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RE: a question about the mixture distribution

From: Joseph Standing <joseph.standing>
Date: Tue, 20 Oct 2009 09:24:07 +0200

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: owner-nmusers
Behalf Of Fidler,Matt,FORT WORTH,R&D
Sent: den 13 oktober 2009 21:33
To: wu kehua
Cc: nmusers
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: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
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 20 2009 - 03:24:07 EDT

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