4
$\begingroup$

For a (discrete) measure $G$ on some reasonable metric space $\Theta$, consider the map $G \mapsto f_G$ defined as $$ f_G := f*G(dx) := \int f(dx|\theta) G(d\theta) $$ for some nice kernel function $f(dx|\theta)$ on $X\times \Theta$ where $X$ is some other reasonable metric space. We think of $f_G$ as a measure on $X$, which is a smoothed version of $G$. For simplicity one can think of $X = \mathbb{R}^d$ and $f(dx|\theta) = f(x|\theta) dx$ for some density $f(x|\theta)$ with respect to Lebesgue measure. We can even restrict to the case $f(x|\theta) = f(x-\theta)$ assuming $X =\Theta = \mathbb{R}^d$.

Let us equip both the spaces of measures on $\Theta$ and $X$ with some Wasserstein distance, say $W_2$. The question is how to obtain a bound of the form $$ \alpha(W_2(G,G')) \le W_2(f_G,f_{G'}), \quad \forall G,G' $$ for some $\alpha : [0,\infty) \to [0,\infty)$. In other words, we are interested in the modulus of continuity of the "inverse" of $G \mapsto f_{G}$. One can even consider the simplest case where $G = \delta_{\theta}$ and $G' = \delta_{\theta'}$ so that $W_2(G,G') = d_{\Theta}(\theta,\theta')$.

$\endgroup$
2
  • 1
    $\begingroup$ What are $dx|\theta$ and $x|\theta$? For $x-\theta$ to make sense, you that $\Theta$ is not just a metric space but has a linear structure. Or do you consider it to be a topological group? $\endgroup$ – Dirk Apr 3 '14 at 9:27
  • $\begingroup$ @Dirk: I agree with you regarding the linear structure. You can assume what is necessary for $x - \theta$ to make sense (e.g. a topological group, vector space, etc.). What I originally meant when writing it is the special case of $X = \Theta = \mathbb{R}^d$. That is the main case I care about. $\endgroup$ – passerby51 Apr 3 '14 at 15:14
4
$\begingroup$

It is intuitively clear that for the simplest case $G = \delta_\theta$ and $G' = \delta_{\theta'}$ in $\mathbb{R}^n$ we have $W_2(G,G') = W_2(f_G,f_{G'})$ for all $f$. The less trivial direction can be seen for instance by considering the orthogonal projection $p$ of a transport $\sigma$ between $f_G$ and $f_{G'}$ to the line $l$ passing through $\theta$ and $\theta'$. Since the orthogonal projection does not increase distances and since on the line the monotone rearrangement is the optimal way to tranport, we get $$ W_2^2(f_G,f_{G'}) = \inf_\sigma\int|x-y|^2\,d\sigma(x,y) \ge \inf_{\sigma}\int|p(x)-p(y)|^2\,d\sigma(x,y) \ge W_2^2(G,{G'}). $$

However, in general it is harder to give useful bounds:

Example Let $X = \Theta = \mathbb{R}$ and $f(x) = \frac12\chi_{[-1,1]}(x)$. Take $n \in \mathbb{N}$ and define $$ G = \sum_{i=1}^n \frac{1}{n}\delta_{2i} $$ and $$ G' = \frac{1}{2n}\delta_{1} + \frac{1}{2n}\delta_{2n+1} + \sum_{i=1}^{n-1} \frac{1}{n}\delta_{2i+1}. $$

Then $W_2(G,G') = 1$, but by looking at the monotone rearrangement for the diffused measures $f_G$ and $f_{G'}$ we see that $$ W_2(f_G,f_{G'}) = \sqrt{\frac1n\int_0^1t^2\,dt} = \frac{1}{\sqrt{3n}}. $$ Thus the modulus $\alpha$ has the constraint $\alpha(1) = 0$ and consequently $\alpha(t) = 0$ for all $t \in (0,1]$.

$\endgroup$
2
  • $\begingroup$ Thanks. Regarding the simplest case, I can see that $W_2(f_G,f_{G'}) \le W_2(G,G')$ if $G$ and $G'$ are point masses, but I can't see the reverse inequality in general. It seems the diffused measures get closer in general. Regarding your example, thanks for the nice construction. What can be assumed to rule out this? It seems that assuming $G$ and $G'$ to have support in a compact set rules this out. $\endgroup$ – passerby51 Apr 23 '14 at 17:32
  • $\begingroup$ I added a way to see the reverse inequality. I'll think about your followup question later when I have more time. $\endgroup$ – Tapio Rajala Apr 23 '14 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.