Chuanpu, In all stable problems that I tried, parametrization
ETA() $OMEGA 0.1 ; estimated was equivalent (in terms of the estimated value and objective function) to THETA(*)*ETA() $OMEGA 1 FIXED Also, H0: THETA=0, vs. H1: THETA<>0 is the same as H0: OMEGA=0, vs. H1: OMEGA>0 since OMEGA=THETA^2 In theta-form, the problem has two identical solution THETA()=SQRT(OMEGA) and THETA()= -SQRT(OMEGA) Leonid -------------------------------------- Leonid Gibiansky, Ph.D. President, QuantPharm LLC web: www.quantpharm.com e-mail: LGibiansky at quantpharm.com tel: (301) 767 5566 On 8/30/2010 1:41 PM, Hu, Chuanpu [CNTUS] wrote:
*From:* Hu, Chuanpu [CNTUS] *Sent:* Monday, August 30, 2010 8:46 AM *To:* 'Mark Sale' *Cc:* 'nmusers' *Subject:* RE: [NMusers] Block versus diagonal omega Mark, Nice thought – the test can be conducted, but the devil is in the details. This has to do with the intricacies of the role alternative hypothesis plays in hypothesis testing: For the original parameterization testing OMEGA, the hypothesis test is H0: OMEGA=0, vs. H1: OMEGA>0 For the THETA parameterization testing OMEGA, the hypothesis test is H0: THETA=0, vs. H1: THETA<>0 So without getting into the math, the intuitive argument is that the alternative hypotheses in the 2 situations are different, therefore it is logical that the testing criteria must change. The world of math does not contain contradictions even though it may appear so at times. J Chuanpu *From:* Mark Sale [mailto:[email protected]] *Sent:* Sunday, August 29, 2010 9:19 AM *To:* Hu, Chuanpu [CNTUS] *Cc:* nmusers *Subject:* RE: [NMusers] Block versus diagonal omega Chuanpu, Do I extrapolate correctly then that: V = THETA(1)*EXP(THETA(2)*ETA(1)) . . . $OMEGA (1,FIXED). Can be tested (THETA(2) <> 0), since it is not a truncated distribution? might be an interesting exercise to do this with LRT and compare to the randomization test with the usual specification. Mark --- On *Fri, 8/27/10, Hu, Chuanpu [CNTUS] /<[email protected]>/* wrote: From: Hu, Chuanpu [CNTUS] <[email protected]> Subject: RE: [NMusers] Block versus diagonal omega To: "Mark Sale - Next Level Solutions" <[email protected]>, "Eleveld,DJ" <[email protected]> Cc: "nmusers" <[email protected]> Date: Friday, August 27, 2010, 4:33 PM Theoretically, the NONMEM objective function drop for adding a diagonal element follows a mixture chi-square distribution, from which follows that using the “usual” chi-square distribution would be conservative. This has to do with 0 being on the boundary of possible values. (See Pinheiro and Bates, Mixed Effects Models in S and S-PLUS, Springer, 2000.) As this boundary issue does not apply to off-diagonal elements, the “usual” chi-square distribution should be fine (with the usual statistical asymptotic caveats). I’d like to mention that, while the “find the best fit” mindset may be suitable for the typical exploratory setting, the p-values from repeated (e.g., stepwise) tests are not statistically interpretable. To have valid p-values, confirmatory analyses would be needed, which in my mind deserves a wider use. J Chuanpu ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~* Chuanpu Hu, Ph.D. Director, Pharmacometrics Pharmacokinetics Biologics Clinical Pharmacology Janssen Pharmaceutical Companies of Johnson & Johnson C-3-3 200 Great Valley Parkway Malvern, PA 19355 Tel: 610-651-7423 Fax: (610) 993-7801 E-mail: [email protected] </mc/[email protected]> ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*
