证据理论入门笔记（2）—— 多种合成公式

Yager 合成公式

  Yager 合成公式去掉了 DS 合成公式中的归一化因子 $\frac{1}{1-k}$ ，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： $$k=\sum _{A_1\cap A_2\cap A_3\cdots =\varnothing }{m}_1(A_1)m_2(A_2)m_3(A_3)\cdots$$ m ( ∅ ) = 0 m ( A ) = ∑ A 1 ∩ A 2 ∩ A 3 ⋯ = A m 1 ( A 1 ) m 2 ( A 2 ) m 3 ( A 3 ) ⋯   , A ≠ ∅ , Ω m ( Ω ) = ∑ A 1 ∩ A 2 ∩ A 3 ⋯ = Ω m 1 ( A 1 ) m 2 ( A 2 ) m 3 ( A 3 ) ⋯ + k \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} m(∅)m(A)m(Ω)​=0=A1​∩A2​∩A3​⋯=A∑​m1​(A1​)m2​(A2​)m3​(A3​)⋯,A​=∅,Ω=A1​∩A2​∩A3​⋯=Ω∑​m1​(A1​)m2​(A2​)m3​(A3​)⋯+k​

Example

Ω = { A , B , C } m 1 : m 1 ( { A } ) = 0.98 , m 1 ( { B } ) = 0.01 , m 1 ( { C } ) = 0.01 m 2 : m 2 ( { A } ) = 0 , m 2 ( { B } ) = 0.01 , m 2 ( { C } ) = 0.99 m 3 : m 3 ( { A } ) = 0.9 , m 3 ( { B } ) = 0 ,   m 3 ( { C } ) = 0.1 \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} ​Ω={ A,B,C}m1​:m1​({ A})=0.98,m2​:m2​({ A})=0,m3​:m3​({ A})=0.9,​m1​({ B})=0.01,m2​({ B})=0.01,m3​({ B})=0,​m1​({ C})=0.01m2​({ C})=0.99m3​({ C})=0.1​ 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 × 0 × 0.9 + 0.01 × 0.01 × 0 + 0.01 × 0.99 × 0.1 ] = 0.99901 m ( { A } ) = m 1 ( { A } ) m 2 ( { A } ) m 3 ( { A } ) = 0.98 × 0 × 0.9 = 0 m ( { B } ) = 0 m ( { C } ) = m 1 ( { C } ) m 2 ( { C } ) m 3 ( { C } ) 1 − k = 0.01 × 0.99 × 0.1 = 0.00099 m ( Ω ) = 0 + k = 0.99901 \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\})&=\frac{m_1(\{C\})m_2(\{C\})m_3(\{C\})}{1-k} \\ &=0.01\times0.99\times0.1 \\ &=0.00099 \\ m(\Omega)&=0+k=0.99901 \\ \end{aligned} km({ A})m({ B})m({ C})m(Ω)​=1−[m1​({ A})m2​({ A})m3​({ A})+m1​({ B})m2​({ B})m3​({ B})+m1​({ C})m2​({ C})m3​({ C})]=1−[0.98×0×0.9+0.01×0.01×0+0.01×0.99×0.1]=0.99901=m1​({ A})m2​({ A})m3​({ A})=0.98×0×0.9=0=0=1−km1​({ C})m2​({ C})m3​({ C})​=0.01×0.99×0.1=0.00099=0+k=0.99901​

