From: XIA LI <*lix4*>

Date: Mon, 17 Nov 2008 00:28:25 -0500 (EST)

Leonid,

Sorry, I did make myself clear.

CL=THETA(1)*EXP(ETA(1)) (1)

where ETA(1) is Normal( 0, omega^2) or

log Normal(Eta_bar,omega^2)

Adding one more stage means giving some functions for the MEAN and VARIANCE of ETA(1), say:

Eta_bar=THETA(2)

omega^= THETA(3)*EXP(ETA(2)) (2)

Sorry for any confusion!

Best,

Xia

---- Original message ----

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

*>From: Leonid Gibiansky <LGibiansky *

*>Subject: Re: [NMusers] Very small P-Value for ETABAR
*

*>To: Xia Li <lix4 *

*>Cc: "'Nick Holford'" <n.holford *

*>
*

*>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
*

*>>>
*

*>>
*

======================================

Xia Li

Mathematical Science Department

University of Cincinnati

Received on Mon Nov 17 2008 - 00:28:25 EST

Date: Mon, 17 Nov 2008 00:28:25 -0500 (EST)

Leonid,

Sorry, I did make myself clear.

CL=THETA(1)*EXP(ETA(1)) (1)

where ETA(1) is Normal( 0, omega^2) or

log Normal(Eta_bar,omega^2)

Adding one more stage means giving some functions for the MEAN and VARIANCE of ETA(1), say:

Eta_bar=THETA(2)

omega^= THETA(3)*EXP(ETA(2)) (2)

Sorry for any confusion!

Best,

Xia

---- Original message ----

======================================

Xia Li

Mathematical Science Department

University of Cincinnati

Received on Mon Nov 17 2008 - 00:28:25 EST