Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 26K (Probability and Measure)

Assume $X_n\to X$ in $L^2$. The reverse triangle inequality for the $L^2$ norm givesSquaring the norms yields

Now assume $X_n\to X$ almost surely and $\mathbb{E}[X_n^2]\to \mathbb{E}[X^2]$. In particular, $(X_n)$ is bounded in $L^2$, so it is uniformly integrable in $L^1$. Almost-sure convergence and uniform integrability therefore give

We next show that $X_n$ converges weakly to $X$ in the L2 space. For $Y\in L^2$, let $Y_M=Y{1}_{\{|Y|\le M\}}$. Thenwhile the Cauchy-Schwarz inequality gives, uniformly in $n$,Letting first $n\to \infty$ and then $M\to \infty$ proves $\mathbb{E}[X_{n}Y]\to \mathbb{E}[XY]$. Taking $Y=X$ and expanding the square now givesConsequentlyThis proves the almost-sure convergence and convergence of second moments criterion.

The quantile function $X_{\mu }=F_{\mu }^{-1}$ is nondecreasing, hence a Borel measurable function. More explicitly, for every $x\in \mathbb{R}$,Thus $X_{\mu }$ is a random variable, and because Lebesgue measure is being used on $(0,1)$,Its cumulative distribution function is therefore $F_{\mu }$, soThis is inverse transform sampling.

Suppose first thatThis 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 weakly to $\nu$. Part (a)(i) also gives

Conversely, suppose that ${\mu }_n$ converges weakly to $\nu$ and the displayed second moments converge. By the stated quantile coupling for convergence in distribution,Part (i) identifies their second moments with those of the measures, so part (a)(ii) applies and gives convergence in $L^2$. ThereforeThis is the one-dimensional second-Wasserstein convergence criterion realized by the common-quantile coupling.