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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.

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  • $\begingroup$ 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). $\endgroup$ Sep 17, 2014 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – bdeonovic
    Sep 18, 2014 at 0:02

1 Answer 1

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Here is one way to do it:

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

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