<ol>
<li><p>I'm not trying to be picky, but just in case, there is no such thing as <em>the</em> consistent but inefficient estimator. Perhaps you mean <em>a</em> consistent but inefficient estimator?</p>
</li>
<li><p>Your estimator is the OLS estimator. See <span class="math-container">$X'X$</span> cancelled.</p>
</li>
<li><p>If you only use the moment conditions <span class="math-container">$E(x_t u_t)=0$</span>, then OLS is the only GMM (or actually MM) estimator that can follow. As the moment conditions are exactly identifying, there are no other GMM estimators. The OLS estimator is efficient if the error term is Gaussian (in which case OLS = MLE). Otherwise, the OLS estimator is inefficient.</p>
</li>
<li><p>If you are <em>allowed</em> to change the moment conditions to something like <span class="math-container">$E(A_t x_t u_t)=0$</span> for some nonrandom and nonsingular <span class="math-container">$A_t$</span>, then you can have other estimators by choosing <span class="math-container">$A_t$</span> in an inefficient way. (WLS with wrong weights is an example: <span class="math-container">$A_t = tI$</span>.) But we usually regard <span class="math-container">$E(A_t x_t u_t)=0$</span> as different moment conditions from <span class="math-container">$E(x_t u_t)=0$</span> although one implies the other, vice versa.</p>
</li>
<li><p>It's really a matter of definitions. What do you mean by <em>inefficient</em>? Does it mean asymptotic inefficiency? Note that the OLS estimator is inefficient unless the error distribution is Gaussian although it is a BLUE. Also, can you use moment conditions implied by <span class="math-container">$E(x_t u_t)=0$</span>, not just <span class="math-container">$E(x_t u_t)=0$</span>. Clarification is required.</p>
</li>
</ol>
<p>Perhaps it's just the OLS estimator, which is inefficient under the given conditions (i.e., without normality). The OLS estimator is, however, asymptotically efficient.</p>

CC BY-SA 4.0
2021-06-26T09:09:48.413Z
2021-06-26T09:09:48.413Z
chan1142
<p>Consider the following linear model</p>
<p><span class="math-container">$$y_t = x_t' \beta +u_t$$</span></p>
<p>where <span class="math-container">$t =1,...,T$</span> and <span class="math-container">$x_t = (x_{1t} x_{2t} ... x_{kt})'$</span> , <span class="math-container">$ \beta$</span> is <span class="math-container">$k \times 1$</span> vector of unknown coefficients, <span class="math-container">$u_t$</span> is an iid disturbance term with the variance <span class="math-container">$\sigma^2$</span> and <span class="math-container">$E(x_tu_t)=0$</span> for all t.</p>
<p>Find the consistent but inefficient GMM estimator.</p>
<hr />
<p>My solution:</p>
<p>I know that <span class="math-container">$E(x_tu_t)= \frac{1}{T} \sum^T_{t=1} [x_t(y_t-x_t'\beta)]=0$</span></p>
<p>Define the Jacobian matrix</p>
<p><span class="math-container">$$J(B)= g(B)' W g(B)$$</span></p>
<p>where <span class="math-container">$g(B)=\frac{1}{T} \sum^T_{t=1} [x_t(y_t-x_t'\beta)]$</span> and <span class="math-container">$W=I_k$</span></p>
<p>Here, I define W as an identity matrix, because efficiency  depends on W matrix and when W=I, I guess that this estimator become inefficient . (Maybe wrong, I don't know exactly)</p>
<p>Then, the Jacobean matrix <span class="math-container">$J(B)$</span> in the matrix form is written as</p>
<p><span class="math-container">$$J(B)=[\frac{1}{T} X'(y-X\beta]' I_k [\frac{1}{T} X'(y-X\beta]$$</span></p>
<p>Let's minimize J(B) w.r.t <span class="math-container">$\beta$</span></p>
<p>That's, <span class="math-container">$\partial J(B) / \partial \beta =0 $</span></p>
<p>Then, <span class="math-container">$$\hat{\beta} = (X'XX'X)^{-1} X'XX'y$$</span></p>
<p>This result seems odd to me.</p>
<p>How do you solve for this question ? Where I'm wrong? Please share your ideas with me.</p>

