[Sebastian Kripfganz]

The test statistic produced by your estat overid onelevel command is calculated as the difference of the individual Hansen test statistics from the two models, as you already observed. The right comparison in the output of estat overid, difference would be the very last row, where the difference test statistic is shown as 8.1042. These numbers do not coincide exactly, because for the estat overid, difference table the reduced model (without the level moment conditions) was computed using a restricted weighting matrix, taking the weighting matrix of the full model and leaving out the rows and columns for the level model. This weighting matrix differs from the one computed for the first model. The two test statistics are asymptotically equivalent, but not identical in finite samples.

The benefit of using the restricted weighting matrix is that it does not require estimating both models, and that the test statistic is guaranteed to be non-negative. The advantage of comparing the Hansen tests from two separately estimated models is that it gives you full flexibility about which moment conditions you want to test without getting lost in the larger table produced by estat overid, difference.

[Mugi Jang]

Dear professor Sebastian Kripfganz
You recommended me to try the manual difference-in-Hansen test outlined on slide 49 of your 2019 London Stata Conference presentation
So I tried as follows

*comparison of one and two level equation
1.nested model
xtdpdgmm L(0/1).n w k, model(diff) collapse gmm(n, lag(2 4)) gmm(w k, lag(1 3)) gmm(n, lag(1 1)diff model(level)) two vce(r) overid // gmm(w k, lag(0 0) diff model(level)) dropped
estat overid
estimates store onelevel

2.nesting full model
xtdpdgmm L(0/1).n w k, model(diff) collapse gmm(n, lag(2 4)) gmm(w k, lag(1 3)) gmm(n, lag(1 1) diff model(level)) gmm(w k, lag(0 0) diff model(level)) two vce(r) overid
estat overid
estat overid,difference
estat overid onelevel

From the 1.nested model the output is as follows
1, model(diff):
L2.n L3.n L4.n
2, model(diff):
L1.w L2.w L3.w L1.k L2.k L3.k
3, model(level):
L1.D.n

. estat overid

Sargan-Hansen test of the overidentifying restrictions
H0: overidentifying restrictions are valid
2-step moment functions, 2-step weighting matrix chi2(7) = 8.2384
Prob > chi2 = 0.3120
2-step moment functions, 3-step weighting matrix chi2(7) = 6.6284
Prob > chi2 = 0.4686


From the 2.nesting full model the output is as follows

1, model(diff):
L2.n L3.n L4.n
2, model(diff):
L1.w L2.w L3.w L1.k L2.k L3.k
3, model(level):
L1.D.n
4, model(level):
D.w D.k
5, model(level):
_cons

. estat overid
Sargan-Hansen test of the overidentifying restrictions
H0: overidentifying restrictions are valid

2-step moment functions, 2-step weighting matrix chi2(9) = 16.1962
Prob > chi2 = 0.0629

2-step moment functions, 3-step weighting matrix chi2(9) = 13.8077
Prob > chi2 = 0.1293

estat overid,difference
Sargan-Hansen (difference) test of the overidentifying restrictions
H0: (additional) overidentifying restrictions are valid

2-step weighting matrix from full model

| Excluding | Difference
Moment conditions | chi2 df p | chi2 df p
------------------+-----------------------------+-----------------------------
1, model(diff) | 14.6666 6 0.0230 | 1.5296 3 0.6754
2, model(diff) | 4.0234 3 0.2590 | 12.1728 6 0.0582
3, model(level) | 15.8404 8 0.0447 | 0.3558 1 0.5509
4, model(level) | 12.0861 7 0.0978 | 4.1102 2 0.1281
model(diff) | 0.0000 0 . | 16.1962 9 0.0629
model(level) | 8.0920 6 0.2314 | 8.1042 3 0.0439


Finally I utilized the xtdpdgmm postestimation command estat overid to compute the difference of two nested overidentification test statistics


estat overid onelevel

Sargan-Hansen difference test of the overidentifying restrictions
H0: additional overidentifying restrictions are valid
2-step moment functions, 2-step weighting matrix chi2(2) = 7.9578
Prob > chi2 = 0.0187
2-step moment functions, 3-step weighting matrix chi2(2) = 7.1793
Prob > chi2 = 0.0276


Here are my questions
why does not the values of the last Sargan-Hansen difference test (right above)match to Sargan-Hansen (difference) test of the overidentifying restrictions of nesting model

7.9578 != 4.1102

But it matches to the difference value of the two model's Sargan-Hansen test of the overidentifying restrictions


16.1962 - 8.2384 = 7.9578

why I can't find this value(7.9578) in the 2.nesting full model's "| Excluding | Difference output" table
and also after excluding the model 4 the chi2 value of nesting model is 12.0861 not 8.2384 in "| Excluding | Difference output" table

