Xia,
I could be missing something but this
ETA(1)= THETA(2)*exp(ETA(2)) (Eq. 1)
does not make sense to me. In the original definition, ETA(1) is the
random variable with normal distribution. Even if posthoc ETAs are not
normal, they are still random. For example, it can be either positive or
negative (unlike ETA1 given by (1)). If I the understood intentions
correctly, this is an attempt to describe a transformation of the random
effects to make it normal:
CL = THETA(1) exp(ETA(1)) is replaced by
CL = THETA(1) exp(THETA(2)*exp(ETA(1))) (2)
But not every transformation is reasonable. I hardly can imagine the
case when you may want to use (2). Could you give some more realistic
examples, please, and situation when they were useful?
On the separate note, mean of THETA(2)*exp(ETA(2)) is not equal to
THETA(2): geometric mean of THETA(2)*exp(ETA(2)) is equal to THETA(2)
Thanks
Leonid
--------------------------------------
Leonid Gibiansky, Ph.D.
President, QuantPharm LLC
web: www.quantpharm.com
e-mail: LGibiansky at quantpharm.com
tel: (301) 767 5566
Xia Li wrote:
Hi Nick,
My pleasure!
This is a topic from Bayesian Hierarchical Model(BHM). If we look at the
simplest PK statement: CL=THETA(1)*EXP(ETA(1)), where ETA(1) is the between
subject random effect. We assume the "similarity" among the subjects may be
modeled by THETA(1) and ETA(1).
Now here, if we observe that there is an underlying pattern between
ETA(1)'s, i.e. deviation from zero or no longer normal and we assume that
there is a similarity among those patterns.
Since ETA(1)'s are assumed similar, it is reasonable to model the
"similarity" among the ETA(1)'s by THETA(2) and ETA(2): ETA(1)=
THETA(2)*exp(ETA(2)). Hence we have one more stage, ETA(1) now is
lognormal(nonsymmetrical) with mean THETA(2) (doesnt have to be zero).
We will not say the variance of ETA(1) is confounded with the variance of
ETA(2), we say it is a function of variance of ETA(2).In statistics,
confounding means hard to distinguish from each other. Here, it is a direct
causation.
Sorry I don't have a NM-TRAN code for this now. I usually use SAS and Win
bugs to do modeling and haven't tried this BHM in NONMEM. I will figure out
can I do it in NONMEM later.
Best,
Xia
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Nick Holford
Sent: Friday, November 14, 2008 3:34 PM
To: nmusers
Subject: Re: [NMusers] Very small P-Value for ETABAR
Jakob, Mats,
Thanks very much for your careful explanations of how asymmetric EBE
distributions can arise. That is very helpful for my understanding.
Xia,
I am intrigued by your suggestion for how to estimate and account for
the bias in the mean of the EBE distribution.
In the usual ETA on EPS model I might write:
; SD of residual error for mixed proportional and additive random effects
PROP=THETA(1)*F
ADD=THETA(2)
SD=SQRT(PROP*PROP + ADD*ADD)
Y=F + EPS(1)*SD*EXP(ETA(1))
where EPS(1) is distributed mean zero, variance 1 FIXED
and ETA(1) is the between subject random effect for residual error
You seem to be suggesting:
ETABAR=THETA(3)
Y=F + EPS(1)*SD*EXP(ETA(1)) * ETABAR*EXP(ETA(2))
It seems to me that the variance of ETA(1) will be confounded with the
variance of ETA(2). Would you please explain more clearly (with an
explicit NM-TRAN code fragment if possible) what you are suggesting?
Best wishes,
Nick
Xia Li wrote:
Hi Jakob,
Thank you very much for the information adding an "eta on epsilon". This
is
what I did in my research and I am glad to see people in Pharmacometrics
is
using it.
And in Bayesian analysis, adding one more stage for ETA, i.e
ETA=ETABAR*exp(eta2), eta2~N(0,omega2) will allow the deviation from zero
and shrinkage of ETA.
Again, thanks all for your input.:)
Best Regards,
Xia
Xia Li
Mathematical Science Department
University of Cincinnati
