Let $\lambda_i(t),S_i(t)$ be the hazard and survival functions of two populations for $i=1,2$ and satisfy that: $\frac{S_2(t)}{1-S_2(t)}=\phi\frac{S_1(t)}{1-S_1(t)}$ (1) I want to proof that $\lim_{t\to\infty }\frac{\lambda_2(t)}{\lambda_1(t)}=1$

Using that $\lambda(t)=\frac{f(t)}{1-F(t)}$ where $f(·)$ is a pdf and $F(·)$ their cdf and the fact that (1) can be expressed as $\frac{1-F_2(t)}{F_2(t)}=\phi\frac{1-F_1(t)}{F_1(t)}$ then: $$\lim_{t \to \infty}\left [ \frac{f_2(t)}{1-F_2(t)}\div \frac{f_1(t)}{1-F_1(t)} \right ]=\lim_{t \to \infty}\left [ \frac{f_2(t)}{\frac{\phi (1-F_1(t))F_2(t)}{F_1(t)}}\div \frac{f_1(t)}{1-F_1(t)} \right ]=\lim_{t \to \infty}\frac{f_2(t)F_1(t)}{\phi F_2(t)f_1(t)}=\frac{1}{\phi}\lim_{t \to \infty}\frac{f_2(t)}{f_1(t)}$$

  • $\begingroup$ What is your question? $\endgroup$ – whuber Dec 29 '18 at 18:30
  • $\begingroup$ @whuber how to prove that the limit tends to 1. $\endgroup$ – willy Dec 29 '18 at 19:24
  • $\begingroup$ Okay, thank you. Perhaps you might be able to show $$\lambda_2/\lambda_1=(1-S_2)/(1-S_1)$$ and go on from there. One approach is through the expressions $$\lambda_i = -\frac{d}{dt}\log S_i(t).$$ $\endgroup$ – whuber Dec 29 '18 at 23:20
  • $\begingroup$ @whuber Thank you, So this means that the limit of the quotient doesn't need to satifsfy the condition (1). $\endgroup$ – willy Dec 31 '18 at 0:36
  • $\begingroup$ If you solved this you can answer the Q yourself! Please do so! $\endgroup$ – kjetil b halvorsen Dec 31 '18 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.