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How do I interpret the j-test result in this result from 'gmm' command from 'gmm' package?

Does it mean that I am safe to use my gmm (generalized method of moments) model?

 Call:
 gmm(g = Y ~ X + X_lag1 + Y_lag1, x = Rh)


 Method:  twoStep 

 Kernel:  Quadratic Spectral

 Coefficients:
                   Estimate     Std. Error   t value      Pr(>|t|)   
 (Intercept)        3.1606e-03  4.3265e-03   7.3052e-01  4.6507e-01
 X                  2.5763e-03  2.1111e-03   1.2204e+00  2.2232e-01
 X_lag1            -2.57E-03    2.13E-03    -1.20E+00    2.28E-01
 Y_lag1             2.35E-01    4.20E-02     5.60E+00    2.17E-08

 J-Test: degrees of freedom is 0 
                 J-test               P-value            
 Test E(g)=0:    4.9684798424935e-27  *******        
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  • $\begingroup$ I vote to leave open as the question does not seem to stem from an issue with using the package, but rather related to the logic of the underlying test. $\endgroup$ – Christoph Hanck Mar 2 '18 at 9:33
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No. You have as many instruments in Rh as endogenous variables, i.e., exact identification. In that case, the J-statistic is by construction zero.

Here is a proof for linear GMM estimators. The notation follows Hayashi, i.e., instruments $x$, regressors $z$. Sample moment vectors and matrices are denoted by $s_{ij}$ and $S_{ij}$.

The GMM criterion function, which for the $J$-test is evaluated at the GMM estimator $\widehat{\delta}(\hat S^{-1})$ and the efficient weighting matrix $\hat W=\hat S^{-1}$ is given by $$ J(\tilde{\delta},\widehat{W})=n(s_{xy}-S_{xz}\tilde{\delta})'\widehat{W}(s_{xy}-S_{xz}\tilde{\delta}). $$ The linear GMM estimator is $$ \widehat{\delta}(\widehat{W})=(S_{xz}'\widehat{W}S_{xz})^{-1}S_{xz}'\widehat{W}s_{xy} $$ Under exact identification, $S_{xz}$ is square, so that $$ (S_{xz}'\widehat{W}S_{xz})^{-1}=S_{xz}^{-1}\widehat{W}^{-1}S_{xz}'^{-1} $$ and hence \begin{eqnarray*} s_{xy}-S_{xz}\widehat{\delta}(\widehat{W})&=&s_{xy}-\underbrace{S_{xz}S_{xz}^{-1}}_{=I}\underbrace{\widehat{W}^{-1}\underbrace{S_{xz}'^{-1}S_{xz}'}_{=I}\widehat{W}}_{=I}s_{xy}\\ &=&s_{xy}-s_{xy}=0 \end{eqnarray*} Thus, under exact identification, the criterion function is zero for any admissible weighting matrix $\hat W$, and thus, the $J$-statistic is identically zero.

The J-test is also called test for overidentifying restrictions - i.e., if you have more instruments than you need, you can exploit that overidenfication to test the joint validity of all instruments. If you don't have that, you cannot use the test (whence gmm appropriately does not return a p-value).

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    $\begingroup$ OK, in that case you will have as many variables in Rh as regressors, so still no overidentification, and my point stays the same. $\endgroup$ – Christoph Hanck Mar 2 '18 at 11:01
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    $\begingroup$ No. My point was and is that the J-test does not work under exact identification, and hence provides no information on whether or not you can trust your GMM model. $\endgroup$ – Christoph Hanck Mar 2 '18 at 11:02
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    $\begingroup$ No. My point is: when you have exact identification, the J-statistic is zero whether your instruments are valid or not. So you cannot use it to test their validity in this case. $\endgroup$ – Christoph Hanck Mar 2 '18 at 11:04
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    $\begingroup$ So the baseline is my j-statistics do not provide any help nor information to check the validity of my model because J-stats is close to zero (exact identification). $\endgroup$ – Eric Mar 2 '18 at 11:09
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    $\begingroup$ Yes. I now also added a proof of this result for the linear case. $\endgroup$ – Christoph Hanck Mar 2 '18 at 11:26

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