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()
}



________________________________________
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


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