Re:constant CV

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