Here's a simple but interesting coupling / measure theory question:
Say we have measures $P_1, P_2$ and probabilities $Q_1, Q_2$ on a nice measurable space $(\mathcal{X},\mathscr{F})$, with $P_i(A) \leq Q_i(A)$ for all $A \in \mathscr{F}$ and $i=1,2$.
Let $\bar P$ be a coupling of $P_1$ and $P_2$, meaning that $\bar P$ is a measure on $(\mathcal{X} \times \mathcal{X}, \mathscr{F} \otimes \mathscr{F})$ with $\bar P(A \times \mathcal{X}) = P_1(A)$ and $\bar P(\mathcal{X} \times A) = P_2(A)$ for all $A \in \mathscr{F}$.
Does there exist a coupling $\bar Q$ of $Q_1$ and $Q_2$ such that $\bar P(B) \leq \bar Q(B)$ for
any $B \in \mathscr{F} \otimes \mathscr{F}$?
Thank you very much!
Best Answer
Taking $A=\mathcal{X}$ in the definition of the coupling $\bar P$, we infer that $P_1(\mathcal{X})=P_2(\mathcal{X})$. Call this number $p$. If $p=1$ then $Q_i=P_i$ so we may assume that $p<1$. Write $M_i=Q_i-P_i$ for $i=1,2$ and define $$\bar Q=\bar P+\frac{M_1 \times M_2}{1-p} \,,$$ where $M_1 \times M_2$ is the product measure. Then for all $A \in \mathscr{F}$ we have $$\bar Q(A \times \mathcal{X}) = P_1(A)+\frac{M_1(A)M_2(\mathcal{X})}{1-p}=Q_1(A)$$ and similarly $\bar Q(\mathcal{X} \times A) = Q_2(A)$.