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

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: 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 - 09:30:55 EST