孙全提出的合成公式

  有 n 个证据源，对的应基本概率分配函数为 $m_1,m_2,\cdots ,m_n$。设证据源 $i$ 和 $j$ 之间的冲突系数为 $k_{ij}$：$$k_{ij}=\sum _{A_1\cap A_2=\varnothing }{m}_i(A_1)m_j(A_2)=1-\sum _{A_1\cap A_2\ne \varnothing }{m}_i(A_1)m_j(A_2)$$ $\epsilon$ 为证据的可信度：$$\epsilon =e^{-\hat{k}},\hat{k}=\frac{2}{n(n-1)}\sum _{i<j}{k}_{ij}$$ 公式定义如下：$$m(\varnothing )=0$$ $$m(A)=p(A)+k\times \epsilon \times q(A),A\ne \varnothing ,\Omega$$ $$m(\Omega )=p(\Omega )+k\times \epsilon \times q(\Omega )+k(1-\epsilon )$$ $$p(A)=\sum _{A_1\cap A_2\cap \cdots \cap A_n=A}{m}_1(A_1)m_2(A_2)\cdots m_n(A_n)$$ $$q(A)=\frac{1}{n}\sum _{i=1}^n{m}_i(A)$$ $m(A)$ 又可写成如下形式：$$m(A)=(1-k)\frac{p(A)}{1-k}+k\times \epsilon \times q(A)$$ 式中第一项的 $\frac{p(A)}{1-k}$ 是 DS 合成公式。

Example

（本示例采用前面 Yager 合成公式示例给出的数据）
k 12 = ∑ A 1 ∩ A 2 = ∅ 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 − ∑ A 1 ∩ A 3 ≠ ∅ 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 k ^ = 2 n ( n − 1 ) ∑ i < j k i j = 2 3 × ( 3 − 1 ) ( 0.99 + 0.117 + 0.901 ) = 0.6693 ε = e − k ^ = 0.5120 m ( { A } ) = ( 1 − k ) p ( { A } ) 1 − k + k × ε × q ( { A } ) = 0 + 0.99901 × 0.5120 × 1 3 ( 0.98 + 0 + 0.9 ) = 0.3205 m ( { B } ) = ( 1 − k ) p ( { B } ) 1 − k + k × ε × q ( { B } ) = 0 + 0.99901 × 0.5120 × 1 3 ( 0.01 + 0.01 + 0 ) = 0.0034 m ( { C } ) = ( 1 − k ) p ( { C } ) 1 − k + k × ε × q ( { C } ) = ( 1 − 0.99901 ) × 1 + 0.99901 × 0.5120 × 1 3 ( 0.01 + 0.99 + 0.1 ) = 0.1885 m ( Ω ) = ( 1 − k ) p ( Ω ) 1 − k + k × ε × q ( Ω ) + k ( 1 − ε ) = 0 + 0 + 0.99901 × ( 1 − 0.5120 ) = 0.4875 \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} k12​k13​k23​k^εm({ A})m({ B})m({ C})m(Ω)​=A1​∩A2​=∅∑​m1​(A1​)m2​(A2​)=m1​({ A})m2​({ B})+m1​({ A})m2​({ C})+m1​({ B})m2​({ A})+m1​({ B})m2​({ C})+m1​({ C})m2​({ A})+m1​({ C})m2​({ B})=0.99=1−A1​∩A3​​=∅∑​m1​(A1​)m3​(A3​)=1−[m1​({ A})m3​({ A})+m1​({ B})m3​({ B})+m1​({ C})m3​({ C})]=0.117=0.901=n(n−1)2​i<j∑​kij​=3×(3−1)2​(0.99+0.117+0.901)=0.6693=e−k^=0.5120=(1−k)1−kp({ A})​+k×ε×q({ A})=0+0.99901×0.5120×31​(0.98+0+0.9)=0.3205=(1−k)1−kp({ B})​+k×ε×q({ B})=0+0.99901×0.5120×31​(0.01+0.01+0)=0.0034=(1−k)1−kp({ C})​+k×ε×q({ C})=(1−0.99901)×1+0.99901×0.5120×31​(0.01+0.99+0.1)=0.1885=(1−k)1−kp(Ω)​+k×ε×q(Ω)+k(1−ε)=0+0+0.99901×(1−0.5120)=0.4875​

Smets 合成公式

  Smets 认为证据源之间的冲突只能来自于对辨识框架的错误定义。这样，Smets 将冲突系数 $k$ 保留不用，而不用于归一化。（论文[2]中用 $m(\varnothing )$ 表示 $k$，根据论文中给的示例来看 $m(\varnothing )$ 不表示空集的 bpa，空集的 bpa 依然是0，论文中的 $\varnothing$ 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：$$\begin{array}{rl}m(A)& =\sum _{A_1\cap A_2\cap A_3\cdots =A}{m}_1(A_1)m_2(A_2)m_3(A_3)\cdots ,A\ne \varnothing \\ & \\ k& =\sum _{A_1\cap A_2\cap A_3\cdots =\varnothing }{m}_1(A_1)m_2(A_2)m_3(A_3)\cdots \end{array}$$

