7
$\begingroup$

What is the critical differences between Diff-in-diff and regression discontinuity? I understood how those worked but still not clear with the differences. I googled and looked up some papers which used either of them or even both for their analysis but the authors didn't mention why they were using both. Does anyone can help with this with an intuitive explanation?

$\endgroup$

2 Answers 2

10
$\begingroup$

RD is about comparing two groups that are very similar except for the treatment because the treatment depends discontinuously on some cutoff. For example, those with a test score of 1499 don't get to go college and those with 1501 do. The underlying ability of these two groups is probably similar enough, so the wage comparison for those on either side of 1500 has high internal validity because the only difference between the two is the college education. Everything else is constant. Unfortunately, RD does not have very much to say about what happens at a score of 1357 or 1600.

DID is about comparing two groups that could have some pre-existing difference on top of treatment, but the effect of that difference is assumed to be constant over time. When you take the first difference of the outcome for each group over time, the time-invariant effect is subtracted out and doesn't contaminate the comparison in the second difference.

So RD requires different assumptions and less data that DID, but it estimates a more local effect around the cutoff. DID requires panel data and is more global in some sense.

In the extreme case when the number of periods before and after the treatment is very large, we could do an RDD with time as the running variable and the difference between treatment and control groups as the outcome. I don't think it is possible to go in the other direction.

$\endgroup$
0
0
$\begingroup$

From Regression and Other Stories, by Aki Vehtari, Andrew Gelman, and Jennifer Hill:

Regression discontinuity versus difference-in-differences. There may be a temptation to use a difference-in-differences approach in situations where the data arise from a prospective or retrospective regression discontinuity design. After all, don’t such designs often lead to situations in which there is an exposed and an unexposed group and measures of a response of interest both before and after the treatment group is exposed? Why wouldn’t this be ideal?

As it turns out, this design is particularly poorly suited to a difference-in-differences approach because, by design, a simple application of difference-in-differences is likely to suffer from bias arising from regression to the mean. For example, consider our hypothetical regression discontinuity example with grade retention imposed based on student test scores falling below a given threshold. The regression discontinuity design is capitalizing on the fact that, for students with scores close to the threshold, falling above or below the cutoff is for all intents and purposes randomly assigned. For instance you might imagine that students with test scores in that range all have scores that come from a probability distribution centered at the threshold value.

If we accept this premise, we realize that if nothing changed for those students over time, that is, if there were no treatment effect, then the next time we administered a similar test there would be a high probability that the students who scored in the lower part of this distribution (below the cutoff) the first time would be likely to score higher the next time and that the students who score relatively high (above the cutoff) the first time would be likely to score lower the next time. Therefore, if there were truly no treatment effect then we would still (repeatedly across samples) estimate a negative treatment effect due to regression to the mean.

The regression discontinuity analysis avoids this problem by conditioning on the assignment variable. This sets up direct comparisons between units with the same value of the assignment variable (confounder) and highlights the role of specification of the models for E(y1|x) and E(y0|x).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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