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：

$$
\begin{aligned}
P \left( A \right) & = \sum_{i=1}^{n} P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
P \left( B \right) & = \sum_{i=1}^{n} P \left( A_{i} \right) P \left( B \mid A_{i} \right)
\end{aligned}
$$

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：

$$
\textcolor{yellow}{
\sum_{i=1}^{n} B_{i} = \Omega
}
$$

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：

$$
\textcolor{yellow}{
B_{i} B_{j} = \phi
}
$$

证明 $P \left( A \right)$ 的全概率公式

根据 条 件 概 率 ：

首先：

$$
\begin{aligned}
\textcolor{gray}{ P \left( A \right) P \left( B_{\textcolor{pink}{1}} \mid A \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( A \right) \frac{P \left( A B_{\textcolor{pink}{1}} \right)}{P \left( A \right)} = P \left( A B_{\textcolor{pink}{1}} \right) } \\ \\
\textcolor{gray}{ P \left( A \right) P \left( B_{\textcolor{pink}{2}} \mid A \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( A \right) \frac{P \left( A B_{\textcolor{pink}{2}} \right)}{P \left( A \right)} = P \left( A B_{\textcolor{pink}{2}} \right) } \\ \\
& \textcolor{gray}{ \vdots } \\ \\
\textcolor{springgreen}{ P \left( A \right) P \left( B_{\textcolor{pink}{n}} \mid A \right) } & = \textcolor{gray}{ P \left( A \right) \frac{P \left( A B_{\textcolor{pink}{n}} \right)}{P \left( A \right)} = } \textcolor{yellow}{ P \left( A B_{\textcolor{pink}{n}} \right) }
\end{aligned}
$$

又因为：

$$
\begin{aligned}
\textcolor{gray}{ P \left( B_{\textcolor{pink}{1}} \right) P \left( A \mid B_{\textcolor{pink}{1}} \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( B_{\textcolor{pink}{1}} \right) \frac{P \left( A B_{\textcolor{pink}{1}} \right)}{P \left( B_{\textcolor{pink}{1}} \right)} = P \left( A B_{\textcolor{pink}{1}} \right) } \\ \\
\textcolor{gray}{ P \left( B_{\textcolor{pink}{2}} \right) P \left( A \mid B_{\textcolor{pink}{2}} \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( B_{\textcolor{pink}{2}} \right) \frac{P \left( A B_{\textcolor{pink}{2}} \right)}{P \left( B_{\textcolor{pink}{2}} \right)} = P \left( A B_{\textcolor{pink}{2}} \right) } \\ \\
& \textcolor{gray}{ \vdots } \\ \\
\textcolor{springgreen}{ P \left( B_{\textcolor{pink}{n}} \right) P \left( A \mid B_{\textcolor{pink}{n}} \right) } & \textcolor{gray}{ = P \left( B_{\textcolor{pink}{n}} \right) \frac{P \left( A B_{\textcolor{pink}{n}} \right)}{P \left( B_{\textcolor{pink}{n}} \right)} = } \textcolor{yellow}{ P \left( A B_{\textcolor{pink}{n}} \right) }
\end{aligned}
$$

于是：

$$
\begin{align}
\textcolor{gray}{P \left( A \right) P \left( B_{\textcolor{pink}{1}} \mid A \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( B_{\textcolor{pink}{1}} \right) P \left( A \mid B_{\textcolor{pink}{1}} \right) } \notag \\ \notag \\
\textcolor{gray}{ P \left( A \right) P \left( B_{\textcolor{pink}{2}} \mid A \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( B_{\textcolor{pink}{2}} \right) P \left( A \mid B_{\textcolor{pink}{2}} \right) } \notag \\ \notag \\
& \textcolor{gray}{ \vdots } \notag \\ \notag \\
\textcolor{springgreen}{ P \left( A \right) P \left( B_{\textcolor{pink}{n}} \mid A \right) } & = \textcolor{springgreen}{ P \left( B_{\textcolor{pink}{n}} \right) P \left( A \mid B_{\textcolor{pink}{n}} \right) } \tag{1}
\end{align}
$$

对上面的式子 $(1)$ 等号两端同时求和，可得：

$$
\begin{aligned}
& P \left( A \right) P \left( B_{n} \mid A \right) = P \left( B_{n} \right) P \left( A \mid B_{n} \right) \\ \\
\Rightarrow \ & \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( A \right) P \left( B_{i} \mid A \right) = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \textcolor{orange}{ \sum_{i = 1}^{n} } \textcolor{tan}{ P \left( B_{i} \mid A \right) } = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \textcolor{orange}{ \sum_{i = 1}^{n} } \textcolor{tan}{ \frac{P \left( A B_{i} \right)}{P \left( A \right)} } = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \textcolor{tan}{ \frac{P \left[ A \left( \textcolor{orange}{ \sum_{i = 1}^{n} } B_{i} \right) \right]}{P \left( A \right)} } = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \frac{P \left[ A \textcolor{red}{ \left( \sum_{i = 1}^{n} B_{i} \right) } \right]}{P \left( A \right)} = \sum_{i = 1}^{n} P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \frac{P \left( A \textcolor{red}{ \Omega } \right)}{P \left( A \right)} = \sum_{i = 1}^{n} P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \textcolor{pink}{ \frac{P \left( A \Omega \right)}{P \left( A \right)} } = \sum_{i = 1}^{n} P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & P \left( A \right) \textcolor{pink}{ 1 } = \sum_{i = 1}^{n} P \left( B_{i} \right) P \left( A \mid B_{i} \right) \\ \\
\Rightarrow \ & \textcolor{springgreen}{ P \left( A \right) = \sum_{i = 1}^{n} P \left( B_{i} \right) P \left( A \mid B_{i} \right) }
\end{aligned}
$$

