Re: distribution assumption of Eta in NONMEM

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