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.

  • $\begingroup$ 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. $\endgroup$ Jun 25, 2021 at 20:13
  • $\begingroup$ 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 $\endgroup$
    – studentp
    Jun 25, 2021 at 20:56
  • $\begingroup$ What is the property of this new instrument? It is chosen randomly? @JesperHybel $\endgroup$
    – studentp
    Jun 25, 2021 at 21:18
  • $\begingroup$ 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. $\endgroup$ Jun 25, 2021 at 22:45
  • $\begingroup$ Also note that you are wrong when you say $W=I$ results in inconsistency. $\endgroup$ Jun 25, 2021 at 22:49

1 Answer 1

  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.


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.