证明 $P \left( B \right)$ 的全概率公式

根据 条 件 概 率 ：

首先：

$$
\begin{aligned}
\textcolor{gray}{ P \left( B \right) P \left( A_{\textcolor{pink}{1}} \mid B \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( B \right) \frac{P \left( B A_{\textcolor{pink}{1}} \right)}{P \left( B \right)} = P \left( B A_{\textcolor{pink}{1}} \right) } \\ \\
\textcolor{gray}{ P \left( B \right) P \left( A_{\textcolor{pink}{2}} \mid B \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( B \right) \frac{P \left( B A_{\textcolor{pink}{2}} \right)}{P \left( B \right)} = P \left( B A_{\textcolor{pink}{2}} \right) } \\ \\
& \textcolor{gray}{ \vdots } \\ \\
\textcolor{springgreen}{ P \left( B \right) P \left( A_{\textcolor{pink}{n}} \mid B \right) } & \textcolor{gray}{ = P \left( B \right) \frac{P \left( B A_{\textcolor{pink}{n}} \right)}{P \left( B \right)} = } \textcolor{yellow}{ P \left( B A_{\textcolor{pink}{n}} \right) }
\end{aligned}
$$

又因为：

$$
\begin{aligned}
\textcolor{gray}{ P \left( A_{\textcolor{pink}{1}} \right) P \left( B \mid A_{\textcolor{pink}{1}} \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( A_{\textcolor{pink}{1}} \right) \frac{P \left( B A_{\textcolor{pink}{1}} \right)}{P \left( A_{\textcolor{pink}{1}} \right)} = P \left( B A_{\textcolor{pink}{1}} \right) } \\ \\
\textcolor{gray}{ P \left( A_{\textcolor{pink}{2}} \right) P \left( B \mid A_{\textcolor{pink}{2}} \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( A_{\textcolor{pink}{2}} \right) \frac{P \left( B A_{\textcolor{pink}{2}} \right)}{P \left( A_{\textcolor{pink}{2}} \right)} = P \left( B A_{\textcolor{pink}{2}} \right) } \\ \\
& \textcolor{gray}{ \vdots } \\ \\
\textcolor{springgreen}{ P \left( A_{\textcolor{pink}{n}} \right) P \left( B \mid A_{\textcolor{pink}{n}} \right) } & \textcolor{gray}{ = P \left( A_{\textcolor{pink}{n}} \right) \frac{P \left( B A_{\textcolor{pink}{n}} \right)}{P \left( A_{\textcolor{pink}{n}} \right)} = } \textcolor{yellow}{ P \left( B A_{\textcolor{pink}{n}} \right) }
\end{aligned}
$$

于是：

$$
\begin{align}
\textcolor{gray}{P \left( B \right) P \left( A_{\textcolor{pink}{1}} \mid B \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( A_{\textcolor{pink}{1}} \right) P \left( B \mid A_{\textcolor{pink}{1}} \right) } \notag \\ \notag \\
\textcolor{gray}{ P \left( B \right) P \left( A_{\textcolor{pink}{2}} \mid B \right) } & \textcolor{gray}{=} \textcolor{gray}{ P \left( A_{\textcolor{pink}{2}} \right) P \left( B \mid A_{\textcolor{pink}{2}} \right) } \notag \\ \notag \\
& \textcolor{gray}{ \vdots } \notag \\ \notag \\
\textcolor{springgreen}{ P \left( B \right) P \left( A_{\textcolor{pink}{n}} \mid B \right) } & = \textcolor{springgreen}{ P \left( A_{\textcolor{pink}{n}} \right) P \left( B \mid A_{\textcolor{pink}{n}} \right) } \tag{2}
\end{align}
$$

对上面的式子 $(2)$ 等号两端同时求和，可得：

$$
\begin{aligned}
& P \left( B \right) P \left( A_{n} \mid B \right) = P \left( A_{n} \right) P \left( B \mid A_{n} \right) \\ \\
\Rightarrow \ & \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( B \right) P \left( A_{i} \mid B \right) = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \textcolor{orange}{ \sum_{i = 1}^{n} } \textcolor{tan}{ P \left( A_{i} \mid B \right) } = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \textcolor{orange}{ \sum_{i = 1}^{n} } \textcolor{tan}{ \frac{P \left( B A_{i} \right)}{P \left( B \right)} } = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \textcolor{tan}{ \frac{P \left[ B \left( \textcolor{orange}{ \sum_{i = 1}^{n} } A_{i} \right) \right]}{P \left( B \right)} } = \textcolor{orange}{ \sum_{i = 1}^{n} } P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \frac{P \left[ B \textcolor{red}{ \left( \sum_{i = 1}^{n} A_{i} \right) } \right]}{P \left( B \right)} = \sum_{i = 1}^{n} P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \frac{P \left( B \textcolor{red}{ \Omega } \right)}{P \left( B \right)} = \sum_{i = 1}^{n} P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \textcolor{pink}{ \frac{P \left( B \Omega \right)}{P \left( B \right)} } = \sum_{i = 1}^{n} P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & P \left( B \right) \textcolor{pink}{ 1 } = \sum_{i = 1}^{n} P \left( A_{i} \right) P \left( B \mid A_{i} \right) \\ \\
\Rightarrow \ & \textcolor{springgreen}{ P \left( B \right) = \sum_{i = 1}^{n} P \left( A_{i} \right) P \left( B \mid A_{i} \right) }
\end{aligned}
$$

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