From: "Luciane Velasque"
Subject:[NMusers] CV Date:Tue, 4 Feb 2003 15:48:17 -0200 Dear NM-users, I am a statistician and I am working with NONMEM in my master thesis. In the NONMEM manual the eta estimated is expressed as CV= sqrt(ETA)*100. I would like to know why is the Eta expressede like this ? I know CV as: CV = sqrt(x)/ average(x)*100. Luciane Velasque _________________________________________ From:Leonid Gibiansky Subject:Re: [NMusers] CV Date:Tue, 04 Feb 2003 13:19:13 -0500 Luciane Most likely, the model that you refer looks similar to CL=THETA*EXP(ETA) For the FO method, this is equivalent to CL=THETA*(1+ETA) Variance of CL is THETA^2*OMEGA where OMEGA is the variance of ETA Standard deviation of CL is THETA*SQRT(OMEGA) Mean of CL is THETA CV = SD/ MEAN*100=SQRT(OMEGA)*100 (and it should be OMEGA, not eta in that expression that you mentioned) Hope this helps, Leonid _________________________________________ From:"Bachman, William" Subject:RE: [NMusers] CV Date:Tue, 4 Feb 2003 13:27:29 -0500 The expression for CV is dependent on the parameterization of the error model. If the error model is proportional, e.g. V = TVV+ TVV*ETA(n) (or the exponential equivalent, V = TVV * EXP(ETA(n)) ) then the CV = sqrt(omega(n))*100 If the error model is additive, eg. V= TVV+ETA(n), then the CV becomes CV = sqrt(omega(n))/theta(n) * 100 in both cases omega(n) is the variance of ETA(n) and theta(n) is the population estimate of the parameter (mean). email@example.com GloboMax LLC 7250 Parkway Drive, Suite 430 Hanover, MD 21076 Voice: (410) 782-2205 FAX: (410) 712-0737 _________________________________________ From:firstname.lastname@example.org Subject:RE: [NMusers] CV for exponential model Date:Tue, 4 Feb 2003 17:18:49 -0500 I guess Luciane's confusion is about the exponential model. Basically, it is because of the FO(first order) method (first order Taylor expansion). If we assume V=theta*exp(eta) with eta~N(0, omega**2), V follows an exact lognormal distribution. But Taylor expansion at zero will yield V=theta+theta*eta approximately. In other words, V approximately follows a normal distribtuion with mean of theta and variance of theta**2*omega**2. Then CV of y is approximately sqrt(var(y))/mean(y)*100=theta*omega/theta*100=omega*100. I remember we learned this in two of our statistic courses, survival analysis and matrix algebra, something about the first order Delta method. If Y=f(X), Y=f(X0)+(X-X0)*f'(X0) and E(Y)=E(X) and Var(Y)=f'(X0)**2*Var(X) by Delta method which is based on first order Taylor expansion. Of course, if we assuem Y=theta*(1+eta), it is straightforward. Hope this helps Yaning Wang Department of Pharmaceutics College of Pharmacy University of Florida _________________________________________ From:Matthew Riggs Subject:RE: [NMusers] CV for exponential model Date:Wed, 5 Feb 2003 06:19:36 -0800 (PST) Luciane, As Bill stated, the CV = sqrt(omega(n))*100 for a proportional structure: V= TVV*(1+ETA(n)) and for an additive structure (V= TVV+ETA(n)), the CV = sqrt(omega(n))/theta(n)* 100. And as Yaning stated, the FO method applies a first order Taylor expansion about an assumed mean eta of zero; mathematically collapsing the exponential error form into a proportional form = same calculation for CV. However, it's been my impression that when using FOCE or LAPLACIAN with an exponential structure, the math does not "collapse" so easily and you need to consider calculating the CV% = 100*sqrt(exp(omega(n))-1), particularly with "large" values of omega (e.g., >=0.16). It's easy enough to see the deviation you'd get by using the two methods (proportional vs. exponential), especially at higher omegas, by comparing the CV% values across a range of omegas in a spreadsheet. This was addressed back in 1997 by Stuart Beal, see: http://www.cognigencorp.com/nonmem/nm/98sep261997.html -Matt _________________________________________