## θ2 = 1 / θ1 [PK / PD]

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

Or the other way ’round: You would have to claim

Since you are talking about PK, most metrics follow a lognormal distribution,

With such an approach you are not aiming at equivalence any more, but at

For background see there.

» 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 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.

—

Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮

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*Dif-tor heh smusma*🖖_{}Helmut Schütz

The quality of responses received is directly proportional to the quality of the question asked. 🚮

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### Complete thread:

- Asymmetric margins in PK study msmnainar 2022-05-06 07:22 [PK / PD]
- θ2 = 1 / θ1Helmut 2022-05-06 09:32