# Expected value of Cumulative Hazard

Define $T=\min(T^0,C)$ where $T^0$ is the failure time and $C$ is the censoring time. Define the failure indicator $$\delta = \begin{cases} 1 & \text{if T^0\leq C}\\ 0 & \text{if T^0> C}\end{cases}$$ Furthermore, let $\Lambda(t)$ be the cumulative hazard function for $T^0$. Assume the random censorship model. Show that $\text{E}[\delta] = \text{E}[\Lambda(T)]$.

First approach

\begin{align*} \text{E}[\delta] &= \text{Pr}\left\{ T^0 \leq C\right\}\\ &= \int_0^\infty \int_0^c f_{T^0,C}(t,c)dtdc\\ &= \int_0^\infty \int_0^c f_{T^0}(t)f_C(c)dtdc\\ &= \int_0^\infty \int_t^\infty f_{T^0}(t)f_C(c)dcdt\\ &= \int_0^\infty f_{T^0}(t)\int_t^\infty f_C(c)dcdt\\ &= \end{align*} I get stuck here. Another approach is \begin{align*} \text{E}[\Lambda(T)] &= \int_0^\infty \Lambda(t)f_{T^0}(t)dt\\ &= \int_0^\infty \int_0^t\dfrac{f_{T^0}(x)}{S_{T^0}(x)}f_{T^0}(t)dxdt\\ &= \int_0^\infty \int_0^t\dfrac{f_{T^0}(x)}{S_{T^0}(x)}f_{T^0}(t)dxdt\\ &= \int_0^\infty \dfrac{f_{T^0}(x)}{S_{T^0}(x)}\int_x^\infty f_{T^0}(t)dtdx\\ &= \int_0^\infty \dfrac{f_{T^0}(x)}{S_{T^0}(x)}S_{T^0}(x)dx\\ &= 1 \end{align*} This does not seem right...so I am lost. Can anyone help? Let me know if you need clarification on any of the terms.

• What's the point in mentioning your dinner in a question posted on this forum??? Besides, we already know what you have after dinner (full stomach and a strong urge to go to bed). Sep 17, 2014 at 23:03
• Sorry, for self-learning problems its standard for the poster to provide atleast an initial attempt at the problem. I was too busy to type up all of my work, but still wanted to post the problem in hopes that it would give somebody some time to think about it. Sep 18, 2014 at 0:02

\begin{align*} \text{E}[\Lambda_T(T)] &= \int_0^\infty \Lambda_T(t)f_T(t)dt\\ &= \int_0^C \Lambda_T(t)f_T(t)dt + \int_C^\infty \Lambda_T(C)f_T(t)dt\\ &= \int_0^C -\log{S_T(t)}f_T(t)dt + \Lambda_T(C)\int_C^\infty f_T(t)dt\\ &= \int_0^{-\log{S_T(C)}} ue^{-u}du + \Lambda_T(C)S_T(C)\\ &= \left[ -ue^{-u} - e^{-u}\right]_0^{-\log S_T(C)} + \Lambda_T(C)S_T(C)\\ &= (\log S_T(C))S_T(C) - S_T(C) + 1 + \Lambda_T(C)S_T(C)\\ &= -\Lambda_T(C)S_T(C) - S_T(C) + 1 + \Lambda_T(C)S_T(C)\\ &= 1-S_T(C)\\ &= F_T(C)\\ &= \text{Pr}\left\{ T^0 \leq C\right\}\\ &= \text{E}[\delta] \end{align*}