Scaling property of the Wasserstein metric

Question

I would need help with this example.

Let $(S, ||\cdot||)$ denote a normed vector space over $K =\mathbb R$ or $K =\mathbb C$. Let $X$ and $Y$ be $S$-valued random vectors with $E~[~||X||~] < \infty$ and $E~[~||Y||~] < \infty$.

Prove that, for every $c \in K $\ $ \{0\}$: $d_W(\mathcal{L}(cX), \mathcal{L}(cY )) = |c| d_W(\mathcal{L}(X), \mathcal{L}(Y))$.

Hint: For $f : S → R$ with $Lip(f) ≤ 1$ consider $f_c(x) := \frac{1}{|c|} f(cx)$ for $x \in S$.

$d_W(\mathcal{L}(X),\mathcal{L}(Y)) = \sup_{f \in F_W} \int_S f d\mathcal{L}(X) - \int_S f d\mathcal{L}(Y)$ and $F_W$ denote the set of all functions $f : S → R$, which are Lipschitz continuous with constant at most $1$.

Proof: First I show that $f_c \in F_W$:

$Lip(f_c) = \frac{1}{|c|} sup_{x,y \in S, x \neq y}\frac{f(cx)-f(cy)}{d(x,y)} \leq \frac{Lip(f)}{|c|} \leq \frac{1}{|c|}$ -> $f_c \in F_W $.

$d_W(\mathcal{L}(cX), \mathcal{L}(cY)) = \sup_{f \in F_W} \int_S f(cx) d\mathcal{L}(Xc)(x) - \int_S f(xc) d\mathcal{L}(Yc)(x) = \sup_{f \in F_W} \int_S f(cx) d\mathcal{L}(X)(\frac{x}{c}) - \int_S f(xc) d\mathcal{L}(Y)(\frac{x}{c}) = ?$

Have I got everything right so far? If so, how should I proceed now?

Thanks for the help!

Answer 1

You've made a few little mistakes.
First, the Lipschitz constant of $f_c$ is bounded by $1$ rather than $1/c$ : $$\frac{1}{|c|}\frac{|f(cx)-f(cy)|}{d(x,y)} \leq\frac{1}{|c|} \frac{Lip(f)\cdot d(cx,cy)}{d(x,y)} \leq \frac{1}{|c|}\frac{Lip(f)\cdot |c|d(x,y)}{d(x,y)} \le 1 $$ Second for any $f$, you have $$\int_S fd\mathcal L(cX) \stackrel{(1)}{=} \int_S f(\color{red}{x})d\mathcal L(cX) (x) \stackrel{(2)}{=}\int_S f({x})d\mathcal L(X) \left(\frac x c\right) \stackrel{(3)}{=}\int_S f(c{z})d\mathcal L(X) (z)$$ Where $(1)$ is the definition of the integral of $f$ against the distribution of $cX$, $(2)$ follows from the observation that $``\{cX \in dx\} = \left\{X\in \frac{dx}{c}\right\}" $ (which can be made rigorous) and $(3)$ follows from letting $z:=x/c$.

From there we get that $$\begin{align*} d_W(\mathcal{L}(cX), \mathcal{L}(cY)) &= \sup_{f \in F_W} \int_S f(c{z})d\mathcal L(X) (z) -\int_S f(c{z})d\mathcal L(Y) (z)\\ &= \sup_{f \in F_W} \int_S |c|\cdot\frac{f(c{z})}{|c|}d\mathcal L(X) (z) - \int_S |c|\cdot\frac{f(c{z})}{|c|}d\mathcal L(Y) (z)\\ &=|c|\left[\sup_{f_c \in F_W^c} \int_S f({z})d\mathcal L(X) (z) -\int_S f({z})d\mathcal L(Y) (z)\right] \end{align*}$$

Where $F_W^c := \left\{z\mapsto\frac{f(c{z})}{|c|}\mid f\in F_W\right\}$.

Can you finish the proof now ?