# What is the intuitive difference between Wasserstein-1 distance and Wasserstein-2 distance?

What is the intuitive difference between Wasserstein-1 distance and Wasserstein-2 distance, and how to know which one to use?

The answer boils down to the difference between the distance and its squared value. Assume we use the Euclidean distance, and imagine we want to transform a histogram $$P_S$$ whose bins are $$x\in \mathbb{R}^m$$ to another histogram $$P_T$$ with bins $$y \in \mathbb{R}^n$$. The $$W_1$$ and $$W_2$$ distances between the two histograms are defined as follows:
$$W_1 (P_S, P_T) = \underset{\gamma \in \Pi}{\text{min }} \sum_{x,y} |x-y|\gamma(x,y)$$. $$W_2 (P_S, P_T) = \left(\underset{\gamma \in \Pi}{\text{min }} \sum_{x,y} (x-y)^2\gamma(x,y)\right)^{1/2}$$ where $$\gamma$$ denotes the transportation plan and $$\Pi$$ denotes all plans, whose marginals are $$P_T$$ and $$P_S$$. One example, in which the distances will be very different is dealing with outliers. Imagine that, 95% of the density of the source distribution is in the range [0,50] and the rest is in the range $$[1000,1100] \sim$$ outliers. The $$L^2$$ distance between the outlier and histogram $$P_T$$ will be way higher than the $$L^1$$ distance. Hence, in the $$W_2$$ case, the best transportation plan is very much dependent on the outlier, i.e., moving the outlier's mass effectively is very important even though it forms only 5% of the total data. While the $$L^1$$ distance penalises way less such outliers and in a sense it is more robust.
This depends on the application and what you really want to measure with the Wasserstein distance. If you are working with measures that might contain outliers, then understanding how each distance deals with outliers is useful. In another context, imagine you are interested in the quantity $$\frac{\partial W^{P}}{\partial (x_i-y_j)}$$ (i.e., how much the distance between the two histograms change as a function of the difference between two bins) and note that it's different between $$p=1$$ and $$p=2$$.