Proof of property of Mallows/Wasserstein metric

Question

Let $\mathcal F_p=\{H\text{ is a cumulative distribution function}:\int|x|^pdH<\infty\}$. Define on $\mathcal F_p,$ Mallows' metric ($p$ Wasserstein metric), $d_p,p\ge1$ for two random variables $X,Y$ as $$d_p(X,Y)=\inf_{(X,Y)\sim\pi}\left[E_\pi|X-Y|^p\right]^\frac{1}{p}.$$ If $\{X_j\},\{Y_j\}$ are two sequences of independent random variables with $EX_j^2<\infty~\forall j$, $EY_j^2<\infty~\forall j$ and $EX_j=EY_j~\forall j$, then show that $$\left[d_2\left(\sum_1^nX_j,\sum_1^nY_j\right)\right]^2\le\sum_1^n[d_2(X_j,Y_j)]^2$$

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Attempt. $$d_2\left(\sum_1^nX_j,\sum_1^nY_j\right)=\inf_\pi\left[E_\pi\sum_1^n|X_j-Y_j|^2\right]^\frac{1}{2}\\\le\inf_\pi\sum_1^n\left[E_\pi|X_j-Y_j|^2\right]^\frac{1}{2}~\text{(Minkowski's inequality)}\\\le\sum_1^n\inf_\pi\left[E_\pi|X_j-Y_j|^2\right]^\frac{1}{2}~{(\inf(A+B)\le\inf A+\inf B)}\\=\sum_1^nd_2(X_j,Y_j)$$ However, if I were to square both sides, this is a looser bound than the one I need to prove. Moreover, I haven't really used the independence and equal mean assumption anywhere in the proof. Where am I going wrong?

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Note that I need this property to show "closeness" of sampling distribution of the bootstrapped mean to the sample mean using Mallows' metric.

Answer 1

[I was in the middle of this post when you posted your answer. For future referential purpose, I am completing the draft.]

Subadditivity of Mallows distance has been well-documented in many papers.

The first thing is to note that the distance is indeed a metric and attains the infimum in $\mathcal F_p:$ this is possible by considering $X^*:=F_X^{-1}(U), ~Y^*:=F_Y^{-1}(U), ~U\sim\mathrm U(0, 1) .$ See $\rm[I,~lemma ~2.3]$ which uses the Fréchet-Hoeffding bound to deduce this.

Now (cf. $\rm[II,~lemma ~8.6]$), using this, WLOG, we assume the the $B\times B$-valued $(X_j, Y_j) $ are independent and $d_p(X_j, Y_j) =\mathbf E[\Vert X_j-Y_j\Vert^p]^{\frac 1p}.$ Using Minkowski's inequality, one can conclude $$ d_p\left(\sum_j X_j, \sum_j Y_j\right)\leq \sum_jd_p(X_j, Y_j).$$

When the space is equipped with orthogonality, one can improve the result above (cf. $\rm[II,~lemma ~8.7]$). Take $p=2, $ with $B$ being a Hilbert space with inner product $\langle\cdot, \cdot\rangle$ and assuming $\mathbf E[X_j]=\mathbf E[Y_j],$ one gets $$\mathbf E[\langle X_j-Y_j, X_k-Y_k\rangle]=\begin{cases}d_2(X_j,Y_j)^2&k=j\\0&k\ne j\end{cases}.$$ So \begin{align} d_2\left(\sum_j X_j, \sum_j Y_j\right)^2&\leq \mathbf E\left[\left\langle \sum_j(X_j-Y_j), \sum_j(X_j-Y_j)\right\rangle\right]\\&=\sum_jd_2(X_j,Y_j)^2.\end{align}

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References:

$\rm[I]$ Central Limit Theorem and convergence to stable laws in Mallows distance, Oliver Johnson, Richard Samworth, $2005, $ url: https://arxiv.org/abs/math/0406218.

$\rm[II]$ Some Asymptotic Theory for the Bootstrap, Peter J. Bickel, David A. Freedman, Ann. Statist. $ 9 ~(6) ~1196$–$ 1217, $ November, $1981. $url: https://doi.org/10.1214/aos/1176345637.

Answer 2

Figured it on my own. Here are some key steps.

1. $[d_2(\sum X_j,\sum Y_j)]^2=(\inf_\pi [E_\pi \{\sum(X_j-Y_j)\}^2]^{1/2})^2=\inf_\pi ([E_\pi \{\sum(X_j-Y_j)\}^2]^{1/2})^2$ as $\inf a_i^2=(\inf a_i)^2$ for $a_i\ge0$.
2. $\inf_\pi E_\pi \{\sum(X_j-Y_j)\}^2\le \inf_\pi E_\pi\sum(X_j-Y_j)^2$ by expansion using assumptions.
3. $\inf_\pi\sum E_\pi(X_j-Y_j)^2=\sum\inf_\pi E_\pi(X_j-Y_j)^2$ as $\inf (A+B) =\inf A +\inf B$.