= Solution
Define
$$
F_+(t)=\frac12\{\Phi(t)+\Phi(t-2b_1)\},
\qquad
F_-(t)=F_+(t+2b_1)
=\frac12\{\Phi(t)+\Phi(t+2b_1)\}.
$$
Thus $F_+$ is the equal mixture of $N(0,1)$ and $N(2b_1,1)$, while $F_-$ is the equal mixture of $N(0,1)$ and $N(-2b_1,1)$. They have finite variance and everywhere positive densities. Normal symmetry shows that $F_+$ is symmetric about $b_1$ and $F_-$ about $-b_1$.
For every $t$,
$$
0\leq\Phi(t)-\Phi(t-2b_1)
\leq\Phi(b_1)-\Phi(-b_1)=2\varepsilon,
$$
where the maximum occurs at $t=b_1$. Consequently
$$
|F_+(t)-\Phi(t)|
=\frac12|\Phi(t-2b_1)-\Phi(t)|
\leq\varepsilon.
$$
The same argument, shifted and reflected, applies to $F_-$. Hence both distributions belong to $\mathcal P_\varepsilon^K(\Phi)\cap\mathcal M$ and satisfy the required translation relation.
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