# RE: Very small P-Value for ETABAR

From: Xia Li <lix4>
Date: Mon, 17 Nov 2008 19:25:26 -0500

Dear All,
We begin with

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

where ETA(1) is N( 0, omega^2),
So the p-value for testing mean of ETA(1)=0 would be "expected" to be
large.

When (1) doesn't look like the right model, say b/c mean of ETA(1) looks
non-zero,

I said: modify (1) by letting

ETA(1)= THETA(2)*exp(ETA(2)) (2a).

with ETA(2) being N(0.v), so that ETA(1) now has possibly non-zero mean.

Jakob you are right, combing (1) and (2a) we see THETA(1) and THETA(2)
appear together making both THETA's non-identifiable.

What I probably thought was, instead of (2a), set

EXP(ETA(1))= exp(W)*exp(ETA(2))= exp(W+ETA(2)) (2b)

for a covariate W. Combining (1) and (2b)

CL=THETA(1)*EXP(ETA(1))= THETA(1)*exp(W)*exp(ETA(2))

giving mean of ETA(1) non-zero, and dependent on covariate W. We are back to
the point that 'good' covariates help modeling...

Regarding the question how equation 2c affects equation 1 in my previous
email.
omega^2= THETA(3)*EXP(ETA(2)) (2c)

Instead of assuming ETA(1) is normally distributed with same variance
omega^2, we say different omega^2 for different characteristic groups.
Omega^2_j=theta3_j*exp(eta2_ij)

By doing so, ETA(1) is a mixture of j normals with different variance. We
know a mixture of normals (having identical mean and variance) is normal,
whereas, a mixture of normals (if means are identical, but variances are
not) may be non-normal.

Now, 2(b) may help explaining the nonzero mean of ETA(1) and 2(c) may help
explaining the asymmetric shape of ETA(1).

I tried to attach a doodles graph to help my explanation but it seems
attachment is not allowed...
http://i36.tinypic.com/152hpwy.jpg
Again, thanks for all input and let me know if I missed something or
misunderstood the original problem.

Best,
Xia

-----Original Message-----
From: owner-nmusers
Behalf Of Ribbing, Jakob
Sent: Monday, November 17, 2008 2:27 AM
To: XIA LI; nmusers
Subject: RE: [NMusers] Very small P-Value for ETABAR

Xia,

I must admit, I am still confused. In my mind, you can not estimate
THETA(2) in your code, since it is completely confounded with THETA(1).
Moreover, if you fix THETA(2) to a non-zero value, THETA(1) will no
longer be the typical value of CL (or the population typical value of
CL), meaning that the interpretability of THETA(1) is lost.

Regarding your definition of omega^ I think this is an attempt to allow
a semi-parametric model. Can you please explain how equation 2 affects
equation 1, using code acceptable in the nonmem NONMEM program?
Currently, I am not clear on how many random effects you are estimating
for CL.

Thanks

Jakob

-----Original Message-----
From: owner-nmusers
On Behalf Of XIA LI
Sent: 17 November 2008 05:28
To: Leonid Gibiansky
Cc: 'Nick Holford'; 'nmusers'
Subject: Re: [NMusers] Very small P-Value for ETABAR

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
<nmusers
>
>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
[mailto: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
>>>
>>
======================================
Xia Li
Mathematical Science Department
University of Cincinnati
Received on Mon Nov 17 2008 - 19:25:26 EST

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