For $P_0=N(-\theta_0,1)$ and $P_1=N(\theta_0,1)$, the nominal log-likelihood ratio for one observation is $2\theta_0x$. Under sufficiently small <epsilon-contamination neighborhoods>, the least-favourable pair clips this likelihood ratio between two constants. Symmetry turns its log into a positive multiple of the winsorized score
$$
q_c(x)=\operatorname{clip}(x,-c,c).
$$
The robust form of the <Neyman-Pearson lemma> therefore rejects for large values of
$$
\sum_{i=1}^nq_c(X_i),
$$
with boundary randomization and threshold chosen so that the worst-case null rejection probability is $\alpha$. Extreme observations contribute only $\pm c$, preventing a few contaminants from dominating the test.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 223 / 3 / d / Solution

{title2=$\mathcal P_\epsilon(P)$}

The epsilon-contamination neighborhood of $P$ is the class $\{(1-\epsilon)P+\epsilon Q:Q\text{ is a probability distribution}\}$.


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 Incoming links: Epsilon-contamination neighborhood