Thanks lot in advance

[Sebastian Kripfganz]

Filip Novinc

1. If the difference-in-Hansen test fails for one set of instruments, it might be reasonable to just drop this set of instruments and continue with all of the other instruments. You might want to test for potential underidentification problems after dropping a set of instruments using the underid postestimation command.
2. Which tests you report depends on the message you want to conveye. A good strategy might be to report the overall Hansen test and a single difference-in-Hansen test for all level instruments jointly. If you want to be more specific about different sets of instruments, you can also report all of the individual circled difference-in-Hansen tests.
3. (a) is a robustness test to potential problems with estimating the optimal weighting matrix; this could be useful. (b) is unclear to me what kind of robustness you are assessing; depends on your research question. (c) it is not sure what you will achieve with these comparisons. (d) it might be insightful to check the robustness to different lag lengths for the instruments.

Mugi Jang

1. Note that with model(fod), the contemporaneous variables (lag 0) already qualify as instruments (for exogenous and predetermined variables). The test results here look all fine.
2. It is hard to say what is going on here. You might want to try keeping the instruments for the indicator variables for the first-differenced model, at least for testing purposes. You could also try the manual difference-in-Hansen test outlined on slide 49 of my 2019 London Stata Conference presentation; make sure that the smaller model is nested in the larger one!

[Mugi Jang]

Dear professor Kripfganz
I am trying to measure the rate of return to R&D , so I build a model dyanmic panel data model
dlntfp(firm performance) = L.dlntfp + L.dlnrnd(firm RnD investment) + L.dinvest(firm physical invest) + L.dlnrcapital(firm captal sotck) + i.type_num(firm tipe)
I first implemented model(fod) collapse nolevel and second stage I added the level equation
then at the second stage the excluding differenc hansen test value changed form insignificant to significant
what does this phenomina happen?
If the results of difference gmm does not satisfy anyone of test(Arellano-Bond test for autocorrelation of the first-differenced residuals, Sargan-Hansen test of the overidentifying restrictions ,Sargan-Hansen (difference) test of the overidentifying restrictions) then can't we move to sys-gmm.
Thanks lot!



1. only model(fod) not level equation

. xi:xtdpdgmm L(0/1).dlntfp L(1/1).dlnrnd L(1/1).dinvest L(1/1).dlnrcapital i.type_num , model(fod
> ) collapse iv( i.type ) gmmiv(L.dlntfp, lag(1 3 ) ) gmmiv(L.dlnrnd, lag(1 3 )) gmmiv(L.dinvest,
> lag(1 3 )) gmmiv(L.dlnrcapital,lag(1 3 )) vce(robust) teffects overid two
i.type_num _Itype_num_1-4 (naturally coded; _Itype_num_1 omitted)
i.type _Itypea1-4 (_Itypea1 for type==대기업 omitted)

Generalized method of moments estimation


Group variable: firmid Number of obs = 5690
Time variable: year Number of groups = 815

Moment conditions: linear = 22 Obs per group: min = 4
nonlinear = 0 avg = 6.981595
total = 22 max = 7

(Std. err. adjusted for 815 clusters in firmid)
------------------------------------------------------------------------------
| WC-Robust
dlntfp | Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
dlntfp |
L1. | .2366397 .1872553 1.26 0.206 -.130374 .6036533
|
dlnrnd |
L1. | -.0956772 .0571342 -1.67 0.094 -.2076581 .0163037
|
dinvest |
L1. | .0007188 .0010524 0.68 0.495 -.0013438 .0027815
|
dlnrcapital |
L1. | -.0430342 .0297595 -1.45 0.148 -.1013618 .0152934
|
_Itype_num_2 | .0423446 .1142619 0.37 0.711 -.1816045 .2662937
_Itype_num_3 | .0406287 .0307536 1.32 0.186 -.0196472 .1009046
_Itype_num_4 | .087836 .0908781 0.97 0.334 -.0902819 .2659538
|
year |
2016 | -.0333049 .0093917 -3.55 0.000 -.0517122 -.0148976
2017 | -.0495779 .0079012 -6.27 0.000 -.065064 -.0340919
2018 | -.0027143 .0064672 -0.42 0.675 -.0153898 .0099612
2019 | -.0495071 .0129707 -3.82 0.000 -.0749292 -.024085
2020 | .0116514 .009511 1.23 0.221 -.0069897 .0302926
2021 | -.054457 .0174896 -3.11 0.002 -.0887359 -.0201781
|
_cons | -.0319817 .1018606 -0.31 0.754 -.2316247 .1676614
------------------------------------------------------------------------------
Instruments corresponding to the linear moment conditions:
1, model(fodev):
L1.L.dlntfp L2.L.dlntfp L3.L.dlntfp
2, model(fodev):
L1.L.dlnrnd L2.L.dlnrnd L3.L.dlnrnd
3, model(fodev):
L1.L.dinvest L2.L.dinvest L3.L.dinvest
4, model(fodev):
L1.L.dlnrcapital L2.L.dlnrcapital L3.L.dlnrcapital
5, model(fodev):
_Itypea2 _Itypea3 _Itypea4
6, model(level):
2016bn.year 2017.year 2018.year 2019.year 2020.year 2021.year
7, model(level):
_cons

