Dempster证据合成公式 证据组合规则

Yager 合成公式

$\frac{1}{1-k}$ ，并将冲突系数 $k$ 分配给了辨识框架 $\Omega$ 对应的基本概率分配函数 $m(\Omega)$。公式定义如下： $k=\sum_{A_1\cap A_2\cap A_3\cdots=\emptyset}m_1(A_1)m_2(A_2)m_3(A_3)\cdots$ $\begin{aligned} m(\emptyset)&=0 \\ \\ m(A)&=\sum_{A_1\cap A_2\cap A_3\cdots=A}m_1(A_1)m_2(A_2)m_3(A_3)\cdots,\quad A \neq\emptyset,\Omega \\ \\ m(\Omega)&=\sum_{A_1\cap A_2\cap A_3\cdots=\Omega}m_1(A_1)m_2(A_2)m_3(A_3)\cdots+k \\ \end{aligned}$

Example

$\begin{aligned} &\Omega=\{A,B,C\} &\quad &\quad \\ &m_1:m_1(\{A\})=0.98,& m_1(\{B\})=0.01,\quad & m_1(\{C\})=0.01 \\ &m_2:m_2(\{A\})=0, & m_2(\{B\})=0.01,\quad & m_2(\{C\})=0.99 \\ &m_3:m_3(\{A\})=0.9, & m_3(\{B\})=0, \:\qquad&m_3(\{C\})=0.1 \\ \end{aligned}$ $\begin{aligned} k&=1-[m_1(\{A\})m_2(\{A\})m_3(\{A\})+m_1(\{B\})m_2(\{B\})m_3(\{B\})+m_1(\{C\})m_2(\{C\})m_3(\{C\})] \\ &=1-[0.98\times0\times0.9+0.01\times0.01\times0+0.01\times0.99\times0.1] \\ &=0.99901 \\ m(\{A\})&=m_1(\{A\})m_2(\{A\})m_3(\{A\}) \\ &=0.98\times0\times0.9 \\ &=0 \\ m(\{B\})&=0 \\ m(\{C\})&=m_1(\{C\})m_2(\{C\})m_3(\{C\}) \\ &=0.01\times0.99\times0.1 \\ &=0.00099 \\ m(\Omega)&=0+k=0.99901 \\ \end{aligned}$

孙全提出的合成公式

$m_1,m_2,\cdots,m_n$。设证据源 $i$ 和 $j$ 之间的冲突系数为 $k_{ij}$：$k_{ij}=\sum_{A_1\cap A_2=\emptyset}m_i(A_1)m_j(A_2)=1-\sum_{A_1\cap A_2\neq\emptyset}m_i(A_1)m_j(A_2)$ $\varepsilon$ 为证据的可信度：$\varepsilon=e^{-\hat{k}}, \quad \hat{k}=\frac{2}{n(n-1)}\sum_{i<j}k_{ij}$ 公式定义如下：$m(\emptyset)=0$ $m(A)=p(A)+k\times\varepsilon\times q(A),\quad A \neq\emptyset,\Omega$ $m(\Omega)=p(\Omega)+k\times\varepsilon\times q(\Omega)+k(1-\varepsilon)$ $p(A)=\sum_{A_1\cap A_2\cap\cdots\cap A_n=A}m_1(A_1)m_2(A_2)\cdots m_n(A_n)$ $q(A)=\frac{1}{n}\sum_{i=1}^{n}m_i(A)$ $m(A)$ 又可写成如下形式：$m(A)=(1-k)\frac{p(A)}{1-k}+k\times\varepsilon\times q(A)$ 式中第一项的 $\frac{p(A)}{1-k}$

