Question on 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, $y_t$ is an iid disturbance term with the variance $\sigma^2$ and $E(x_tu_t)=0$ for all t.
Suppose that the linear regression satisfies the following restrictions:
$$\beta_1 + \beta_2 =0$$
$$2 \beta_4 - \beta_3 =0$$
Find a consistent and efficient GMM estimator for $\beta$.
My solution attempt is very trivial. Thus, I didn't write it down. Please shear your ideas about this question with me. Thank you

The restrictions you gave are linear and so we will find that this GMM specification is equivalent to restricted OLS as is most natural. We can express the restrictions as,
$$R\beta = r$$
Where $r = \begin{bmatrix}0\\0\end{bmatrix}$ and $R=\begin{bmatrix}1 & 1 & 0 & 0\\ 0 & 0 & -1 & 2 \end{bmatrix}$.
Since this will be an overidentified GMM we will need to estimate the weight matrix,
$$\hat\Sigma = \sigma^2 n^{-1} X^TX$$
Next let us consider the optimization problem,
$$\min_\beta [\frac{1}{n}\sum_i(Y_i-\beta X_i)]^T\hat\Sigma^{-1}[\frac{1}{n}\sum_i(Y_i-\beta X_i)] \text{ s.t. } R\beta = \boldsymbol{0}$$
Yielding the Langrangian problem,
$$\min_\beta [\frac{1}{n}\sum_i(Y_i-\beta X_i)]^T\hat\Sigma^{-1}[\frac{1}{n}\sum_i(Y_i-\beta X_i)] + \lambda R\beta$$
However, since this is OLS and we know what our choice of $\hat\Sigma$ is we can expand the criterion to,
$$Q(\beta)=\frac{1}{2n\sigma^2}(y-X\beta)^TX(X^TX)^{-1}X^T(y-X\beta)=\frac{1}{2n\sigma^2}(y-X\beta)^T(y-X\beta)$$
Which will yields the exact same FOCs as the usual restricted OLS estimator.
So we let us just derive the estimator for restricted OLS (dropping the constant up front right now for notation). We have the Lagrangian,
$$\mathcal{L}(\beta,\lambda)=(y-X\beta)^T(y-X\beta) + \lambda R\beta=y^Ty-2y^Tx\beta+\beta^TX^TX\beta + \lambda\beta R$$
This yields the FOCs,
$$\frac{\partial \mathcal{L}}{\partial \hat\beta} = -2X^Ty+2\hat\beta X^TX + \lambda R$$
$$\frac{\partial \mathcal{L}}{\partial \lambda} = R\beta$$
So setting equal to zero we have the system,
$$2\hat\beta X^TX + \lambda R=2X^Ty$$
$$R\beta = 0$$
Solving for $\hat\beta$ in the first equation yields,
$$\hat\beta = (X^TX)^{-1}X^Ty - \frac{1}{2}(X^TX)^{-1}R^T\lambda$$
Also multiplying both sides by $R$ yields,
$$R\hat\beta = 0 = R\hat\beta_{OLS} - \frac{1}{2}R(X^TX)^{-1}R^T\lambda$$
So,
$$\lambda = R\hat\beta_{OLS}(\frac{1}{2}R(X^TX)^{-1}R^T)^{-1}$$
Thus we have determined our estimator for the restricted OLS (GMM),
$$\hat\beta = \hat\beta_{OLS}-(X^TX)^{-1}R^T(\frac{1}{2}R(X^TX)^{-1}R^T)^{-1}R\hat\beta_{OLS}$$
Now let's consider consistency and efficiency (these properties are actually immediate from our work with the GMM version of the estimator but it is useful to see that they hold).
Consistency:
Note that from the assumption $\hat\beta_{OLS}$ is consistent. So if the restriction is correctly specified so that $R\hat\beta_{OLS}=0$ then this zeros out the restricted least squares adjustment so that,
$$\hat\beta = \hat\beta_{OLS} \to \beta$$
Efficiency:
Consider that $Var(\hat\beta_{OLS})=\sigma^2 (X^TX)^{-1}$. Thus, we know that $Var(\hat\beta)=\sigma^2 L (X^TX)^{-1}$ for $L=I-(X^TX)^{-1}R^T(\frac{1}{2}R(X^TX)^{-1}R^T)^{-1}R)$. Thus we have, $Var(\hat\beta)\leq Var(\hat\beta_{OLS})$. Yielding efficiency. We can also justy skip this computation becasue we are using an optimal weighting matrix with known variance.
