Fixed effects in regression discontinuity design

Question

I want to do a non parametric RDD type analysis to know the impact of an intervention (a single dummy variable) on an outcome variable. I have several 'boundaries' (which are actually different geographical locations) around which I will be picking observations.

Consider that I have only 2 observations at each boundary- one with the intervention and one without it. I can assume that all other relevant variables are the same for the pair of observations at each boundary.

Can I then regress the outcome variable on the intervention dummy along with boundary fixed effects? My main concern is regarding use of fixed effects with just 2 observations at each boundary — I am not sure if that is legit.

I will be taking observations only close to the boundary, as in a non-parametric RDD. — bitter-sweet, Oct 03 '15 at 16:42
@gung I didn't find the OPs response enlightening, but I assume [this](https://en.wikipedia.org/wiki/Regression_discontinuity_design#Non-parametric_estimation) is the intent — Glen_b, Oct 03 '15 at 17:32
Someone please help! Let me know if the question isn't clear enough! — bitter-sweet, Oct 04 '15 at 14:05
If your outcome is the difference in outcomes across the boundary, $LATE = \frac{y^+-y^-}{x^+-x^-}$, where $x$ is expected level of treatment, any fixed effect should get differenced out. That is, your non-parametric method is implicitly using a FE approach. — dimitriy, Oct 08 '15 at 01:36
@DimitriyV.Masterov Thanks. Agreed that is one way to do it. But what if I want to use the outcome itself as the dependent variable? Then I have 2 observations at each boundary (i'm not taking any differences). Can I add boundary dummies then? — bitter-sweet, Oct 11 '15 at 11:16

dimitriy · Accepted Answer · 2015-10-13T00:49:46.060

For reasons explained in my comment, you will get identical estimates for the treatment coefficient.

Here's a numerical example of "hard" RDD using Stata. We will use a experimental dataset of 12 cars. Each car was run once without a beneficial fuel additive (condition 1) and once with (condition 2). The outcome is miles per gallon. This setup is similar to your cross-border pairs, where one member is treated, but they are otherwise similar.

. use http://www.stata-press.com/data/r14/fuel, clear

. gen diff = mpg2-mpg1

. list, clean noobs

    mpg1   mpg2   diff  
      20     24      4  
      23     25      2  
      21     21      0  
      25     22     -3  
      18     23      5  
      17     18      1  
      18     17     -1  
      24     28      4  
      20     24      4  
      24     27      3  
      23     21     -2  
      19     23      4  

. reg diff

      Source |       SS           df       MS      Number of obs   =        12
-------------+----------------------------------   F(0, 11)        =      0.00
       Model |           0         0           .   Prob > F        =         .
    Residual |       80.25        11  7.29545455   R-squared       =    0.0000
-------------+----------------------------------   Adj R-squared   =    0.0000
       Total |       80.25        11  7.29545455   Root MSE        =     2.701

------------------------------------------------------------------------------
        diff |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
       _cons |       1.75   .7797144     2.24   0.046     .0338602     3.46614
------------------------------------------------------------------------------

. gen pair_id = _n

. reshape long mpg, i(pair_id) j(treat)
(note: j = 1 2)

Data                               wide   ->   long
-----------------------------------------------------------------------------
Number of obs.                       12   ->      24
Number of variables                   4   ->       4
j variable (2 values)                     ->   treat
xij variables:
                              mpg1 mpg2   ->   mpg
-----------------------------------------------------------------------------

. reg mpg i.treat  i.pair_id 

      Source |       SS           df       MS      Number of obs   =        24
-------------+----------------------------------   F(12, 11)       =      4.03
       Model |       176.5        12  14.7083333   Prob > F        =    0.0139
    Residual |      40.125        11  3.64772727   R-squared       =    0.8148
-------------+----------------------------------   Adj R-squared   =    0.6127
       Total |     216.625        23  9.41847826   Root MSE        =    1.9099

