[R-sig-ME] lmer (lme4): % total variance explained by random effect

[Jarrod Hadfield]

Hi Katharina,

The difficulty with random intercept-slope models is that they  
(usually) give rise to a non-constant variance across the range of the  
covariate. This means that the percentage of variance explained will  
depend on which value of the covariate you evaluate.

If you extract the (co)variance matrix for the intercept-slopes:

V<-VarCorr(name_of_model)$name_of_random_effect

and then set up a design matrix with ones in the first column, and a  
set of covariate values in the second:

Z<-cbind(rep(1,100), seq(-1,1,length=100))

then

M<-diag(Z%*%V%*%t(Z))

is equal to the variance explained by the random effect for each value  
of the covariate.  If there are no other effects M/(M+Ve) gives you  
the proportion explained where Ve is the residual variance.

Hope this helps,

Jarrod

On 2 Aug 2009, at 16:23, Katharina May wrote:

> Hi,
>
> just out of curiosity because nobody is answering:
> is it not not possible to calculate the variance described by a random
> effect on slope and intercept as percentage of the total variance
> (variance of random effect + unexplained variance)?
>
> Would be more than happy if somebody can help me...
>
> Thanks,
>
>       Katharina
>
>
> 2009/7/24 Katharina May <may.katharina at googlemail.com>:
>> Hello,
>>
>> just to say sorry if this questions may be somewhat  
>> "inappropriate": I'm a
>> bachelor student,
>> recently started with R and with trying to understand mixed models,  
>> but I'm
>> somewhat stuck with
>> the following problem and hope somebody might be able to help me  
>> finding a
>> solution:
>>
>> How can I get the variance (in % of the total variance) which is  
>> explained
>> by the random effect (both on slope
>> and intercept together)?
>> My aim is to say something like xx% of the variance is explained by  
>> the
>> random effect...
>>
>> As I'm not sure how to deal with this I would be more than happy  
>> for any
>> hints...
>>
>> Thank you very much and With Best Wishes from Freising/Germany,
>>
>>                         Katharina
>>
>>
>>
>> here an example output of a mixed model I use with 1 random effect  
>> on both
>> slope and intercept,
>> fitted with method=ML:
>>
>>
>> Linear mixed model fit by maximum likelihood
>> Formula: log(AGB) ~ log(BM_roots) + (log(BM_roots) |
>> as.factor(biomass_data[which(biomass_data$woody ==      1), 2]))
>>    Data: biomass_data[which(biomass_data$woody == 1), ]
>>    AIC   BIC logLik deviance REMLdev
>>  588.6 619.6 -288.3    576.6     583
>> Random effects:
>>  Groups                                                     Name
>>                          Variance   Std.Dev.   Corr
>>  as.factor(biomass_data[which(biomass_data$woody == 1), 2])  
>> (Intercept)
>> 1.7568529  1.325463
>>                                                              
>> log(BM_roots)
>>                           0.0071313  0.084447  -0.393
>>  Residual
>> 0.0809467 0.284511
>> Number of obs: 1282, groups:  
>> as.factor(biomass_data[which(biomass_data$woody
>> == 1), 2]), 22
>>
>> Fixed effects:
>>               Estimate Std. Error t value
>> (Intercept)    1.33062    0.29669    4.48
>> log(BM_roots)  0.93182    0.02441   38.17
>>
>> Correlation of Fixed Effects:
>>             (Intr)
>> log(BM_rts) -0.446
>>
>>
>
>
>
> -- 
> Time flies like an arrow, fruit flies like bananas.
>
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