[R-sig-ME] lmer (lme4): % total variance explained by random effect
may.katharina at googlemail.com
Sun Aug 9 22:53:51 CEST 2009
Oh yes, now I get the picture, thanks a lot!
as the variance explained by the random effect for each value of the
covariate is (luckely) more or less similar within my data, I will
take the mean to have an approximation..
Thank you very much,
2009/8/3 Jarrod Hadfield <j.hadfield at ed.ac.uk>:
> 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:
> 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))
> 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,
> On 2 Aug 2009, at 16:23, Katharina May wrote:
>> 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...
>> 2009/7/24 Katharina May <may.katharina at googlemail.com>:
>>> just to say sorry if this questions may be somewhat "inappropriate": I'm
>>> bachelor student,
>>> recently started with R and with trying to understand mixed models, but
>>> somewhat stuck with
>>> the following problem and hope somebody might be able to help me finding
>>> How can I get the variance (in % of the total variance) which is
>>> 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
>>> Thank you very much and With Best Wishes from Freising/Germany,
>>> here an example output of a mixed model I use with 1 random effect on
>>> 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
>>> 0.0071313 0.084447 -0.393
>>> 0.0809467 0.284511
>>> Number of obs: 1282, groups:
>>> == 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:
>>> log(BM_rts) -0.446
>> Time flies like an arrow, fruit flies like bananas.
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