Nick, Mats
I would guess that nonmem should inflate variance (for this example)
trying to fit the observed uniform (-0.5, 0.5) into some normal N(0, ?).
This example (if I read it correctly) shows that Nonmem somehow
estimates variance without making distribution assumption.
Nick, you mentioned:
"the mean estimate of OMEGA(1) was 0.0827"
does it mean that Nonmem-estimated OMEGA was close to 0.0827 or you
refer to the variances of estimated ETAs?
Thanks
Leonid
--------------------------------------
Leonid Gibiansky, Ph.D.
President, QuantPharm LLC
web: www.quantpharm.com
e-mail: LGibiansky at quantpharm.com
tel: (301) 767 5566
Mats Karlsson wrote:
Nick,
It has been showed over and over again that empirical Bayes estimates,
when individual data is rich, will resemble the true individual
parameter regardless of the underlying distribution. Therefore I don’t
understand what you think this exercise contributes.
Best regards,
Mats
Mats Karlsson, PhD
Professor of Pharmacometrics
Dept of Pharmaceutical Biosciences
Uppsala University
Box 591
751 24 Uppsala Sweden
phone: +46 18 4714105
fax: +46 18 471 4003
*From:* [email protected]
[mailto:[email protected]] *On Behalf Of *Nick Holford
*Sent:* Monday, May 31, 2010 6:05 PM
*To:* [email protected]
*Cc:* 'Marc Lavielle'
*Subject:* Re: [NMusers] distribution assumption of Eta in NONMEM
Hi,
I tried to see with brute force how well NONMEM can produce an empirical
Bayes estimate when the ETA used for simulation is uniform. I attempted
to stress NONMEM with a non-linear problem (the average DV is 0.62). The
mean estimate of OMEGA(1) was 0.0827 compared with the theoretical value
of 0.0833.
The distribution of 1000 EBEs of ETA(1) looked much more uniform than
normal.
Thus FOCE show no evidence of normality being imposed on the EBEs.
$PROB EBE
$INPUT ID DV UNIETA
$DATA uni1.csv ; 100 subjects with 1 obs each
$THETA 5 ; HILL
$OMEGA 0.083333333 ; PPV_HILL = 1/12
$SIGMA 0.000001 FIX ; EPS1
$SIM (1234) (5678 UNIFORM) NSUB=10
$EST METHOD=COND MAX=9990 SIG=3
$PRED
IF (ICALL.EQ.4) THEN
IF (NEWIND.LE.1) THEN
CALL RANDOM(2,R)
UNIETA=R-0.5 ; U(-0.5,0.5) mean=0, variance=1/12
HILL=THETA(1)*EXP(UNIETA)
Y=1.1**HILL/(1.1**HILL+1)
ENDIF
ELSE
HILL=THETA(1)*EXP(ETA(1))
Y=1.1**HILL/(1.1**HILL+1) + EPS(1)
ENDIF
REP=IREP
$TABLE ID REP HILL UNIETA ETA(1) Y
ONEHEADER NOPRINT FILE=uni.fit
I realized after a bit more thought that my suggestion to transform the
eta value for estimation wasn't rational so please ignore that senior
moment in my earlier email on this topic.
Nick
--
Nick Holford, Professor Clinical Pharmacology
Dept Pharmacology & Clinical Pharmacology
University of Auckland,85 Park Rd,Private Bag 92019,Auckland,New Zealand
tel:+64(9)923-6730 fax:+64(9)373-7090 mobile:+64(21)46 23 53
email: [email protected] <mailto:[email protected]>
http://www.fmhs.auckland.ac.nz/sms/pharmacology/holford
