From: "clement" <clement@ha.mc.ntu.edu.tw>
Subject:constant CV
Date: Thu, 26 Apr 2001 19:21:04 +0800

Hello, nmusers
As a beginner for NONMEM, I wonder would someone kindly told me what's the point to consider in selecting the following parameter or error models. In addition to statistical consideration, is there any physiological meaning for the selection?

(If body weight is a factor to consider)
TVCL = THETA(1)*WT
TVCL= THETA(1)*WT*EXP(THETA(2))
TVCL=THETA(1)*(WT/60)*EXP(THETA(2))

CL=TVCL+ETA(1)
CL=TVCL*(1+ETA(1))
CL=TVCL*EXP(ETA(1))
LOG(CL)=LOG(TVCL)+ETA(1)

Any help will be appreciated.

Clement

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From: "Sale, Mark" <ms93267@GlaxoWellcome.com>
Subject: RE:constant CV
Date: Thu, 26 Apr 2001 08:14:12 -0400

Clement,

In my view there is a continuum of criteria for selection of any models for empiric pk/pd modeling. The ends of the spectrum are the purely statistical (what best fits the data/can be statistically defended) and the physiologic (what makes sense based on our understanding of they physiology/pharmacology). Most of us are somewhere in between, insisting on some (in my view limited) statistical rigor, but based in our understanding of the physiology. Among nonmem users, I tend to lean toward the statistical end, among statisticians, I'm far toward the physiologic end.

So, the issues to address are:
What do the data tell us the model should be:
What makes sense physiologically.

The first is easy to address, just run all of them and see which has the lowest objective function (corrected for degrees of freedom, also will successfully complete a covariance and have an estimation correlation matrix with all off-diagonal element with absolute value < 0.95). Of course, trying 4 different model for each parameter/covariate model and each interindividual error, and all the possible combinations leads to literally millions of models.

The second is harder.
First, I think that the second and third models are overparameterized. THETA(1) and THETA(2) are indistinguishable. I usually use something like.

TVCL = THETA(1)*EXP(THETA(2)*(WT-60)/11)

Where the standard deviation of WT is 11, and the mean is 60. (Don't know if I mean geometric of arithmetic mean, probably geometric) First, note that WT is centered and scaled. This improves the numerical stability, and give THETA(1) the nice property of being the value for a typical individual.

Note however, that in this model, there will be a positive value for someone with a WT of 0. Personally, this doesn't bother me, since I've never seen someone with a WT of zero. But it does bother some people. However, the model

TVCL = THETA(1)*WT

Is a different model, you are making a different statement about physiology, that TVCL is proportion not to EXP(WT), but to WT. Generally, in adults we don't have sufficient range in WT to distinguish these models, so they can often be used interchangeable. However, if there is a wide range of WT (e.g., peds and adults), you may find that one is "better" (lower objective function) than the other. In this case, I'd run both and choose the one with the lower objective function.

The same approach applies to the selection of the interindividual error, a balance between statistical and physiologic criteria. We often find that population parameters are log normally distributed. In this case the model
CL = TVCL*EXP(ETA(1)) is appropriate. This model also has the nice property of never having a negative value for CL (if TVCL is positive). This is useful if you are using METHOD = 1, or POSTHOC. This model is usually my default. The other two (+ETA and *(1+ETA)) can be tried if you have reason to believe that the parameter is normally distributed. We recently used this error for a model of Hemoglobin A1c, where we knew that baseline value was normally distributed. However, in general, we use the log normal distribution for pretty much everything, since we always run POSTHOC and often use METHOD = 1.

Mark

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From: SMITH_BRIAN_P@Lilly.com
Subject: Re:constant CV
Date: Thu, 26 Apr 2001 08:59:14 -0500

As a statistician, I cannot explain physiological meaning. I can interpret for you what these mathematical equations imply. You then should be able to judge whether or not these are physiologically meaningful.

1) TVCL = THETA(1)*WT

Mathematically, this simply implies that a doubling of weight will double the clearance. This would imply that weight normalized dosing would produce equivalent AUCs.

2)TVCL = THETA(1)*WT*EXP(THETA(2))

In a mathematical since one cannot distinguish between THETA(1) and THETA(2) with this model. It is incorrect. I think you meant to consider the model

TVCL = THETA(1)*WT**(THETA(2))

This is called a power model. When THETA(2) = 1 it becomes the first model that you stated. Thus, it allows you to examine whether model 1) is tenable. Notice, that when THETA(2) = 0, then the model implies that clearance is independent of weight. Thus, it also allows for you to judge whether the independence of weight and clearance is tenable. Mathematically, a doubling of weight will increase clearance by a factor of 2**THETA(2). Thus, for instance if THETA(2) = 0.8 then a doubling of weight (a 100% increase in weight) will cause a 74% increase in clearance.

3)TVCL = THETA(1)*(WT/60)*EXP(THETA(2))

Again, I believe that you did not mean to consider this model since THETA(1) and THETA(2) are indistinguishable. I think you wanted to consider

TVCL = THETA(1)*(WT/60)**(THETA(2))

Statistically this model is exactly the same as model 2). Everything that I said about model 2) could be said for model 3). Additionally, this model makes THETA(1) easier to interpret. THETA(1) becomes the estimate of an individual's clearance that has a weight of 60. Both model 2) and model 3) will give the exact same estimate for THETA(2).

