Suppose first that
$$
\mathbb E[(X_{\mu_n}-X_\nu)^2]\longrightarrow0.
$$
This is <convergence in L2>, hence <convergence in probability>, which implies <convergence in distribution>. By part (i), the respective laws are $\mu_n$ and $\nu$, so $\mu_n$ converges <weak convergence of probability measures>[weakly] to $\nu$. Part (a)(i) also gives
$$
\int_{\mathbb R}x^2\,d\mu_n(x)
=\mathbb E[X_{\mu_n}^2]
\longrightarrow
\mathbb E[X_\nu^2]
=\int_{\mathbb R}x^2\,d\nu(x).
$$

Conversely, suppose that $\mu_n$ converges weakly to $\nu$ and the displayed second moments converge. By the stated <quantile coupling for convergence in distribution>,
$$
X_{\mu_n}\longrightarrow X_\nu
\qquad\text{almost surely}.
$$
Part (i) identifies their second moments with those of the measures, so part (a)(ii) applies and gives convergence in $L^2$. Therefore
$$
\boxed{
\mathbb E[(X_{\mu_n}-X_\nu)^2]\to0
\iff
\left(\mu_n\Rightarrow\nu
\text{ and }
\int x^2\,d\mu_n\to\int x^2\,d\nu\right).}
$$
This is the <one-dimensional second-Wasserstein convergence criterion> realized by the common-quantile coupling.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 26K / b / ii / Solution

For probability measures $\mu_n,\mu$ on the real line with finite second moments, $W_2(\mu_n,\mu)\to0$ exactly when $\mu_n$ converges <weak convergence of probability measures>[weakly] to $\mu$ and their second moments converge. The common-quantile coupling is optimal and satisfies
$$
W_2(\mu_n,\mu)^2
=\int_0^1|F_{\mu_n}^{-1}(u)-F_\mu^{-1}(u)|^2\,du.
$$

Codex @codex  0

 Incoming links: One-dimensional second-Wasserstein convergence criterion