CC BY-SA 4.0
2021-06-25T20:13:05.737Z
You must have arrived at $X^\top XX^\top (y-Xb) =0$ where premultiplication of $X^\top X$ is redundant so the solution is standard OLS. Quite efficient under the conditions you describe. Try choosing another set of instruments perhaps $x^2 = (x_1^2,..,x_K^2)$ and see how that goes.
Jesper Hybel
CC BY-SA 4.0
2021-06-25T20:56:41.070Z
I don’t understand what you said in the last sentence. I have one more question, for this given question, does application of  method of moment estimation create an inefficient estimate as well? @JesperHybel
studentp
CC BY-SA 4.0
2021-06-25T21:18:51.470Z
What is the property of this new instrument? It is chosen randomly? @JesperHybel
studentp
CC BY-SA 4.0
2021-06-25T22:45:48.747Z
Not entirely random. I was implicitly assuming that $\mathbb E[x_tu_t] = 0$ was a consequence of $\mathbb E[u_t\lvert x_t] = 0$ thereby implying $\mathbb E[f(x_t) u_t] = 0$ in general and $\mathbb E[x_t^2 u_t] = 0$ in particular.
Jesper Hybel
CC BY-SA 4.0
2021-06-25T22:49:35.753Z
Also note that you are wrong when you say $W=I$ results in inconsistency.
Jesper Hybel
CC BY-SA 4.0
2021-06-26T03:46:47.500Z
Okay I see. Well, if I use method of moment estimation instead of GMM, is it possible to generate inefficient estimate? (Without using any new instrument) @JesperHybel
studentp
CC BY-SA 4.0
2021-06-26T03:48:48.370Z
Sorry,there is typo.  I correct it.  When W=I, I try to get inefficient GMM. Is this wrong? @JesperHybel
studentp
CC BY-SA 4.0
2021-06-26T03:51:44.797Z
I want to correct who’s question. I want to find consistent but INEFFICIENT GMM estimate. Sorry for typo @JesperHybel
studentp
CC BY-SA 4.0
2021-06-26T08:07:38.197Z
Dear @JesperHybel I will be happy if you help me
studentp
CC BY-SA 4.0
2021-06-26T08:14:32.593Z
I'm not really sure I understand what you mean by "inefficient". Is this a term you actually understand or is it something that is written in an assignment?
Jesper Hybel
CC BY-SA 4.0
2021-06-26T08:17:12.873Z
Yes, from problem set. Thank you so much dear @JesperHybel
studentp
CC BY-SA 4.0
2021-06-26T08:20:51.437Z
Maybe they are just referring to the estimator using the identity matrix as inefficient because it is not using the optimal weight matrix. Nevertheless, in this particular case this does not really result in an inefficient estimator in the sense that if you had chosen another weight matrix you would have had a smaller variance.
Jesper Hybel
CC BY-SA 4.0
2021-06-26T08:21:41.880Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/126882/discussion-between-jesper-hybel-and-1190).
Jesper Hybel
<econometrics><applied-econometrics><self-study><time-series><gmm>
Finding consistent but inefficient GMM estimate
studentp
# Finding consistent but inefficient GMM estimate

(Asked by: username_0 on 2021-06-25)

Consider the following linear model

$$y\_t = x\_t' \beta +u\_t$$

where $t =1,...,T$ and $x\_t = (x\_{1t} x\_{2t} ... x\_{kt})'$ , $ \beta$ is $k \times 1$ vector of unknown coefficients, $u\_t$ is an iid disturbance term with the variance $\sigma^2$ and $E(x\_tu\_t)=0$ for all t.

Find the consistent but inefficient GMM estimator.

---

My solution:

I know that $E(x\_tu\_t)= \frac{1}{T} \sum^T\_{t=1} [x\_t(y\_t-x\_t'\beta)]=0$

Define the Jacobian matrix

$$J(B)= g(B)' W g(B)$$

where $g(B)=\frac{1}{T} \sum^T\_{t=1} [x\_t(y\_t-x\_t'\beta)]$ and $W=I\_k$

Here, I define W as an identity matrix, because efficiency depends on W matrix and when W=I, I guess that this estimator become inefficient . (Maybe wrong, I don't know exactly)

Then, the Jacobean matrix $J(B)$ in the matrix form is written as

$$J(B)=[\frac{1}{T} X'(y-X\beta]' I\_k [\frac{1}{T} X'(y-X\beta]$$

Let's minimize J(B) w.r.t $\beta$

That's, $\partial J(B) / \partial \beta =0 $

Then, $$\hat{\beta} = (X'XX'X)^{-1} X'XX'y$$

This result seems odd to me.

