 Solution

For $P_0=N(-{\theta }_0,1)$ and $P_1=N({\theta }_0,1)$, the nominal log-likelihood ratio for one observation is $2{\theta }_{0}x$. 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 scoreThe robust form of the Neyman-Pearson lemma therefore rejects for large values ofwith 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.