I have tried that in the past and I changed it to: Data test2; Set test; .. Array subx(*) subx1-subx34; .. Subj(i)=1/(1+exp(-subx(i))); But my answer is Test 1 Param sub1 treat subn subj1 Logit -1.11 ates -0.03 0.24 Test2 Param sub1 treat subn subj1 Logit -1.11 ates -0.03 . The test2 should have the result from the calculation Subj(i)=1/(1+exp(-subx(i))); However I get a (.) for subj1 (i.e., no calculation value missing), but the same code does give the correct answer for Test1. That is what I need to know why so Test2 can be corrected. ... View more

Yes, that was the issue and thanks for your feedback. ... View more

The result that I would like to have and I will use only subject #4 as an example but it would be done for all subjects. Original data Subject Time DV IPRED eps1 eps2 4 10 5 3 0.01 0.1 4 12 21 1 0.01 0.1 4 20 20 15 0.01 0.1 4 24 30 25 0.01 0.1 Recalculation of values at 20 and 24 hrs which will result in a number at 24 hrs less than the value at 20 hrs 4 10 5 3 0.01 0.1 4 12 21 1 0.01 0.1 4 20 (new value) 15 0.01 0.1 4 24 (value lt 20) 25 0.01 0.1 The log is: 1 OPTIONS NONOTES NOSTIMER NOSOURCE NOSYNTAXCHECK; 72 73 74 75 76 data test; 77 input Subject Time dv ipred eps1 eps2; 78 datalines; NOTE: SAS went to a new line when INPUT statement reached past the end of a line. NOTE: The data set WORK.TEST has 15 observations and 6 variables. NOTE: DATA statement used (Total process time): real time 0.01 seconds cpu time 0.02 seconds 96 ; 97 run; 98 99 100 101 data old; 102 set test; 103 retain subject dv lagdv dv eps1 eps2; 104 if time in (20 24); 105 lagdv=lag(dv); 106 %let eps1=0.0025; 107 %let eps2=0.096; 108 eps1=rand('LOGNORMAL')*sqrt(&eps1); 109 eps2=ranD('LOGNORMAL')*sqrt(&eps2); 110 dv=ipred*(1+eps1) + eps2; 111 array subs subject; 112 do i=1 to dim(subs); 113 if dv>lag(dv) then do until dv __ 22 76 ERROR 22-322: Syntax error, expecting one of the following: (, =. ERROR 76-322: Syntax error, statement will be ignored. 114 output; 115 keep subject time lagdv dv eps1 eps2; 116 end; 117 end; 118 run; NOTE: The SAS System stopped processing this step because of errors. WARNING: The data set WORK.OLD may be incomplete. When this step was stopped there were 0 observations and 6 variables. NOTE: DATA statement used (Total process time): real time 0.00 seconds cpu time 0.00 seconds 119 ... View more

The resulting Table20 did not have values for 24 hr.  Where can I find that table? Obs Subject Time DV123 1 20 10 2 20 10 3 20 20 ... View more

data new; input subject time dv; datalines; 1 20 10 1 24 5 2 20 10 2 24 15 3 20 10 3 24 20 ; run; data old; set new; retain subject dv lagdv x; do subject=1 to 3; if time in (20 24); lagdv=lag(dv); if dv>lag(dv) then x=0.5*lag(dv); end; output; keep subject dv x; run; I ran this code and the result had an issue with outputting the correct subject's number.  Can someone correct the code for me so that subjects 1-3 will be output?    Obs subject dv x123456 4 10 . 4 5 . 4 10 5.0 4 15 5.0 4 10 5.0 4 20 7.5 ... View more

For each subject there will be only one value at 20 hr. and at 24 hr. ... View more

Ok, I will write some code using the lag function to resolve the issue and if successful, I will post the results. ... View more

Yes that does give an explicit example and also addresses the issue of the seed which I failed to mention in the original post. ... View more

I received the following reply: Is your concern about controlling the seed of the Rand('Normal') ? If not the equivalent code would be : eps1 = rand('Normal') * sqrt(feps1); Which is the same as yours (without the mean), but the issue for me which I failed to initially clearly state would be to control the seed as indicated in the response. ... View more

My concern is to control the seed. The sqrt value is a value determined from a separate analysis and I only want to add random error and it is not a parameter of the normal distribution. ... View more