4) CL = TVCL + ETA(1)

This implies that clearance has a normal distribution. With this assumption there is a positive probability that a clearance could take any value from negative infinity to infinity. Obviously, physiologically this is impossible since clearance has to be a positive value. In my mind this makes this a non-usable error assumption. It is used, however. Additionally, this model assumes that the variability in the absolute since is the same for all clearances. Since there tends to be a larger variance with larger clearances this also should make this error structure suspect.

5) CL=TVCL*(1+ETA(1))

This is often called a constant coefficient of variation model. ETA(1) is normally distributed with mean 0 and unknown variance. Call this unknown variance omegasquared. Call the square root of this value omega. This model implies that the standard deviation of clearance is omega*TVCL. Or that the coefficient of variation, which is the standard deviation divided by the mean is omega*TVCL/TVCL = omega. Physiologically it often makes since to assume that the coefficient of variation is constant. However, this model does have a similar detriment as model 4). ETA(1) has positive probability for any value from negative infinity to infinity. Thus, it allows for ETA(1) to be less than 1, which implies that it allows clearance to be negative. Again, this is physiologically impossible.

6) CL = TVCL*EXP(ETA(1))

This model also implies constant coefficient of variation. It implies that clearance has a log-normal distribution. It is mathematically equivalent to model 7), so I will defer my discussion of this model.

7) LOG(CL) = LOG(TVCL) + ETA(1)

First, notice when we exponentiate each side we get

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

which is model 6). Mathematically these models are identical. They imply that LOG(CL) has a normal distribution. That is that LOG(CL) has positive probability for any value from negative infinity to infinity. This implies that clearance has positive probability for any value from 0 to infinity. This is a tenable assumption since clearance has to be positive. Statistics fact: the mean and variance of a log normal distribution are exp(log(TVCL) + 0.5*omegasquared) and (exp(omegasquared) - 1)*exp(2*log(TVCL) + omegasquared) respectively. This implies that the coefficient of variation, standard deviation divided by mean, is sqrt(exp(omegasqrared) - 1). As you can see the coefficient of variation is then constant (it does not change with the mean). Physiologically then, models 6) and 7), which are equivalent mathematically, are the best choices for analysis.

Models 5), 6), and 7) and the first order approximation in NONMEM.

Numerically neither model 5), 6), or 7) are easy to solve. Different approximations to models are provided by NONMEM. With the first order method, models 6) and 7), which are mathematically equivalent, provide different solutions. Interestingly, models 5) and 6) provide exactly the same solutions!! It has been a while since I was an active user of NONMEM. I did not realize that model 7) was even possible to specify in NONMEM. I hope others will verify this. If model 7) is possible, then I believe it should be preferred when using first order (or any other approximation provided by NONMEM). Even though each error term in a NONMEM model is assumed to have a normal distribution, the resulting model does not. Numerically it would be just about impossible to find estimates for the "true" model. Most approximation methods approximate the stated model with one that is normally distributed. That is it uses a model where the error terms are additive not multiplicative or exponential. At this point numerical solutions are possible. My theory, unproven, is that you would get a closer approximation with model 7) since the error term is additive.

I hope this helps.

Sincerely,

Brian Smith

 

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From: Thierry Buclin <Thierry.Buclin@chuv.hospvd.ch>
Subject: constant CV
Date: Fri, 27 Apr 2001 09:39:41 +0200

Brian

Thank you for the light-clear explanation about Clement's question. I have a perplexing interrogation about the intuitive meaning of the coefficient ov variation estimated in the "constant CV" error model (e.g. CL=TVCL*EXP(ETA(1)). You state that the mean and variance of a log normal distribution are exp(log(TVCL) + 0.5*omegasquared) and exp(omegasquared) - 1)*exp(2*log(TVCL) + omegasquared) respectively, implying that the coefficient of variation is sqrt(exp(omegasqrared) - 1).

Consider a "constant" error model giving three individual estimates of CL, say 2, 4 and 6 L/h. The omegasquared is 4 and the SD is 2 L/h : this value describes the typical variation among individual estimates, this is intuitively convincing, OK ?

Now consider a "constant CV" error model giving individual CL estimates of 2, 4 and 8 L/h. Intuitively, I would consider that the Logs display a typical variation of Log(2), and therefore the CL estimates a variation by a factor of 2, hence a CV of 100%. However, the calculation you mention gives a CV of 78.5%. I am convinced that it is statistically founded, but how could you reconcile this result with common sense intuition ?

In other words, if I fit a model assuming constant CV error, is it better to report CV values calculated according to sqrt(exp(omegasqrared) - 1) or to exp(sqrt(omegasquared)) -1 ?

Thierry BUCLIN, MD
Lecturer, consulting physician and clinical researcher
Division of Clinical Pharmacology
University Hospital CHUV - Beaumont 633
CH 1011 Lausanne - SWITZERLAND
Tel: +41 21 314 42 61 - Fax: +41 21 314 42 66