Example

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

Dubois and Prade 合成公式

  对于有两个证据源的问题，证据源1有子集（命题）$A_1$，证据源2有子集 $A_2$，当 $A_1\cap A_2=\varnothing$ 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：$$m(A)=\sum _{A_1\cap A_2=A}{m}_1(A_1)m_2(A_2)+\sum _{A_1\cup A_2=A,A_1\cap A_2=\varnothing }{m}_1(A_1)m_2(A_2)$$

Example

Ω = { A , B , C } m 1 : m 1 ( { A } ) = 0.1 , m 1 ( { B } ) = 0.1 , m 1 ( { C } ) = 0.8 m 2 : m 2 ( { A } ) = 0.8 , m 2 ( { B } ) = 0.1 , m 2 ( { C } ) = 0.1 \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} ​Ω={ A,B,C}m1​:m1​({ A})=0.1,m2​:m2​({ A})=0.8,​m1​({ B})=0.1,m2​({ B})=0.1,​m1​({ C})=0.8m2​({ C})=0.1​ 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 × 0.1 + 0.1 × 0.1 + 0.1 × 0.8 + 0.1 × 0.1 + 0.8 × 0.8 + 0.8 × 0.1 = 0.83 m ( { A } ) = m 1 ( { A } ) m 2 ( { A } ) = 0.1 × 0.8 = 0.08 m ( { B } ) = 0.01 m ( { C } ) = 0.08 m ( { A , B } ) = ∑ A 1 ∩ A 2 = { A , B } m 1 ( A 1 ) m 2 ( A 2 ) + ∑ A 1 ∪ A 2 = { A , B } , A 1 ∩ A 2 = ∅ m 1 ( A 1 ) m 2 ( A 2 ) = 0 + m 1 ( { A } ) m 2 ( { B } ) + m 1 ( { B } ) m 2 ( { A } ) = 0 + 0.1 × 0.1 + 0.1 × 0.8 = 0.09 m ( { A , C } ) = 0 + m 1 ( { A } ) m 2 ( { C } ) + m 1 ( { C } ) m 2 ( { A } ) = 0 + 0.1 × 0.1 + 0.8 × 0.8 = 0.65 m ( { B , C } ) = 0 + m 1 ( { B } ) m 2 ( { C } ) + m 1 ( { C } ) m 2 ( { B } ) = 0 + 0.1 × 0.1 + 0.1 × 0.8 = 0.09 m ( { A , B , C } ) = 0 \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} km({ A})m({ B})m({ C})m({ A,B})m({ A,C})m({ B,C})m({ A,B,C})​=m1​({ A})m2​({ B})+m1​({ A})m2​({ C})+m1​({ B})m2​({ A})+m1​({ B})m2​({ C})+m1​({ C})m2​({ A})+m1​({ C})m2​({ B})=0.1×0.1+0.1×0.1+0.1×0.8+0.1×0.1+0.8×0.8+0.8×0.1=0.83=m1​({ A})m2​({ A})=0.1×0.8=0.08=0.01=0.08=A1​∩A2​={ A,B}∑​m1​(A1​)m2​(A2​)+A1​∪A2​={ A,B},A1​∩A2​=∅∑​m1​(A1​)m2​(A2​)=0+m1​({ A})m2​({ B})+m1​({ B})m2​({ A})=0+0.1×0.1+0.1×0.8=0.09=0+m1​({ A})m2​({ C})+m1​({ C})m2​({ A})=0+0.1×0.1+0.8×0.8=0.65=0+m1​({ B})m2​({ C})+m1​({ C})m2​({ B})=0+0.1×0.1+0.1×0.8=0.09=0​

