[R-sig-ME] trouble using the lmer function

[Ken Beath]

The problem is in the calculation of the derivatives, this

grMod(donnees$DOSE,log(5),log(500),log(30),log(2))

will show that some are NaN. It probably means that underflow has resulted
in a 0/0 in the gradient formula.

Ken

On Sun, August 23, 2009 9:41 am, Julie Bertrand wrote:
> Dear All,
>
> I meet trouble using the lmer function on a data set simulated with a
> gamma Emax Model.
>
> I use the following code, where initial conditions are set to the true
> values :
> GammaEmaxmodel<-function(D,lE0,lE50,lEmax,lHill)exp(lE0)+exp(lEmax)*D^exp(lHill)/(exp(lE50)^exp(lHill)+D^exp(lHill))
> grMod <- deriv(body(GammaEmaxmodel), namevec =
> c("lE0","lE50","lEmax","lHill"), func = GammaEmaxmodel)
>
> fit.lme4<-nlmer(DV~grMod(DOSE,lE0,lE50,lEmax,lHill)~(lE0+lEmax+lE50)|ID,#
>         donnees,
>         start = c(lE0=log(5),lE50=log(500),lEmax=log(30),lHill=log(2))
>         )
> and obtain the error message : "Erreur dans asMethod(object) : matrix is
> not symmetric [1,2]"
>
> I am on Windows XP, with R version 2.9.1 (2009-06-26) and
> lme4_0.999375-31.
>
> When I do not try to estimate the Hill parameter I obtain no error message
> :
>  >
> GammaEmaxmodel<-function(D,lE0,lE50,lEmax)exp(lE0)+exp(lEmax)*D/(exp(lE50)+D)
>  > grMod <- deriv(body(GammaEmaxmodel), namevec =
> c("lE0","lE50","lEmax"), func = GammaEmaxmodel)
>  >
>  >
>  > fit.lme4<-nlmer(DV~grMod(DOSE,lE0,lE50,lEmax)~(lE0+lEmax+lE50)|ID,#
> + donnees,
> + start = c(lE0=log(5),lE50=log(500),lEmax=log(30))
> + )
>  > fit.lme4
> Nonlinear mixed model fit by the Laplace approximation
> Formula: DV ~ grMod(DOSE, lE0, lE50, lEmax) ~ (lE0 + lEmax + lE50) | ID
>    Data: donnees
>   AIC  BIC logLik deviance
>  2235 2273  -1107     2215
> Random effects:
>  Groups   Name  Variance Std.Dev. Corr
>  ID       lE0    1.7172  1.3104
>           lEmax 18.8214  4.3384   -0.171
>           lE50  27.4499  5.2393   -0.128  0.954
>  Residual        4.2807  2.0690
> Number of obs: 324, groups: ID, 81
>
> Fixed effects:
>       Estimate Std. Error t value
> lE0     1.2769     0.1591   8.024
> lE50    7.9102     0.6817  11.603
> lEmax   4.6933     0.5867   7.999
>
> Correlation of Fixed Effects:
>       lE0    lE50
> lE50  -0.058
> lEmax -0.098  0.962
>
> Further, I succeed in estimating the Hill parameter using the nlme
> function (nlme_3.1-93) :
>
>  > fit.nlme2<-nlme(DV~GammaEmaxmodel(DOSE,lE0,lE50,lEmax,lHill),
> + data=dat,
> + fixed=lE0+lEmax+lE50+lHill~1,
> + random=lE0+lEmax+lE50~1,
> +             start =
> c(lE0=log(5),lE50=log(500),lEmax=log(30),lHill=log(2)),
> + weights=NULL
> +      )
>  > fit.nlme2
> Nonlinear mixed-effects model fit by maximum likelihood
>   Model: DV ~ GammaEmaxmodel(DOSE, lE0, lE50, lEmax, lHill)
>   Data: dat
>   Log-likelihood: -1007.409
>   Fixed: lE0 + lEmax + lE50 + lHill ~ 1
>       lE0     lEmax      lE50     lHill
> 1.4964002 3.2601144 6.0436652 0.4885341
>
> Random effects:
>  Formula: list(lE0 ~ 1, lEmax ~ 1, lE50 ~ 1)
>  Level: ID
>  Structure: General positive-definite, Log-Cholesky parametrization
>          StdDev       Corr
> lE0      7.642895e-02 lE0  lEmax
> lEmax    5.686393e-01 1.00
> lE50     3.690756e-05 0.01 0.01
> Residual 3.906402e+00
>
> Number of Observations: 324
> Number of Groups: 81
>
> The data are simulated with Hill=2, please find enclose the data set
> that leads to these outputs (simdat.txt).
> I also simulated a correlation between the lEmax and lE50 parameters
> however I coul not figure out how to code such a random effects matrix
> in lmer.
>
> The first error message is due to lmer trouble in estimating the Hill
> parameter on this data set, or must I change something in my code or
> data set ?
>
> Thank you in advance.
>
> Julie Bertrand
>
> --
> UMR 738, INSERM, Université Paris Diderot
>
>
>
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