证据理论（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 11−k\frac{1}{1-k}1−k1 ，并将冲突系数 kkk 分配给了辨识框架 Ω\OmegaΩ 对应的基本概率分配函数 m(Ω)m(\Omega)m(Ω)。公式定义如下： k=∑A1∩A2∩A3⋯=∅m1(A1)m2(A2)m3(A3)⋯k=\sum_{A_1\cap A_2\cap A_3\cdots=\emptyset}m_1(A_1)m_2(A_2)m_3(A_3)\cdotsk=A1∩A2∩A3⋯=∅∑m1(A1)m2(A2)m3(A3)⋯ m(∅)=0m(A)=∑A1∩A2∩A3⋯=Am1(A1)m2(A2)m3(A3)⋯,A≠∅,Ωm(Ω)=∑A1∩A2∩A3⋯=Ωm1(A1)m2(A2)m3(A3)⋯+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}m1:m1({A})=0.98,m1({B})=0.01,m1({C})=0.01m2:m2({A})=0,m2({B})=0.01,m2({C})=0.99m3:m3({A})=0.9,m3({B})=0,m3({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−[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]=0.99901m({A})=m1({A})m2({A})m3({A})=0.98×0×0.9=0m({B})=0m({C})=m1({C})m2({C})m3({C})=0.01×0.99×0.1=0.00099m(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\})&=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} 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=m1({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 m1,m2,⋯,mnm_1,m_2,\cdots,m_nm1,m2,⋯,mn。设证据源 iii 和 jjj 之间的冲突系数为 kijk_{ij}kij：kij=∑A1∩A2=∅mi(A1)mj(A2)=1−∑A1∩A2≠∅mi(A1)mj(A2)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)kij=A1∩A2=∅∑mi(A1)mj(A2)=1−A1∩A2=∅∑mi(A1)mj(A2) ε\varepsilonε 为证据的可信度：ε=e−k^,k^=2n(n−1)∑i<jkij\varepsilon=e^{-\hat{k}}, \quad \hat{k}=\frac{2}{n(n-1)}\sum_{i<j}k_{ij}ε=e−k^,k^=n(n−1)2i<j∑kij 公式定义如下：m(∅)=0m(\emptyset)=0m(∅)=0 m(A)=p(A)+k×ε×q(A),A≠∅,Ωm(A)=p(A)+k\times\varepsilon\times q(A),\quad A \neq\emptyset,\Omegam(A)=p(A)+k×ε×q(A),A=∅,Ω m(Ω)=p(Ω)+k×ε×q(Ω)+k(1−ε)m(\Omega)=p(\Omega)+k\times\varepsilon\times q(\Omega)+k(1-\varepsilon)m(Ω)=p(Ω)+k×ε×q(Ω)+k(1−ε) p(A)=∑A1∩A2∩⋯∩An=Am1(A1)m2(A2)⋯mn(An)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)p(A)=A1∩A2∩⋯∩An=A∑m1(A1)m2(A2)⋯mn(An) q(A)=1n∑i=1nmi(A)q(A)=\frac{1}{n}\sum_{i=1}^{n}m_i(A)q(A)=n1i=1∑nmi(A) m(A)m(A)m(A) 又可写成如下形式：m(A)=(1−k)p(A)1−k+k×ε×q(A)m(A)=(1-k)\frac{p(A)}{1-k}+k\times\varepsilon\times q(A)m(A)=(1−k)1−kp(A)+k×ε×q(A) 式中第一项的 p(A)1−k\frac{p(A)}{1-k}1−kp(A) 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）
k12=∑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})=0.99k13=1−∑A1∩A3≠∅m1(A1)m3(A3)=1−[m1({A})m3({A})+m1({B})m3({B})+m1({C})m3({C})]=0.117k23=0.901k^=2n(n−1)∑i<jkij=23×(3−1)(0.99+0.117+0.901)=0.6693ε=e−k^=0.5120m({A})=(1−k)p({A})1−k+k×ε×q({A})=0+0.99901×0.5120×13(0.98+0+0.9)=0.3205m({B})=(1−k)p({B})1−k+k×ε×q({B})=0+0.99901×0.5120×13(0.01+0.01+0)=0.0034m({C})=(1−k)p({C})1−k+k×ε×q({C})=(1−0.99901)×1+0.99901×0.5120×13(0.01+0.99+0.1)=0.1885m(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} k12k13k23k^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)2i<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 kkk 保留不用，而不用于归一化。（论文[2]中用 m(∅)m(\emptyset)m(∅) 表示 kkk，根据论文中给的示例来看 m(∅)m(\emptyset)m(∅) 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：m(A)=∑A1∩A2∩A3⋯=Am1(A1)m2(A2)m3(A3)⋯,A≠∅k=∑A1∩A2∩A3⋯=∅m1(A1)m2(A2)m3(A3)⋯\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}m(A)k=A1∩A2∩A3⋯=A∑m1(A1)m2(A2)m3(A3)⋯,A=∅=A1∩A2∩A3⋯=∅∑m1(A1)m2(A2)m3(A3)⋯

