Interchanging the order of the finite sums gives
$$
\begin{aligned}
\sum_{i=1}^n\widehat M_i
&=\sum_{i=1}^nv_i
-\sum_{i=1}^n\sum_{j=1}^i\frac{v_j}{n-j+1}\\
&=\sum_{j=1}^nv_j
-\sum_{j=1}^n\frac{v_j}{n-j+1}
\sum_{i=j}^n1\\
&=\sum_{j=1}^nv_j-\sum_{j=1}^nv_j=0.
\end{aligned}
$$
Each hazard increment is counted once for every individual exposed to it, exactly reproducing its event count.


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 207 / 4 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-207/4/c/ii/solution