1
$\begingroup$

Let $R_1,...,R_K$ be competing risks and put $T=\min(R_1,...,R_K)$ and $\delta = i$ if $T=R_i$. The cumulative incidence function is then defined as $F_i(t):=P(T\le t,\delta = i)$. The cause-specific hazard rate is then defined as $$h_i(t)=\lim_{\Delta t\to 0}\frac{P(t\le T<t+\Delta t,\delta = i|T\ge t)}{\Delta t}.$$It is then stated that $$F_i(t)=\int_0^th_i(u)\exp(-H_T(u))du,$$where $H_T(u)=\sum_{j=1}^K\int_0^th_j(u)du$ (the cumulative hazard rate of $T$). However I can not see how this follows and could use some help on how to show it. If you need more information let me know. Thanks!

$\endgroup$

1 Answer 1

1
$\begingroup$

Wikipedia has some good pages on the relationships between the event-time CDF, survival function, and cumulative hazard.

$h(t)$ can be seen as $f(t)/S(t)$, the pdf divided by the the survival function. Additionally, $\text{exp}[-H(t)]=S(t)$. It should then be clear that

$$\int_0^t h(u)\cdot \text{exp}[-H(u)] du=\int_0^t \frac{f(u)}{S(u)}S(u) du=\int_0^t f(u) du=F(t). $$

For cause-specific risks it appears we are not conditioning on a risk but incorporating the risk type within the probability statement, i.e. $h_i(t)=f_i(t)/S(t)=f(t,\delta)/S(t)=h(t,\delta)$. Then

\begin{eqnarray} \int_0^t h_i(u)\cdot \text{exp}[-H_T(u)] du&=&\int_0^t h(u,\delta)\cdot \text{exp}[-H_T(u)] du\\ &=&\int_0^t \frac{f(u,\delta)}{S(u)}S(u) du\\ &=&\int_0^t f(u,\delta) du\\ &=&\int_0^t f_i(u) du\\ &\overset{\text{by definition}}{=}&F_i(t). \end{eqnarray}

Some of the steps are unnecessary but serve as a reminder that we aren't conditioning on a particular type of risk. Often times a subscript can suggest a conditional probability. Let me know if your question is regarding something else.

$\endgroup$
5
  • $\begingroup$ I see why these equalities are true however (I assume you mean $F=$CDF) I can't see how $F(t)=F_i(t)$. $\endgroup$
    – statman
    Dec 16, 2021 at 16:47
  • $\begingroup$ Thank you for the edit. However I think we are taking past each other. It is the first equality , $F_i=\int_0^th_i(u)\exp(-H_T(u))du$, that I am unsure of. It seems like you are using that or maybe I'm misunderstanding. $\endgroup$
    – statman
    Dec 16, 2021 at 21:17
  • $\begingroup$ Ah, I see I wasn't as clear as I could have been. I have revised my answer. $\endgroup$ Dec 16, 2021 at 21:39
  • $\begingroup$ Ah so $f_i$ is the marginal pdf? $\endgroup$
    – statman
    Dec 16, 2021 at 21:43
  • $\begingroup$ Based on the $i$ subscript we could call it the marginal cause-specific pdf. $\endgroup$ Dec 16, 2021 at 21:46

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.