I am using SAS for Windows 9.4. Below is the log for what I thought was simple SAS/IML code. But I cannot understand why I am receiving an error. Can anyone explain why and how to fix? proc iml;     a1={1 2 3 4 0 0};     a2={1 1 1 1 1 1};     a3=j(1,6,0);     a4=j(1,6,0);     n_cols=ncol(a1);     PRINT a1 a2 a3 a4 n_cols;     do i=1 to n_cols;     if a1[i] ^= 0 then do;        * Condition [1] *;         a3[i]=1;         a4[i]=a2/a1; *Why is this line processed when Condition [1] not met *;     end;     else do;         a3[i]=9;         a4[i]=9;     end; end; PRINT a3 a4; quit;   The log file is attached. ... View more

I need some help with a problem related to a previous post. Here's the situation: I have a huge dataset (N=10000) that contains simulated event times (with right censored data) and I would like to generate a similar simulated dataset by using the estimated Weibull parameters from that dataset. First I ran PROC LIFEREG with the following model statement:      MODEL day*cnsr(1)= / dist=Weibull; This generated the following ML estimates: Intercept=7.7, Scale=4.5, Weibull_Scale=2300, Weibull_Shape=0.22.   Then, to simulate time values in a new huge dataset, I used the SAS statements:      ...      call streaminit(72131);      Day=rand("Weibull", 0.22, 2300);      ... But the distribution of times on this new dataset is very different. For example, the proportion with day<=365 in this new dataset is 48%, as compared to 58% in the original dataset. This discrepancy persists, even if I change the randomization seed or increase the number of simulated records.      I found the documentation for PROC LIFEREG and the RAND function too confusing to help me reconcile this problem.  If anyone can offer me any help, I would greatly appreciate it.   ... View more

Why doesn't PROC LIFREG Weibull parameter estimates match those that I used in the RAND function?      Below is SAS code that generates subject survival data from a Weibull distribution using the RAND function. I then analyze the survival data with PROC LIFEREG, which outputs an estimate of the shape and scale parameters. Although the estimated shape parameter from PROC LIFEREG is approximately the same as used to generate the subject data, the scale parameter is not even close. So what exactly is “scale” estimate from PROC LIFEREG? Is there a way to estimate the Weibull parameters from PROC LIFEREG (or another SAS procedure)? Any help is appreciated   data X;   call streaminit(2468);     a=1.6; *Shape;   b=4.0; *Scale;     TRT=1;   do Subject=1 to 10000;     T=rand("Weibull", a, b);     Event=(rand("Uniform")<0.20);     output;   end; run;   proc lifereg data=X;   model T*Event(0)=TRT / dist=Weibull; run; ... View more