<p>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.</p>

<p>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. </p>

<p>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.</p>

CC BY-SA 3.0
2017-10-13T14:58:07.353Z
2017-10-13T14:58:07.353Z
chan1142
<p>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? </p>

<p>Thanks!</p>

CC BY-SA 3.0
2017-02-20T16:47:46.227Z
I think this is not properly an econ question. Maybe you should think about posting it on stats.stackexchange.com
PostDocing
CC BY-SA 3.0
2017-02-21T00:21:19.790Z
$\bar{g}(\theta)'W\bar{g}(\theta)$ 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. :)
chan1142
CC BY-SA 3.0
2017-02-21T11:22:40.803Z
@chan1142 Please post answers as answers so they can be voted on by the community.
Giskard
CC BY-SA 3.0
2017-10-13T14:59:24.970Z
@denesp Did what you suggested. Thanks and sorry for such a delay.
chan1142
<econometrics>
GMM - when a model is identified
Neta_1990
# GMM - when a model is identified

(Asked by: username_0 on 2017-02-19)

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?

Thanks!

**Tags:** econometrics

> **Comments:**
> **username_1**: I think this is not properly an econ question. Maybe you should think about posting it on stats.stackexchange.com  
> **username_2**: $\bar{g}(\theta)'W\bar{g}(\theta)$ 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. :)  
> **username_3**: @username_2 Please post answers as answers so they can be voted on by the community.  
> **username_2**: @denesp Did what you suggested. Thanks and sorry for such a delay.  

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## 1 answers

### Answer by username_2 on 2017-10-13. Score: 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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