How do you solve for this question ? Where I'm wrong? Please share your ideas with me.

**Tags:** econometrics, applied-econometrics, self-study, time-series, gmm
> **Comments:**
> **username_1**: You must have arrived at $X^\top XX^\top (y-Xb) =0$ where premultiplication of $X^\top X$ is redundant so the solution is standard OLS. Quite efficient under the conditions you describe. Try choosing another set of instruments perhaps $x^2 = (x_1^2,..,x_K^2)$ and see how that goes.  
> **username_0**: I don’t understand what you said in the last sentence. I have one more question, for this given question, does application of  method of moment estimation create an inefficient estimate as well? @JesperHybel  
> **username_0**: What is the property of this new instrument? It is chosen randomly? @JesperHybel  
> **username_1**: Not entirely random. I was implicitly assuming that $\mathbb E[x_tu_t] = 0$ was a consequence of $\mathbb E[u_t\lvert x_t] = 0$ thereby implying $\mathbb E[f(x_t) u_t] = 0$ in general and $\mathbb E[x_t^2 u_t] = 0$ in particular.  
> **username_1**: Also note that you are wrong when you say $W=I$ results in inconsistency.  
> **username_0**: Okay I see. Well, if I use method of moment estimation instead of GMM, is it possible to generate inefficient estimate? (Without using any new instrument) @JesperHybel  
> **username_0**: Sorry,there is typo.  I correct it.  When W=I, I try to get inefficient GMM. Is this wrong? @JesperHybel  
> **username_0**: I want to correct who’s question. I want to find consistent but INEFFICIENT GMM estimate. Sorry for typo @JesperHybel  
> **username_0**: Dear @JesperHybel I will be happy if you help me  
> **username_1**: I'm not really sure I understand what you mean by "inefficient". Is this a term you actually understand or is it something that is written in an assignment?  
> **username_0**: Yes, from problem set. Thank you so much dear @JesperHybel  
> **username_1**: Maybe they are just referring to the estimator using the identity matrix as inefficient because it is not using the optimal weight matrix. Nevertheless, in this particular case this does not really result in an inefficient estimator in the sense that if you had chosen another weight matrix you would have had a smaller variance.  
> **username_1**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/126882/discussion-between-jesper-hybel-and-1190).  

---

## 1 answers

### Answer by username_2 on 2021-06-26. Score: 1 | Accepted answer

1. I'm not trying to be picky, but just in case, there is no such thing as *the* consistent but inefficient estimator. Perhaps you mean *a* consistent but inefficient estimator?
2. Your estimator is the OLS estimator. See $X'X$ cancelled.
3. If you only use the moment conditions $E(x\_t u\_t)=0$, then OLS is the only GMM (or actually MM) estimator that can follow. As the moment conditions are exactly identifying, there are no other GMM estimators. The OLS estimator is efficient if the error term is Gaussian (in which case OLS = MLE). Otherwise, the OLS estimator is inefficient.
4. If you are *allowed* to change the moment conditions to something like $E(A\_t x\_t u\_t)=0$ for some nonrandom and nonsingular $A\_t$, then you can have other estimators by choosing $A\_t$ in an inefficient way. (WLS with wrong weights is an example: $A\_t = tI$.) But we usually regard $E(A\_t x\_t u\_t)=0$ as different moment conditions from $E(x\_t u\_t)=0$ although one implies the other, vice versa.
5. It's really a matter of definitions. What do you mean by *inefficient*? Does it mean asymptotic inefficiency? Note that the OLS estimator is inefficient unless the error distribution is Gaussian although it is a BLUE. Also, can you use moment conditions implied by $E(x\_t u\_t)=0$, not just $E(x\_t u\_t)=0$. Clarification is required.

Perhaps it's just the OLS estimator, which is inefficient under the given conditions (i.e., without normality). The OLS estimator is, however, asymptotically efficient.


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