Why Multivariate L1 Distance from Coarsened Exact Matching (CEM) is high compared to Univariate Imbalance for each matching covariate?

Question

I was looking at a paper link here by Blackwell et al. (2010) on CEM in Stata.

In one example using an example data set, the authors ran CEM using the matching covariates such as age, education, black, nodegree, and re74. For the imbalance measure for each covariate (univariate imbalance), all the measures became below 0.1 (where 0 means fully balanced and 1 means not at all balanced). But why Multivariate L1 distance, which takes account for all the imbalance measures at a time, is relatively so high (nearly 0.51)?

I read the definition of how Multivariate L1 distance is caculated. Is the above reason due to the fact that Multivariate L1 distance is calculated based on absolute difference of frequencies over all the matching covairates between treatment and control group?
In other words, does Multivariate L1 distance can be high because the range of the values for each matching covariate can differ from each other (e.g. age ranges 17-55 whereas black and nodegree have binary values) so that calculating absolute difference for all naturally generates higher Multivariate L1 distance?

I am really interested in to understand this and I could not find a good explanation for this on the internet. If I am wrong, I am happy to hear someone's comment on this.

Answer 1

Just to clarify, those balance statistics were reported prior to matching. When using CEM, the L1 statistics are equal to zero for all covariates and in the joint distribution when the same coarsening procedure is used to calculate the L1 metric and specify the bins for matching (King, Lucas, & Nielsen, 2017). This is the touted strength of CEM. Of course, the problem with CEM is that you have to discard many units to do so, which changes your target population and can reduce precision. The only explanation for why the L1 statistics after CEM are not equal to zero is that different coarsening structures were used for calculating the L1 metric and for matching.

Back to your question. Consider a dataset with 4 people in it, 2 treated and 2 control. Let the 2 treated units be a white man and a black woman. Let the 2 control units be a black man and a white woman. The proportion of black people in each treatment group is the same (.5), and the proportion of women in each treatment group is the same (.5), so the univariate L1 metrics for race and gender would both be equal to 0. But if you consider the multidimensional histogram that includes both gender and race, you will see that there is not a single bin with both a treated unit and a control unit, meaning the L1 metric will be equal to 1, indicating complete imbalance.

The point is that the marginal distributions do not correspond to the joint distribution. In the joint distribution, it is much harder for two units to fall into the same bin because they have to be exactly the same on every single one of the (coarsened) covariates. It is much easier for two units to fall into the same bin in a univariate distribution, though; they only need to be similar on one covariate.