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Re: Linear VS LTBS

From: Leonid Gibiansky <LGibiansky>
Date: Fri, 21 Aug 2009 08:45:55 -0400

Large RSE, inability to converge, failure of the covariance step are
often caused by the over-parametrization of the model. If you already
have bootstrap, look at the scatter-plot matrix of parameters versus
parameters (THATA1 vs THETA2, .., THETA1 vs OMEGA1, ...), these are very
  informative plots. If you have over-parametrization on the population
level, it will be seen in these plots as strong correlations of the
parameter estimates.

Also, look on plots of ETAs vs ETAs. If you see strong correlation
(close to 1) there, it may indicate over-parametrization on the
individual level (too many ETAs in the model).

For random effect with a very large RSE on the variance, I would try to
remove it and see what happens with the model: often, this (high RSE) is
the indication that the error effect is not needed.

Also, try combined error model (on log-transformed variables):

   Y = LOG(IPRED) + W1*EPS(1)


Why concentrations were on LOQ? Was it because BQLs were inserted as
LOQ? Then this is not a good idea.

Leonid Gibiansky, Ph.D.
President, QuantPharm LLC
e-mail: LGibiansky at
tel: (301) 767 5566

Indranil Bhattacharya wrote:
> Hi Joachim, thanks for your suggestions/comments.
> When using LTBS I had used a different error model and the error block
> is shown below
> IPRED = -5
> IF (F.GT.0) IPRED = LOG(F) ;log transforming predicition
> W=1
> IWRES=IRES/W ;Uniform Weighting
> Y = IPRED + ERR(1)
> I also performed bootsrap on both LTBS and non-LTBS models and the
> non-LTBS CI were much more tighter and the precision was greater than
> non-LTBS.
> I think the problem plausibly is with the fact that when fitting the
> non-transformed data I have used the proportional + additive model while
> using LTBS the exponential model (which converts to additional model due
> to LTBS) was used. The extra additive component also may be more
> important in the non-LTBS model as for some subjects the concentrations
> were right on LOQ.
> I tried the dual error model for LTBS but does not provide a CV%. So I
> am currently running a bootstrap to get the CI when using the dual error
> model with LTBS.
> Neil
> On Fri, Aug 21, 2009 at 3:01 AM, Grevel, Joachim
> <Joachim.Grevel
> wrote:
> Hi Neil,
> 1. When data are log-transformed the $ERROR block has to change:
> additive error becomes true exponential error which cannot be
> achieved without log-transformation (Nick, correct me if I am wrong).
> 2. Error cannot "go away". You claim your structural model (THs)
> remained unchanged. Therefore the "amount" of error will remain the
> same as well. If you reduce BSV you may have to "pay" for it with
> increased residual variability.
> 3. Confidence intervals of ETAs based on standard errors produced
> during the covariance step are unreliable (many threads in NMusers).
> Do bootstrap to obtain more reliable C.I..
> These are my five cents worth of thought in the early morning,
> Good luck,
> Joachim
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> -----Original Message-----
> *From:* owner-nmusers
> <mailto:owner-nmusers
> [mailto:owner-nmusers
> <mailto:owner-nmusers
> Bhattacharya
> *Sent:* 20 August 2009 17:07
> *To:* nmusers
> *Subject:* [NMusers] Linear VS LTBS
> Hi, while data fitting using NONMEM on a regular PK data set
> and its log transformed version I made the following observations
> - PK parameters (thetas) were generally similar between
> regular and when using LTBS.
> -ETA on CL was similar
> -ETA on Vc was different between the two runs.
> - Sigma was higher in LTBS (51%) than linear (33%)
> Now using LTBS, I would have expected to see the ETAs unchanged
> or actually decrease and accordingly I observed that the eta
> values decreased showing less BSV. However the %RSE for ETA on
> VC changed from 40% (linear) to 350% (LTBS) and further the
> lower 95% CI bound has a negative number for ETA on Vc (-0.087).
> What would be the explanation behind the above observations
> regarding increased %RSE using LTBS and a negative lower bound
> for ETA on Vc? Can a negative lower bound in ETA be considered
> as zero?
> Also why would the residual vriability increase when using LTBS?
> Please note that the PK is multiexponential (may be this is
> responsible).
> Thanks.
> Neil
> --
> Indranil Bhattacharya
> --
> Indranil Bhattacharya
Received on Fri Aug 21 2009 - 08:45:55 EDT

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