I am currently running computations through a "Fuzzy" Regression discontinuity Design. Suppose my data are in the following form:

• $Z$: assignment variable; if $Z > Z_0$ then the person is assigned to the treatment with a certain probability $p_D$ (since we are in the "fuzzy" RDD framework, $p_D<1$).
• $D$: treatment status; $D=1$ if the person is treated, 0 otherwise.
• $X$: set of exogenous variables.
• $Y$: Binary outcome variable.

To my knowledge - see e.g. [1] - running a fuzzy RDD is equivalent to apply Instrumental Variables using $Z$ as instrument (hence at the first stage we should have $D$ regressed on $Z$ and $X$).

In order to estimate the model through Stata I used the following code:

biprobit (Y = X D) (D = X Z)


According to some research I have done - see Nichols' pdf at [2] - the -biprobit- package should be required because of the binary nature of the endogenous variable ($D$).

Do you find the above codes correct? Is it also possible to use a simple linear probability model like this?

ivregress 2sls Y X (D=Z)


Thanks fo any help,

Stefano

[1] Angrist, J. D., Pischke, J. (2008). Mostly Harmless Econometrics: An Empiricist's Companion. Princeton University Press.

This is partial answer. I think you should probably use both the biprobit and the ivreg/ivreg2 commands to check how robust your effects are. I like the biprobit approach given your data, but it does make some strong assumptions (no heteroskedasticity, no hetrogenous effects, normality of errors).* However, there's also a dedicated RD command in Stata called rdrobust. It can handle the fuzzy design and may be installed with:
net install rdrobust, from(http://www-personal.umich.edu/~cattaneo/rdrobust) replace

• In theory, it's problematic since you're dividing $\beta_i$ by $\sigma_i=\sigma$ to get to the standard normal. However, according to Austin, biprobit is remarkably robust to variability in the treatment effect (random coefficients) in simulation. Oct 23 '13 at 18:16