Re: IIV on %CV-Scale - Standard Error (SE)

Jakob I am not sure that the formula that you present is correct: 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 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() > } > > ________________________________________