@FreelanceReinh Thanks for your further explanation. I guess if we have more than 3 time-varying variables, this approach would be very laborious. I addition, the PROC TRANSPOSE might experience difficulty when transferring a wide dataset to a narrow dataset, given there is more than one time-varying variables (e.g., A_wk1, A_wk2, ... A_wkm, and B_wk1, B_wk2, ... B_wkn). ... View more

Thanks for the notice, @FreelanceReinh    Great! And I have one more question. If the two time-varying variables change on the same day (tie), how do we count them? Only two rows with a same start and stop time? Thank you. ... View more

Hi, @FreelanceReinh    Thank you for your suggestion. Hmmm, it's a great idea to deal with multiple time-dependent variables. Please let me reproduce your idea.   First, via the counting process method. Say, we have two time-varying variables named A and B, and there are other fixed variables which is represented in union by U. If there is an individual who was followed for 14 days, for whom the A changed on day 7 from 0 to 1. In addition, the B changed on day 4 from 1 to 0. According to my understanding of your idea, there should be three rows for this individual in the overall table and they are:   row one: A=0, B=1, start=0, stop=4 row two: A=0, B=0, start=4, stop=7 row three A=1, B=0, start=7, stop=14   Therefore, every change in the time-varying variable produces an extra row for the individual, given that the time-varying variables do not change on the same day (i.e., tie).   I have not yet thought about the programming method and if I have ideas about it I would update this post. Again, thank you for your feedback.   Tom ... View more

Hi, Gris   Thank you very much for the suggestion. I see, and I think I should paste part of my code below.   After the PROC LIFEREG procedure, I got the estimation of parameters. Then, I used PROC DATA to simulate projections before I used these data to plot the CDF.   The shape parameter is rho, along with its confidence interval of rho_l and rho_u.   As you can see in the PROC DATA procedure, I did not use the confidence interval of model coefficients to simulate, instead, I only used that of the shape parameter. I don't know if this is the standard of art, or if I should have used the confidence interval of coefficients too.   data hazard; do t = 0 to 14 by 0.01; /* Adjust the range and increment as appropriate for your data */ lambda = exp( - 0.0105*65 + 0.49*1.75 + 0.04*45 + 0.29*0 + 0.70*0 + 0.29*1); /* Replace `intercept` with your actual intercept value */ rho = 2.3531; /* Shape parameter from PROC LIFEREG output */ rho_l = 2.0850; rho_u = 2.6557; h_t = rho/lambda * ((t/lambda)**(rho - 1)); /* Calculate hazard */ loght = log(rho/lambda * ((t/lambda)**(rho - 1))); lower_h_t = rho_u/lambda * ((t/lambda)**(rho_u - 1)); upper_h_t = rho_l/lambda * ((t/lambda)**(rho_l - 1)); suriv = 1 - exp(-(t/lambda) ** rho); suriv_u = 1 - exp(-(t/lambda) ** rho_u); suriv_l = 1 - exp(-(t/lambda) ** rho_l); output; end; *drop lambda rho rho_l rho_u; run; data hazard2; do t = 0 to 14 by 0.01; /* Adjust the range and increment as appropriate for your data */ lambda = exp( - 0.0105*65 + 0.49*1.75 + 0.04*45 + 0.29*0 + 0.70*0 + 0.29*0); /* Replace `intercept` with your actual intercept value */ rho = 2.3531; /* Shape parameter from PROC LIFEREG output */ rho_l = 2.0850; rho_u = 2.6557; h_t = rho/lambda * ((t/lambda)**(rho - 1)); /* Calculate hazard */ loght = log(rho/lambda * ((t/lambda)**(rho - 1))); lower_h_t = rho_u/lambda * ((t/lambda)**(rho_u - 1)); upper_h_t = rho_l/lambda * ((t/lambda)**(rho_l - 1)); suriv = 1 - exp(-(t/lambda) ** rho); suriv_u = 1 - exp(-(t/lambda) ** rho_u); suriv_l = 1 - exp(-(t/lambda) ** rho_l); output; end; *drop lambda rho rho_l rho_u; run; ... View more

Hi, Ksharp   Thanks for your feedback. However, I don't see it that way. The stddev is the standard deviation of the predictor and the predictor can only be 1 or 0, given the predictor represents the trial with only 2 arms (e.g., drug vs placebo). A predictor of binomial distribution does not have to be only either 1 or 0, it can be any integer. Therefore, the distribution of the predictor is a  Bernoulli distribution, not a binomial distribution. My first post made the mistake and it should have been Bernoulli distribution.   Regards, ... View more

Sorry, it should be a Bernoulli distribution. Each observation in this case either receives the drug or not. Therefore, the random variable is either 1 or 0. The random variable of a binomial distribution can have numbers other than 0 or 1. It should have been Bernoulli distribution. ... View more

Yes. And in your case, the maximum value of the square root is 0.50. ... View more

Hello, Freelance   I think you're right. The default value of stddev is 0.5, which is the standard deviation of a binary 1:1 match grouping variable, i.e., square root of 0.5(1 - 0.5). Thank you. ... View more

Hello, Rick. I am not sure about the parameter stddev. Is it the standard error of a coefficient estimate from a PROC PHREG procedure, or something different? Thanks. ... View more

Hi, ballardw. Thank you for your feedback. I understand where you are coming from, but I need to clarify that my intention is to estimate the actual power of the study after we have the actual events observed. This helps to evaluate the risk of type II error of this study. ... View more

The formula has a big error. %let se = %sysevalf((&lnHrupper - &lnHrLower)/2*1.96); This is wrong. The correct should be, %let se = %sysevalf((&lnHrupper - &lnHrLower)/(2*1.96)); I am not 100% sure about what the parameter of stddev should be. I think it is the standard deviation of logHR which approximately has a normal distribution. Based on this understanding, I constructed the formula to calculate the value, by (UpperCI - LowerCI)/(2 x 1.96) ... View more