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

Hello Jakob,

I remember I developed code at one point to calculate a SE on a typical =
clearance which was a function of four covariates..

The general method is to use formula incorporating partial derivatives =
with respect to each parameter.

http://www.itl.nist.gov/div898/handbook/mpc/section5/mpc55.htm this =
link gives good background information.


The S-plus piece of code that can be used is :


OM1 <- 0.16
SE1 <- 0.04

# PERCENTOM1<- sqrt(OM1) # the expression to calculate a standard error =
on it

DDDDD <- deriv(~ sqrt(OM1) ,
c("OM1"),function(OM1) NULL, formal=T)


DERIV <- DDDDD(OM1) # .grad[, "OM1"] <- 0.5 * =
(OM1^-0.5) which is 1/ 2 . SQRT(OM1)
PARDERIV <- attr(DERIV,"gradient")

PERCENTOM1SE<- sqrt ( (PARDERIV[1]*SE1)^2)


For this simple example computing partial derivative by hand is easy but =
for more complex formulas things can be tricky and S-plus can calculate =
this for you.

http://statpages.org/erpropgt.html this link calculates SE of function =
of up to two variables.

Hope this helps,



Samer Mouksassi











-----Original Message-----
From: owner-nmusers_at_globomaxnm.com on behalf of Ribbing, Jakob
Sent: Fri 2/15/2008 04:03
To: NONMEM users forum
Cc: varun goel; pwestwood02_at_qub.ac.uk
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:
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 - 08:40:09 EST