From: Leonid Gibiansky <*LGibiansky*>

Date: Fri, 14 Nov 2008 18:37:22 -0500

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: 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
*

*> 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
*

*>>
*

*> *

Received on Fri Nov 14 2008 - 18:37:22 EST

Date: Fri, 14 Nov 2008 18:37:22 -0500

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:

Received on Fri Nov 14 2008 - 18:37:22 EST