topic Re: The Weibull accelerated failure time model parameters estimated but how to use them for predicti in Statistical Procedures

The Weibull accelerated failure time model parameters estimated but how to use them for predicting?

TomHsiung — Thu, 09 May 2024 11:50:52 GMT

Hello, everyone

I managed to estimate the parameters for a Weibull accelerated failure time model from the PROC LIFEREG procedure. I have the point and interval estimations for predictor coefficients. In addition, I have the estimations for the Weibull shape parameter.

After I get these estimations, I would like to simulate the cumulative incidence curve/survival curve. I know I should use the point estimation of these coefficients and the shape parameter in the prediction model. But I am not sure whether to use the interval estimation of coefficients, or the shape parameter.

You answer is much appreciated!

Re: The Weibull accelerated failure time model parameters estimated but how to use them for predicti

OsoGris — Thu, 09 May 2024 13:56:39 GMT

Tom: You can get the survival probability in this manner: use the OUTPUT statement to request the CDF= keyword and post-process this as Survival = 1 - CDF in a subsequent DATA step. I'm not sure how to get a cumulative incidence value since PROC LIFEREG doesn't model any competing risks scenario for which the CIF is appropriate.

Re: The Weibull accelerated failure time model parameters estimated but how to use them for predicti

sbxkoenk — Thu, 09 May 2024 14:05:54 GMT

What you CANNOT do:

use PROC SCORE
use the STORE statement (PROC LIFEREG) to store the fitted model in an item store. Then use PROC PLM to create predictions on new data. I think the PLM scoring possibility is disabled for models coming from LIFEREG. If not, don't trust the predictions (they are wrong).

What you CAN do:

specify a missing value for the response variable for each observation that you want to score but not use for fitting the model. Since such observations are not used to fit the model, it does not matter whether those observations contain a non-zero weight.
specify a weight of 0 for observations that you want score but not use for fitting the model. (PROC LIFEREG has a WEIGHT statement)

I agree it's not optimal since every time you want to score another set of data ... you need to fit your model again.

Re: The Weibull accelerated failure time model parameters estimated but how to use them for predicti

sbxkoenk — Thu, 09 May 2024 14:27:25 GMT

You can also use this formula to get predictions on new data (use a data step where t_value is your input-time-value to be scored):

predicted_target = 1/(1+exp(-((t_value - intercept) / scale_parameter)));

Please verify by scoring the training data and comparing with proc lifereg predictions.

I won't put my hand up to it, but I think that's the right formula.

Koen

Re: The Weibull accelerated failure time model parameters estimated but how to use them for predicti

TomHsiung — Fri, 10 May 2024 08:57:25 GMT

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;