Asking this for a colleague:

They have two groups of people with various characteristics and to address the issue of comparability of outcomes they would like to reference a paper or book that discusses the concept and problems of common support.

  • $\begingroup$ I think I know what you mean, David, but I'm not quite sure because "support" means several things in statistics. Among them that might be relevant here are (a) values that a random variable can attain and (b) the physical extent or amount of any sample of a substance (such as a volume of soil or water) represented by a measurement. However, neither seem exactly right for this question. Could you perhaps indicate what you mean by "common support"? $\endgroup$
    – whuber
    Mar 10, 2011 at 1:33
  • $\begingroup$ I mean support in the sense of where a function is non-zero. I think for the example at hand, one has a binary variable s and some other background variables gathered on people, and the potential problem is that for people in the group with s=1 and people in the group with s=2, their range of measured values on the background variables might not overlap much. In this instance all is ok, but a referee wants a discussion that says why there is no problem about finding a region of common support for these two groups. My colleague is not a statistician so needs a pointer even if only to a textbook $\endgroup$ Mar 10, 2011 at 2:38
  • 3
    $\begingroup$ From your comment above and as you've tagged this with propensity-scores, I believe 'common support' means the same as 'overlap in the distributions of covariates' as discussed by Rubin 2008. My reading is that this paper says lack of such overlap is a problem. I'm no expert on propensity score analysis, but I understand Rubin to be saying that this is a problem for estimating causal effects from observational data whether or not you use propensity scores, but using propensity scores make it more obvious. $\endgroup$
    – onestop
    Mar 10, 2011 at 10:55
  • $\begingroup$ @onestop - this is it. I'll have a look at Rubin, it may be just what my colleague needs. $\endgroup$ Mar 10, 2011 at 21:44
  • $\begingroup$ @David @onestop @whuber, it has nothing to do with support of a distribution ? my answer is totally out of the point? what are propensity score ? $\endgroup$ Mar 11, 2011 at 12:22

2 Answers 2


Matching and Balance

The general issue falls under 'matching', which is the process of sampling units to create subsamples that are maximally 'balanced' i.e as similar as possible on uninteresting (e.g. non-treatment) covariates while differing on an interesting one, with an eye to making causal claims about the latter. Ideally, matched samples have the same distribution of uninteresting covariates. Less stringently, their covariates have overlapping support.

Propensity scores are one approach to the balance problem, but there are a more, not all of which are so dependent on building another model to make a matching 'score' and implicitly relying on its assumptions. For example, a recent alternative would be 'coarsened exact matching'.


This field is getting pretty big, but here is an unsystematic sample from my shelf and hard disk, that might prove helpful.

Gelman and Hill ch. 9 and 10 are very good, I think, assuming that you're operating in a regression framework to start with. For just a reference to the issue and its possible solutions, this might be a good one. You might also find Imai et al. (2008) helpful, mostly for the error decomposition.

More generally, there are a bunch of article length reviews available from Stuart's papers page or from King's causal inference page. Morgan and Winship is a thorough book length treatment of causal inference that puts the common support issue in a bigger picture.


If I understand, you have a classification problem where you observe $(X_1,Y_1),\dots,(X_n,Y_n)$. Denoting by $\mathcal{X}$ the space where $X$ takes values, you assume the existance of $\mathcal{X}_{10}$ the smallest closed subset of $\mathcal{X}$ such that:

$\mathcal{X}\setminus \mathcal{X}_{10}$ can be partinioned into $\mathcal{X}_{1\setminus 0}\cup \mathcal{X}_{0\setminus 1}$ with $P_{0}(X\in \mathcal{X}_{1\setminus 0})=0$, $P_{1}(X\in \mathcal{X}_{0\setminus 1})=0$ (for $i=0,1$ $P_i$ is the distribution of $(X|Y=i)$)

I think there are two problems that may arise from your question:

  1. How do you justify a decomposition of the distribution to be able to solve the problem only on $\mathcal{X}_{10}$ (Theoretical problem)
  2. How do you estimate $\mathcal{X}_{1\setminus 0}$, $\mathcal{X}_{0\setminus 1}$ and$\mathcal{X}_{0 1}$ from data (practical estimation problem)

Theoretical problem In this case (and under suitables conditions) you can decompose $P_{1}$ and $P_0$ using the Radon-Nikodym theorem (This version) into:

$$P_{1}=P_{1,\mathcal{X}_{1\setminus 0}}+P_{1,\mathcal{X}_{0 1}}$$

$$P_{0}=P_{0,\mathcal{X}_{0\setminus 1}}+P_{0,\mathcal{X}_{0 1}}$$

with $P_{0,\mathcal{X}_{0 1}}\sim P_{1,\mathcal{X}_{0 1}}$ (mutually absolutly continuous)

$P_{1,\mathcal{X}_{0 1}} \bot P_{1,\mathcal{X}_{1\setminus 0}} \bot P_{0,\mathcal{X}_{0\setminus 1}}$ (mutually singular) Note that distributions in the decomposition are not probabilities, and you need to renormalize (i.e. compute $P_{i}(X\in A)$ when $A$ is one of the three above mentionned sets). This renormalization is also required from a practical point of view.

Once you have done the renormalization you can construct a classification rule on $\mathcal{X}_{0 1}$ using renormalized version of $P_1$ and $P_0$ 's restrictions. You can extend this classification rule in the obvious way to the whole $\mathcal{X}$.

Estimation problem If you want to "learn" the above mentionned rule from the data, you need to estimate the corresponding sets I guess there is a large litterature in "support" estimation but in view of the classification problem at the end, it is not a big deal since the places where you might be wrong in terms of support estimation are does where you don't have much "mass" and hence does that are not really important for the classification errors at the end... at least from a bayesian perspective of which error is important... (see my answer here)


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