Assume first that $0<\epsilon<1/2$, set $k=\lfloor\epsilon n\rfloor$, and replace the last $k$ observations by $100$. Among the remaining $m=n-k$ observations, let
$$
Y_i=\mathbf1_{\{X_i>\epsilon\}},
\qquad
p_\epsilon=\mathbb P(Z>\epsilon)=1-\Phi(\epsilon).
$$
The contaminated median exceeds $\epsilon$ whenever
$$
\sum_{i=1}^mY_i\geq\frac{n+1}{2}-k.
$$
The right-hand threshold divided by $m$ tends to
$$
q_\epsilon=\frac{1/2-\epsilon}{1-\epsilon}<p_\epsilon.
$$
Therefore, for all sufficiently large $n$, it is at most $p_\epsilon-\delta_\epsilon$ for some $\delta_\epsilon>0$. The <Hoeffding inequality> gives
$$
\mathbb P\!\left(
\sum_{i=1}^mY_i<\frac{n+1}{2}-k
\right)
\leq e^{-2m\delta_\epsilon^2}
\leq e^{-c_\epsilon n}.
$$
The supremum over adversarial perturbations is at least this explicit construction, proving the claim. When $\epsilon\geq1/2$, replacing at least half the sample makes the conclusion immediate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 223 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-223/4/a/solution