HI All,I am trying to simulate a AFT Weibull model in R using a log-linear mod
log(Y)=?0+?1X1+?2X2+cW
Where W is a extreme value distribution. My R code is below. I can not understand why the intercept estimate is 2.94, not lose to 1?? Thanks,?
Vincent
install.packages("extRemes")
library(extRemes)
n<-500000
x1<-rnorm(n,0,1)
x2<-rnorm(n,0,1)
error<-revd(n, loc = 0, scale = 1)
b0<- 1
b1<-0.5
b2<--1.2
c<-1
time<-exp(b0+b1*x1+b2*x2+c*error)
status<-rep(1,n)
survreg(Surv(time, status==1) ~ x1+x2,dist="exponential")
####output
Coefficients:
(Intercept) x1 x2
2.9390956 0.4931199 -1.2061369
Scale fixed at 1
#####
Simulating a Weibull regression model
2 messages · duo wan, Mark Clements
From ?survival::survreg.distributions, we find that the CDF for an extreme value distribution is defined as F=1-e^{-e^t}, whereas ?extRemes::revd gives G(x) = exp(-exp((x - location)/scale)) (or F=e^{-e^t}) when shape=0.
If you use time<-exp(b0+b1*x1+b2*x2-c*error) # or time<-rexp(n, rate=1/exp(b0+b1*x1+b2*x2)) # or time<-rweibull(n, shape=c, scale=exp(b0+b1*x1+b2*x2)) then you could model using survreg(Surv(time, status==1) ~ x1+x2,dist="exponential") # or survreg(Surv(time, status==1) ~ x1+x2,dist="weibull") -- Mark
From: R-sig-Epi <r-sig-epi-bounces at r-project.org> on behalf of duo wan <duo_wan at yahoo.com>
Sent: 11 October 2022 20:05
To: r-sig-epi at r-project.org <r-sig-epi at r-project.org>
Subject: [R-sig-Epi] Simulating a Weibull regression model
Sent: 11 October 2022 20:05
To: r-sig-epi at r-project.org <r-sig-epi at r-project.org>
Subject: [R-sig-Epi] Simulating a Weibull regression model
[Some people who received this message don't often get email from duo_wan at yahoo.com. Learn why this is important at https://aka.ms/LearnAboutSenderIdentification ] HI All,I am trying to simulate a AFT Weibull model in R using a log-linear mod log(Y)=?0+?1X1+?2X2+cW Where W is a extreme value distribution. My R code is below. I can not understand why the intercept estimate is 2.94, not lose to 1? Thanks, Vincent install.packages("extRemes") library(extRemes) n<-500000 x1<-rnorm(n,0,1) x2<-rnorm(n,0,1) error<-revd(n, loc = 0, scale = 1) b0<- 1 b1<-0.5 b2<--1.2 c<-1 time<-exp(b0+b1*x1+b2*x2+c*error) status<-rep(1,n) survreg(Surv(time, status==1) ~ x1+x2,dist="exponential") ####output Coefficients: (Intercept) x1 x2 2.9390956 0.4931199 -1.2061369 Scale fixed at 1 ##### [[alternative HTML version deleted]] _______________________________________________ R-sig-Epi at r-project.org mailing list https://eur01.safelinks.protection.outlook.com/?url=https%3A%2F%2Fstat.ethz.ch%2Fmailman%2Flistinfo%2Fr-sig-epi&data=05%7C01%7Cmark.clements%40ki.se%7C9224c78aa653458e9ca808daabb3324e%7Cbff7eef1cf4b4f32be3da1dda043c05d%7C0%7C0%7C638011083709550655%7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C3000%7C%7C%7C&sdata=GIx6eSNqJs4NPSC1MhXz%2FgfoKrO5nSQQM8gzc8BaPXI%3D&reserved=0 Nar du skickar e-post till Karolinska Institutet (KI) innebar detta att KI kommer att behandla dina personuppgifter. Har finns information om hur KI behandlar personuppgifter<https://ki.se/medarbetare/integritetsskyddspolicy>. Sending email to Karolinska Institutet (KI) will result in KI processing your personal data. You can read more about KI's processing of personal data here<https://ki.se/en/staff/data-protection-policy>.