θ2 = 1 / θ1 [PK / PD]

posted by Helmut Homepage – Vienna, Austria, 2022-05-06 09:32 (10 d 11:27 ago) – Posting: # 22964
Views: 155

Hi msmnainar,

» Is it possible to use asymmetric margin (e.g., 80-140%) in the PK study?

Possible: yes. Acceptable: no.

» If yes, what are the pros and cons of using asymmetric margin?

If you use untransformed data (i.e., evaluate the study by an additive model), the limits should be symmetrical in the linear domain. Say, your clinically not relevant difference \(\small{\Delta=20\%}\), then $$\left\{\theta_1,\theta_2\right\}=\left\{1-\Delta,1+\Delta\right\}=\left\{80\%,120\%\right\}.\tag{1}$$You get maximum power for no difference in treatments or \(\theta_0=100\%\) because $$\theta_0=(\theta_1+\theta_2)/2=100\,(0.80+1.20)=100\%\,.\tag{2}$$ With your margins you get maximum power at \(\small{\theta_0=110\%}\) and hence, perfectly matching treatments (\(\small{\theta_0=100\%}\)) would have lower power for any given sample size. That’s not what we want.
Or the other way ’round: You would have to claim and justify that you have two not clinically relevant differences \(\small{\left\{\Delta_1=20\%,\Delta_2=40\%\right\}}\). Such a justification will be difficult at least.

Since you are talking about PK, most metrics follow a [image] lognormal distribution, i.e., we assume a multiplicative model. In order to evaluate the study by a linear model (requiring additive effects), we \(\small{\log_{e}\text{-}}\)transform data. With any given \(\small{\Delta}\) the limits are symmetrical in the \(\small{\log_{e}\text{-}}\)domain and asymmetrical back-transformed. With \(\small{\Delta=20\%}\) again, we get in the \(\small{\log_{e}\text{-}}\)domain $$\log_{e}\left\{\theta_1,\theta_2\right\}=\log_{e}\left\{1-\Delta,1+\Delta\right\}=\left\{-0.2231,+0.2231\right\}\tag{2}$$ and in the linear domain $$\left\{\theta_1,\theta_2\right\}=\left\{1-\Delta,\left(1-\Delta\right)^{-1}\right\}=\left\{80\%,125\%\right\}.\tag{3}$$ Similarly to above, with your limits you get maximum power at $$100\,\exp\big{(}\log_{e}(0.80)+\log_{e}(1.40)\big{)}/2=105.83\%\,.\tag{4}$$
With such an approach you are not aiming at equivalence any more, but at suprabioavailability. Your chance that such an approach will be accepted by any authority is close to nil.

For background see there.

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