ssussu ☆ China, 20191206 10:58 (253 d 03:51 ago) Posting: # 20937 Views: 3,815 

Dear guys, When we conducted a BE study designed with two way crossover, one subject dropped out at the second period, so we just got one period data.At this situation, should this subject's data with one period be included in the BE assessment? Why? Best regards 
Beholder ★ Russia, 20191206 12:36 (253 d 02:13 ago) (edited by Beholder on 20191206 13:06) @ ssussu Posting: # 20940 Views: 3,121 

Dear ssussu! » At this situation, should this subject's data with one period be included in the BE assessment? No, you could not. » Why? Because you need comparison between T and R for each subject. This is the aim  to compare. In your case you have either T or R (I dont know exactly which drug was taken on period 2). At least, you could use Cmax value from period 2 if the subject withdraw after Cmax was reached. AUC  no way. — Best regards Beholder 
ElMaestro ★★★ Belgium?, 20191206 13:14 (253 d 01:35 ago) @ Beholder Posting: # 20941 Views: 3,113 

Hello, both, » » At this situation, should this subject's data with one period be included in the BE assessment? » » No, you could not. We do not need a TR comparison in each subject to do an analysis. We are just stuck in that paradigm. We could in principle fit a model with missing period data. And all data could be used for the calculation of the CI. I do not know why there is a regulatory tradition for only using completers  I would say there is even a potential scientific and ethical advantage of using all period data. — I could be wrong, but... Best regards, ElMaestro "Pass or fail" (D. Potvin et al., 2008) 
Helmut ★★★ Vienna, Austria, 20191206 14:47 (253 d 00:02 ago) @ ElMaestro Posting: # 20942 Views: 3,093 

Ahoy my capt’n and welcome back to this side of the pond! » I do not know why there is a regulatory tradition for only using completers … Those days when studies were evaluated by a paired ttest with a pocket calculator? » – I would say there is even a potential scientific and ethical advantage of using all period data. Correct. If we would be allowed (pun!) to use a mixedeffects model. Patterson and Jones argued against this doubtful practice and the ethical implications of discarding data.*
— Diftor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
PharmCat ★ Russia, 20191206 23:46 (252 d 15:03 ago) @ Helmut Posting: # 20954 Views: 3,067 

Hi all! » Correct. If we would be allowed (pun!) to use a mixedeffects model. Patterson and Jones argued against this doubtful practice and the ethical implications of discarding data. Really, there is no problem in mixed model approach to use all data if model constructed correctly. May be not all regulatory bodies can understand how it works  it is not rare situations when I get issue like "Give me ANOVA!!! I don't want to see your GLM or MIXED...", pfff... or "In guideline ***(any ancient as mammoth's excrement guideline) written ratio of geometric means  please just divide one to another and give me CI"... So here two answers  theoretically there is no problem to use all data, but regulatory authorities can bann you for it. (IMHO) 
wienui ★ Germany, Oman, 20191207 04:01 (252 d 10:48 ago) (edited by wienui on 20191207 06:47) @ PharmCat Posting: # 20955 Views: 3,060 

Dear all, » So here two answers  theoretically there is no problem to use all data, but regulatory authorities can bann you for it. (IMHO) I also agree about this as a previous acadmic person, but as a regulator, which regulatory supportive Guidelines (EMA & FDA) reasons, can I rely on it to be able to use data from study subjects who didn't completed all study periods? Best regards, Osama — Cheers, Osama 
mittyri ★★ Russia, 20191207 20:58 (251 d 17:51 ago) @ wienui Posting: # 20959 Views: 3,029 

Dear Osama, » I also agree about this as a previous acadmic person, but as a regulator, which regulatory supportive Guidelines (EMA & FDA) reasons, can I rely on it to be able to use data from study subjects who didn't completed all study periods? I doubt that something exists at the moment. Please also see the related discussion and cites of EMA guideline. — Kind regards, Mittyri 
wienui ★ Germany, Oman, 20191209 05:25 (250 d 09:24 ago) @ mittyri Posting: # 20963 Views: 2,927 

Dear Mittyri, » I doubt that something exists at the moment. » Please also see the related discussion and cites of EMA guideline. Thank you very much, it is really helpful. Best regards, Osama — Cheers, Osama 
ElMaestro ★★★ Belgium?, 20191208 01:31 (251 d 13:18 ago) @ PharmCat Posting: # 20961 Views: 3,008 

