Let $I_0=B^c$ and $m_0=|I_0|$. Each $q_j$ with $j\in I_0$ is a valid <p-value> by the argument in part c. If the <Holm step-down procedure> selects any index from $I_0$, let $r$ be the rank of the first such index. All $r-1$ earlier selections belong to $B$, so
$$
r\leq|B|+1=p-m_0+1,
\qquad p-r+1\geq m_0.
$$
Selection through rank $r$ implies
$$
q_{\tau(r)}\leq\frac\alpha{p-r+1}\leq\frac\alpha{m_0}.
$$
Consequently
$$
\mathbb P(\widehat B\not\subseteq B)
\leq\mathbb P\!\left(\min_{j\in I_0}q_j\leq\frac\alpha{m_0}\right)
\leq\sum_{j\in I_0}\mathbb P\!\left(q_j\leq\frac\alpha{m_0}\right)
\leq\alpha.
$$
Thus the procedure controls the <familywise error rate> without requiring independence among the p-values.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 205 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-205/4/d/solution