Omega Block off Diagonals
From: Brett Houk <bhouk@amgen.com>
Subject: [NMusers] Omega Block off Diagonals
Date: 6/27/2003 12:33 PM
Users,
I am currently running a placebo effect mixture model (responders and
non-responders) with Omega Block (3) and getting off diagonal elements
coming up as negative. Has anyone seen this result previously, and/or does
anyone have any insight as to the meaning of such a result? I will past the
control stream and final parameter est. below.
Thanks,
Brett
$PROB PLACEBO MIXTURE MODEL
$INPUT C ID TIME DV
$DATA data.csv IGNORE=C
$PRED
CALLFL = 1
EST = MIXEST
TVE0=THETA(1)
E0=TVE0*EXP(ETA(1))
IF (MIXNUM.EQ.2) THEN
EMAX = THETA(3)*EXP(ETA(3))
ELSE
EMAX = THETA(2)*EXP(ETA(2))
ENDIF
TVIT=THETA(4)
IT50=TVIT*EXP(ETA(4))
TVHL=THETA(5)
HILL=TVHL
F=E0*(1-(EMAX*(TIMW**HILL))/((IT50**HILL)+(TIMW**HILL)))
IPRED=F
W=(THETA(6)**2+F*F*THETA(7)**2)**0.5
IRES=DV-IPRED
IWRES=IRES/W
Y=F+W*EPS(1)
$MIX
NSPOP= 2
P(1) = THETA(8)
P(2) = 1 - THETA(8)
$OMEGA BLOCK (3)
0.2
0.01 0.2
0.01 0.01 0.2
$OMEGA BLOCK (1)
0.2
$THETA
(18.1) ;THETA(1) TVE0 typical value of E0
(-0.02) ;THETA(2) Emax for population 1
(0.135) ;THETA(3) Emax for population 2
(0, 5.28) ;THETA(4) TVIT typical value for IT50
(0, 1.98) ;THETA(5) TVHL typical value of Hill Factor
(0, 2.42) ;THETA(6) Additive error term
(0, 0.001) ;THETA(8) Proportional error term
(0, 0.765, 1) ;THETA(7) Proportion of population in Pop1
$SIGMA
1 FIX; Eps(1)
$COV
$EST NOABORT MAXEVAL=9999 POSTHOC
****************************************************************************
******************************
THETA - VECTOR OF FIXED EFFECTS PARAMETERS *********
TH 1 TH 2 TH 3 TH 4 TH 5 TH 6 TH 7
TH 8
1.84E+01 1.08E-02 8.80E-02 5.33E+00 1.98E+00 2.43E+00
4.89E-08 7.47E-01
OMEGA - COV MATRIX FOR RANDOM EFFECTS - ETAS ********
ETA1 ETA2 ETA3 ETA4
ETA1
+ 1.88E-01
ETA2
+ -3.30E+00 2.56E+02
ETA3
+ 1.04E+00 -4.19E+01 6.84E+01
ETA4
+ 0.00E+00 0.00E+00 0.00E+00 9.37E+01