Hi all, » » Correct. If we would be allowed (pun!) to use a mixedeffects model. Patterson and Jones argued against this doubtful practice and the ethical implications of discarding data. » » Really, there is no problem in mixed model approach to use all data if model constructed correctly. But would we really need a mixed model ?? I seem to recall a thread about it some time ago. I believe the normal linear model fits without trouble to a 222BE dataset with a missing period, so why would it be necessary to use a mixed model for that situation at all? As I recall it mixed model done with ML are proven to give unbiased estimates, but it has never been proven that REML fits are unbiased (and REML is generally used in BE, whenever a discussion of the mixed model comes into play), hence I can't imagine that bias would be a reason for picking the mixed model over a linear model when we have a simple missing period situation?!? Am curious to learn more about this. Here's an example in R: rm(list=ls(all=TRUE)) Note e.g. that the two fits have different residuals and residual df's, which to me means incomplete subjects are not deleted (R does not know and is not being told something is incomplete; the full rank design matrix is still invertible and so on). Many thanks for any light you can shed onto this. Edit: I added set.seed(123456) to your code in order to get reproducible results. [Helmut]— I could be wrong, but... Best regards, ElMaestro "Pass or fail" (D. Potvin et al., 2008) 
Shuanghe ★★ Spain, 20191209 12:00 (250 d 02:49 ago) @ ElMaestro Posting: # 20964 Views: 2,867 

Hi ElMaestro, » Note e.g. that the two fits have different residuals and residual df's, which to me means incomplete subjects are not deleted (R does not know and is not being told something is incomplete; the full rank design matrix is still invertible and so on). What about the following?
X3 < Xm[7,]
Xm has missing period (1) for subject 4, X3 has no subject 4. compare anova(M2) and anova(M3) , residual and df of residual are same. 90% CI also same. So wouldn't it mean that R deleted the extra period (2) of subject 4 in Xm automatically when doing BE evaluation? Lsmeans are different, so subject 4 period 2 was kept for that calculation. I woulds say that this behaviour is the same as SAS.— All the best, Shuanghe 
PharmCat ★ Russia, 20191209 15:13 (249 d 23:36 ago) @ ElMaestro Posting: # 20966 Views: 2,851 

» But would we really need a mixed model ?? Hello, ElMaestro! Look at residuals: resid(M1) This is how I understand: For observation 8 we have 3.608225e16, I think, that mean, that this subject affect on "intravariation" and make it less, and you can see SE is smaller. "Intraindividual part" went to coefficient and began part of interindividual. Brr.. I don't know how to say with сlever words. In lm / glm we see each observation as at statistically independent. And with lm we make smart trick when "storing" interindividual variation in model coefficients, and exclude it from model error. Missing observation violate this system. In mixed model we have another situation  all observation of subject is one statistically independent observation  realization of multidimentional variable. In the case of the ML / REML estimation, such a significant transition of one component of the variation to another does not occur in the case when the data contains missing values. ML and REML variation estimates all biased, but less biased than lm with missing data. 
ElMaestro ★★★ Belgium?, 20191221 15:02 (237 d 23:47 ago) @ PharmCat Posting: # 21012 Views: 2,685 

Hi PharmCat, » For observation 8 we have 3.608225e16, I think, This. I think, is around the "effective zero" for fits in R at default settings on 64 and 32bit systems. » I don't know how to say with сlever words. Very unfortunate, because I did not understand what was being said. I would like to get the insight. It is at the limits of my conception. » ML and REML variation estimates all biased, but less biased than lm with missing data. Is this a fact? How do we actually know this? Do you have a reference I coud learn from (not Pinheiro and Bates, I don't understand a word of it). Does "less biased" apply to both the fixed effects and to the variance components? — I could be wrong, but... Best regards, ElMaestro "Pass or fail" (D. Potvin et al., 2008) 
PharmCat ★ Russia, 20191222 01:12 (237 d 13:37 ago) @ ElMaestro Posting: # 21013 Views: 2,674 

Hi ElMaestro! » This. I think, is around the "effective zero" for fits in R at default settings on 64 and 32bit systems. Anyway residual "can't" be zero. » Is this a fact? How do we actually know this? Do you have a reference I coud learn from (not Pinheiro and Bates, I don't understand a word of it). For ML: Yes, it's a fact. ML biased always "by definition". proof For REML: this is a more difficult question. We can see following: * link 1 * link 2 * link 3 * link 4 * link 5 * link 6 Some links behind the paywall, but this problem can be solved with scihub. What can we say: "REML does effectively reduce bias.", "The REML estimates are typically less biased than the ML methods." REML does not always eliminate all of the bias in parameter estimation, since many methods for obtaining REML estimates cannot return negative estimates of a variance component. However, this source of bias also exists with ML, so REML is clearly the preferred method for analyzing large data sets with complex structure. Hm... I understand that REML not always unbiased. May be I'm wrong, but all above make me think this. » Does "less biased" apply to both the fixed effects and to the variance components? Mmm. If we call model coefficients as fixed effects  they are unbiased, no problems with them. Variance component in LM if say strictly is a model error. LM have analytical solution and ubiased estimator of variance. But model should be fitted correctly: each level of any factor should have at least 2 observations. 
Helmut ★★★ Vienna, Austria, 20191222 10:37 (237 d 04:12 ago) @ ElMaestro Posting: # 21016 Views: 2,698 

