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

Dear All,
 
 Correction to my previous posting
 
 E(Y)=Var(X)*(2*theta)^2, but calculations in the example have been done correctly.
 
 This comes from delta method which is if y is a function of x, such that
 
 y=g(x), then E(y) and Var(Y) can be obtained by taking a first order taylor series expansion around E(x).
 
 y=E(g(x))+g'(x)*(x-E(x)) ; where g'(x) is the first derivative of g(x) with respect to x
 
 then
 E(y)=E(g(x))
 Var(y)=g'(x)^2*Var(x)
 
 Hope it helps
 
 Varun Goel
 PhD Candidate, Pharmacometrics
 Experimental and Clinical Pharmacology
 University of Minnesota

"Ribbing, Jakob" <Jakob.Ribbing_at_pfizer.com> 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()
}



________________________________________
From: owner-nmusers_at_globomaxnm.com [mailto:owner-nmusers_at_globomaxnm.com] On Behalf Of varun goel
Sent: 14 February 2008 23:07
To: pwestwood02_at_qub.ac.uk; 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


       
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Received on Fri Feb 15 2008 - 09:48:55 EST