martin Senior Austria, 20170508 19:50 Posting: # 17322 Views: 1,073 

Dear all, I would like to open again a new thread regarding endogenous compounds in BE assessments. Simple case:  assuming no circadian rhythm / temporarily constant  baseline subtraction method (e.g. median of at several predose samples) Several guidelines mention that a negative predose corrected value should be set to zero. However, what about subsequent levels? I think setting also all concentrations after the first negative to zero would make some sense as otherwise we just have “noise” in the AUC (i.e. increase variability) and it’s rather impossible to get a reliable estimate for lambda_z (for calculation of AUC0inf). I think this should be also inline with guidelines as the method for baseline subtraction needs to be specified apriori. Best regards and looking forward to thoughts from more experienced members in this forum Martin PS.: premise of stable endogenous could be verified by looking at adjusted concentrations after the first negative (inclusive) adjusted concentration: should be symmetrically distributed around zero. 
Helmut Hero Vienna, Austria, 20170508 21:10 @ martin Posting: # 17324 Views: 887 

Hi Martin, » Several guidelines mention that a negative predose corrected value should be set to zero. However, what about subsequent levels? » I think setting also all concentrations after the first negative to zero would make some sense […] this should be also inline with guidelines as the method for baseline subtraction needs to be specified apriori. Well, this is exactly what I have done for ages. never got a deficiency letter. » PS.: premise of stable endogenous could be verified by looking at adjusted concentrations after the first negative (inclusive) adjusted concentration: should be symmetrically distributed around zero. Really? You explained that to me so may times; excuse my walnutsized brain. Baseline (C_{0}) = lognormal. Postdose (C_{1}) = lognormal. C_{0}–C_{1} = normal?
— Regards, Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. ☼ Science Quotes 
martin Senior Austria, 20170508 21:19 @ Helmut Posting: # 17325 Views: 877 

Hi Helmut, Thank you very much for your response  much appreciated. Regarding distributions  thank you for the reminder & sorry for not being clear We look at the end of the profile assuming C0=C1 this would lead to
C0 < rlnorm(n=1e6, meanlog=log(1)0.5*log(0.3^2+1), sdlog=sqrt(log(0.3^2+1))) best regards Martin 