. estat serial, ar(1/3)

Arellano-Bond test for autocorrelation of the first-differenced residuals
H0: no autocorrelation of order 1 z = -2.6375 Prob > |z| = 0.0084
H0: no autocorrelation of order 2 z = -0.5151 Prob > |z| = 0.6065
H0: no autocorrelation of order 3 z = 0.4108 Prob > |z| = 0.6812

. estat overid

Sargan-Hansen test of the overidentifying restrictions
H0: overidentifying restrictions are valid

2-step moment functions, 2-step weighting matrix chi2(8) = 9.4444
Prob > chi2 = 0.3062

2-step moment functions, 3-step weighting matrix chi2(8) = 11.7550
Prob > chi2 = 0.1625

. estat overid, difference

Sargan-Hansen (difference) test of the overidentifying restrictions
H0: (additional) overidentifying restrictions are valid

2-step weighting matrix from full model

| Excluding | Difference
Moment conditions | chi2 df p | chi2 df p
------------------+-----------------------------+-----------------------------
1, model(fodev) | 6.0013 5 0.3061 | 3.4431 3 0.3282
2, model(fodev) | 7.4374 5 0.1901 | 2.0070 3 0.5709
3, model(fodev) | 6.1205 5 0.2947 | 3.3239 3 0.3443
4, model(fodev) | 5.8152 5 0.3246 | 3.6292 3 0.3044
5, model(fodev) | 8.3917 5 0.1359 | 1.0527 3 0.7885
6, model(level) | 4.0922 2 0.1292 | 5.3523 6 0.4995
model(fodev) | . -7 . | . . .




2. adding sys-gmm

. xtdpdgmm L(0/1).dlntfp L(1/1).dlnrnd L(1/1).dinvest L(1/1).dlnrcapital i.type_num , model(fod) collapse gmmiv(L.dlntfp, lag(1 . ) )
>gmmiv(L.dlnrnd, lag(1 . )) gmmiv(L.dinvest, lag(1 . )) gmm
> iv(L.dlnrcapital,lag(1 . )) gmmiv(L.dlntfp, model(l)lag(0 0) ) gmmiv(L.dlnrnd, model(l)lag(0 0))
> gmmiv(L.dinvest, model(l)lag(0 0)) gmmiv(L.dlnrcapital,model(l)lag(0 0)) iv( i.type ,model(l) )
> vce(robust) teffects overid
i.type_num _Itype_num_1-4 (naturally coded; _Itype_num_1 omitted)
i.type _Itypea1-4 (_Itypea1 for type==대기업 omitted)

Generalized method of moments estimation



Group variable: firmid Number of obs = 5690
Time variable: year Number of groups = 815

Moment conditions: linear = 34 Obs per group: min = 4
nonlinear = 0 avg = 6.981595
total = 34 max = 7

