From: "Luciane Velasque" 
Subject:[NMusers] CV
Date:Tue, 4 Feb 2003 15:48:17 -0200

Dear NM-users,

I am a statistician and I am working with NONMEM in my master thesis.

In the NONMEM manual  the eta estimated is expressed as CV= sqrt(ETA)*100.

I would like to know why  is the Eta expressede like this ?

I know CV as:

CV = sqrt(x)/ average(x)*100.

Luciane Velasque

From:Leonid Gibiansky 
Subject:Re: [NMusers] CV
Date:Tue, 04 Feb 2003 13:19:13 -0500

Most likely, the model that you refer looks similar to 
For the FO method, this is equivalent to 
Variance of CL is THETA^2*OMEGA where OMEGA is the variance of ETA 
Standard deviation of CL is THETA*SQRT(OMEGA) 
Mean of CL is THETA 
CV = SD/ MEAN*100=SQRT(OMEGA)*100 

(and it should be OMEGA, not eta in that expression that you mentioned) 

Hope this helps, 

From:"Bachman, William" 
Subject:RE: [NMusers] CV
Date:Tue, 4 Feb 2003 13:27:29 -0500

The expression for CV is dependent on the parameterization of the error model.
If the error model is proportional, e.g. V = TVV+ TVV*ETA(n) (or the exponential 
equivalent, V = TVV * EXP(ETA(n))  )
then the CV = sqrt(omega(n))*100
If the error model is additive, eg. V= TVV+ETA(n),
then the CV becomes  CV = sqrt(omega(n))/theta(n) * 100
in both cases omega(n) is the variance of ETA(n) and theta(n) is the population
estimate of the parameter (mean). 
GloboMax LLC 
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Subject:RE: [NMusers] CV for exponential model
Date:Tue, 4 Feb 2003 17:18:49 -0500

I guess Luciane's confusion is about the exponential model. Basically, it is 
because of the FO(first order) method (first order Taylor expansion). 

If we assume V=theta*exp(eta) with eta~N(0, omega**2), V follows an exact 
lognormal distribution. But Taylor expansion at zero will yield 
V=theta+theta*eta approximately. In other words, V approximately follows a 
normal distribtuion with mean of theta and variance of theta**2*omega**2. Then 
CV of y is approximately 

I remember we learned this in two of our statistic courses, survival analysis 
and matrix algebra, something about the first order Delta method. 
If Y=f(X), Y=f(X0)+(X-X0)*f'(X0) and E(Y)=E(X) and Var(Y)=f'(X0)**2*Var(X) by 
Delta method which is based on first order Taylor expansion. 

Of course, if we assuem Y=theta*(1+eta), it is straightforward.

Hope this helps

Yaning Wang
Department of Pharmaceutics
College of Pharmacy
University of Florida

From:Matthew Riggs 
Subject:RE: [NMusers] CV for exponential model
Date:Wed, 5 Feb 2003 06:19:36 -0800 (PST)

Luciane, As Bill stated, the CV = sqrt(omega(n))*100
for a proportional structure: V= TVV*(1+ETA(n)) and
for an additive structure (V= TVV+ETA(n)), the CV =
sqrt(omega(n))/theta(n)* 100.

And as Yaning stated, the FO method applies a first
order Taylor expansion about an assumed mean eta of
zero; mathematically collapsing the exponential error
form into a proportional form = same calculation for

However, it's been my impression that when using FOCE
or LAPLACIAN with an exponential structure, the math
does not "collapse" so easily and you need to consider
calculating the CV% = 100*sqrt(exp(omega(n))-1),
particularly with "large" values of omega (e.g.,
>=0.16).  It's easy enough to see the deviation you'd
get by using the two methods (proportional vs.
exponential), especially at higher omegas, by
comparing the CV% values across a range of omegas in a

This was addressed back in 1997 by Stuart Beal, see: