因果推断笔记1

R
因果推断
笔记
Author

kili

Published

2025-01-02

Tower property

给定随机变量或向量A,B,C我们有:(事实上整本书的推导将大量使用该定律)

\[ \begin{align} E(A) &= E\{E(A|B)\} ,\\ E(A|C) &= E\{E(A|B,C)|C\} \\ \end{align} \]

由条件概率公式\(P(A|B)P(B)=f(A,B),P(A|B,C)P(B|C)=P(A,B|C)\)我们有如下证明: \[ \begin{align*} \text{proof: } E(A) &= \int_{-\infty }^{\infty }Af(A) \,dA \\ &= \int_{-\infty }^{\infty }A\left(\int_{-\infty}^{\infty} f(A,B) \,dB \right) \,dA \\ &= \iint Af(A|B)f(B)dAdB \\ &= \int_{}^{} E(A|B)f(B) \,dB \\ &= E\{E(A|B)\} \\ E(A|C) &= \int_{-\infty }^{\infty }Af(A|C) \,dA \\ &= \int_{-\infty }^{\infty }A\left(\int_{-\infty}^{\infty} f(A,B|C) \,dB \right) \,dA \\ &= \iint Af(A|B,C)f(B|C)dAdB \\ &= \int_{}^{} E(A|B,C)f(B|C) \,dB \\ &= E\{E(A|B,C)|C\} \\ \end{align*} \]

\[ \begin{align} \end{align} \]

给定随机变量A,随机变量或向量B,C,我们有 \[ \begin{align} \end{align} \]

\[ \begin{align} \end{align} \]

\[ \begin{align} \end{align} \]