logistic regression

From: "Charlotte van Kesteren" <Apcks@slz.nl> Subject: logistic regression Date: Fri, 14 Sep 2001 16:14:58 +0200 Dear NONMEM-users, We have AUC-values and toxicity data of a total of 143 individuals, originating from 4 studies with different treatment schedules. The toxicity data are dichotomous, i.e. the adverse effect either occurs (1) or it does not (0). We have one data point for each patient. With logistic regression in NONMEM, we have tried to model the relation between exposure and the chance of toxicity. Furthermore, we want to investigate a possible schedule dependency in this relation. However, we are not sure whether it is appropriate to estimate interindividual variability with logistic regression with only one observation per individual. Furthermore, how can we judge goodness of fit with such a data set? Does anyone have experience with these kind analyses? Thank you in advance for your help. Best regards, Charlotte van Kesteren

From: Lewis B Sheiner <lewis@c255.ucsf.edu> Subject: Re: logistic regression Date: Fri, 14 Sep 2001 08:41:29 -0700 Can't be done. -- _/ _/ _/_/ _/_/_/ _/_/_/ Lewis B Sheiner, MD (lewis@c255.ucsf.edu) _/ _/ _/ _/_ _/_/ Professor: Lab. Med., Biophmct. Sci., Med. _/ _/ _/ _/ _/ Box 0626, UCSF, SF, CA, 94143-0626 _/_/ _/_/ _/_/_/ _/ 415-476-1965 (v), 415-476-2796 (fax)

From: Michael.J.Fossler@dupontpharma.com Subject: Re: logistic regression Date: Fri, 14 Sep 2001 13:08:52 -0400 I have performed logistic regression with similar types of data. You could simply perform the analysis in either S-plus or SAS. BTW, an excellent text on LR is Hosmer and Lemeshow, Applied Logistic Regression. This text covers many of the questions that you have asked (none of which are really simple) and is very readable Mike *********************************************************************** Michael J. Fossler Associate Director Drug Metabolism and Pharmacokinetics, DuPont Pharmaceuticals (302) 366-6445 Cell: (302) 584-5495 michael.j.fossler@dupontpharma.com

From: "Piotrovskij, Vladimir [JanBe]" <VPIOTROV@janbe.jnj.com> Subject: RE: logistic regression Date: Mon, 17 Sep 2001 14:05:46 +0200 Charlotte, It is possible to solve some of your problems in NONMEM. However, the best way is to apply generalized linear regression using one of the statistical packages. With NONMEM, try the following control stream: $PROB dichotomous response: fixed effect of schedule $DATA nmd.ssc $INPUT ID AUC SCHD DV ; schedule coded as 1,2,3, etc. $PRED SCHD1 = 0 SCHD2 = 0 SCHD3 = 0 SCHD4 = 0 IF (SCHD.EQ.1) SCHD1 = 1 IF (SCHD.EQ.2) SCHD2 = 1 IF (SCHD.EQ.3) SCHD3 = 1 IF (SCHD.EQ.4) SCHD4 = 1 SLOPE = THETA(1) E50 = SCHD1*THETA(2)+SCHD2*THETA(3)+SCHD3*THETA(4)+SCHD4*THETA(5) INT = -LOG(E50) * SLOPE LOGIT = INT + SLOPE * LOG(AUC) + ETA(1) A=EXP(LOGIT) P=A/(1+A) IF (DV.EQ.1) Y=P IF (DV.EQ.0) Y=1-P $THETA (2 5 7); 1 SIGM (0 30 50); 2 E50 SCHD=1 (20 50 70); 3 E50 SCHD=2 (40 70 90); 4 E50 SCHD=3 (60 100 200); 5 E50 SCHD=4 $OMEGA .0001 $EST METHOD=COND LAPLACE LIKE MAX=500 PRINT=10 $COV $TABLE ID AUC SCHD DV FILE=tab.ssc ONEHEADER NOPRINT I tested it using simulation-fitting. Note that you need sufficient number of individuals per schedule to identify all the parameters with sufficient precision. In my simulation I included 20 individuals per schedule and it was OK. Best regards, Vladimir ------------------------------------------------------------------------ Vladimir Piotrovsky, Ph.D. Research Fellow Global Clinical Pharmacokinetics and Clinical Pharmacology (ext. 5463) Janssen Research Foundation B-2340 Beerse Belgium Email: vpiotrov@janbe.jnj.com

From: "James Bailey" <James_Bailey@EmoryHealthCare.org> Subject: logistic regression Date: Tue, 18 Sep 2001 16:24:59 -0500 I believe the difficulty with logistic regression for sparse dichotomous data can be well appreciated by considering the case of binary data (for example, loss of responsiveness with an intravenous anesthetic) with one data point per patient. The probability of a positive drug effect is given by P = C**gamma/(C**gamma + C50**gamma) (1) This is equivalent to a model which postulates an underlying continuous drug effect E given by E = gamma*ln(C/C50) + epsilon (2) where epsilon is a random variable with a logistic distribution. It is further postulated that a positive binary drug effect is observed if E > 0 The probability of positive binary drug effect is equal to the probability that epsilon is greater than -gamma*ln(C/C50). and using the definition of the logistic distribution one can easily derive equation (1). Now consider interpatient variability and assume that ln(C50) =ln(<C50>) + eta where <C50> is the "typical value" and eta is normally distributed. Then E = gamma*ln(C/<C50>) + gamma*eta + epsilon In this case the probability of a positive binary drug effect is equal to the probability that the random variable gamma*eta + epsilon is greater than -gamma*ln(C/<C50>). However, consider the situation where epsilon conforms to a normal distribution instead of a logistic distribution. Then gamma*eta + epsilon also has a normal distribution and it is impossible to determine the relative contributions of eta and epsilon to the overall variance. In this situation it is impossible to do a complete analysis of binary data with one data point per patient. This, of course, corresponds to probit analysis but it makes the difficulty apparent. The normal and logistic distributions are not that different. Doing a population analysis of sparse binary data depends on the ability to distinguish between the two distributions and will be almost impossible. Furthermore, it rests on the assumption of an underlying logistic distribution for the intrapatient variability (in epsilon), and there is little basis for this assumption. I and my colleague Wei Lu have done some simulations and our results indicate that from 5-10 data points per patient are necessary to estimate <C50> or gamma with any degree of reliability. Jim Bailey