Let $X$ be a Polish space endowed with a bounded metric $\rho_X$. Let $\mu, \mu'$ be two probability measures, and $\kappa, \kappa'$ be two stochastic kernels on $X$. Assume that $\kappa, \kappa'$ are Lipschitz in the following sense: $$ \bar\rho_X(\kappa(x), \kappa'(x')) \leq \gamma\cdot \rho_X(x, x') $$ for some $\gamma > 0$ and all $x,x'\in X$. Here $$ \bar\rho_X:(\nu,\nu')\to\inf_{N \in \mathcal C(\nu, \nu')}\int_{X^2} \rho_X(x, x')N(\mathrm dx\times \mathrm dx') $$ is the Wasserstein distance on measures induced by the original metric $\rho_X$ on points. By $\mathcal C(\nu, \nu')$ I denote the set of all coupling of probability measures $\nu, \nu'$.
Consider now the product measures $P = \mu\otimes\kappa$ and $P' = \mu'\otimes \kappa'$, and let their right marginal be $\nu = \mu\kappa$ and $\nu' = \mu'\kappa'$. I think I can show that $\bar\rho_X(\nu, \nu') \leq \gamma \bar\rho_X(\mu, \mu')$ (if there is a textbook proof of this, I'd be interested in seeing it). Thus, if we assume the product metric to be $$ \rho_{X^2}(x, x', y, y') = \rho_X(x,x') + \rho_X(y,y') \tag{1} $$ we get $$ \bar\rho_{X^2}(P, P') \leq (1 + \gamma)\cdot \bar\rho_X(\mu, \mu'). $$ The same bounds would of course apply to the case $$ \rho_{X^2}(x, x', y, y') = \max(\rho_X(x,x'), \rho_X(y,y')), \tag{2} $$ however I think that they are too conservative in this case. Is there a way to come with better bounds on the Wasserstein distance for the produce measure in case (2)?
Notation-wise: $P = \mu\otimes\kappa$ is the product measure, a unique solution for $$ P(A\times B) = \int_A \kappa(B|x)\mu(\mathrm dx). $$ Its marginals are $\mu$ and $\nu = \mu\kappa$, the latter define by $$ \mu\kappa(B) = \int_X \kappa(B|x)\mu(\mathrm dx). $$