Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 38 / 2 / b / iii / Solution

{title2=$\mathbb EH(\min(T,c))=\mathbb P(T\le c)$}

For a nonnegative continuous event time with <survival function> S, <hazard function> h and H the <cumulative hazard>, split the expectation at c. The censored contribution is S(c)H(c). The event contribution is $\int_0^cH(t)f(t)dt=1-S(c)-S(c)H(c)$ by <integration by parts>, since $dH=hdt$ and $f=hS$. Adding gives the displayed identity. Linearity extends it to the expected observed event count in a cohort, without requiring independence between subjects.

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 Incoming links: Mean accumulated hazard before fixed censoring