(Std. err. adjusted for 815 clusters in firmid)
------------------------------------------------------------------------------
| Robust
dlntfp | Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
dlntfp |
L1. | .6732203 .0246499 27.31 0.000 .6249074 .7215332
|
dlnrnd |
L1. | .0140929 .0018071 7.80 0.000 .0105511 .0176346
|
dinvest |
L1. | 5.76e-06 7.07e-06 0.82 0.415 -8.10e-06 .0000196
|
dlnrcapital |
L1. | -.0123889 .0017524 -7.07 0.000 -.0158236 -.0089542
|
_Itype_num_2 | -.0007039 .0083043 -0.08 0.932 -.0169801 .0155723
_Itype_num_3 | .0167453 .0080707 2.07 0.038 .000927 .0325635
_Itype_num_4 | .0119466 .0081141 1.47 0.141 -.0039568 .0278499
|
year |
2016 | -.0458018 .005804 -7.89 0.000 -.0571774 -.0344262
2017 | -.0513295 .0055171 -9.30 0.000 -.0621428 -.0405162
2018 | .0014926 .0048898 0.31 0.760 -.0080912 .0110764
2019 | -.0680147 .0056909 -11.95 0.000 -.0791686 -.0568607
2020 | .0193935 .0058651 3.31 0.001 .0078981 .0308889
2021 | -.0765423 .0055467 -13.80 0.000 -.0874137 -.0656709
|
_cons | .0185271 .0081254 2.28 0.023 .0026016 .0344526
------------------------------------------------------------------------------
Instruments corresponding to the linear moment conditions:
1, model(fodev):
L1.L.dlntfp L2.L.dlntfp L3.L.dlntfp L4.L.dlntfp L5.L.dlntfp
2, model(fodev):
L1.L.dlnrnd L2.L.dlnrnd L3.L.dlnrnd L4.L.dlnrnd L5.L.dlnrnd
3, model(fodev):
L1.L.dinvest L2.L.dinvest L3.L.dinvest L4.L.dinvest L5.L.dinvest
4, model(fodev):
L1.L.dlnrcapital L2.L.dlnrcapital L3.L.dlnrcapital L4.L.dlnrcapital
L5.L.dlnrcapital
5, model(level):
L.dlntfp
6, model(level):
L.dlnrnd
7, model(level):
L.dinvest
8, model(level):
L.dlnrcapital
9, model(level):
_Itypea2 _Itypea3 _Itypea4
10, model(level):
2016bn.year 2017.year 2018.year 2019.year 2020.year 2021.year
11, model(level):
_cons

. estat serial, ar(1/3)

Arellano-Bond test for autocorrelation of the first-differenced residuals
H0: no autocorrelation of order 1 z = -10.9716 Prob > |z| = 0.0000
H0: no autocorrelation of order 2 z = 0.4605 Prob > |z| = 0.6451
H0: no autocorrelation of order 3 z = 1.2161 Prob > |z| = 0.2239

. estat overid

Sargan-Hansen test of the overidentifying restrictions
H0: overidentifying restrictions are valid

1-step moment functions, 1-step weighting matrix chi2(20) = 68.5536
note: * Prob > chi2 = 0.0000

1-step moment functions, 2-step weighting matrix chi2(20) = 55.3697
note: * Prob > chi2 = 0.0000

* asymptotically invalid if the one-step weighting matrix is not optimal

. estat overid, difference

Sargan-Hansen (difference) test of the overidentifying restrictions
H0: (additional) overidentifying restrictions are valid

1-step weighting matrix from full model
note: asymptotically invalid if the one-step weighting matrix is not optimal

| Excluding | Difference
Moment conditions | chi2 df p | chi2 df p
------------------+-----------------------------+-----------------------------
1, model(fodev) | 56.1563 15 0.0000 | 12.3972 5 0.0297
2, model(fodev) | 58.8869 15 0.0000 | 9.6666 5 0.0853
3, model(fodev) | 53.7500 15 0.0000 | 14.8036 5 0.0112
4, model(fodev) | 62.2399 15 0.0000 | 6.3137 5 0.2769
5, model(level) | 55.4676 19 0.0000 | 13.0860 1 0.0003
6, model(level) | 62.7131 19 0.0000 | 5.8405 1 0.0157
7, model(level) | 61.7584 19 0.0000 | 6.7951 1 0.0091
8, model(level) | 64.4376 19 0.0000 | 4.1159 1 0.0425
9, model(level) | 59.0935 17 0.0000 | 9.4600 3 0.0238
10, model(level) | 56.8654 14 0.0000 | 11.6881 6 0.0693
model(fodev) | 0.3769 0 . | 68.1766 20 0.0000
model(level) | 11.5289 7 0.1172 | 57.0247 13 0.0000

[Filip Novinc]

Dear professor Kripfganz,

Regarding difference-in-Hansen test after xtdpdgmm estimation, may I ask two questions please:

1) If moment conditions for just one variable in model(level) do not pass the test, is it ok not to use instruments for model(level) just for that variable? Do reviewers see this as a cherry-picking or is it legit?

2) estat overid, difference gave me the following output:

Is it ok to report overall Hansen test + p-values circled in red (since these test the Blundell-Bond assumption, if my understading is correct)?

I am asking this since I've got a lot of models to report, but cannot figure out how to extract the diff-in-Hansen test in a table to word document. Any suggestions here?

3) For the purpose of testing robustness of dynamic panel data SYS-GMM estimation, what of the following do you suggest: a) comparing twostep and igmm estimations; b) using another (similar) dependent variable; c) reporting FE and OLS estimations; d) something else?

[Mugi Jang]

Thank you very much Sebastian Kripfganz
and sorry about my misnumbering
2.Second Q corresponds to 3.Replacing with higher lagged
and
3. Third Q corresponds to 2.including higher lagged differenced instruments

Anyway I clearly understand what redundant means
Thanks lot again!