Hi ElMaestro, » » For observation 8 we have 3.608225e16, I think, » I think, is around the "effective zero" for fits in R at default settings on 64 and 32bit systems. Yes, it is.
— Diftor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
ElMaestro ★★★ Belgium?, 20191223 14:37 (236 d 00:12 ago) @ Helmut Posting: # 21021 Views: 2,585 

Hi Hötzi, » » » For observation 8 we have 3.608225e16, I think, » » I think, is around the "effective zero" for fits in R at default settings on 64 and 32bit systems. » » Yes, it is. »
This comparison in your context is just a test if the difference is less than about 10^{8 }since there is an implied tolerance argument for all.equal , the square root of .Machine$double.eps Effective zero residuals will be somewhat better than 10^{8} in practice. They will depend on the approach used to find the solution; in lm I believe the approach is via a qr decomposition of the model matrix, and R by defualt has a tol argument in that function of 10^{7} which lm may be leaning on. Here's an example of a perfect fit, therefore having effective zero residuals: a=c(rep(1,5), rep(2,5), rep(3,5)) It may actually not be the best example since the dependents are all representable inernally in R's (and computer's) binary. Perhaps this makes a better point: a=c(rep(pi,5), rep(sin(1.5+pi),5), rep(log(pi),5)) — I could be wrong, but... Best regards, ElMaestro "Pass or fail" (D. Potvin et al., 2008) 
PharmCat ★ Russia, 20191224 14:18 (235 d 00:31 ago) @ Helmut Posting: # 21025 Views: 2,583 

» » » For observation 8 we have 3.608225e16, I think, » » I think, is around the "effective zero" for fits in R at default settings on 64 and 32bit systems. Hi Helmut! I suppose hat this value come from QR decomposition with pivoting of rank deficient X matrix. So real value = 0. Can residual of random variable be equal zero?  No => bias. 
Helmut ★★★ Vienna, Austria, 20191224 14:54 (234 d 23:55 ago) @ PharmCat Posting: # 21026 Views: 2,517 

Hi PharmCat » I suppose hat this value come from QR decomposition with pivoting of rank deficient X matrix. So real value = 0. Can residual of random variable be equal zero?  No => bias. I disagree. In the model ε = 0. However in the fit, the sum of residuals is only asymptotically 0. We shouldn’t speak of bias when we obtain something sufficiently close to the numeric resolution of the machine. — Diftor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
ElMaestro ★★★ Belgium?, 20191224 15:10 (234 d 23:39 ago) @ Helmut Posting: # 21027 Views: 2,525 

Hi both, » I disagree. In the model ε = 0. However in the fit, the sum of residuals is only asymptotically 0. We shouldn’t speak of bias when we obtain something sufficiently close to the numeric resolution of the machine. Perhaps I am getting it wrong; the sentence above sounds a little off and I have a feeling you may not be discussing the same thing? The sum of residuals for a fitted normal linear model will be zero. Not asymptotically. If you sum them in R or any other software you will get zero, be it either like zerozero or effectively zero, depending on the implementation. This is because the underlying assumption is that epsilon be distributed with mean zero. If we end up with a nonzero sum, I'd say we have screwed up somewhere. Try sum(resid(M)) — I could be wrong, but... Best regards, ElMaestro "Pass or fail" (D. Potvin et al., 2008) 
Helmut ★★★ Vienna, Austria, 20191228 13:55 (231 d 00:54 ago) @ ElMaestro Posting: # 21032 Views: 2,431 

Hi ElMaestro, » The sum of residuals for a fitted normal linear model will be zero. Not asymptotically. Sorry, I did not mean asymptotically (central limit theorem ) but approximately (due to numeric issues). Theoretically the sum should be zero, of course. IIRC, in old versions of lm() fitting data without errors (like your first example) threw an error and there was even a warning in the manpage.— Diftor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes 
PharmCat ★ Russia, 20191224 18:40 (234 d 20:09 ago) @ Helmut Posting: # 21028 Views: 2,528 

» I disagree. In the model ε = 0. However in the fit, the sum of residuals is only asymptotically 0. We shouldn’t speak of bias when we obtain something sufficiently close to the numeric resolution of the machine. Hi Helmut! I don't mean that we have bias because 0 not equal eps(). Numericaly we can make calculation with values lower that eps() like: (1E40 + 2E40) * 3E+40 = 9.0 . Problems can start with things like this: 1 + 1E40, for example: (1E+40  1E+40 + 1 == 1E+40 + 1  1E+40) == false .  this not what about a try to say.I try to say, that zero residual (or effective zero residual) can't be at all, "by definition" in linear model. 