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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 11:44:20 -0500

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
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 - 12:44:20 EDT

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