[Sebastian Kripfganz]

1. Redundant means that these instruments are perfectly collinear with the already included instruments. Therefore, you do not need to include them because they would be dropped during the estimation anyway (just as any estimation command drops regressors which are perfectly collinear with others). However, strictly speaking, these additional lagged instruments for the level model are only redundant if ALL lags for the first-differenced model are used WITHOUT collapsing (and with balanced panel data). As soon as you collapse the instruments or limit their lag order, there is a good chance that the extra lags for the level model are no longer redundant, and therefore could be used in the estimation.
2. Yes, that would exactly be the recommended approach.
3. In your example, the higher-order lags are not strictly redundant because the lag limit for the first-differenced model is restricted to 3 and the instruments are also collapsed.

Here is an example of that redundancy, where adding the first lag of the instruments for the model in level neither changes the number of instruments actually used nor the estimation results:

. webuse psidextract

. xtdpdgmm L(0/1).lwage wks, gmm(L.lwage wks, l(1 .) m(d)) gmm(L.lwage wks, d l(0 0)) vce(r)

Generalized method of moments estimation

Fitting full model:
Step 1         f(b) =  .02160153

Group variable: id                           Number of obs         =      3570
Time variable: t                             Number of groups      =       595

Moment conditions:     linear =      47      Obs per group:    min =         6
                    nonlinear =       0                        avg =         6
                        total =      47                        max =         6

                                   (Std. err. adjusted for 595 clusters in id)
------------------------------------------------------------------------------
             |               Robust
       lwage | Coefficient  std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       lwage |
         L1. |   .8324395   .0321538    25.89   0.000     .7694192    .8954597
             |
         wks |  -.0000475   .0014533    -0.03   0.974     -.002896    .0028009
       _cons |   1.209187   .2559019     4.73   0.000     .7076281    1.710745
------------------------------------------------------------------------------
Instruments corresponding to the linear moment conditions:
 1, model(diff):
   3:L1.L.lwage 4:L1.L.lwage 5:L1.L.lwage 6:L1.L.lwage 7:L1.L.lwage
   4:L2.L.lwage 5:L2.L.lwage 6:L2.L.lwage 7:L2.L.lwage 5:L3.L.lwage
   6:L3.L.lwage 7:L3.L.lwage 6:L4.L.lwage 7:L4.L.lwage 7:L5.L.lwage 3:L1.wks
   4:L1.wks 5:L1.wks 6:L1.wks 7:L1.wks 3:L2.wks 4:L2.wks 5:L2.wks 6:L2.wks
   7:L2.wks 4:L3.wks 5:L3.wks 6:L3.wks 7:L3.wks 5:L4.wks 6:L4.wks 7:L4.wks
   6:L5.wks 7:L5.wks 7:L6.wks
 2, model(level):
   3:D.L.lwage 4:D.L.lwage 5:D.L.lwage 6:D.L.lwage 7:D.L.lwage 2:D.wks 3:D.wks
   4:D.wks 5:D.wks 6:D.wks 7:D.wks
 3, model(level):
   _cons

. xtdpdgmm L(0/1).lwage wks, gmm(L.lwage wks, l(1 .) m(d)) gmm(L.lwage wks, d l(0 1)) vce(r)

Generalized method of moments estimation

Fitting full model:
Step 1         f(b) =  .02160153

Group variable: id                           Number of obs         =      3570
Time variable: t                             Number of groups      =       595

Moment conditions:     linear =      47      Obs per group:    min =         6
                    nonlinear =       0                        avg =         6
                        total =      47                        max =         6

                                   (Std. err. adjusted for 595 clusters in id)
------------------------------------------------------------------------------
             |               Robust
       lwage | Coefficient  std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
       lwage |
         L1. |   .8324395   .0321538    25.89   0.000     .7694192    .8954597
             |
         wks |  -.0000475   .0014533    -0.03   0.974     -.002896    .0028009
       _cons |   1.209187   .2559019     4.73   0.000     .7076281    1.710745
------------------------------------------------------------------------------
Instruments corresponding to the linear moment conditions:
 1, model(diff):
   3:L1.L.lwage 4:L1.L.lwage 5:L1.L.lwage 6:L1.L.lwage 7:L1.L.lwage
   4:L2.L.lwage 5:L2.L.lwage 6:L2.L.lwage 7:L2.L.lwage 5:L3.L.lwage
   6:L3.L.lwage 7:L3.L.lwage 6:L4.L.lwage 7:L4.L.lwage 7:L5.L.lwage 3:L1.wks
   4:L1.wks 5:L1.wks 6:L1.wks 7:L1.wks 3:L2.wks 4:L2.wks 5:L2.wks 6:L2.wks
   7:L2.wks 4:L3.wks 5:L3.wks 6:L3.wks 7:L3.wks 5:L4.wks 6:L4.wks 7:L4.wks
   6:L5.wks 7:L5.wks 7:L6.wks
 2, model(level):
   3:D.L.lwage 5:D.L.lwage 6:D.L.lwage 7:D.L.lwage 5:L1.D.L.lwage 2:D.wks
   4:D.wks 6:D.wks 7:D.wks 4:L1.D.wks 6:L1.D.wks
 3, model(level):
   _cons

