# Proving $d_W(X,Y) \leq \vert \lambda_1-\lambda_2\vert$ for the Wasserstein metric

Suppose, that $$X \sim \text{Pois}(\lambda_1)$$ and $$Y \sim \text{Pois}(\lambda_2)$$, and consider the Wasserstein metric

$$d_W(X,Y) = \sup_{h \in \text{Lip}(1)} \vert E[h(X)]-E[h(Y)] \vert,$$ where $$\text{Lip}(1)$$ denotes the real functions which are Lipschitz with constant atmost 1.

I want to prove the following upper bound: $$d_W(X,Y) \leq \vert \lambda_1-\lambda_2\vert$$

I am not sure whether or not this holds for a more general class of distribution (if we replace the $$\lambda$$'s with expectations), and by extension whether or not we need to use properties of the Poisson distribution.

As a generic attempt, the taking an arbitrary $$h \in \text{Lip}(1)$$, we get

$$\vert E[h(X)]-E[h(Y)] \vert \leq E[\vert h(X) - h(Y)\vert ] \leq E[\vert X - Y \vert ],$$ but this is not useful. We would need to not move the absolute value inside the integral, but then I am coming up blank for ways to do this estimate.

Also I tried googling for any nice results for the Lipschitz functions and the Poisson distribution but could not find anything. I guess we could try to use that the Poisson distribution is discrete, but I don't think sums are preferable to $$E$$ here.

• Why do you think it's not useful? The difference of Poisson variables is somehow calculated. Mar 13, 2023 at 12:21
• Yes but do we know the distribution of $X-Y$? This is not a new Poisson variable as it can take negative values Mar 13, 2023 at 13:12
• Okay, assuming $X,Y$ independent, there is something called the Skellam distribution. However we are still looking at the absolute value of this, and from what I can gather from other posts on here, it is a quite complicated expression for the expectation, and it is not clear to me that this is bounded by what we want. Mar 13, 2023 at 13:46

The inequality you have already deduced is useful. Now it is a matter of coming up with the right coupling. You want to find a pair $$(X,Y)$$ such that the marginals are Poisson distributed with rates $$\lambda_1$$ and $$\lambda_2$$ and such that $$E[\lvert X-Y\rvert]\leq\lvert\lambda_1-\lambda_2\rvert.$$ To do this we can use the fact that the sum of independent Poisson variables is again Poisson distributed.
Assume w.l.o.g. that $$\lambda_1\geq\lambda_2$$ and let $$A\sim\text{Pois}(\lambda_2),B\sim\text{Pois}(\lambda_1-\lambda_2)$$ be independent. Then we define $$Y=A$$ and $$X=A+B$$ and notice that $$X\sim\text{Pois}(\lambda_1),Y\sim\text{Pois}(\lambda_2)$$. Since $$X-Y=B\sim\text{Pois}(\lambda_1-\lambda_2)$$ we have $$E[\lvert X-Y\rvert]=E[X-Y]=\lambda_1-\lambda_2=\lvert\lambda_1-\lambda_2|.$$