Re: a question about the mixture distribution

2009/10/20 Joseph Standing <[email protected]> > Dear Kehua, > > > > We discussed your question in the Uppsala pharmacometrics meeting > yesterday. If I understood your original question correctly, it was that > there were two possible conditions for ALPHA and BASE, and that each > combination was possible. A general principle for coding this: > > > > …. > > $MIX > > > > P(1) = Theta(alph) * Theta(Base) > > P(2) = (1 – Theta(alph)) * Theta(Base) > > P(3) = Theta(alph) * (1 - Theta(base)) > > P(4) = (1 – Theta(alph)) * (1-Theta(base)) ; = 1 – P(1) – P(2) – > P(3) > > > > … > > > > $THETA (0,x,1) ; Theta(alph) > > $THETA (0,y,1) ; Theta(base) > > > > > > You should be able to extend this general principle to suit your specific > case. > > Best wishes, > > > > Joe Standing > > > ------------------------------ > > *From:* [email protected] [mailto:[email protected]] > *On Behalf Of *Fidler,Matt,FORT WORTH,R&D > *Sent:* den 13 oktober 2009 21:33 > *To:* wu kehua > *Cc:* [email protected] > *Subject:* RE: [NMusers] a question about the mixture distribution > > > > Kehua, > > > > The number of subgroups depend on how you view the subgroups. > > > > There are three subgroups for ALPH, and there are two subgroups for BASE. > Therefore, there could be > > > > 3*2 = 6 subgroups > > > > or > > > > 3+2 = 5 subgroups -- 3 for ALPH and 2 for BASE. > > > > > > While technically there are 6 groups, you only code for 5, and each of > these five have sufficient information to determine which group each > individual is a member of. > > > > Matt. > > > > > ------------------------------ > > *From:* wu kehua [mailto:[email protected]] > *Sent:* Tuesday, October 13, 2009 11:58 AM > *To:* Fidler,Matt,FORT WORTH,R&D > *Cc:* [email protected] > *Subject:* Re: [NMusers] a question about the mixture distribution > > Hi Matt, > > Thank you very much for your kindly reply. It is very helpful. > > I still have another question about this issue. I have tried the first > option. I applied three subgroups on ALPH and two subgroups on BASE. I think > there should be six subgroups in total, right? But in the resulting data, > there are just five subgroups. The code follows, > > $PRED > > CALLFL =1 > EST = MIXEST > IF (MIXNUM.EQ.3) THEN > ALPH = (THETA(3)) > END IF > IF (MIXNUM.EQ.2) THEN > ALPH = (THETA(2)) > END IF > IF (MIXNUM.EQ.1) THEN > ALPH = (THETA(1)) > ENDIF > > > IF (MIXNUM.EQ.4) THEN > BASE=THETA(4) > ELSE > BASE=THETA(5) > END IF > > > > $MIX > > P(1) = THETA(6) > P(2)=THETA(7) > P(3) = 1-THETA(7)-THETA(6) > P(4)=THETA(8) > P(5)=1-THETA(8) > NSPOP = 5 > > Sorry to bother you and thank you very much. I will try that one mentioned > in your mail and let you know the results. > I appreciate your helps. > > Best regards, > > Kehua > > 2009/10/13 Fidler,Matt,FORT WORTH,R&D <[email protected]> > > Kehua, > > > > Option 1 is definitely better. This states that there is a possibility > that a population falls into ALPH1 or ALPH2. Within that same population > there are two populations for BASE. > > > > The other option states that each person in the has distinct parameters > that four populations fall into: ALPH - Pop1, ALPH - Pop 2, Base -Pop 1, or > Base -Pop2. Therefore, if you selected ALPH - pop1, you wouldn't have the > parameter base. (You require this by having P1 - P4 to add up to be one - > the total probability). > > > > A third option you may consider is if you have reason to believe that the > populations that have ALPH1 and Base1 are the same: > > > > $PRED > IF (MIXNUM.EQ.2) THEN > ALPH =THETA(1) > > ELSE > ALPH = THETA(2) > ENDIF > > IF (MIXNUM.EQ.2) THEN > > > BASE=THETA(3) > ELSE > BASE=THETA(4) > END IF > > $MIX > > P(1) = THETA(5) > P(2) = 1-THETA(5) > > NSPOP = 2 > > > > I haven't run anything like Option 1, and am unsure if NONMEM supports two > separate populations for ALPH and BASE. Has anyone tried this? > > > > Matt. > > > ------------------------------ > > *From:* [email protected] [mailto:[email protected]] > *On Behalf Of *wu kehua > *Sent:* Tuesday, October 13, 2009 10:58 AM > *To:* [email protected] > *Subject:* [NMusers] a question about the mixture distribution > > Hi, > > I am a new NONMEM user. I have a question about mixture distribution. > > I have two parameters. How to apply mixture distribution on the both > parameters? I should use the first one or the second one? > > *First,* > $PRED > IF (MIXNUM.EQ.2) THEN > ALPH =THETA(1) > END IF > IF (MIXNUM.EQ.1) THEN > ALPH = THETA(2) > ENDIF > > IF (MIXNUM.EQ.3) THEN > BASE=THETA(3) > ELSE > BASE=THETA(4) > END IF > > $MIX > > P(1) = THETA(5) > P(2) = 1-THETA(5) > P(3)=THETA(6) > P(4)=1-THETA6) > NSPOP = 4 > > > *Second,* > > IF (MIXNUM.EQ.1) THEN > ALPH =THETA(1) > BASE=THETA(3) > END IF > IF (MIXNUM.EQ.2) THEN > ALPH = THETA(1) > BASE=THETA(4) > ENDIF > IF (MIXNUM.EQ.3) THEN > ALPH = THETA(2) > BASE=THETA(3) > ELSE > ALPH = THETA(2) > BASE=THETA(4) > END IF > > $MIX > > P(1) = THETA(5) > P(2) = THETA(6) > P(3)=THETA(7) > P(4)=1-THETA(5)-THETA(6)-THETA(7) > NSPOP = 4 > > Thank you very much! > > Best regards, > > Kehua > > > This e-mail (including any attachments) is confidential and may be > legally privileged. 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