Suppose we are using GMM to estimate the model $y=m(x;\theta_0)+u$ (for some function $m$ and $\theta_0 \in\mathbb{R^p}$) and we are using the moment condition: $\mathbb{E}[g(\theta_0)]=0$ where $g:\mathbb{R^p}\rightarrow\mathbb{R^p}$ so the model is just identified. My question is how do we know that our estimator $\hat\theta$ is not dependent on the weight matrix?


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    $\begingroup$ I think this is not properly an econ question. Maybe you should think about posting it on stats.stackexchange.com $\endgroup$
    – PhDing
    Commented Feb 20, 2017 at 16:47
  • $\begingroup$ $\bar{g}(\theta)'W\bar{g}(\theta)$ is minimized if and only if $\bar{g}(\theta)=0$ in that case, where $W$ is symmetric and positive definite. For whatever $W$, you always attain $\bar{g}(\theta)=0$, where the root is the GMM estimator. By the way, GMM is relevant more to econometrics than to statistics. :) $\endgroup$
    – chan1142
    Commented Feb 21, 2017 at 0:21
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    $\begingroup$ @chan1142 Please post answers as answers so they can be voted on by the community. $\endgroup$
    – Giskard
    Commented Feb 21, 2017 at 11:22
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    $\begingroup$ @denesp Did what you suggested. Thanks and sorry for such a delay. $\endgroup$
    – chan1142
    Commented Oct 13, 2017 at 14:59

1 Answer 1


In the exactly identified case, it is natural to assume that a unique $\theta$ satisfies $\bar{g}(\theta) = 0$, because the number of parameters is equal to the number of equations. Let $\hat\theta$ denote such $\theta$ value.

When $W$ is positive definite, $\bar{g}(\theta)' W \bar{g}(\theta) \ge 0$ (because $W$ is positive definite), and $\bar{g}(\theta)' W \bar{g}(\theta)$ attains zero if and only if $\bar{g}(\theta)=0$. That is, the global minimizer of $\bar{g}(\theta)' W \bar{g}(\theta)$ equals the solution to $\bar{g}(\theta)=0$, which is $\hat\theta$, for whatever positive definite $W$ matrix. Thus, $W$ is irrelevant.

Note that this argument does not hold if $W$ is not positive definite. For example, if $W$ is positive semi-definite but not positive definite, then the GMM estimator may not be unique.


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