2. The equation of interest is yt = m(Xt, β) + et with E[Zt...

2. The equation of interest is $y_t = m(X_t, \beta) + e_t$ with $E[Z_t e_t] = 0$, where $m(x, \beta)$ is a known nonlinear function, $\beta^T = (\beta_1^T, \beta_2^T)$ is a $k \times 1$ vector of unknown parameters and $Z$ is an $l \times 1$ vector of instrumental variables. (a) Show all steps on how to construct the GMM estimator for $\beta$, denoted by $\hat{\beta}_{gmm}$. (b) Show that $\hat{\beta}_{gmm}$ is efficient in the GMM sense. (c) By considering testing $H_0 : \beta_2 = 0$, where $\beta_2$ is a vector of $k_2 \times 1$ parameters, please give all steps on using Langrange Multiplier (LM) or Rao test to this testing problem.

Transcript

00:01 Hello students in this question given that xt is equal to x1, t x2, t so on up to xn, t and also given yt so that is equal to y1, t y2, t so on up to yn, t.

00:25 So linear regression model is given by yi is equal to xi beta plus epsilon i so in matrix notation y is equal to x beta plus epsilon and sb is the sum of squared residual so that is summation i running from 1 to n yi minus xi t t whole square so here also we have xi t so this can be written as y minus xb power t into y minus x into b so the value of b the value of b which minimizes this sum is called ols estimator for beta so the function s of b is quadratic in b with positive definition definite and therefore this function possesses a unique value minimum at b is equal to beta the minimum can be calculated using matrix calculation so here o is equal to ds divided by db of beta cap so that is t divided by db of yty minus bt xt y minus yt xb plus bt xt into xb given b is equal to beta cap so now differentiating will be getting minus 2 xt y plus 2 xt x beta cap so using the matrix calculus matrix s has full column matrix full column rank therefore xt x invertible and the least square estimator for beta beta is given by least square estimator for beta is given by beta cap is equal to xt x whole inverse xt y so x equal to xt is equal to x 1 comma t so on up to xn comma t and next year x 1 comma t so on up to xn comma t so now multiplying these two x 1 dash t x 1 comma t plus x 2 dash t comma x 2 comma t plus so on up to x n dash comma t into x n comma t so that can be written as summation s running from 1 to n xs of t dash into xs of t similarly x dash y so that is x 1 dash t x 2 dash t so on up to x n dash t into here y 1 t so on up to y n comma t so multiplying these two similarly like above will be getting summation s from 1 to n x dash s comma t into ys comma t so now combining the terms beta cap is equal to summation s running from 1 to n xs dash comma t into xs comma t whole inverse into summation xs comma t dash ys comma t...

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