2
$\begingroup$

I am trying to understand how IV/just identified GMM and overidentified GMM compare when it comes to efficiency.

The way I understand it, we are able to identify the vector of coefficients in IV and hence set our orthogonality conditions exactly to 0. In the overidentified case, we have to use a distance metric J with a weighting matrix that measures how far away from zero our orthogonality conditions are. We can find an optimal weighting matrix S that minimises this distance giving us efficient GMM.

My question is now whether I will be able to reduce my asymptotic variance by adding new instruments to my IV estimation and estimating it with efficient GMM. Intuitively, I would say the IV case is less efficient since it uses a subset of orthogonality conditions of the efficient GMM case which is equivalent to using a non efficient weighting matrix, hence it can’t be as efficient. At the same time, this distance logic doesn’t really apply to IV since we can find coefficients such that J is 0.

If there is a reference that explains this I would be happy to read about it since I don’t manage to prove which asymptotic variance is smaller.

(The logic and all mentioned above is based on Hayashi Econometrics)

$\endgroup$
1
  • $\begingroup$ I found the answer in Hayashi Chapter 3 Analytical Question 4 for which solutions are provided online. $\endgroup$
    – Blo4d
    Sep 19, 2021 at 15:58

1 Answer 1

2
$\begingroup$

Hayashi's exercise mentioned in the comments (see http://fhayashi.fc2web.com/hayashi_econometrics.htm) is a good reference, yes. He considers the case where the set of instruments $x_i$ consists of the regressors $z_i$ (exploited by OLS) plus overidentifying instruments, call them $o_i$ (i.e., GMM exploits all of $x_i$). Thus, $$ x_i=\begin{pmatrix} z_i\\ o_i \end{pmatrix}. $$

He spells out that GMM with an efficient choice of weighting matrix making use of additional overidentifying moment conditions beyond those exploited by OLS when all regressors are predetermined is still efficient (in the sense that the asymptotic covariance matrix of GMM is weakly smallest in the matrix sense).

He does so by showing that the OLS asymptotic covariance matrix can be cast as a special case of the general covariance matrix of possibly inefficient GMM, concretely one that puts zero weight on the overidentifying moments and that selects the efficient weighting matrix for the moments actually exploited (the regressors $z_i$).

Given that efficiency of GMM is a general result, this may not come as a surprise.

That said, it may be worth noting that, under homoskedasticity, GMM and OLS are actually equally efficient, i.e., while GMM is not outperformed, it also does not provide any asymptotic efficiency gains over OLS.

To see this, recall that the covariance matrix of efficient GMM is $$ Avar(\hat\delta_{GMM})=(\Sigma_{xz}'S^{-1}\Sigma_{xz})^{-1}, $$ where $S^{-1}$ is the limit of the efficient weighting matrix (notation is as in Hayashi). In particular, under homoskedasticity, $S=\sigma^2\Sigma_{xx}$, i.e., the covariance of the product of instruments and errors has a particular product form.

Partition $\Sigma_{xz}$, the covariance of instruments and regressors, into that of instruments with regressors and with overidentifying instruments, $$ \Sigma_{xz}=\begin{pmatrix} \Sigma_{zz}\\ \Sigma_{oz} \end{pmatrix} $$ Likewise, partition $$ S=\sigma^2\begin{pmatrix} \Sigma_{zz}&\Sigma_{oz}'\\ \Sigma_{oz}&\Sigma_{oo} \end{pmatrix} $$ Plugging into the formula for the GMM covariance matrix gives ($\Sigma_{zz}$ is of course symmetric) $$ Avar(\hat\delta_{GMM})=\sigma^2\left(\begin{pmatrix} \Sigma_{zz}&\Sigma_{oz}' \end{pmatrix}\begin{pmatrix} \Sigma_{zz}&\Sigma_{oz}'\\ \Sigma_{oz}&\Sigma_{oo} \end{pmatrix}^{-1}\begin{pmatrix} \Sigma_{zz}\\ \Sigma_{oz} \end{pmatrix}\right)^{-1}, $$ We now make use of the result for partitioned inverses (see, e.g., Proof: Adding additional regressor and the influence on the adjusted R^2). Letting $\tilde\Sigma=\Sigma_{oo}-\Sigma_{oz}\Sigma_{zz}^{-1}\Sigma_{oz}'$, we obtain that $\sigma^{2}S^{-1}$ equals $$ \begin{pmatrix} \Sigma_{zz}^{-1}+\Sigma_{zz}^{-1}\Sigma_{oz}'\tilde\Sigma^{-1}\Sigma_{oz}\Sigma_{zz}^{-1}&-\Sigma_{zz}^{-1}\Sigma_{oz}'\tilde\Sigma^{-1}\\ -\tilde\Sigma^{-1}\Sigma_{oz}\Sigma_{zz}^{-1}&\tilde\Sigma^{-1} \end{pmatrix} $$

Plugging back into the previous equation and multiplying out gives that $$ \begin{pmatrix} \Sigma_{zz}^{-1}+\Sigma_{zz}^{-1}\Sigma_{oz}'\tilde\Sigma^{-1}\Sigma_{oz}\Sigma_{zz}^{-1}&-\Sigma_{zz}^{-1}\Sigma_{oz}'\tilde\Sigma^{-1}\\ -\tilde\Sigma^{-1}\Sigma_{oz}\Sigma_{zz}^{-1}&\tilde\Sigma^{-1} \end{pmatrix}\begin{pmatrix} \Sigma_{zz}\\ \Sigma_{oz} \end{pmatrix} = \begin{pmatrix} I\\ O \end{pmatrix}, $$ so that $$ Avar(\hat\delta_{GMM})=\sigma^2\Sigma_{zz}^{-1}, $$ which is exactly the covariance matrix of OLS under homoskedasticity.

[Of course, when we do not just have homoskedasticity in the data, but also compute the GMM estimator with the homoskedastic weighting matrix ($\hat W=\hat\sigma^2X'X/n$), it reduces to 2SLS, which, if the instruments contain the regressors, reduces to OLS, so that GMM and OLS would even be numerically identical.]

It may also be relevant in practice to note that these are first-order asymptotic results, i.e., the case in which we have a finite number of strong instruments and infinitely many observations. The big literature on many and weak instruments suggests that, in practically relevant situations, such first-order asymptotics often do not reflect finite-sample situations very well.

In particular, this literature shows that GMM may be rather biased and imprecise in finite samples, so that throwing in any moment conditions may not be sound advice with small $n$ even if the above first-order asymptotics suggest you stand nothing to lose with this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.