# Is this correct for (generalised) method of moments?

I am reviewing a MOOC and realised... I didn't grasp the method of moments, so I did try to get it back "from scratch".

Are the above equations/sentences correct ? In the MOOC the analysis is made on the mapping between $$\theta$$ and the moments, and then using the inverse of this function, but I find it confusing and prefer to think directly in terms of the moments $$\mapsto \theta$$ mapping. $$\newcommand\array{\begin{bmatrix}#1\end{bmatrix}}$$

### My interpretation of (generalised) method of moments

Case #1: the estimator can be expressed as the expected value of a function of X

• $$\theta = E[g(X)]$$, $$G = g(X)$$, $$\bar G_n = \frac{1}{n} \sum_{i=1}^n g(x_i)$$
• by LLN applied on $$g(X)$$ we have: $$\hat \theta_n := \bar G_n \xrightarrow{(p/d)} \theta := E[G]$$
• by CLT applied to $$g(X)$$ we have: $$\hat \theta_n := \bar G_n \xrightarrow{(d)} \sim N(E[G],\frac{var(G)}{n})$$

Case #2: the estimator can be expressed as a function of the expected value of a function of X

• $$\theta = h(E[g(X)])$$, $$G = g(X)$$, $$\bar G_n = \frac{1}{n} \sum_{i=1}^n g(x_i)$$
• by LLN applied on $$g(X)$$ and the CMT we have: $$\hat \theta_n := h(\bar G_n) \xrightarrow{(p/d)} \theta := h(E[G])$$
• by CLT applied to $$g(X)$$ and the delta method we have: $$\hat \theta_n := h(\bar G_n) \xrightarrow{(d)} \sim N \left(h(E[G]),\frac{\left( \frac{dh}{dE[G]} \right)^2 var(G)}{n}\right)$$

Case #3: the estimators can be expressed as D functions of K multiple expected values of functions of X

• $$\theta = \{h_1(E[g_1(X)],E[g_2(X)],...,E[g_K(X)]), h_2(E[g_1(X)],E[g_2(X)],...,E[g_K(X)]),...,h_D(\cdot)\}$$ with $$1 \leq K \leq D$$
• $$G_k = g_k(X)$$, $$\bar G_{k,n} = \frac{1}{n} \sum_{i=1}^n g_k(x_i)$$, $$J(H)$$ is the $$D \times K$$ matrix of the $$\frac{dh_d}{dE[g_d(X)]}$$ derivatives, and $$Cov(G)$$ is the $$D \times D$$ covariance matrix between the $$g_d(x)$$ random variables
• by LLN applied on $$g_k(X)$$ and the CMT we have: $$\hat \theta_n := \begin{bmatrix} h_1(\bar G_{1,n}, \bar G_{2,n}, ..., \bar G_{K,n}) \\ h_2(\bar G_{1,n}, \bar G_{2,n}, ..., \bar G_{K,n}) \\ ... \\ h_D(\bar G_{1,n}, \bar G_{2,n}, ..., \bar G_{K,n}) \\ \end{bmatrix} \xrightarrow{(p/d)} \theta := \begin{bmatrix} h_1(E[G_1], E[G_2], ..., E[G_K]) \\ h_2(E[G_1], E[G_2], ..., E[G_K]) \\ ... \\ h_D(E[G_1], E[G_2], ..., E[G_K]) \\ \end{bmatrix}$$
• by CLT applied to $$g(X)$$ and the delta method we have: $$\hat \theta_n := \begin{bmatrix} h_1(\bar G_{1,n}, \bar G_{2,n}, ..., \bar G_{K,n}) \\ h_2(\bar G_{1,n}, \bar G_{2,n}, ..., \bar G_{K,n}) \\ ... \\ h_D(\bar G_{1,n}, \bar G_{2,n}, ..., \bar G_{K,n}) \\ \end{bmatrix} \xrightarrow{(d)} \sim N_D \left( \begin{bmatrix} h_1(E[G_1], E[G_2], ..., E[G_K]) \\ h_2(E[G_1], E[G_2], ..., E[G_K]) \\ ... \\ h_D(E[G_1], E[G_2], ..., E[G_K]) \\ \end{bmatrix} ,\frac{ J(H) ~Cov(G) ~(J(H))' }{n}\right)$$