.scr(x1)  -0.051                                   
z.s(x2)     0.038  0.259                            
z.scr(x3)   0.045  0.156  0.113                     
z.(x4      -0.040  0.144 -0.052  0.044              
z.(x5       0.026 -0.368 -0.339 -0.072 -0.073       
z.scor(x6) -0.031 -0.020  0.002 -0.143 -0.004  0.004

Here is some sample data, to be fit with glmer(y ~ x1 + (1|RE), data=d, family=poisson).

d <- data.frame(
  y  =  c(3, 5, 2, 6, 3, 7, 2, 3, 0, 4, 0,10, 1, 4, 0, 4, 2, 3, 0, 6, 
          3, 4, 2, 3, 2, 3, 3, 4, 0, 5, 6, 5, 4, 4, 0, 3, 1, 6, 0, 3, 2, 
          2, 1, 6, 2, 7, 0, 2, 0, 4, 0, 6, 4, 5, 1, 5, 1, 4, 1, 2, 3, 6, 
          6, 7, 0, 5, 0, 9, 1, 4, 5, 6, 1, 7, 1, 4, 1, 4, 0, 4, 1, 6, 1, 
          4, 0, 7, 1, 4, 0, 6, 0, 7, 2, 6, 0, 6, 1, 5, 0, 4, 1, 7, 2, 4, 
          1, 5, 1, 7, 2, 5, 0, 4, 3, 5, 1, 4, 0, 3, 0, 6, 0, 8, 3, 9, 0, 
          2, 3, 8, 0, 1, 0, 3, 0, 5, 0, 4, 4, 5, 0, 5, 1, 5, 3, 5, 1, 4, 
          3, 4, 4, 4, 4, 4, 4, 7, 1, 8, 1, 4, 0, 2, 2, 5, 1, 4, 1, 5, 1, 
          4, 2, 4, 2, 4, 0, 6, 1, 6, 0, 6, 1, 2, 1, 3, 1, 8, 1, 6, 1, 6, 
          0, 6, 1, 6, 2, 6, 2, 4, 0, 1, 1, 1, 1, 6, 5, 5, 1, 5, 2, 4, 2, 
          6, 1, 7, 1, 8, 2, 8, 1, 8, 2, 4, 1, 7, 3, 6, 4, 7, 3, 7, 1, 6, 
          3, 5, 1,10, 1, 7, 2, 5, 1, 5, 0, 6, 1, 8, 4, 7, 1, 6, 1, 9, 
          0, 9, 1, 3, 2, 5, 2, 9, 3, 5, 0, 2, 2, 3, 0, 5, 0, 5, 0, 4, 3, 
          6, 1,10, 2, 8, 0, 6, 0, 4, 2, 6, 2, 4, 2, 6, 1, 4, 0, 5, 2, 
          6, 1, 5, 2, 5, 1, 5, 1, 5),
  x1 = rep(c(0.1008, 0.0511, 0.1792, 1.0345), c(80, 80, 80, 60)),
  RE = rep(c(37, 88, 139, 190, 241, 292, 343, 394, 91, 142, 193, 244, 295, 
             346, 397, 40, 94, 145, 43, 196, 247, 298, 349, 400, 301, 352, 
             403, 250, 148, 199, 46, 97, 355, 406, 253, 304, 49, 100, 151, 
             202, 37, 88, 139, 190, 241, 292, 343, 394, 91, 142, 193, 244, 
             295, 346, 397, 40, 43, 94, 145, 247, 298, 349, 196, 400, 199, 
             250, 301, 352, 406, 46, 97, 148, 403, 49, 100, 151, 202, 253, 
             304, 355, 37, 88, 139, 190, 241, 292, 343, 394, 193, 244, 346, 
             397, 295, 40, 91, 142, 43, 94, 145, 196, 46, 97, 148, 151, 247, 
             400, 298, 349, 352, 199, 250, 301, 403, 253, 304, 355, 202, 406, 
             49, 100, 37, 88, 139, 190, 241, 292, 343, 394, 346, 397, 193, 
             244, 295, 40, 91, 142, 43, 94, 145, 196, 247, 298, 349, 400, 
             97, 148, 46, 199, 250, 301), each=2)
)
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    $\begingroup$ To answer your question (now edited away) about sample data: No, there's not a great way to include sample data; using dput, as you did, is the best you can usually do, though it can usually be made cleaner by setting the rownames to NULL first, and in some cases (as here) using rep can help tremendously. $\endgroup$ Mar 27 '12 at 1:23
  • $\begingroup$ The other guess I can suggest you look at is your model was fit by LaPlace rather than by REML. I do not know whether that might be an issue. I see you do have a random effect after all. $\endgroup$ Mar 27 '12 at 5:32
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    $\begingroup$ I don't think such a value exists for a GLMM. The model you show that does have a Residual component is a LMM not a GLMM. In a GLMM there is a known mean-variance relationship and there isn't a parameter sigma to estimate. You can compute the residual deviance but this doesn't fit into the scheme of being a variance parameter (and hence can not be squared to give a standard deviation). That would be sufficient for it not to be shown in the output from the GLMM. $\endgroup$ Mar 27 '12 at 9:47
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Such a value doesn't exists for a GLMM. The model you show that does have a Residual component is a LMM not a GLMM. In a GLMM there is a known mean-variance relationship and there isn't a parameter $\sigma$ to estimate. You can compute the residual deviance but this doesn't fit into the scheme of being a variance parameter (and hence can not be squared to give a standard deviation). That would be sufficient for it not to be shown in the output from the GLMM.

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    $\begingroup$ I'm re-posting this as an answer here as whilst I wasn't 100% sure, Ben Bolker (one of the lme4 developers) comment on the Q on SO that my Answer was correct/explained the lack of Residual component in the output from lmer() that the OP was discussing. Not sure why the Answer turned into a comment, but I do believe this answers the software implementation bit of the Q. $\endgroup$ Mar 27 '12 at 16:30

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