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.
