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I understand that regression discontinuity determines changes in the fitted line (or coefficient) at the point in a variable which defines whether the person received the treatment/intervention.

It has been said that, as such, RD is a form of instrumental variable regression? Does this mean it is a special case of regression of which heckman (or models which predict treatment) are examples?

Is regression discontinuity a form of instrumental variable regression? If yes, how so?

dimitriy
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user163367
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    Possible duplicate of [Fuzzy RDD issue](https://stats.stackexchange.com/questions/121750/fuzzy-rdd-issue) – dimitriy Oct 19 '17 at 14:39

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Yes and no. Regression discontinuity comes in two forms: sharp and fuzzy.

Sharp RD is not a form of instrumental variables regression. Rather, it is a very special case of matching. Please see page 289 of Lee and Lemieux (2010) for an explanation:

https://www.aeaweb.org/articles?id=10.1257/jel.48.2.281

The gist is that sharp RD trivially satisfies ignorability (or unconfoundedness) assumption but violates the overlap assumption, which necessitates local continuity as a replacement assumption for overlap. Less technically, if you look at page 318 of Lee and Lemieux, you can see that in the sharp RD case, you only need one equation estimated with OLS.

The other case, fuzzy RD, is a form of instrumental variables regression. See pages 299 and 300 of Lee and Lemieux for the technical explanation, and pages 327 and 328 for the two equations to be estimated with 2SLS. The gist is that the ignorability assumption is no longer satisfied, but we can use the expected discontinuity at the cutoff as an instrumental variable for the actual discontinuity at the cutoff.

Peter
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