------------------------------------------------------------------------------
         mpg |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
     2.treat |       1.75   .7797144     2.24   0.046     .0338602     3.46614
             |
     pair_id |
          2  |          2   1.909902     1.05   0.317    -2.203667    6.203667
          3  |         -1   1.909902    -0.52   0.611    -5.203667    3.203667
          4  |        1.5   1.909902     0.79   0.449    -2.703667    5.703667
          5  |       -1.5   1.909902    -0.79   0.449    -5.703667    2.703667
          6  |       -4.5   1.909902    -2.36   0.038    -8.703667   -.2963331
          7  |       -4.5   1.909902    -2.36   0.038    -8.703667   -.2963331
          8  |          4   1.909902     2.09   0.060    -.2036669    8.203667
          9  |   1.03e-15   1.909902     0.00   1.000    -4.203667    4.203667
         10  |        3.5   1.909902     1.83   0.094    -.7036669    7.703667
         11  |   1.15e-15   1.909902     0.00   1.000    -4.203667    4.203667
         12  |         -1   1.909902    -0.52   0.611    -5.203667    3.203667
             |
       _cons |     21.125    1.40565    15.03   0.000     18.03118    24.21882
------------------------------------------------------------------------------

. xtreg mpg i.treat, fe

    Fixed-effects (within) regression               Number of obs     =         24
    Group variable: pair_id                         Number of groups  =         12

    R-sq:                                           Obs per group:
         within  = 0.3141                                         min =          2
         between =      .                                         avg =        2.0
         overall = 0.0848                                         max =          2

                                                    F(1,11)           =       5.04
    corr(u_i, Xb)  = 0.0000                         Prob > F          =     0.0463

    ------------------------------------------------------------------------------
             mpg |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
    -------------+----------------------------------------------------------------
         2.treat |       1.75   .7797144     2.24   0.046     .0338602     3.46614
           _cons |         21   .5513413    38.09   0.000     19.78651    22.21349
    -------------+----------------------------------------------------------------
         sigma_u |  2.6809513
         sigma_e |  1.9099024
             rho |  .66334557   (fraction of variance due to u_i)
    ------------------------------------------------------------------------------
    F test that all u_i=0: F(11, 11) = 3.94                      Prob > F = 0.015

All 3 methods yield an estimated marginal improvement of 1.75 miles per gallon, with a standard error of 0.78.

Got it. I also have control variables that can differ b/w the 2 observations at a location. I think then in the diff approach I should include (as an explanatory variable) the diff in control variables. But I also want my sample to have locations where both observations are treatment or both of them are non treatment (because they still contribute to determining coefficients of the control variables, right?). Then how do I go about the diff approach? And would that be equivalent to the FE approach? — bitter-sweet, Oct 13 '15 at 13:33
@bitter-sweet Covariates are not required for identification, but they can added to increase the precision of the estimated treatment effect by absorbing some residual variance. A good robustness check is to do RD with each of the covariates as the outcome (or to just graph them against the running variable). An effect/jump will detect violations of the assumptions underlying the RD model. I would probably exclude locations where both observations are either treatment or control since you don't have an apples-to-apples comparison that can be made in such places. — dimitriy, Oct 13 '15 at 17:14
Including covariates would be important in my case because they could in fact be quite different between the 2 obs at a location (I am aware this creates a problem for the RDD). If I'm including covariates, should I still exclude locations where there is no change in treatment status? Do I not lose much by selectively excluding such locations? — bitter-sweet, Oct 14 '15 at 08:37
If you're willing to give up on identification, you can certainly run any regression you want! More seriously, my understanding is that throwing away these unmatched data away is analogous to enforcing the common support assumption in propensity score matching: nothing can be said about some observations if there's no valid counterfactual for them. To quote John Tukey, "The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data." — dimitriy, Oct 14 '15 at 18:08
Ok sounds good. Now assume that covariates don't differ enough to make my RDD invalid but I still want to include them (as is done in many papers). Also I want to include locations with both non-treatment or both treatment observations in order to better determine coefficients of covariates. Then are the difference and FE approaches equivalent? Of course I won't be getting the same coefficients now, but conceptually are they the same or is one method preferable to another? To be clear, in the diff approach I would be looking at a dummy for change in treatment status. — bitter-sweet, Oct 15 '15 at 04:19

Fixed effects in regression discontinuity design

1 Answers1