The sample median attains the equivariant upper bound:
$$
\varepsilon^*(\widehat\theta_{\mathrm{med}})
=\frac1n\left\lfloor\frac{n-1}{2}\right\rfloor.
$$
For the Huber estimator, fewer than half the observations cannot overpower the bounded scores of the uncontaminated majority. Evaluating the estimating equation below $x_{(1)}-k$ or above $x_{(n)}+k$ makes every uncontaminated score have the same sign. A contaminating majority can balance these scores arbitrarily far away, so
$$
\varepsilon^*(\widehat\theta_{\mathrm{Hub},k})
=\frac1n\left\lfloor\frac{n-1}{2}\right\rfloor.
$$
A $\gamma$-trimmed mean remains bounded while at most $\lfloor\gamma n\rfloor$ arbitrary observations are removed by each tail trim; one more arbitrarily large replacement survives. Hence
$$
\varepsilon^*(\widehat\theta_{\mathrm{trim},\gamma})
=\frac{\lfloor\gamma n\rfloor}{n}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 223 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-223/1/c/solution