# 中介效应：有序因果中介分析的半参数估计A-理论

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New！ `lianxh` 命令发布了：

`. ssc install lianxh`

`. help lianxh`

⛳ Stata 系列推文：

Source: Xiang Zhou, 2021, Semiparametric estimation for causal mediation analysis with multiple causally ordered mediators. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 84(3), 794– 821. -Link--Data-

## 2. 符号、定义和识别

### 2.1 两个因果有序中介的情况

#### 2.1.1 定义PSEs

(a) $A$ $\to$ $Y$; (b) $A$ $\to$ ${M}_{2}$ $\to$ $Y$; (c) $A$ $\to$ ${M}_{1}$ $\to$ $Y$; (d) $A$ $\to$ ${M}_{1}$ $\to$ ${M}_{2}$ $\to$ $Y$

${\tau }_{A\to Y}\left({a}_{1},{a}_{2},{a}_{12}\right)$ $=$ $\mathbb{E}\left[Y\left(1,{M}_{1}\left({a}_{1}\right),{M}_{2}\left({a}_{2},{M}_{1}\left({a}_{12}\right)\right)\right)-Y\left(0,{M}_{1}\left({a}_{1}\right),{M}_{2}\left({a}_{2},{M}_{1}\left({a}_{12}\right)\right)\right)\right]$

${\tau }_{A\to {M}_{2}\to Y}\left(a,{a}_{1},{a}_{12}\right)$ $=$ $\mathbb{E}\left[Y\left(a,{M}_{1}\left({a}_{1}\right),{M}_{2}\left(1,{M}_{1}\left({a}_{12}\right)\right)\right)-Y\left(a,{M}_{1}\left({a}_{1}\right),{M}_{2}\left(0,{M}_{1}\left({a}_{12}\right)\right)\right)\right]$

${\tau }_{A\to {M}_{1}\to Y}\left(a,{a}_{1},{a}_{12}\right)$ $=$ $\mathbb{E}\left[Y\left(a,{M}_{1}\left(1\right),{M}_{2}\left({a}_{2},{M}_{1}\left({a}_{12}\right)\right)\right)-Y\left(a,{M}_{1}\left(0\right),{M}_{2}\left({a}_{2},{M}_{1}\left({a}_{12}\right)\right)\right)\right]$

${\tau }_{A\to {M}_{1}\to {M}_{2}\to Y}\left(a,{a}_{1},{a}_{2}\right)$ $=$ $\mathbb{E}\left[Y\left(a,{M}_{1}\left({a}_{1}\right),{M}_{2}\left({a}_{2},{M}_{1}\left(1\right)\right)\right)-Y\left(a,{M}_{1}\left({a}_{1}\right),{M}_{2}\left({a}_{2},{M}_{1}\left(0\right)\right)\right)\right]$

${\tau }_{A\to {M}_{1}⇝Y}\left(a,{a}_{2}\right)$ $=$ $\mathbb{E}\left[Y\left(a,{M}_{1}\left(1\right),{M}_{2}\left({a}_{2},{M}_{1}\left(1\right)\right)\right)-Y\left(a,{M}_{1}\left(0\right),{M}_{2}\left({a}_{2},{M}_{1}\left(0\right)\right)\right)\right]$

#### 2.1.2 识别PSEs

• 对任一单位及任何 $a,{m}_{1},{m}_{2}$ , 若 $A=a$ 则 ${M}_{1}={M}_{1}\left(a\right)$ ; 若 $A=a$ 且 ${M}_{1}={m}_{1}$ 则 ${M}_{2}={M}_{2}\left(a\right)$ ; 若 $A=a$${M}_{1}={m}_{1}$ 且 ${M}_{2}={m}_{2}$ 则 $Y=Y\left(a,{m}_{1},{m}_{2}\right)$

• 对任何 $a,{a}_{1},{a}_{2},{m}_{1},{m}_{1}^{\ast },{m}_{2}$, 有 $\left({M}_{1}\left({a}_{1}\right),{M}_{2}\left({a}_{2},{m}_{1}\right),Y\left(a,{m}_{1},{m}_{2}\right)\right)\perp \phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\perp A|X$ ;  $\left({M}_{2}\left({a}_{2},{m}_{1}\right),Y\left(a,{m}_{1},{m}_{2}\right)\right)\perp \phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\perp {M}_{1}\left({a}_{1}\right)|X,A$$Y\left(a,{m}_{1},{m}_{2}\right)\perp \phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\perp {M}_{2}\left({a}_{2},{M}_{1}^{\ast }\right)|X,A,{M}_{1}$

• 只要 ${p}_{X}\left(x\right)>0$ , 则 ${p}_{A|X}\left(a|x\right)>\epsilon >0$ ; 只要 ${p}_{X,{M}_{1}}\left(x,{m}_{1}\right)>0$ , 则 ${p}_{A|X,{M}_{1}}\left(a|x,{m}_{1}\right)>\epsilon >0$ ; 只要 ${p}_{X,{M}_{1},{M}_{2}}\left(x,{m}_{1},{m}_{2}\right)>0$ , 则 ${p}_{A|X,{M}_{1},{M}_{2}}\left(a|x,{m}_{1},{m}_{2}\right)>\epsilon >0$ 。此处 $p\left(\cdot \right)$ 表示概率密度函数。