Let $\hat{\pi}^N$ be an AW-consistent estimator of $\pi$ (i.e., $\hat{\pi}^N$ is a strongly consistent estimator of $\pi$ under adapted (or called nested) Wasserstein distance $AW(\pi, \hat{\pi}^N)\to 0 $ a.s.).


How to prove that $W(\hat{\pi}^N)$ is a consistent estimator of $W(\pi)$?



How to prove that is a consistent estimator?
Take $\pi^N$ with $AW(\pi^N, \pi) \leq \frac{1}{N}$, where we denote by $\mu^N$ and $\nu^N$ the marginals of $\pi^N$.


Note that by the backward induction for $AW$ (cf. [here](https://epubs.siam.org/doi/pdf/10.1137/16M1080197)), it holds
$$
AW(\pi, \pi^N) = \inf\_{\kappa\_1 \in \Pi(\mu, \mu^N)} d\_{X\_1}(x\_1, y\_1) + W\_1(\pi\_{x\_1}, \pi^N\_{y\_1}) \kappa\_1(dx\_1, dy\_1),
$$
and thus we can choose $\kappa\_1^N \in \Pi(\mu, \mu^N)$ such that
$$
\int W\_1(\pi\_{x\_1}, \pi^N\_{y\_1}) \kappa\_1^N(dx\_1, dy\_1) \leq \frac{1}{N},
$$
and further clearly the second marginal converges, i.e., $W\_1(\nu^N, \nu) \leq \frac{1}{N}$.


By applying twice the triangle inequality, we get
\begin{align}
\int W\_1(\pi\_{x\_1}, \nu) \mu(dx\_1) - \int W\_1(\pi^N\_{y\_1}, \nu^N) \mu^N(dy\_1) &= \int W\_1(\pi\_{x\_1}, \nu) - W\_1(\pi^N\_{y\_1}, \nu^N) \kappa\_1(dx\_1, dy\_1) \\
&\leq \int W\_1(\nu, \nu^N) + W\_1(\pi\_{x\_1}, \pi\_{y\_1}^N) \kappa\_1^N(dx\_1, dy\_1) \\
&\leq \frac{2}{N}
\end{align}
and vice versa.


Since the denominator of $W(\pi)$ should be strictly larger than zero (and converges as well since $W\_1(\nu^N, \nu)$ goes to zero), we get that the estimator is consistent.


