[Sarah Magd]

For the IV-GMM, I am also confused about the naming of this method. As an example from the literature, Acheampong et al.(2021) mention that they use the instrumental variable generalized method of moment (IV-GMM) based on the Baum et al. (2003).
In this case, would IV-GMM in ivreg2 be equivalent to the one-step sys-GMM in xtdpdgmm?

[Sebastian Kripfganz]

• Yes, in a static model with predetermined/endogenous variables, the two-step GMM estimator can still be useful.
• I do not know what is meant by IV-GMM here.

[Sarah Magd]

Thanks a lot for the constructive answer.

- As far as I understood, if our model is specified as a static model, and we want to control for the fixed effects and obtain more efficient estimates, we can still use the two-step GMM estimator. Am I right?
- Would you please clarify the difference between these methods: IV-GMM & sys-GMM. Because I found a paper in the literature where they use the IV-GMM to estimate a static model. They also claim that "the IV-GMM controls for the endogeneity problem and variable omission bias, and produces consistent estimates".
- How can we estimate the IV-GMM using the xtdpdgmm?

[Sebastian Kripfganz]

• Yes, the one-step estimator with the default weighting matrix yields the 2SLS estimator.
• The two-step system GMM estimator would be generally more efficient because it accounts for the extra variance coming from the unobserved fixed effects. It would be equivalent to the 2SLS estimator only if there were no unobserved fixed effects (in other words, if their variance was zero).
• Checking for serial correlation after a static model could still be useful. If you have predetermined or endogenous regressors, serial correlation would still affect the first admissible lag for the instruments. Serial correlation might also indicate that a dynamic model may be more appropriate to avoid an omitted-variable bias.

[Sarah Magd]

#########################################
# Estimating a static model using the xtdpdgmm
#########################################

Dear Prof. Kripfganz,

If we specify a model without the lagged dependent variable as a regressor (i.e., a static model), and this model suffers from the endogeneity problem (i.e., due to the reverse causality). Also, the specification should account for the unobserved fixed effects.
- As far as I know, using the system GMM estimator with the default weighting matrix in xtdpdgmm would be equivalent to the 2SLS estimator. Am I right?
- If we use the two-step GMM estimator with this specification, would it still be more efficient in this case than the one-step system GMM?
- Do we also need to check the serial correlation after estimating the static model with GMM?

[Sebastian Kripfganz]

From the information you have provided, it is not clear how exactly you specified the model. The full command syntax would help. Assuming that you are referring to the first-differenced model, the starting lag 0 is not valid for the lagged dependent variable because it is correlated with the first-differenced error term. You need to start from lag 1 instead. For an endogenous variable, the respective starting lag would be 2 (assuming no serial correlation of the idiosyncratic level errors).

If you were using the forward-orthogonal transformation for the model instead of the first-difference transformation, the first admissible lag would be 0 for a predetermined variable and 1 for an endogenous variable.

[Neyati Ahuja]

Hello Prof. Sebastian

I have a query in respect of predeterminant variable.
When I took lag of dependent variable as the predetermined variable, with lag (0 0), (0 1),.....(0 .), the resulting coefficient obtained for the lag depenedent variable after running the xtdpdgmm command is positive but not significant. However, when the same is repeated with lag starting from 1 the results for the coefficient is positive as well as significant.
I have checked for instrument validity using Sargan and Hansen Test also.
1. What should be done in such situation.
2. Is there a problem in model specification or we can take lag of predeterminant variable lag starting from 1.
3. If ans to previous question is yes, what should be the starting of lag of endogenous variables. Can it still start from 1.

Regards

[Sebastian Kripfganz]

I am afraid the last xtdpdgmm update (version 2.3.11) was premature and did more harm than good. I "fixed" a bug that actually wasn't one. I apologize for this mishap.

With the now available latest version 2.4.0, the correct computations have been restored for estat serial and estat hausman. Furthermore, a minor bug in option auxiliary has been fixed which was introduced in version 2.3.10.

