基于条件概率详解全概率公式的证明

一、前言

在另一篇文章中，「荒原之梦考研数学」通过图解的方式证明了全概率公式，在本文中，「荒原之梦考研数学」将使用传统的证明方法实现对全概率公式的证明：

$$\begin{array}{rl}P\left(A\right)& =\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ P\left(B\right)& =\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\end{array}$$

二、正文

完备事件

若 $B_i$ 为完备事件，则事件 $B_1$, $B_2$, $\cdots$, $B_n$ 的和事件一定是必然事件，即：

$$\sum _{i=1}^n{B}_i=\mathrm{\Omega }$$

若 $B_i$ 为不相容事件，则对于任意的属于 $B_i$ 的事件 $B_{i1}$ 和 $B_{i2}$, 一定有：

$$B_i{B}_j=\varphi$$

证明 $P\left(A\right)$ 的全概率公式

根据 条 件 概 率 ：

首先：

$$\begin{array}{rl}P\left(A\right)P(B_1\mid A)& =P\left(A\right)\frac{P\left(A{B}_1\right)}{P\left(A\right)}=P\left(A{B}_1\right)\\ \\ P\left(A\right)P(B_2\mid A)& =P\left(A\right)\frac{P\left(A{B}_2\right)}{P\left(A\right)}=P\left(A{B}_2\right)\\ \\ & ⋮\\ \\ P\left(A\right)P(B_n\mid A)& =P\left(A\right)\frac{P\left(A{B}_n\right)}{P\left(A\right)}=P\left(A{B}_n\right)\end{array}$$

又因为：

$$\begin{array}{rl}P\left(B_1\right)P(A\mid B_1)& =P\left(B_1\right)\frac{P\left(A{B}_1\right)}{P\left(B_1\right)}=P\left(A{B}_1\right)\\ \\ P\left(B_2\right)P(A\mid B_2)& =P\left(B_2\right)\frac{P\left(A{B}_2\right)}{P\left(B_2\right)}=P\left(A{B}_2\right)\\ \\ & ⋮\\ \\ P\left(B_n\right)P(A\mid B_n)& =P\left(B_n\right)\frac{P\left(A{B}_n\right)}{P\left(B_n\right)}=P\left(A{B}_n\right)\end{array}$$

于是：

$$\begin{array}{rl}P\left(A\right)P(B_1\mid A)& =P\left(B_1\right)P(A\mid B_1)\\ \\ P\left(A\right)P(B_2\mid A)& =P\left(B_2\right)P(A\mid B_2)\\ \\ & ⋮\\ \\ \text{(1)}& P\left(A\right)P(B_n\mid A)& =P\left(B_n\right)P(A\mid B_n)\end{array}$$

对上面的式子 $(1)$ 等号两端同时求和，可得：

$$\begin{array}{rl}& P\left(A\right)P(B_n\mid A)=P\left(B_n\right)P(A\mid B_n)\\ \\ \Rightarrow & \sum _{i=1}^{n}P\left(A\right)P(B_i\mid A)=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)\sum _{i=1}^{n}P(B_i\mid A)=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)\sum _{i=1}^n\frac{P\left(A{B}_i\right)}{P\left(A\right)}=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)\frac{P\left[A\left(\sum _{i=1}^n{B}_i\right)\right]}{P\left(A\right)}=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)\frac{P\left[A\left(\sum _{i=1}^n{B}_i\right)\right]}{P\left(A\right)}=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)\frac{P\left(A\mathrm{\Omega }\right)}{P\left(A\right)}=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)\frac{P\left(A\mathrm{\Omega }\right)}{P\left(A\right)}=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)1=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\\ \\ \Rightarrow & P\left(A\right)=\sum _{i=1}^{n}P\left(B_i\right)P(A\mid B_i)\end{array}$$

证明 $P\left(B\right)$ 的全概率公式

根据 条 件 概 率 ：

首先：

$$\begin{array}{rl}P\left(B\right)P(A_1\mid B)& =P\left(B\right)\frac{P\left(B{A}_1\right)}{P\left(B\right)}=P\left(B{A}_1\right)\\ \\ P\left(B\right)P(A_2\mid B)& =P\left(B\right)\frac{P\left(B{A}_2\right)}{P\left(B\right)}=P\left(B{A}_2\right)\\ \\ & ⋮\\ \\ P\left(B\right)P(A_n\mid B)& =P\left(B\right)\frac{P\left(B{A}_n\right)}{P\left(B\right)}=P\left(B{A}_n\right)\end{array}$$

又因为：

$$\begin{array}{rl}P\left(A_1\right)P(B\mid A_1)& =P\left(A_1\right)\frac{P\left(B{A}_1\right)}{P\left(A_1\right)}=P\left(B{A}_1\right)\\ \\ P\left(A_2\right)P(B\mid A_2)& =P\left(A_2\right)\frac{P\left(B{A}_2\right)}{P\left(A_2\right)}=P\left(B{A}_2\right)\\ \\ & ⋮\\ \\ P\left(A_n\right)P(B\mid A_n)& =P\left(A_n\right)\frac{P\left(B{A}_n\right)}{P\left(A_n\right)}=P\left(B{A}_n\right)\end{array}$$

于是：

$$\begin{array}{rl}P\left(B\right)P(A_1\mid B)& =P\left(A_1\right)P(B\mid A_1)\\ \\ P\left(B\right)P(A_2\mid B)& =P\left(A_2\right)P(B\mid A_2)\\ \\ & ⋮\\ \\ \text{(2)}& P\left(B\right)P(A_n\mid B)& =P\left(A_n\right)P(B\mid A_n)\end{array}$$

对上面的式子 $(2)$ 等号两端同时求和，可得：

$$\begin{array}{rl}& P\left(B\right)P(A_n\mid B)=P\left(A_n\right)P(B\mid A_n)\\ \\ \Rightarrow & \sum _{i=1}^{n}P\left(B\right)P(A_i\mid B)=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)\sum _{i=1}^{n}P(A_i\mid B)=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)\sum _{i=1}^n\frac{P\left(B{A}_i\right)}{P\left(B\right)}=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)\frac{P\left[B\left(\sum _{i=1}^n{A}_i\right)\right]}{P\left(B\right)}=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)\frac{P\left[B\left(\sum _{i=1}^n{A}_i\right)\right]}{P\left(B\right)}=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)\frac{P\left(B\mathrm{\Omega }\right)}{P\left(B\right)}=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)\frac{P\left(B\mathrm{\Omega }\right)}{P\left(B\right)}=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)1=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\\ \\ \Rightarrow & P\left(B\right)=\sum _{i=1}^{n}P\left(A_i\right)P(B\mid A_i)\end{array}$$

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