[Mugi Jang]

Dear professor Sebastian Kripfganz
I hope you are well today
Always thank you for your clear answer
This time I have three question as followings
1.First Q
In your London presentation(2019)p.31 you tald that "In combination with the moment conditions for the differenced model, further lags for the level model are redundant" with roodman(2009), and Arellano and Bover(1995)
Here what is the exact meaning of redundant? does it mean that we do not need to include higher order lags or must not use higer lags additionally ?

2.Second Q
another question is if higher order lags are redundant and it is better to use only one moment contion then instead of using the first zero lag can we use higher lag one. in case of predetermined variable

3. Third Q
Last, if higher lags are redundant, then why including the higer lags affect the estimation results


The folliwings are three case
1.using only first differenced instrument in level equation :gmm(n, lag(1 1) diff model(level))
xtdpdgmm L(0/1).n w k, model(diff) collapse gmm(n, lag(2 4)) gmm(w k, lag(1 3)) gmm(n, lag(1 1) diff model(level)) gmm(w k, lag(0 0) diff model(level)) two vce(r) // p.36

------------------------------------------------------------------------------
| WC-Robust
n | Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
n |
L1. | .5117523 .1208484 4.23 0.000 .2748937 .7486109
|
w | -1.323125 .2383451 -5.55 0.000 -1.790273 -.855977
k | .1931365 .0941343 2.05 0.040 .0086367 .3776363
_cons | 4.698425 .7943584 5.91 0.000 3.141511 6.255339
------------------------------------------------------------------------------

2.including higher lagged differenced instruments in level equation :gmm(n, lag(1 3) diff model(level))
xtdpdgmm L(0/1).n w k, model(diff) collapse gmm(n, lag(2 4)) gmm(w k, lag(1 3)) gmm(n, lag(1 3) diff model(level)) gmm(w k, lag(0 0) diff model(level)) two vce(r) // p.36

------------------------------------------------------------------------------
| WC-Robust
n | Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
n |
L1. | .4951825 .0955246 5.18 0.000 .3079577 .6824073
|
w | -1.2974 .192342 -6.75 0.000 -1.674383 -.9204167
k | .2078625 .0951428 2.18 0.029 .0213861 .394339
_cons | 4.649591 .5885837 7.90 0.000 3.495989 5.803194
------------------------------------------------------------------------------


3.Replacing with higher lagged differenced instruments in level equation :gmm(n, lag(3 3) diff model(level))
xtdpdgmm L(0/1).n w k, model(diff) collapse gmm(n, lag(2 4)) gmm(w k, lag(1 3)) gmm(n, lag(3 3) diff model(level)) gmm(w k, lag(0 0) diff model(level)) two vce(r) // p.36

------------------------------------------------------------------------------
| WC-Robust
n | Coefficient std. err. z P>|z| [95% conf. interval]
-------------+----------------------------------------------------------------
n |
L1. | .4765705 .0994517 4.79 0.000 .2816488 .6714922
|
w | -1.412922 .2631131 -5.37 0.000 -1.928614 -.8972297
k | .1916385 .1015619 1.89 0.059 -.0074192 .3906962
_cons | 5.016399 .8148121 6.16 0.000 3.419396 6.613401
------------------------------------------------------------------------------

Thank you so much again!!!

[Sebastian Kripfganz]

Because they are are used for different model transformations - here: forward-orthogonal deviations and mean deviations - these instruments (more precisely: these moment conditions) carry different informational content. If that was not the case, the command would automatically drop redundant instruments.

Let F be the forward-orthogonal deviations transformation matrix, M be the mean deviations transformation matrix, Z be the untransformed instrument, and e be the untransformed errors. The respective moment conditions are
E [Z' (F e)] = 0 for the forward-orthogonally deviated errors
E [Z' (M e)] = 0 for the mean deviated errors

Now redefine the Zf = F' Z and Zm = M' Z, then the moment conditions can be written as
E [Zf' e] = 0 for the level errors
E [Zm' e] = 0 for the level errors

Clearly, Zf and Zm are different instruments (even though Z was initially the same). See also slide 33 of my presentation.

