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
Received on Fri Feb 15 2008 - 04:03:02 EST
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