Put $D_j=\widetilde H(x_j)-\widetilde H(x_{j-1})$, with $\widetilde H(x_0)=0$. Then
$$
\sum_{i=1}^n\widetilde H(x_i)
=\sum_{j=1}^n(n-j+1)D_j.
$$
The natural local calibration is therefore
$$
(n-j+1)D_j=v_j,
\qquad
D_j=\frac{v_j}{n-j+1}.
$$
Taking $\widetilde H$ to be the right-continuous step function with these increments gives
$$
\widetilde H(t)=\sum_{j:x_j\leq t}\frac{v_j}{n-j+1}.
$$
Because $n-j+1$ is exactly the risk-set size $r_j$, this is the estimator from part c and automatically satisfies $\sum_i\{v_i-\widetilde H(x_i)\}=0$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 207 / 4 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-207/4/e/solution