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]<mailto:[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]> 
[mailto:[email protected]<mailto:[email protected]>] On 
Behalf Of wu kehua
Sent: Tuesday, October 13, 2009 10:58 AM
To: [email protected]<mailto:[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




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