广义矩估计GMM（Generalized Method of Moments）

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.title[
# 广义矩估计GMM（Generalized Method of Moments）
]
.author[
### 文旷宇
]
.institute[
### 华中科技大学
]

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---

### 主要内容

- 矩条件

- 2SLS

- 有效GMM

- GMM框架下的假设检验

---

###矩条件

- 工具变量法要寻找一个工具变量 `$Z_i$`与 `$X_i$`高度相关而与 `$e_i$`无关，则有 `$E[Z_ie_i]=0 \Rightarrow E[Z_i(Y_i-X_i'\beta)]=0$`

- 对于GMM，有 `$E[g(w_i,\beta)]=0$`，这里 `$g:R^p\times R^k\rightarrow R^l$`（ `$p$` 个数据， `$k$` 个参数， `$l$` 个矩条件）

- `$E[g(w_i,\beta)]=0,\forall \beta'\not=\beta,E[g(w_i,\beta')]\not=0$`，我们就说模型是可识别的；但如果多个参数都能使这个成立，那么矩条件就不能被识别。

- `$l=k$`时为恰好识别， `$\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\hat{\beta}_{MM})=0$`
 
 - `$l>k$`时为过度识别， `$\hat{\beta}_{GMM}=\arg\min\limits_{\beta}\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]'W_n\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]$`

---

###广义矩估计

- 证明估计量的一致性

- 由大数定律可知，
 
 `$\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]'W_n\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]\rightarrow^p_{n\rightarrow \infty} \Big[E[g(w_i,\beta)]\Big]'W\Big[E[g(w_i,\beta)]\Big]$`
 
 - 根据定义， `$\hat{\beta}_{GMM}$`最小化了 `$\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]'W_n\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]$`，因此如果我们想要证明 `$\hat{\beta}_{GMM}\to\beta$`，只需要证明 `$\beta$`最小化了 `$\Big[E[g(w_i,\beta)]\Big]'W\Big[E[g(w_i,\beta)]\Big]$`。
 
---

###广义矩估计（续）

- 一阶条件为 `$\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\beta)}{\partial \beta}\Big]'W_n\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)\Big]=0$`，

- 即 `$\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\Big]'W_n\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\hat{\beta}_{GMM})\Big]=0$`

- 泰勒展开，在 `$\beta$` 附近对 `$\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\hat{\beta}_{GMM})$`进行泰勒展开，

`$$\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\hat{\beta}_{GMM})=\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)+(\frac{1}{n}\sum^n_{i=1}\frac{\partial g(\omega_i,\beta^{\ast})}{\partial\beta})(\hat{\beta}_{GMM}-\beta)$$`

- 此时一阶条件变为

`$$\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\Big]'W_n\Big[\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)+(\frac{1}{n}\sum^n_{i=1}\frac{\partial g(\omega_i,\beta^{\ast})}{\partial\beta})(\hat{\beta}_{GMM}-\beta)\Big]=0$$`

---

###广义矩估计（续）

`$$\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\Big]'W_n(\frac{1}{n}\sum^n_{i=1}\frac{\partial g(\omega_i,\beta^{\ast})}{\partial\beta})(\hat{\beta}_{GMM}-\beta)\\=-\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\Big]'W_n\frac{1}{n}\sum\limits_{i=1}^ng(w_i,\beta)$$`
因此， `$\sqrt{n}(\hat{\beta}_{GMM}-\beta)$`可由下式表示：

`$$\small-\Bigg[\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\Big]'W_n(\frac{1}{n}\sum^n_{i=1}\frac{\partial g(\omega_i,\beta^{\ast})}{\partial\beta})\Bigg]^{-1}\Big[\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\Big]'W_n\frac{1}{\sqrt{n}}\sum\limits_{i=1}^n g(w_i,\beta)$$`

令 `$Q=E[\frac{\partial g(\omega_i,\beta)}{\partial \beta}]$`， `$\Sigma=E[g(\omega_i,\beta)g(\omega_i,\beta)']$`

---

###广义矩估计（续）

`$\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\hat{\beta}_{GMM})}{\partial \beta}\to_pQ$`, `$W_n\to_pW$`

`$\frac{1}{n}\sum\limits_{i=1}^n \frac{\partial g(w_i,\beta^{\ast})}{\partial \beta}\to_pQ$`, `$\frac{1}{\sqrt{n}}\sum^n_{i=1}g(\omega_i,\beta)\to_dN(0,\Sigma)$`

因此， `$\sqrt{n}(\hat{\beta}_{GMM}-\beta)\to_d -\big[Q'WQ\big]^{-1}Q'WN(0,\Sigma)$`

`$$\sqrt{n}(\hat{\beta}_{GMM}-\beta)\to_d N(0,\big[Q'WQ\big]^{-1}Q'W\Sigma WQ\big[Q'WQ\big]^{-1}$$`

---

###2SLS

- 矩条件为 `$g(\omega_i,\beta)=Z_i(Y_i-X_i'\beta)$`，则 `$\frac{\partial g(\omega_i,\beta)}{\partial \beta}=-Z_iX_i'$`

- `$\hat{\beta}_{GMM}=arg min\Big[\frac{1}{n} \sum\limits^n_{i=1}Z_i(Y_i-X_i'\beta)\Big]'W_n\Big[\frac{1}{n} \sum\limits^n_{i=1}Z_i(Y_i-X_i'\beta)\Big]$`

