Wasserstein distance of linear combination Given three measures $\mu,\nu,\xi$ on a common space and $d$ be a distance on such space. Let also pick an $\alpha \in (0,1)$
is there some result connecting $W_p(\alpha\mu + (1-\alpha)\nu, \xi)$ and $W_p(\mu,\xi)$ and $W_p(\nu,\xi)$, were $W_p$ denotes the $p-$Wasserin distance, i.e. $W_p(\mu,\nu)=\min_\pi\mathbb{E}_{\pi}[d^p(X,Y)]$, where $X\sim\mu$, $Y\sim\nu$ and the minimization is over couplings of $\mu$ and $\nu$?
I could not find anything anywhere, but maybe there's a quick trick I fail to see?
 A: Nice question. I don't think there is, but I don't have a full answer. $W_p$ makes the space of probability measures into a geodesic metric space, and it looks like your question is about the curvature of this space. More precisely, if you put $\phi_\alpha = W_p(\alpha \mu + (1-\alpha) \nu,\xi)$ for the interpolation between $\phi_0=\nu$ and $\phi_1=\mu$, then you have a "triangle" with vertex $\xi$ and opposite side $\phi_\alpha$, $0 \leq \alpha \leq 1$, and the it seems like you're for the curvature of this opposite side. I don't know of general results for curvature in the Wasserstein space. Also, this vector space interpolation may have very little to do with the Wasserstein geodesic from $\mu$ to $\nu$. So I can't see how this is possible.
Perhaps there is something if you include $W_p(\mu,\nu)$, or a geometric condition on the underlying metric space (like hyperbolicity) ... but maybe that's just pointing back to triangle inequality.
Letting $\mu,\nu,\xi$ be translations of a simple measure seems to rule out any of the basic relations that occur to me. I'd be very interested to hear if there is a neat relation here that I'm not seeing.