Example

（本示例采用前面 Yager 合成公式示例给出的数据）
$\begin{aligned} k_{12}&=\sum_{A_1\cap A_2=\emptyset}m_1(A_1)m_2(A_2) \\ &=m_1(\{A\})m_2(\{B\})+m_1(\{A\})m_2(\{C\})+m_1(\{B\})m_2(\{A\})+m_1(\{B\})m_2(\{C\})+m_1(\{C\})m_2(\{A\})+m_1(\{C\})m_2(\{B\}) \\ &=0.99 \\ k_{13}&=1-\sum_{A_1\cap A_3\neq\emptyset}m_1(A_1)m_3(A_3) \\ &=1-[m_1(\{A\})m_3(\{A\})+m_1(\{B\})m_3(\{B\})+m_1(\{C\})m_3(\{C\})] \\ &=0.117 \\ k_{23}&=0.901 \\ \hat{k}&=\frac{2}{n(n-1)}\sum_{i<j}k_{ij} \\ &=\frac{2}{3\times(3-1)}(0.99+0.117+0.901) \\ &=0.6693 \\ \varepsilon&=e^{-\hat{k}}=0.5120 \\ m(\{A\})&=(1-k)\frac{p(\{A\})}{1-k}+k\times\varepsilon\times q(\{A\}) \\ &=0+0.99901\times0.5120\times\frac{1}{3}(0.98+0+0.9) \\ &=0.3205 \\ m(\{B\})&=(1-k)\frac{p(\{B\})}{1-k}+k\times\varepsilon\times q(\{B\}) \\ &=0+0.99901\times0.5120\times\frac{1}{3}(0.01+0.01+0) \\ &=0.0034 \\ m(\{C\})&=(1-k)\frac{p(\{C\})}{1-k}+k\times\varepsilon\times q(\{C\}) \\ &=(1-0.99901)\times1+0.99901\times0.5120\times\frac{1}{3}(0.01+0.99+0.1) \\ &=0.1885 \\ m(\Omega)&=(1-k)\frac{p(\Omega)}{1-k}+k\times\varepsilon\times q(\Omega)+k(1-\varepsilon) \\ &=0+0+0.99901\times(1-0.5120) \\ &=0.4875 \\ \end{aligned}$

Smets 合成公式

$k$ 保留不用，而不用于归一化。（论文[2]中用 $m(\emptyset)$ 表示 $k$，根据论文中给的示例来看 $m(\emptyset)$ 不表示空集的 bpa，空集的 bpa 依然是0，论文中的 $\emptyset$ 被解释为一个或几个假设没有被考虑在最初的辨识框架中。但是这样合成后的 bpa 累加和会小于1。）公式定义如下：$\begin{aligned} m(A)&=\sum_{A_1\cap A_2\cap A_3\cdots=A}m_1(A_1)m_2(A_2)m_3(A_3)\cdots,\quad A \neq\emptyset \\ \\ k&=\sum_{A_1\cap A_2\cap A_3\cdots=\emptyset}m_1(A_1)m_2(A_2)m_3(A_3)\cdots \end{aligned}$

Example

（本示例采用前面 Yager 合成公式示例给出的数据）
$\begin{aligned} k&=0.99901 \\ m(\{A\})&=0 \\ m(\{B\})&=0 \\ m(\{C\})&=0.00099 \\ m(\Omega)&=0 \\ m(\emptyset)&=0 \\ \end{aligned}$

Dubois and Prade 合成公式

$A_1$，证据源2有子集 $A_2$，当 $A_1\cap A_2=\emptyset$ 时，$m_1(A_1)\cdot m_2(A_2)$ 被分配给子集 $B\cup C$。公式定义如下：$m(A)=\sum_{A_1\cap A_2=A}m_1(A_1)m_2(A_2)+\sum_{A_1\cup A_2=A,A_1\cap A_2=\emptyset}m_1(A_1)m_2(A_2)$

