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RE: a question about the mixture distribution

From: Fidler,Matt,FORT WORTH,R&D <Matt.Fidler>
Date: Tue, 13 Oct 2009 14:32:46 -0500

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:zckehua
Sent: Tuesday, October 13, 2009 11:58 AM
To: Fidler,Matt,FORT WORTH,R&D
Cc: nmusers
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 =
<Matt.Fidler
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: owner-nmusers
[mailto:owner-nmusers
] On Behalf Of wu kehua
Sent: Tuesday, October 13, 2009 10:58 AM
To: nmusers
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




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Received on Tue Oct 13 2009 - 15:32:46 EDT

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