Hi Jakob,

The derivation for your expression comes from a first-order approximation of
the variance often referred to as the "delta method" in the statistical
literature.  The approximation is:

Var(f(x)) ~= [f'(x)^2]Var(x) or equivalently, SE(f(x)) ~= f'(x)SE(x)

If we want SE(omega) but have the SE(omega^2) then x=omega^2 and
f(x)=omega=x^0.5.  Using the first-order approximation we have

SE(omega) = SE(f(x)) 
          = SE(x^0.5)
          = f'(x)SE(x)
          = 0.5(x^-0.5)SE(x) and substituting x=omega^2
          = 0.5(omega^-1)SE(omega^2)
          = SE(omega^2)/(2*omega)

Kind regards,


Kenneth G. Kowalski
President & CEO
A2PG - Ann Arbor Pharmacometrics Group
110 E. Miller Ave., Garden Suite
Ann Arbor, MI 48104
Work:  734-274-8255
Cell:  248-207-5082

-----Original Message-----
Behalf Of Ribbing, Jakob
Sent: Friday, February 15, 2008 4:03 AM
To: NONMEM users forum
Cc: varun goel; [EMAIL PROTECTED]
Subject: [NMusers] IIV on %CV-Scale - Standard Error (SE)

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:

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" <- 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))

Behalf Of varun goel
Sent: 14 February 2008 23:07
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

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%

var.theta.1<-(rse*theta.1/100)^2  ## = 0.51955 

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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