As a major new feature, this latest version can now compute the continously-updating GMM estimator as an alternative to the two-step and iterated GMM estimators. Simply specify the new option cugmm. The CU-GMM estimator updates the weighting matrix simultaneously with the coefficient estimates while minimizing the objective function. This is in contrast to the iterated GMM estimator (of which the two-step estimator is a special case), which iterates back and forth between updating the coefficient estimates and the weighting matrix. As a technical comment: The CU-GMM objective function generally does not have a unique minimum. The estimator therefore can be sensitive to the choice of initial values. By default, xtdpdgmm uses the two-stage least squares estimates, ignoring any nonlinear moment conditions, as starting values for the numerical CU-GMM optimization. This seems to work fine.

The following example illustrates the CU-GMM estimator, and how the xtdpdgmm results can be replicated with ivreg2 (up to minor differences due to the numerical optimization):

. webuse abdata

. xtdpdgmm L(0/1).n w k, gmm(L.n w k, l(1 4) c m(d)) iv(L.n w k, d) cu nofooter

Generalized method of moments estimation

Fitting full model:

Continously updating:
Iteration 0:   f(b) =  .22189289  
Iteration 1:   f(b) =  .08073713  
Iteration 2:   f(b) =  .07655265  
Iteration 3:   f(b) =  .07646044  
Iteration 4:   f(b) =  .07645679  
Iteration 5:   f(b) =  .07645673  

Group variable: id                           Number of obs         =       891
Time variable: year                          Number of groups      =       140

Moment conditions:     linear =      16      Obs per group:    min =         6
                    nonlinear =       0                        avg =  6.364286
                        total =      16                        max =         8

------------------------------------------------------------------------------
           n | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
           n |
         L1. |   .4342625   .1106959     3.92   0.000     .2173024    .6512225
             |
           w |  -2.153388   .3702817    -5.82   0.000    -2.879126   -1.427649
           k |  -.0054155   .1221615    -0.04   0.965    -.2448477    .2340166
       _cons |   7.284639   1.123693     6.48   0.000     5.082241    9.487037
------------------------------------------------------------------------------

. predict iv*, iv
 1, model(diff):
   L1.L.n L2.L.n L3.L.n L4.L.n L1.w L2.w L3.w L4.w L1.k L2.k L3.k L4.k
 2, model(level):
   D.L.n D.w D.k
 3, model(level):
   _cons

. ivreg2 n (L.n w k = iv*), cue cluster(id) nofooter
Iteration 0:   f(p) =  31.065005  (not concave)
Iteration 1:   f(p) =  27.307398  (not concave)
Iteration 2:   f(p) =  26.543788  (not concave)
Iteration 3:   f(p) =  25.047573  (not concave)
Iteration 4:   f(p) =  24.521102  (not concave)
Iteration 5:   f(p) =  24.107293  (not concave)
Iteration 6:   f(p) =  23.931765  (not concave)
Iteration 7:   f(p) =  23.746613  (not concave)
Iteration 8:   f(p) =  23.636564  
Iteration 9:   f(p) =  23.304181  (not concave)
Iteration 10:  f(p) =  23.241277  (not concave)
Iteration 11:  f(p) =  23.178503  (not concave)
Iteration 12:  f(p) =  23.125314  (not concave)
Iteration 13:  f(p) =  23.074408  
Iteration 14:  f(p) =  19.278726  
Iteration 15:  f(p) =  12.160385  (not concave)
Iteration 16:  f(p) =  11.700402  
Iteration 17:  f(p) =   11.03222  (not concave)
Iteration 18:  f(p) =  10.950583  (not concave)
Iteration 19:  f(p) =  10.907663  
Iteration 20:  f(p) =  10.800048  
Iteration 21:  f(p) =  10.704051  
Iteration 22:  f(p) =  10.703945  
Iteration 23:  f(p) =  10.703942  
Iteration 24:  f(p) =  10.703942  