一阶条件为 `$\big[-\frac{1}{n}\sum\limits^n_{i=1}Z_iX_i'\big]'W_n\Big[\frac{1}{n} \sum\limits^n_{i=1}Z_i(Y_i-X_i'\beta)\Big]=0$`

可由此推出 `$\hat{\beta}_{GMMIV}$`的表达式：

`$$\small \begin{aligned}\hat{\beta}_{GMMIV}&=\Bigg[(\frac{1}{n}\sum\limits^n_{i=1}X_iZ_i')W_n(\frac{1}{n}\sum\limits^n_{i=1}Z_iX_i')\Bigg]^{-1}(\frac{1}{n}\sum\limits^n_{i=1}X_iZ_i')W_n(\frac{1}{n}\sum\limits^n_{i=1}Z_iY_i)\\&=\Big[X'ZW_nZ'X\Big]^{-1}(X'Z)W_n(Z'Y)\end{aligned}$$`

当 `$W_n=(Z'Z)^{-1}$`时，  `$\hat{\beta}_{GMMIV}=\hat{\beta}_{2SLS}$`

---

###有效GMM

有效GMM的两步法：

- `$\hat{\Sigma}=\frac{1}{n}\sum\limits_{i=1}^n g(w_i,\hat{\beta})g(w_i,\hat{\beta})'$`，其中 `$\hat{\beta}$`是 `$\beta$`的一致估计量。特别地， `$\hat{\beta}=\arg \min \Big[\frac{1}{n}\sum\limits_{i=1}^n g(w_i,\beta)\Big]'\Big[\frac{1}{n}\sum\limits_{i=1}^n g(w_i,\beta)\Big]$`，

- 令 `$W_n=\hat{\Sigma}^{-1}$`， `$\hat{\beta}_{EGMM}=\arg \min \Big[\frac{1}{n}\sum\limits_{i=1}^n g(w_i,\beta)\Big]'\hat{\Sigma}^{-1}\Big[\frac{1}{n}\sum\limits_{i=1}^n g(w_i,\beta)\Big]$`

- `$\sqrt{n}(\hat{\beta}_{EGMM}-\beta)\to_d N(0,[Q'\Sigma^{-1}Q]^{-1})$`

- 由 `$g(\omega_i,\beta)=Z_i(Y_i-X_i'\beta)=Z_ie_i$` 可推出

`$\Sigma=E\big[g(\omega_i,\beta)g(\omega_i,\beta)'\big]=E\big[Z_ie_i^2Z_i'\big]=E\big[e_i^2Z_iZ_i'\big]$`

---

###有效GMM（续）

- 假定 `$E\big[e_i^2|Z_i\big]=\sigma^2$`

- `$\big[E[e_i^2Z_iZ_i']\big]^{-1}=E\Big[E[e_i^2Z_iZ_i'|Z_i]\Big]^{-1}=E[\sigma^2Z_iZ_i']^{-1}=\sigma^2E[Z_iZ_i']^{-1}$`
 
- 两阶段最小二乘法是GMM的一个特例， `$W_n=\Big[\frac{1}{n}\sum\limits^n_{i=1}Z_iZ_i'\Big]^{-1}$`，且 `$W_n\to_P W=\big[E[Z_iZ_i']\big]^{-1}$`。

- 有效GMM的权重矩阵为 `$W_n$`，且有 `$W_n\to_P W=\big[E[e_i^2Z_iZ_i']\big]^{-1}$`，特别地，同方差假定下（ `$E[e_i^2|Z_i]=\sigma^2$`）的2SLS就是最有效的GMM。

---

###假设检验

- `$H_0 :h(\beta)=0$`； `$H_1 :h(\beta)\not=0$`，其中 `$h(\cdot):R^k\to R^q$` 在 `$\beta$`的某个开区间内连续可导。

- `$\sqrt{n}(h(\hat{\beta}_{GMM})-h(\beta))\to_dN(0,\frac{\partial h(\beta)}{\partial \beta}V(\frac{\partial h(\beta)}{\partial \beta})')$`

- `$h(\hat{\beta}_{GMM})\approx N(0,\frac{1}{n}\frac{\partial h(\beta)}{\partial \beta}V\frac{\partial h(\beta)}{\partial \beta'})$`

- `$h(\hat{\beta}_{GMM})\Big[\frac{1}{n}\frac{\partial h(\beta)}{\partial \beta}V\frac{\partial h(\beta)}{\partial \beta'}\Big]^{-1}h(\hat{\beta}_{GMM})\to _d\chi^2_q$`

- 用 `$\hat{V}$`来估计 `$V$`，则有 `$h(\hat{\beta}_{GMM})\Big[\frac{1}{n}\frac{\partial h(\beta)}{\partial \beta}\hat{V}\frac{\partial h(\beta)}{\partial \beta'}\Big]^{-1}h(\hat{\beta}_{GMM})\to _d\chi^2_q$`

- 因此，Wald统计量 `$nh(\hat{\beta}_{GMM})\Big[\frac{\partial h(\beta)}{\partial \beta}\hat{V}\frac{\partial h(\beta)}{\partial \beta'}\Big]^{-1}h(\hat{\beta}_{GMM})\to _d\chi^2_q$`