[Mugi Jang]

Dear professor Kripfganz
I have a question about your 2019 London Stata Conference presentation p.112

"Again skipping some intermediate steps, we might be willing to treat k as strictly exogenous, using its contemporaneous term as an instrument for the model in mean deviations:"
. xtdpdgmm L(0/2).n L(0/2).w k L(0/3).ys c.w#c.w c.w#c.k, model(fod) collapse gmm(n, lag(1 .)) ///
> gmm(w, lag(0 .)) gmm(k, lag(0 .)) gmm(ys, lag(1 .)) gmm(c.w#c.w, lag(0 .)) gmm(c.w#c.k, lag(0 .)) ///
> gmm(k, lag(0 0) model(md)) teffects two vce(r) overid
Here you use simultaneously gmm(k, lag(0 .)) and gmm(k, lag(0 0) model(md))
My doubt is that althought one is differened and another is mean differenced k,lag(0 0) is used twice as instrument in one equation
then can I repeatably use one instrument variable in different form in one equation
thanks lot you kind reply

[Sebastian Kripfganz]

xtdpd always uses a small-sample adjustment. While the small-sample adjustments differ across commands, none of them are necessarily wrong. I am unable to tell what xtdpd and xtabond2 do differently. In my view, xtdpdgmm follows the usual convention for small-sample standard error correction.

[Mugi Jang]

Thank Professor Sebastian

1.When small option is removed xtdpdgmm and xtabond2 are coincident but xtdpd(stata manual example) is different from the others

xtdpdgmm L(0/1).n L(0/2).(w k) yr1980-yr1984 year, model(diff) iv(L(0/1).(w k) yr1980-yr1984 year,diff ) gmm(L.n, lag(2 .)) gmm(L.n,diff lag(1 1) model(level)) overid w(ind)
L1. | .9603675 .0945063 10.16 0.000 .7751387 1.145596

xtabond2 L(0/1).n L(0/2).(w k) yr1980-yr1984 year, iv(L(0/1).(w k) yr1980-yr1984 year,equation(diff)) gmm(L.n,lag(2 .) equation(diff)) gmm(L.n,equation(level) lag(1 1)) h(2) ar(3)
L1. | .9603675 .0945063 10.16 0.000 .7751387 1.145596

xtdpd L(0/1).n L(0/2).(w k) yr1980-yr1984 year, div(L(0/1).(w k) yr1980-yr1984 year) dgmmiv(n, lag(3 .)) lgmmiv(n, lag(2)) hascons
L1. | .9603675 .095608 10.04 0.000 .7729794 1.147756

2.When small option is added ,xtdpd and xtabond2 are coincident, but not xtdpdgmm, so xtdpdgmm does not replicate the example of stata and xtdpd does not permit small option

xtdpdgmm L(0/1).n L(0/2).(w k) yr1980-yr1984 year, model(diff) iv(L(0/1).(w k) yr1980-yr1984 year,diff ) gmm(L.n, lag(2 .)) gmm(L.n,diff lag(1 1) model(level)) overid w(ind) small
L1. | .9603675 .095335 10.07 0.000 .7732074 1.147528

xtabond2 L(0/1).n L(0/2).(w k) yr1980-yr1984 year, iv(L(0/1).(w k) yr1980-yr1984 year,equation(diff)) gmm(L.n,lag(2 .) equation(diff)) gmm(L.n,equation(level) lag(1 1)) h(2) ar(3) small
L1. | .9603675 .095608 10.04 0.000 .7726711 1.148064

xtdpd L(0/1).n L(0/2).(w k) yr1980-yr1984 year, div(L(0/1).(w k) yr1980-yr1984 year) dgmmiv(n, lag(3 .)) lgmmiv(n, lag(2)) hascons
L1. | .9603675 .095608 10.04 0.000 .7729794 1.147756

This is my conclusion

[Sebastian Kripfganz]

For the first example, add the nolevel option to the xtdpdgmm command. This will give you the same standard errors as the other commands.

In the second example, you generally do not get identical standard errors when you also estimate an intercept. (Note that the other coefficients remain unchanged if you remove the intercept.) Out of the top of my head, I cannot tell you exactly what causes the small difference in standard errors when the intercept is included; it has something to do with the small-sample adjustment. (Without option small, the standard errors between xtabond2 and xtdpdgmm coincide.) I have double checked the code of xtdpdgmm; all it does with the small option is rescaling the variance-covariance matrix by factor N / (N - rank(e(V))) in this case. Interestingly, in your example e(V) is rank deficient; however, this does not seem to explain the differences. Also note that xtabond2 produces different standard errors with and without option nomata.

