Re: Log transformation of concentration

From: Leonid Gibiansky Date: March 26, 2009 technical Source: mail-archive.com
Hi Chenguang, The main reason to do the log transformation is the numerical algorithm used in nonmem for error model. If you try to fit the error model Y=F*EXP(eps) nonmem will take only the first term of the EXP function expansion and will use the error model Y=F*(1+EPS) Therefore, the only way to get true exponential (not proportional) model is to log-transform both parts: LOG(Y)=LOG(F)+EPS Note that this is done on the very last step. All parameters have the same meaning. All differential equations are written and solved for F. Then, after you obtain F, you take the log. In the DV column, you put the log of observed concentrations, so that your actual code is Y=LOG(F)+EPS Last year I compared the performance of FOCE with interaction for models with and without log-transformation, and found the performance to be similar (in terms of bias and precision of parameter estimates): you can find the poster on PAGE web site. Still, for several real data sets, I've seen that the log-transformed model provided slightly better fit, especially for data with large residual error. Leonid -------------------------------------- Leonid Gibiansky, Ph.D. President, QuantPharm LLC web: www.quantpharm.com e-mail: LGibiansky at quantpharm.com tel: (301) 767 5566 Chenguang Wang wrote: > Dear NONMEM users, > > I am working on a PK model and using the log-transformed concentration data. I'v read some discussions in the NONMEM user group about the log-transformed concentration. But I am still not very clear about this. Could anybody give me a reason to do the transform on concentration? Also, I am curious that after the transform, will the fixed effect have the same meaning as that in the untransformed model? For example, theta1 is the clearance, after log-transform of concentration, would the estimation of theta1 still stands for the population clearance? To my simple thinking about the differential equation, > > d(lnc)/dt= (dc/dt)*(1/c). Therefore, a "c" will be multiplied to the right term of the orginal differential equation. I think the solution of that equation might be different from the original one. If it is different, how can I explain the theta1 in the log transformed model? > > Would anyone please give me some explainations or references? > > Thanks a lot! > > Chenguang
Mar 26, 2009 Chenguang Wang Log transformation of concentration
Mar 26, 2009 Leonid Gibiansky Re: Log transformation of concentration
Mar 26, 2009 Jeroen Elassaiss-Schaap RE: Log transformation of concentration
Mar 26, 2009 Jeroen Elassaiss-Schaap RE: Log transformation of concentration
Mar 27, 2009 Joachim Grevel Re: Log transformation of concentration
Mar 27, 2009 Jeroen Elassaiss-Schaap RE: Log transformation of concentration
Mar 27, 2009 Jeroen Elassaiss-Schaap RE: Log transformation of concentration
Mar 27, 2009 Joachim . Grevel Re: Log transformation of concentration
Mar 27, 2009 Bob Leary RE: Log transformation of concentration