Dear Thierry,

thanks for your answer. The same also applies for those two models:

  m1 <- lmer(rt ~ treatment + (1+treatment|subject),data=df,REML=F)
  m2 <- lmer(rt ~ treatment + (1|subject)+ (0+treatment|subject)
,data=df,REML=F)

deviance(m1)
[1] 32769.35

deviance(m2)
[1] 32769.35

anova(m1,m2)
Data: df
Models:
ml3: rt ~ treatment + (1 + treatment | subject)
ml2: rt ~ treatment + (1 | subject) + (0 + treatment | subject)
    Df   AIC   BIC logLik deviance Chisq Chi Df Pr(>Chisq)
ml3 10 32789 32848 -16385    32769
ml2 11 32791 32856 -16385    32769     0      1          1

Can you tell me why this is?

Thanks for your help!


On Sat, Jun 14, 2014 at 12:59 PM, ONKELINX, Thierry <
Thierry.ONKELINX@inbo.be> wrote:

> Dear Ulf,
>
> Treatment is a factor. Hence the random 'slope' is a random intercept for
> each level of treatment. The difference between 1 + treatment and 0 +
> treatment (in case of a factor) is that 1 + treament uses the first level
> as a reference (random 'intercept'). The other levels are coded as the
> difference from the reference levels.
>
> 0 + treatment uses no reference level. All levels are coded as the direct
> effect of the level. Hence you get the same fit but with different
> parametrisation. Note that the variance-covariance matrix of the random
> effects are different.
>
> Best regards,
>
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
> Forest
> team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
> Kliniekstraat 25
> 1070 Anderlecht
> Belgium
> + 32 2 525 02 51
> + 32 54 43 61 85
> Thierry.Onkelinx@inbo.be
> www.inbo.be
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to say
> what the experiment died of. ~ Sir Ronald Aylmer Fisher
> The plural of anecdote is not data. ~ Roger Brinner
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
> ________________________________________
> Van: r-sig-mixed-models-bounces@r-project.org [
> r-sig-mixed-models-bounces@r-project.org] namens Ulf Mertens [
> mertens.ulf@gmail.com]
> Verzonden: zaterdag 14 juni 2014 10:56
> Aan: r-sig-mixed-models@r-project.org
> Onderwerp: [R-sig-ME] Random Intercept + random slope model yields exactly
> the same results as random slope
>
> Hi there,
>
> When I run a random intercept + random slope model, I get exactly the same
> result as when running a model where there is no random intercept. I can't
> figure out why this is.
>
> Model 1:
>
> *m1 <- lmer(rt ~ treatment + (1+treatment|subject),data=df)*
> *m1*
> *Linear mixed model fit by REML ['lmerMod']*
> *Formula: rt ~ treatment + (1 + treatment | subject)*
> *   Data: df*
> *REML criterion at convergence: 32558.52*
> *Random effects:*
> * Groups   Name        Std.Dev. Corr       *
> * subject  (Intercept) 71.64               *
> *          treatment2  54.55    -0.31      *
> *          treatment3  54.79    -0.34  0.89*
> * Residual             97.38               *
> *Number of obs: 2700, groups: subject, 30*
> *Fixed Effects:*
> *(Intercept)   treatment2   treatment3  *
> *   598.6675      -0.9037      50.9770  *
>
> Model 2:
>
> *m2 <- lmer(rt ~ treatment + (0+treatment|subject) ,data=df)*
> *m2*
> *Linear mixed model fit by REML ['lmerMod']*
> *Formula: rt ~ treatment + (0 + treatment | subject)*
> *   Data: df*
> *REML criterion at convergence: 32558.52*
> *Random effects:*
> * Groups   Name       Std.Dev. Corr     *
> * subject  treatment1 71.64             *
> *          treatment2 75.18    0.72     *
> *          treatment3 74.03    0.72 0.94*
> * Residual            97.38             *
> *Number of obs: 2700, groups: subject, 30*
> *Fixed Effects:*
> *(Intercept)   treatment2   treatment3  *
> *   598.6675      -0.9037      50.9770  *
>
> Compare both models:
>
> *anova(m1,m2)*
> *refitting model(s) with ML (instead of REML)*
> *Data: df*
> *Models:*
> *m1: rt ~ treatment + (1 + treatment | subject)*
> *m2: rt ~ treatment + (0 + treatment | subject)*
> *    Df   AIC   BIC logLik deviance Chisq Chi Df Pr(>Chisq)*
> *m1 10 32598 32657 -16289    32578                        *
> *m2 10 32598 32657 -16289    32578     0      0          1*
>
> Thanks in advance
>
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