Solution
= Solution
Without administrative stopping, the probability that one patient experiences $A$ before $B$ is $\theta/(\theta+\phi)$. For $n$ patients the expected event count is therefore
$$
\mathbb E[v_+]=n\frac{\theta}{\theta+\phi}.
$$
= Solution
Without administrative stopping, the probability that one patient experiences $A$ before $B$ is $\theta/(\theta+\phi)$. For $n$ patients the expected event count is therefore
$$
\mathbb E[v_+]=n\frac{\theta}{\theta+\phi}.
$$