# RE: Very small P-Value for ETABAR

From: Xia Li <lix4>
Date: Fri, 14 Nov 2008 17:33:20 -0500

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: owner-nmusers
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
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
>

--
Nick Holford, Dept Pharmacology & Clinical Pharmacology
University of Auckland, 85 Park Rd, Private Bag 92019, Auckland, New Zealand
n.holford
http://www.fmhs.auckland.ac.nz/sms/pharmacology/holford
Received on Fri Nov 14 2008 - 17:33:20 EST

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