With administrative censoring at $c$, one patient is observed to experience $A$ with probability
$$
\int_0^c\theta e^{-(\theta+\phi)t}\,dt
=\frac{\theta}{\theta+\phi}
\left(1-e^{-(\theta+\phi)c}\right),
$$
so
$$
\mathbb E[v_+]
=n\frac{\theta}{\theta+\phi}
\left(1-e^{-(\theta+\phi)c}\right).
$$
The observed time is $\min(T_A,T_B,c)$, whose mean follows from the <tail-sum formula for expectation>:
$$
\mathbb E[\min(T_A,T_B,c)]
=\int_0^ce^{-(\theta+\phi)t}\,dt
=\frac{1-e^{-(\theta+\phi)c}}{\theta+\phi}.
$$
Hence
$$
\mathbb E[x_+]
=n\frac{1-e^{-(\theta+\phi)c}}{\theta+\phi},
\qquad
\frac{\mathbb E[v_+]}{\mathbb E[x_+]}=\theta.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 6 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-207/6/b/iv/solution