Discounting and Dempster 合成公式

  该方法先给每个证据源分配一个表示信任度的系数 $\alpha$，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：$$m_{{\alpha }_i,i}(A)={\alpha }_i{m}_i(A),A\ne \Omega$$ $$m_{{\alpha }_i,i}(\Omega )=1-{\alpha }_i+{\alpha }_i{m}_i(\Omega )$$ 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 $i$ 个证据源。

Example

（本示例采用前面 Dubois and Prade 合成公式示例给出的数据，${\alpha }_1=0.2$，${\alpha }_2=0.8$）
m 0.2 , 1 : m 0.2 , 1 ( { A } ) = 0.02 , m 0.2 , 1 ( { B } ) = 0.02 , m 0.2 , 1 ( { C } ) = 0.16 , m 0.2 , 1 ( Ω ) = 0.8 m 0.8 , 2 : m 0.8 , 2 ( { A } ) = 0.64 , m 0.8 , 2 ( { B } ) = 0.08 , m 0.8 , 2 ( { C } ) = 0.08 , m 0.8 , 2 ( Ω ) = 0.2 \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} ​m0.2,1​:m0.2,1​({ A})=0.02,m0.8,2​:m0.8,2​({ A})=0.64,​m0.2,1​({ B})=0.02,m0.8,2​({ B})=0.08,​m0.2,1​({ C})=0.16,m0.2,1​(Ω)=0.8m0.8,2​({ C})=0.08,m0.8,2​(Ω)=0.2​ 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 × 0.08 + 0.02 × 0.08 + 0.02 × 0.64 + 0.02 × 0.08 + 0.16 × 0.64 + 0.16 × 0.08 = 0.1328 m ( { A } ) = m 0.2 , 1 ( { A } ) m 0.8 , 2 ( { A } ) + m 0.2 , 1 ( { A } ) m 0.8 , 2 ( Ω ) + m 0.2 , 1 ( Ω ) m 0.8 , 2 ( { A } ) 1 − k = 0.02 × 0.64 + 0.02 × 0.2 + 0.8 × 0.64 1 − 0.1328 = 0.6098 m ( { B } ) = m 0.2 , 1 ( { B } ) m 0.8 , 2 ( { B } ) + m 0.2 , 1 ( { B } ) m 0.8 , 2 ( Ω ) + m 0.2 , 1 ( Ω ) m 0.8 , 2 ( { B } ) 1 − k = 0.02 × 0.08 + 0.02 × 0.2 + 0.8 × 0.08 1 − 0.1328 = 0.0803 m ( { C } ) = m 0.2 , 1 ( { C } ) m 0.8 , 2 ( { C } ) + m 0.2 , 1 ( { C } ) m 0.8 , 2 ( Ω ) + m 0.2 , 1 ( Ω ) m 0.8 , 2 ( { C } ) 1 − k = 0.16 × 0.08 + 0.16 × 0.2 + 0.8 × 0.08 1 − 0.1328 = 0.1255 m ( Ω ) = m 0.2 , 1 ( Ω ) m 0.8 , 2 ( Ω ) 1 − k = 0.8 × 0.2 1 − 0.1328 = 0.1845 \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} km({ A})m({ B})m({ C})m(Ω)​=m0.2,1​({ A})m0.8,2​({ B})+m0.2,1​({ A})m0.8,2​({ C})+m0.2,1​({ B})m0.8,2​({ A})+m0.2,1​({ B})m0.8,2​({ C})+m0.2,1​({ C})m0.8,2​({ A})+m0.2,1​({ C})m0.8,2​({ B})=0.02×0.08+0.02×0.08+0.02×0.64+0.02×0.08+0.16×0.64+0.16×0.08=0.1328=1−km0.2,1​({ A})m0.8,2​({ A})+m0.2,1​({ A})m0.8,2​(Ω)+m0.2,1​(Ω)m0.8,2​({ A})​=1−0.13280.02×0.64+0.02×0.2+0.8×0.64​=0.6098=1−km0.2,1​({ B})m0.8,2​({ B})+m0.2,1​({ B})m0.8,2​(Ω)+m0.2,1​(Ω)m0.8,2​({ B})​=1−0.13280.02×0.08+0.02×0.2+0.8×0.08​=0.0803=1−km0.2,1​({ C})m0.8,2​({ C})+m0.2,1​({ C})m0.8,2​(Ω)+m0.2,1​(Ω)m0.8,2​({ C})​=1−0.13280.16×0.08+0.16×0.2+0.8×0.08​=0.1255=1−km0.2,1​(Ω)m0.8,2​(Ω)​=1−0.13280.8×0.2​=0.1845​

