因果推断笔记1

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)

$$\begin{array}{rl}E(A)& =E\{E(A|B)\},\\ E(A|C)& =E\{E(A|B,C)|C\}\end{array}$$

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: $$\begin{array}{rl}\text{proof: }E(A)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}Af(A)dA\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}A({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(A,B)dB)dA\\ & =\iint Af(A|B)f(B)dAdB\\ & ={\int }_{}^{}E(A|B)f(B)dB\\ & =E\{E(A|B)\}\\ E(A|C)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}Af(A|C)dA\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}A({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(A,B|C)dB)dA\\ & =\iint Af(A|B,C)f(B|C)dAdB\\ & ={\int }_{}^{}E(A|B,C)f(B|C)dB\\ & =E\{E(A|B,C)|C\}\end{array}$$

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给定随机变量A,随机变量或向量B,C,我们有 $$\begin{array}{}\end{array}$$

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