CUE estimation
--------------

Estimates efficient for arbitrary heteroskedasticity and clustering on id
Statistics robust to heteroskedasticity and clustering on id

Number of clusters (id) =          140                Number of obs =      891
                                                      F(  3,   139) =    83.84
                                                      Prob > F      =   0.0000
Total (centered) SS     =  1601.042507                Centered R2   =   0.5099
Total (uncentered) SS   =  2564.249196                Uncentered R2 =   0.6940
Residual SS             =  784.7107633                Root MSE      =    .9385

------------------------------------------------------------------------------
             |               Robust
           n | Coefficient  std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
           n |
         L1. |   .4342987   .1003318     4.33   0.000     .2376521    .6309453
             |
           w |  -2.153233   .2986292    -7.21   0.000    -2.738535    -1.56793
           k |  -.0053816   .1162739    -0.05   0.963    -.2332742    .2225111
       _cons |   7.284114   .8901409     8.18   0.000     5.539469    9.028758
------------------------------------------------------------------------------

To update to the new version, type the following in Stata's command window:

net install xtdpdgmm, from(http://www.kripfganz.de/stata) replace

Disclaimer: I have extensively tested this new version. However, due to the complexity of the command, the variety of options, and the lack of alternative software to compare the results for some advanced options, I cannot guarantee that the implementation is error-free. Please let me know if you spot any irregularities.

[Sebastian Kripfganz]

If either X1 or X2 is endogenous, then it usually makes sense to assume that their interaction X3 is endogenous as well. You can then just treat it the same way as any other endogenous variable.

[Sarah Magd]

############################################
#Interaction variables with xtdpdgmm
#############################################
Can you please write a post about how we should instrument the interaction variable X3 (X1*X2) in the following two cases?
Case 1: X1 and X2 are both endogenous
Case 2: X1 is endogenous whereas X2 is predeterminant

[Sebastian Kripfganz]

In the case of system GMM, yes.

[Sarah Magd]

Thanks, Prof. Sebastian Kripfganz
In this case, do you mean that I should ignore the 1-step weighting matrix chi2 test and report the 2-step weighting matrix chi2 only?

[Sebastian Kripfganz]

As the note says, both 1-step tests are asymptotically invalid because the one-step weighting matrix of the system GMM estimator is not optimal. That is especially true for the first of the two reported tests (1-step moment functions, 1-step weighting matrix). You should just ignore that test.

[Sarah Magd]

Sebastian Kripfganz
Dear Prof. Kripfganz,
I have a question regarding the "estat overid" after running the two-step and one-step system GMM. I run the same specification (i.e., with the same number of lags) and get different results for the overidentification test. For the one-step estimation, I never obtain an insignificant P-value for the 1-step moment functions and 1-step weighting matrix. Even if I change the specification of the model. Please find below the details of the output. My sample has 27 cross-section units and 13 years.
What does this conflicting result mean?

##########################
#estat overid after the one step
##########################
Sargan-Hansen test of the overidentifying restrictions
H0: overidentifying restrictions are valid

1-step moment functions, 1-step weighting matrix chi2(21) = 122.9740
note: * Prob > chi2 = 0.0000

1-step moment functions, 2-step weighting matrix chi2(21) = 27.9882
note: * Prob > chi2 = 0.1405

* asymptotically invalid if the one-step weighting matrix is not optimal
####################################
#estat overid after the two step
####################################
Sargan-Hansen test of the overidentifying restrictions
H0: overidentifying restrictions are valid

2-step moment functions, 2-step weighting matrix chi2(21) = 27.3389
Prob > chi2 = 0.1599

2-step moment functions, 3-step weighting matrix chi2(21) = 27.9939
Prob > chi2 = 0.1403

[Sebastian Kripfganz]

Addendum: For a simple example with industry dummies (not using the two-stage approach), see slide 86 of my 2019 London Stata Conference presentation:

• Kripfganz, S. (2019). Generalized method of moments estimation of linear dynamic panel data models. Proceedings of the 2019 London Stata Conference.