Murphy 合成公式

  设有 $n$ 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：$$m^1=f_{DS}(m_{avg},m_{avg})$$ $$m^i=f_{DS}(m^{i-1},m_{avg}),i\ge 2$$

Example

Ω = { 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 \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} ​Ω={ A,B,C}m1​:m1​({ A})=0.5,m2​:m2​({ C})=0.5,​m1​({ B,C})=0.5,m2​({ A,B})=0.5​ m a v g ( { A } ) = m a v g ( { C } ) = m a v g ( { A , B } ) = m a v g ( { B , C } ) = 0.25 m_{avg}(\{A\})=m_{avg}(\{C\})=m_{avg}(\{A,B\})=m_{avg}(\{B,C\})=0.25 mavg​({ A})=mavg​({ C})=mavg​({ A,B})=mavg​({ B,C})=0.25
k = m a v g ( { A } ) m a v g ( { C } ) + m a v g ( { A } ) m a v g ( { B , C } ) + m a v g ( { C } ) m a v g ( { A } ) + m a v g ( { C } ) m a v g ( { A , B } ) + m a v g ( { A , B } ) m a v g ( { C } ) + m a v g ( { B , C } ) m a v g ( { A } ) = 0.25 × 0.25 × 6 = 0.375 m ( { A } ) = m a v g ( { A } ) m a v g ( { A } ) + m a v g ( { A } ) m a v g ( { A , B } ) + m a v g ( { A , B } ) m a v g ( { A } ) 1 − k = 0.25 × 0.25 × 3 1 − 0.375 = 0.3 m ( { B } ) = m a v g ( { A , B } ) m a v g ( { B , C } ) + m a v g ( { B , C } ) m a v g ( { A , B } ) 1 − k = 0.25 × 0.25 × 2 1 − 0.375 = 0.2 m ( { C } ) = m a v g ( { C } ) m a v g ( { C } ) + m a v g ( { C } ) m a v g ( { B , C } ) + m a v g ( { B , C } ) m a v g ( { C } ) 1 − k = 0.25 × 0.25 × 3 1 − 0.375 = 0.3 m ( { A , B } ) = m a v g ( { A , B } ) m a v g ( { A , B } ) 1 − k = 0.25 × 0.25 1 − 0.375 = 0.1 m ( { B , C } ) = m a v g ( { B , C } ) m a v g ( { B , C } ) 1 − k = 0.25 × 0.25 1 − 0.375 = 0.1 \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} km({ A})m({ B})m({ C})m({ A,B})m({ B,C})​=mavg​({ A})mavg​({ C})+mavg​({ A})mavg​({ B,C})+mavg​({ C})mavg​({ A})+mavg​({ C})mavg​({ A,B})+mavg​({ A,B})mavg​({ C})+mavg​({ B,C})mavg​({ A})=0.25×0.25×6=0.375=1−kmavg​({ A})mavg​({ A})+mavg​({ A})mavg​({ A,B})+mavg​({ A,B})mavg​({ A})​=1−0.3750.25×0.25×3​=0.3=1−kmavg​({ A,B})mavg​({ B,C})+mavg​({ B,C})mavg​({ A,B})​=1−0.3750.25×0.25×2​=0.2=1−kmavg​({ C})mavg​({ C})+mavg​({ C})mavg​({ B,C})+mavg​({ B,C})mavg​({ C})​=1−0.3750.25×0.25×3​=0.3=1−kmavg​({ A,B})mavg​({ A,B})​=1−0.3750.25×0.25​=0.1=1−kmavg​({ B,C})mavg​({ B,C})​=1−0.3750.25×0.25​=0.1​

参考文献

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