Hi Jakob, The derivation for your expression comes from a first-order approximation of the variance often referred to as the "delta method" in the statistical literature. The approximation is:

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Var(f(x)) ~= [f'(x)^2]Var(x) or equivalently, SE(f(x)) ~= f'(x)SE(x) If we want SE(omega) but have the SE(omega^2) then x=omega^2 and f(x)=omega=x^0.5. Using the first-order approximation we have SE(omega) = SE(f(x)) = SE(x^0.5) = f'(x)SE(x) = 0.5(x^-0.5)SE(x) and substituting x=omega^2 = 0.5(omega^-1)SE(omega^2) = SE(omega^2)/(2*omega) Kind regards, Ken Kenneth G. Kowalski President & CEO A2PG - Ann Arbor Pharmacometrics Group 110 E. Miller Ave., Garden Suite Ann Arbor, MI 48104 Work: 734-274-8255 Cell: 248-207-5082 [EMAIL PROTECTED] -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ribbing, Jakob Sent: Friday, February 15, 2008 4:03 AM To: NONMEM users forum Cc: varun goel; [EMAIL PROTECTED] Subject: [NMusers] IIV on %CV-Scale - Standard Error (SE) Hi all, I think that Paul stumbled on a rather important issue. The SE of the residual error may not be of primary interest, but the same as discussed under this thread also applies to the standard error of omega. (I changed the name of the subject since this thread now is about omega) I prefer to report IIV on the %CV scale, i.e. sqrt(OMEGAnn) for a parameter with log-normal distribution. It then makes no sense to report the standard error on any other scale. For log-normally distributed parameters the relative SE of IIV then becomes: sqrt(SE.OMEGAnn)/(2*sqrt(OMEGAnn))*100% Notice the factor 2 in the denominator. I got this from Mats Karlsson who picked it up from France Mentré, but I have never seen the actual mathematical derivation for this formula. I think this is what Varun is doing in his e-mail a few hours ago. However, I am not sure; being illiterate I could not understand the derivation. Either way, if we are satisfied with the approximation of IIV as the square root of omega, the factor 2 in the approximation of the SE on the %CV-scale is exact enough. If you would like to convince yourself of that the factor 2 is correct (up to 3 significant digits), you can load the below Splus function and then run with different CV:s, e.g: ratio(IIV=1) ratio(IIV=0.5) Regards Jakob "ratio" <- function (IIV.stdev=1) { ncol <- 1000 #1000 Studies, in which IIV is estimated ETAS <- rnorm(n=1000*ncol, 0, IIV.stdev) ETA <- matrix(data=ETAS, ncol=ncol) IIVs.stds<- colStdevs(ETA) #Estimate of IIV on sd-scale IIVs.vars<- colVars(ETA) #Estimate of IIV on var-scale SE.std <- stdev(IIVs.stds)/sqrt(ncol) SE.var <- stdev(IIVs.vars)/sqrt(ncol) CV.std <- SE.std/IIV.stdev CV.var <- SE.var/(IIV.stdev^2) print(paste("SE on Var scale:", SE.var)) print(paste("SE on Std scale:", SE.std)) print(paste("Ratio CV var, CV std:", CV.var/CV.std)) invisible() } ________________________________________ From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of varun goel Sent: 14 February 2008 23:07 To: [EMAIL PROTECTED]; NONMEM users forum Subject: Re: [NMusers] Combined residual model and IWRES. Dear Paul, You can use the delta method to compute the variance and expected value of a transformation, which is square in your case. given y=theta^2 E(y)=theta^2 Var(y)=Var(theta)+(2*theta)^2 ; the later portion is square of the first derivative of y with respect of theta. In your example theta is the standard deviation whereas error estimate is variance. I did not follow your values very well, so I ran a model with same reparameterization and got following results. theta=2.65, rse=27.2% err=7.04; rse=54.4% theta.1<-2.65 rse<-27.2 var.theta.1<-(rse*theta.1/100)^2 ## = 0.51955 err.1<-7.04 rse.err.1<-54.4#% var.err.1<-(rse.err.1*err.1/100)^2 ## = 14.66 ##now from delta method E(err)=2.65^2 ## 7.025 close to 7.04 var(err)=(2*2.65)^2*0.51955 ## 14.59 close to 14.66 Hope it helps Varun Goel PhD Candidate, Pharmacometrics Experimental and Clinical Pharmacology University of Minnesota