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
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