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
CV
From:Leonid Gibiansky
Subject:Re: [NMusers] CV
Date:Tue, 04 Feb 2003 13:19:13 -0500
Luciane
Most likely, the model that you refer looks similar to
CL=THETA*EXP(ETA)
For the FO method, this is equivalent to
CL=THETA*(1+ETA)
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,
Leonid
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).
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From:yaning@ufl.edu
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
sqrt(var(y))/mean(y)*100=theta*omega/theta*100=omega*100.
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
CV.
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
spreadsheet.
This was addressed back in 1997 by Stuart Beal, see:
http://www.cognigencorp.com/nonmem/nm/98sep261997.html
-Matt
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