# Finding the limit of a quotient of hazard functions

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: $$\lim_{t \to \infty}\left [ \frac{f_2(t)}{1-F_2(t)}\div \frac{f_1(t)}{1-F_1(t)} \right ]=\lim_{t \to \infty}\left [ \frac{f_2(t)}{\frac{\phi (1-F_1(t))F_2(t)}{F_1(t)}}\div \frac{f_1(t)}{1-F_1(t)} \right ]=\lim_{t \to \infty}\frac{f_2(t)F_1(t)}{\phi F_2(t)f_1(t)}=\frac{1}{\phi}\lim_{t \to \infty}\frac{f_2(t)}{f_1(t)}$$

• What is your question? – whuber Dec 29 '18 at 18:30
• @whuber how to prove that the limit tends to 1. – willy Dec 29 '18 at 19:24
• 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 $$\lambda_i = -\frac{d}{dt}\log S_i(t).$$ – whuber Dec 29 '18 at 23:20
• @whuber Thank you, So this means that the limit of the quotient doesn't need to satifsfy the condition (1). – willy Dec 31 '18 at 0:36
• If you solved this you can answer the Q yourself! Please do so! – kjetil b halvorsen Dec 31 '18 at 21:04