In the third example, xtdpdgmm estimates the error variance from the level residuals, while xtdpd and xtabond2 estimate it from the first-differenced residuals. This is because with xtdpdgmm, model(diff) is only specified locally within the gmm() and iv() options; the global default used to compute the standard errors remains model(level). You can either set model(diff) globally, or add vce(, model(diff)) to also set it locally for the VCE. Admittedly, this is a subtle issue easily overlooked; it is a consequence of the flexibility the command provides. (You would still need to remove the small option again for exact replication.)

[Mugi Jang]

Yes here is the example1 of stata manual of xtabond
use https://www.stata-press.com/data/r17/abdata
. xtabond n l(0/1).w l(0/2).(k ys) yr1980-yr1984 year, lags(2) noconstant
(some output omitted)
L1. .6862261 .1486163 4.62 0.000 .3949435 .9775088

xtdpd L(0/2).n L(0/1).w L(0/2).(k ys) yr1980-yr1984 year, noconstant div(L(0/1).w L(0/2).(k ys) yr1980-yr1984 year) dgmmiv(n)
(some output omitted)
L1. | .6862261 .1486163 4.62 0.000 .3949435 .9775088

xtabond2 L(0/2).n L(0/1).w L(0/2).(k ys) yr1980-yr1984 year, gmm(L.n,lag(1 .)) iv( L(0/1).w L(0/2).(k ys) yr1980-yr1984 year ) nolevel small
(some output omitted)
L1. | .6862261 .1486163 4.62 0.000 .3943654 .9780869
this is t value not z value because of small option but the S.E is coincident

xtdpdgmm L(0/2).n L(0/1).w L(0/2).(k ys) yr1980-yr1984 year, model(diff) iv(L(0/1).w L(0/2).(k ys) yr1980-yr1984 year,diff) gmm(L.n, lag(1 .)) nocons small
(some output omitted)
L1. | .6862261 .1482452 4.63 0.000 .3951916 .9772607
this is t value not z value because of small option and the S.E is marginally different

Here is another example of xtdpd of stata nanual Example 5: Allowing for MA(1) errors

xtdpd L(0/1).n L(0/2).(w k) yr1980-yr1984 year, div(L(0/1).(w k) yr1980-yr1984 year) dgmmiv(n, lag(3 .)) hascons
(some output omitted)
L1. | .8696303 .2014473 4.32 0.000 .4748008 1.26446

xtabond2 L(0/1).n L(0/2).(w k) yr1980-yr1984 year, iv(L(0/1).(w k) yr1980-yr1984 year,eq(diff)) gmm(L.n, lag(2 .) eq(diff)) h(2) small
(some output omitted)
L1. | .8696303 .2014473 4.32 0.000 .4741513 1.265109

xtdpdgmm L(0/1).n L(0/2).(w k) yr1980-yr1984 year, model(diff) iv(L(0/1).(w k) yr1980-yr1984 year, diff) gmm(L.n, lag(2 .)) small
(some output omitted)
L1. | .8696303 .2008722 4.33 0.000 .4752813 1.263979

The S.E is delicately different from the first two result

xtdpd L(0/1).n L(0/2).(w k) yr1980-yr1984 year, div(L(0/1).(w k) yr1980-yr1984 year) dgmmiv(n, lag(3 .)) lgmmiv(n, lag(2)) hascons
(some output omitted)
L1. | .9603675 .095608 10.04 0.000 .7729794 1.147756

xtabond2 L(0/1).n L(0/2).(w k) yr1980-yr1984 year, iv(L(0/1).(w k) yr1980-yr1984 year,equation(diff)) gmm(L.n,lag(2 .) equation(diff)) gmm(L.n,equation(level) lag(1 1)) h(2) ar(3) small
(some output omitted)
L1. | .9603675 .095608 10.04 0.000 .7726711 1.148064

xtdpdgmm L(0/1).n L(0/2).(w k) yr1980-yr1984 year, iv(L(0/1).(w k) yr1980-yr1984 year,diff model(diff)) gmm(L.n, lag(2 .)model(diff)) gmm(L.n,diff lag(1 1) model(level)) overid w(ind) small
(some output omitted)
L1. | .9603675 .1256404 7.64 0.000 .7137123 1.207023

In this case the s.e value is very different from the first two results

I just tried to replicate the examples of stata manual (xtabond, xtdpdsys ) with xtdpd, xtabond2 and xtdpdgmm

[Sebastian Kripfganz]

Dear Sebastian Kripfganz
Could you summarize the option about standard error to get the same value among xtdpd, xtabond2 and xtdpdgmm
Thank you in advance

As a starting point, could you please provide a replicable example, say, using the abdata data set, where standard errors are not aligned?