Jakob I am not sure that the formula that you present is correct:

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sqrt(SE.OMEGAnn)/(2*sqrt(OMEGAnn))*100% I think, you do not need to take sqrt(). This is what I would use SE.OMEGAnn/(2*OMEGAnn)*100%

`Note that your S-plus function also does not take a sqrt, so it could be`

`just a typo.`

Factor of 2 is derived from these calculations: Assume that you have random variable X=A+alpha*eps, where eps is the standard normal (mean(eps)=0, var(eps)=1) Then mean(X)=A var(X)=mean( (X-mean(X))**2 ) = alpha**2 sd(X)=sqrt(Var(X))=alpha CV(X)=SD/mean=alpha/A Now, let Y=X**2=(A+alpha*eps)**2 Then mean(Y)=mean(X**2)=mean(A**2+2 alpha A eps + alpha**2 mean(eps**2))= = A**2 + alpha**2 = A**2 ; (neglecting small terms alpha**2) var(Y)=mean( (Y-mean(Y))**2 ) = mean( ((A+alpha*eps)**2-A**2 )**2 ) =

`=mean( (2 alpha A eps)**2) = 4 alpha**2 A**2 (again, keeping only`

`the main term)`

SD(Y) = 2 A alpha CV(Y)=2 alpha/A From here we have CV(X)=CV(Y)/2

`In the context of what we discuss, X is the estimated SD while Y is the`

`estimated variance of OMEGA (or sigma) estimate:`

____________________________ Parameterization 1: PAR=PAR0+THETA(1)*ETA(1) $OMEGA 1 FIXED (X from the discussion above is THETA(1) estimate) --- Parameterization 1: PAR=PAR0+ETA(1) $OMEGA estimated

`Y in this case is the OMEGA estimate, and if the solution does not`

`depend on the parameterization, Y=X**2`

-------------------------

`Another way to derive it is to think in terms of confidence intervals.`

`In SD scale,`

SD.CI = SD +/- 2* SE.SD CV.SD = 2* SE.SD / SD / 2 = SE.SD / SD In OMEGA scale this would result in

`OMEGA.CI = (SD +/- 2* SE.SD)**2=SD**2 +/- 4 * SD * SE.SD (neglecting the`

`term 4 SE.SD**2)`

Then CV.OMEGA = 4 * SD * SE.SD / 2/ SD**2 = 2 SE.SD/SD = 2 CV.SD CV.SD = CV.OMEGA/2 Leonid -------------------------------------- Leonid Gibiansky, Ph.D. President, QuantPharm LLC web: www.quantpharm.com e-mail: LGibiansky at quantpharm.com tel: (301) 767 5566 Ribbing, Jakob wrote:

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 MentrĂ©, 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^2Var(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<-2.65rse<-27.2var.theta.1<-(rse*theta.1/100)^2 ## = 0.51955err.1<-7.04 rse.err.1<-54.4#% var.err.1<-(rse.err.1*err.1/100)^2 ## = 14.66 ##now from delta methodE(err)=2.65^2 ## 7.025 close to 7.04var(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