Example

$\begin{aligned} &\Omega=\{A,B,C\} &\quad &\quad \\ &m_1:m_1(\{A\})=0.1,& m_1(\{B\})=0.1,&\quad m_1(\{C\})=0.8 \\ &m_2:m_2(\{A\})=0.8,& m_2(\{B\})=0.1,&\quad m_2(\{C\})=0.1 \\ \end{aligned}$ $\begin{aligned} k&=m_1(\{A\})m_2(\{B\})+m_1(\{A\})m_2(\{C\})+m_1(\{B\})m_2(\{A\})+m_1(\{B\})m_2(\{C\})+m_1(\{C\})m_2(\{A\})+m_1(\{C\})m_2(\{B\}) \\ &= 0.1\times0.1+0.1\times0.1+0.1\times0.8+0.1\times0.1+0.8\times0.8+0.8\times0.1 \\ &=0.83 \\ m(\{A\})&=m_1(\{A\})m_2(\{A\}) \\ &=0.1\times0.8 \\ &=0.08 \\ m(\{B\})&=0.01 \\ m(\{C\})&=0.08 \\ m(\{A,B\})&= \sum_{A_1\cap A_2=\{A,B\}}m_1(A_1)m_2(A_2)+\sum_{A_1\cup A_2=\{A,B\},A_1\cap A_2=\emptyset}m_1(A_1)m_2(A_2) \\ &=0+m_1(\{A\})m_2(\{B\})+m_1(\{B\})m_2(\{A\}) \\ &=0+0.1\times0.1+0.1\times0.8 \\ &=0.09 \\ m(\{A,C\})&=0+m_1(\{A\})m_2(\{C\})+m_1(\{C\})m_2(\{A\}) \\ &=0+0.1\times0.1+0.8\times0.8 \\ &=0.65 \\ m(\{B,C\})&=0+m_1(\{B\})m_2(\{C\})+m_1(\{C\})m_2(\{B\}) \\ &=0+0.1\times0.1+0.1\times0.8 \\ &=0.09 \\ m(\{A,B,C\})&=0 \end{aligned}$

Discounting and Dempster 合成公式

$\alpha$，用 $m_{\alpha_i,i}$ 表示第 $i$ 个证据源被折扣后的基本概率分配函数。公式定义如下：$m_{\alpha_i,i}(A)=\alpha_im_i(A),\quad A \neq \Omega$ $m_{\alpha_i,i}(\Omega)=1-\alpha_i+\alpha_im_i(\Omega)$ 然后采用 DS 合成公式合成各证据源的 $m_{\alpha_i,i}$。
  当 $\alpha_i=0$ 时，表示第 $i$ 个证据源有问题，不信任它；当 $\alpha_i=1$ 时，表示完全信任第 $i$

Example

（本示例采用前面 Dubois and Prade 合成公式示例给出的数据，$\alpha_1=0.2$，$\alpha_2=0.8$）
$\begin{aligned} &m_{0.2,1}:m_{0.2,1}(\{A\})=0.02,& m_{0.2,1}(\{B\})=0.02,&\quad m_{0.2,1}(\{C\})=0.16,\quad m_{0.2,1}(\Omega)=0.8 \\ &m_{0.8,2}:m_{0.8,2}(\{A\})=0.64,& m_{0.8,2}(\{B\})=0.08,&\quad m_{0.8,2}(\{C\})=0.08,\quad m_{0.8,2}(\Omega)=0.2 \\ \end{aligned}$ $\begin{aligned} k&=m_{0.2,1}(\{A\})m_{0.8,2}(\{B\})+m_{0.2,1}(\{A\})m_{0.8,2}(\{C\})+m_{0.2,1}(\{B\})m_{0.8,2}(\{A\})+m_{0.2,1}(\{B\})m_{0.8,2}(\{C\})+m_{0.2,1}(\{C\})m_{0.8,2}(\{A\})+m_{0.2,1}(\{C\})m_{0.8,2}(\{B\}) \\ &= 0.02\times0.08+0.02\times0.08+0.02\times0.64+0.02\times0.08+0.16\times0.64+0.16\times0.08 \\ &=0.1328\\ m(\{A\})&=\frac{m_{0.2,1}(\{A\})m_{0.8,2}(\{A\})+m_{0.2,1}(\{A\})m_{0.8,2}(\Omega)+m_{0.2,1}(\Omega)m_{0.8,2}(\{A\})}{1-k} \\ &=\frac{0.02\times0.64+0.02\times0.2+0.8\times0.64}{1-0.1328} \\ &=0.6098 \\ m(\{B\})&=\frac{m_{0.2,1}(\{B\})m_{0.8,2}(\{B\})+m_{0.2,1}(\{B\})m_{0.8,2}(\Omega)+m_{0.2,1}(\Omega)m_{0.8,2}(\{B\})}{1-k} \\ &=\frac{0.02\times0.08+0.02\times0.2+0.8\times0.08}{1-0.1328} \\ &=0.0803\\ m(\{C\})&=\frac{m_{0.2,1}(\{C\})m_{0.8,2}(\{C\})+m_{0.2,1}(\{C\})m_{0.8,2}(\Omega)+m_{0.2,1}(\Omega)m_{0.8,2}(\{C\})}{1-k} \\ &=\frac{0.16\times0.08+0.16\times0.2+0.8\times0.08}{1-0.1328} \\ &=0.1255\\ m(\Omega)&=\frac{m_{0.2,1}(\Omega)m_{0.8,2}(\Omega)}{1-k} \\ &=\frac{0.8\times0.2}{1-0.1328} \\ &=0.1845 \\ \end{aligned}$

