Let $Q = $ {$d$-dimensional probability distributions with independent marginals from the exponential family}, i.e. $$Q = \left\{ q(x) = \prod_{i=1}^d q_i(x_i), q_i = a(x) \exp\{\eta(\theta)T(\theta)+a(\theta)\} \;\; \big| \;\; x\in \mathbb{R}^d\right\}.$$
I'm wondering if this is convex in the ambient space $(P(\mathbb{R}^d), W_2)$ (probability distributions on $\mathbb{R}^d$ with 2-Wasserstein distance.).
I believe the exponential family of measures is a convex set in $(P(\mathbb{R}^d), W_2)$ -- is this true? Anyways, I think that $Q$ here is a product of such sets.
A set $A$ is geodesically convex iff the geodesic between every two points in $A$ is contained in $A$. Since the ambient space here is $(P(\mathbb{R}^d), W_2)$, the geodesics between points are really optimal transport plans between them. So the convexity question comes down to "For $p,q\in Q$, does the optimal transport plan between $p$ and $q$ send mass only between marginals, $p_i$ to $q_i$?" I have no idea if this is true or how to prove/disprove it.
edit: one idea I had: since the intersection of convex sets is convex, and $Q$ is the intersection of probability spaces on $\mathbb{R}^d$ where the $i$th component is independent of all others, $$ Q = \cap_{i=1}^d A_i \;\;\;\; \text{ where } A_i = \{q = q_i(x_i)q_{-i}(x_{-i})\}$$ where the subscript $-i$ indicates "all componentx except $i$". This space $A_i$ is the cartesian product of two convex spaces $\{q_i(x_i) \in P(\mathbb{R})\}$ and $\{q_{-i}(x_i) \in P(\mathbb{R}^{d-1})$ so it is convex.
Does this argument hold?
edit literally any thoughts are welcome. No complete answers are needed or expected!
This is not true. Let $O_1 := (0,1)$ and $O_2 := (1,2)$. Then, $q_1(x) = \chi_{O_1}(x_1) \chi_{O_1}(x_2)$ and $q_2(x) = \chi_{O_2}(x_1) \chi_{O_2}(x_2)$ belong to $Q$, but their midpoint does not belong to $Q$.