Search
   >> Home >> Resources & support >> FAQs >> Stata 6: Estimating fixed-effects regression with instrumental variables
Note: This FAQ is for users of Stata 6, an older version of Stata. It is not relevant for more recent versions.

Stata 6: How can I estimate a fixed-effects regression with instrumental variables?

Title   Stata 6: Estimating fixed-effects regression with instrumental variables
Author Vince Wiggins, StataCorp
William Gould, StataCorp
Date November 1999; minor revisions March 2001

Question

Is anyone aware of a routine in Stata to estimate instrumental variable regression for the fixed-effects model? I cannot see that it is possible to do it directly in Stata.

Answer

If we don’t have too many fixed-effects, that is to say the total number of fixed-effects and other covariates is less than Stata's maximum matrix size of 800, and then we can just use indicator variables for the fixed effects. This approach is simple, direct, and always right.

For example, using the auto dataset and rep78 as the panel variable (with missing values dropped) we could estimate a fixed-effects model of mpg on weight and displacement. Let's assume displacement is endogenous and we have gear_ratio and headroom as instruments.

Solution 1

First, generate indicator variables named dr1-dr5, then use ivreg to perform the estimation.

 . tab rep78, gen(dr)

output omitted

 . ivreg mpg weight dr* (displ = gear_ratio headroom)
 
 Instrumental variables (2SLS) regression
 
       Source |       SS       df       MS              Number of obs =      69
 -------------+------------------------------           F(  6,    62) =   20.25
        Model |  1536.20001     6  256.033334           Prob > F      =  0.0000
     Residual |  804.002893    62  12.9677886           R-squared     =  0.6564
 -------------+------------------------------           Adj R-squared =  0.6232
        Total |   2340.2029    68  34.4147485           Root MSE      =  3.6011
 
 ------------------------------------------------------------------------------
          mpg |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
 -------------+----------------------------------------------------------------
 displacement |  -.0158046   .0281621    -0.56   0.577    -.0720999    .0404907
       weight |  -.0038296   .0030457    -1.26   0.213    -.0099178    .0022587
          dr1 |    .093266   2.932757     0.03   0.975    -5.769231    5.955763
          dr2 |  (dropped)
          dr3 |   -.094414   1.444669    -0.07   0.948    -2.982265    2.793437
          dr4 |  -.3131534   1.596628    -0.20   0.845    -3.504767    2.878461
          dr5 |   2.217366   1.895149     1.17   0.246    -1.570984    6.005716
        _cons |   35.79702   4.005494     8.94   0.000     27.79015    43.80389
 ------------------------------------------------------------------------------
 Instrumented:  displacement
 Instruments:   weight dr1 dr2 dr3 dr4 dr5 gear_ratio headroom
 ------------------------------------------------------------------------------

In real examples, you will probably have to increase matsize before estimating the model. You will see

 . ivreg ...
 matsize too small; type -help matsize-
 r(908)

You can set matsize to any number up to 800, assuming you have sufficient memory:

 . set matsize 800

Solution 2

What if we have too many panels to estimate the model directly? In that case, xtdata can be used to transform the data to mean differences, and ivreg can then be used to estimate the fixed-effects model on the transformed data; thus you will not have to reset matsize at all.

We can estimate the same model as above by typing

 . drop if rep78==.
 (5 observations deleted)
 
 . xtdata mpg weight displ gear_ratio headroom, i(rep78) fe clear
 
 . ivreg mpg weight (displ = gear_ratio headroom)
 
 Instrumental variables (2SLS) regression
 
       Source |       SS       df       MS              Number of obs =      69
 -------------+------------------------------           F(  2,    66) =   42.13
        Model |  986.784228     2  493.392114           Prob > F      =  0.0000
     Residual |  804.002893    66   12.181862           R-squared     =  0.5510
 -------------+------------------------------           Adj R-squared =  0.5374
        Total |  1790.78712    68  26.3351047           Root MSE      =  3.4903
 
 ------------------------------------------------------------------------------
          mpg |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
 -------------+----------------------------------------------------------------
 displacement |  -.0158046   .0272954    -0.58   0.565    -.0703016    .0386924
       weight |  -.0038296    .002952    -1.30   0.199    -.0097233    .0020642
        _cons |   36.03048   3.843726     9.37   0.000     28.35623    43.70472
 ------------------------------------------------------------------------------
 Instrumented:  displacement
 Instruments:   weight gear_ratio headroom
 ------------------------------------------------------------------------------

Comparing results

 +----------------------------------------------------------+
 |         |        Solution 1      |        Solution 2     |
 |     mpg |      Coef.   Std. Err. |      Coef.  Std. Err. |
 |---------+------------------------------------------------+
 |   displ |  -.0158046   .0281621  |  -.0158046   .0272954 |
 |  weight |  -.0038296   .0030457  |  -.0038296    .002952 |
 |   _cons |   35.79702   4.005494  |   36.03048   3.843726 |
 +----------------------------------------------------------+

Comparing the estimates, we observe

  1. The coefficients for displ and weight are identical.
  2. The standard errors differ slightly.
  3. The intercepts differ.

The standard errors (SEs) differ by a scale factor and that is easily fixed. The intercept differs because of an unimportant difference in interpretation.

The SEs differ by a scale factor because our estimate of the residual variance, RMSE, also differs between the solutions. In Solution 2, the SEs are not adjusted for the fact that we estimated the fixed effects. We can adjust them:

    SE(soln. 1) = SE(soln. 2) * sqrt( e(dfr) / (e(dfr)-M+1) ) 

where M is the number of panels and M=5 in this case. For instance, the standard error for _b[mpg] can be obtained by typing

 . display _se[displ] * sqrt(e(dfr)/(e(dfr)-5+1))
 .02816214

You do not have to adjust the standard errors—the reported SEs are asymptotically equivalent. Solution 1’s SEs have been adjusted for finite samples, and many researchers prefer that the adjustment be made. With reasonable sample sizes, however, the adjustment will not amount to much.

The intercept differs because of difference in interpretation.

In Solution 1, we did not take any care in computing the intercept and let the indicator for rep78==2 drop out of the equation. Thus the intercept in that equation is the intercept for the case rep78==2.

In Solution 2, xtdata mean-differenced the data and then added back in the overall mean. Thus the reported intercept is essentially the overall intercept for all the data; see the FAQ at stata.com/support/faqs/statistics/intercept-in-fixed-effects-model for a discussion.

Both solutions are equivalent.

The Stata Blog: Not Elsewhere Classified Find us on Facebook Follow us on Twitter LinkedIn Google+ Watch us on YouTube