# Consistent but inefficient GMM

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

As it is seen in the answer, I obtained OLS estimator $$\hat{\beta} = (X'X)^{-1} X’y$$. But OLS estimator is efficient, not inefficient. Thus my answer is wrong.

What should I make an extra assumption to obtain inefficient but consistent estimator? Assuming Weight matrix as I is wrong.

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

Note: the question is as follows. I am asking the part a.

New Update

The efficient GMM estimator depends on the weight matrix.

I firstly define the population moment condition as $$g(B)=\frac{1}{T} \sum^T_{t=1} [x_t(y_t-x_t'\beta)]$$

Based on the Central limit theorem,$$\sqrt{T} g(\beta) \to N(0, P)$$ where $$P=var(\sqrt{T}g(\beta)=\frac{\sigma^2}{T^2}(X'X)$$

Also, Based on Central limit theorem,

$$\sqrt{T} (\hat{\beta}_{GMM} - \beta) \to N(0, V)$$

where $$V=(D'WD)^{-1}D'WPW'D(D'WD)^{-1}$$ for the expected value of the matrix go the first derivates of the moment $$D$$.

I I choose the weight matrix $$W=P^{-1}= (\frac{\sigma^2}{T^2}(X'X))^{-1}$$ instead of choosing any other positive definite matrix $$S$$ for $$W$$, then I can obtain the minimum variance convince matrix.

Well, what is this positive definite matrix $$S$$ which makes the GMM estimator is inefficient.

Can you give me an example for $$S$$ matrix?

Notice that, assuming no multicollinearity, $$X'X$$ has full rank and hence is invertible. Hence, $$(X'XX'X)^{-1}= (X'X)^{-1}(X'X)^{-1}$$ so that $$\hat{\beta} = (X'X)^{-1}\underbrace{(X'X)^{-1}X'X}_{=I}X'y=(X'X)^{-1}X'y.$$