Murphy 合成公式

$n$ 个证据源，该方法先将 $n$ 个证据源的 bpa 取平均得到 $m_{avg}$，再用 DS 合成公式对 $m_{avg}$ 迭代 $(n-1)$ 次得到合成后的 bpa。令 $f_{DS}(S_1,S_2)$ 表示两个证据源的 DS 合成公式，$m^i$ 表示第 $i$ 次迭代后的 bpa，则 Murphy 合成公式定义如下：$m^1=f_{DS}(m_{avg},m_{avg})$ $m^i=f_{DS}(m^{i-1},m_{avg}),\quad i \geq2$

Example

$\begin{aligned} &\Omega=\{A,B,C\} \\ &m_1:m_1(\{A\})=0.5,& m_1(\{B,C\})=0.5, \\ &m_2:m_2(\{C\})=0.5,& m_2(\{A,B\})=0.5 \\ \end{aligned}$ $m_{avg}(\{A\})=m_{avg}(\{C\})=m_{avg}(\{A,B\})=m_{avg}(\{B,C\})=0.25$
$\begin{aligned} k&=m_{avg}(\{A\})m_{avg}(\{C\})+m_{avg}(\{A\})m_{avg}(\{B,C\})+m_{avg}(\{C\})m_{avg}(\{A\})+m_{avg}(\{C\})m_{avg}(\{A,B\})+m_{avg}(\{A,B\})m_{avg}(\{C\})+m_{avg}(\{B,C\})m_{avg}(\{A\}) \\ &=0.25\times0.25\times6 \\ &=0.375 \\ m(\{A\})&=\frac{m_{avg}(\{A\})m_{avg}(\{A\})+m_{avg}(\{A\})m_{avg}(\{A,B\})+m_{avg}(\{A,B\})m_{avg}(\{A\})}{1-k} \\ &=\frac{0.25\times0.25\times3}{1-0.375} \\ &=0.3 \\ m(\{B\})&=\frac{m_{avg}(\{A,B\})m_{avg}(\{B,C\})+m_{avg}(\{B,C\})m_{avg}(\{A,B\})}{1-k} \\ &=\frac{0.25\times0.25\times2}{1-0.375} \\ &=0.2 \\ m(\{C\})&=\frac{m_{avg}(\{C\})m_{avg}(\{C\})+m_{avg}(\{C\})m_{avg}(\{B,C\})+m_{avg}(\{B,C\})m_{avg}(\{C\})}{1-k} \\ &=\frac{0.25\times0.25\times3}{1-0.375} \\ &=0.3 \\ m(\{A,B\})&=\frac{m_{avg}(\{A,B\})m_{avg}(\{A,B\})}{1-k} \\ &=\frac{0.25\times0.25}{1-0.375} \\ &=0.1 \\ m(\{B,C\})&=\frac{m_{avg}(\{B,C\})m_{avg}(\{B,C\})}{1-k} \\ &=\frac{0.25\times0.25}{1-0.375} \\ &=0.1 \\ \end{aligned}$

参考文献

[1] 孙全, 叶秀清, 顾伟康. 一种新的基于证据理论的合成公式[J]. 电子学报, 2000, 28(8):117-119.

[2] Lefevre E , Colot O , Vannoorenberghe P . Belief function combination and conflict management[J]. Information Fusion, 2002, 3(2):149-162.

[3] Murphy C K . Combining belief functions when evidence conflicts[J]. Decision Support Systems, 2000.