Example

（本示例采用前面 Yager 合成公式示例给出的数据）
k=0.99901m({A})=0m({B})=0m({C})=0.00099m(Ω)=0m(∅)=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）A1A_1A1，证据源2有子集 A2A_2A2，当 A1∩A2=∅A_1\cap A_2=\emptysetA1∩A2=∅ 时，m1(A1)⋅m2(A2)m_1(A_1)\cdot m_2(A_2)m1(A1)⋅m2(A2) 被分配给子集 B∪CB\cup CB∪C。公式定义如下：m(A)=∑A1∩A2=Am1(A1)m2(A2)+∑A1∪A2=A,A1∩A2=∅m1(A1)m2(A2)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)m(A)=A1∩A2=A∑m1(A1)m2(A2)+A1∪A2=A,A1∩A2=∅∑m1(A1)m2(A2)

Example

Ω={A,B,C}m1:m1({A})=0.1,m1({B})=0.1,m1({C})=0.8m2:m2({A})=0.8,m2({B})=0.1,m2({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=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=0.83m({A})=m1({A})m2({A})=0.1×0.8=0.08m({B})=0.01m({C})=0.08m({A,B})=∑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=0.09m({A,C})=0+m1({A})m2({C})+m1({C})m2({A})=0+0.1×0.1+0.8×0.8=0.65m({B,C})=0+m1({B})m2({C})+m1({C})m2({B})=0+0.1×0.1+0.1×0.8=0.09m({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 α\alphaα，用 mαi,im_{\alpha_i,i}mαi,i 表示第 iii 个证据源被折扣后的基本概率分配函数。公式定义如下：mαi,i(A)=αimi(A),A≠Ωm_{\alpha_i,i}(A)=\alpha_im_i(A),\quad A \neq \Omegamαi,i(A)=αimi(A),A=Ω mαi,i(Ω)=1−αi+αimi(Ω)m_{\alpha_i,i}(\Omega)=1-\alpha_i+\alpha_im_i(\Omega)mαi,i(Ω)=1−αi+αimi(Ω) 然后采用 DS 合成公式合成各证据源的 mαi,im_{\alpha_i,i}mαi,i。
当 αi=0\alpha_i=0αi=0 时，表示第 iii 个证据源有问题，不信任它；当 αi=1\alpha_i=1αi=1 时，表示完全信任第 iii 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，α1=0.2\alpha_1=0.2α1=0.2，α2=0.8\alpha_2=0.8α2=0.8）
m0.2,1:m0.2,1({A})=0.02,m0.2,1({B})=0.02,m0.2,1({C})=0.16,m0.2,1(Ω)=0.8m0.8,2:m0.8,2({A})=0.64,m0.8,2({B})=0.08,m0.8,2({C})=0.08,m0.8,2(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=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=0.1328m({A})=m0.2,1({A})m0.8,2({A})+m0.2,1({A})m0.8,2(Ω)+m0.2,1(Ω)m0.8,2({A})1−k=0.02×0.64+0.02×0.2+0.8×0.641−0.1328=0.6098m({B})=m0.2,1({B})m0.8,2({B})+m0.2,1({B})m0.8,2(Ω)+m0.2,1(Ω)m0.8,2({B})1−k=0.02×0.08+0.02×0.2+0.8×0.081−0.1328=0.0803m({C})=m0.2,1({C})m0.8,2({C})+m0.2,1({C})m0.8,2(Ω)+m0.2,1(Ω)m0.8,2({C})1−k=0.16×0.08+0.16×0.2+0.8×0.081−0.1328=0.1255m(Ω)=m0.2,1(Ω)m0.8,2(Ω)1−k=0.8×0.21−0.1328=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 nnn 个证据源，该方法先将 nnn 个证据源的 bpa 取平均得到 mavgm_{avg}mavg，再用 DS 合成公式对 mavgm_{avg}mavg 迭代 (n−1)(n-1)(n−1) 次得到合成后的 bpa。令 fDS(S1,S2)f_{DS}(S_1,S_2)fDS(S1,S2) 表示两个证据源的 DS 合成公式，mim^imi 表示第 iii 次迭代后的 bpa，则 Murphy 合成公式定义如下：m1=fDS(mavg,mavg)m^1=f_{DS}(m_{avg},m_{avg})m1=fDS(mavg,mavg) mi=fDS(mi−1,mavg),i≥2m^i=f_{DS}(m^{i-1},m_{avg}),\quad i \geq2mi=fDS(mi−1,mavg),i≥2

Example

Ω={A,B,C}m1:m1({A})=0.5,m1({B,C})=0.5,m2:m2({C})=0.5,m2({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 mavg({A})=mavg({C})=mavg({A,B})=mavg({B,C})=0.25m_{avg}(\{A\})=m_{avg}(\{C\})=m_{avg}(\{A,B\})=m_{avg}(\{B,C\})=0.25mavg({A})=mavg({C})=mavg({A,B})=mavg({B,C})=0.25
k=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=0.375m({A})=mavg({A})mavg({A})+mavg({A})mavg({A,B})+mavg({A,B})mavg({A})1−k=0.25×0.25×31−0.375=0.3m({B})=mavg({A,B})mavg({B,C})+mavg({B,C})mavg({A,B})1−k=0.25×0.25×21−0.375=0.2m({C})=mavg({C})mavg({C})+mavg({C})mavg({B,C})+mavg({B,C})mavg({C})1−k=0.25×0.25×31−0.375=0.3m({A,B})=mavg({A,B})mavg({A,B})1−k=0.25×0.251−0.375=0.1m({B,C})=mavg({B,C})mavg({B,C})1−k=0.25×0.251−0.375=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

[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.