证据理论（1）—— DS证据理论基本理论
证据理论
证据理论 (Theory of Evidence) 是由 Dempster 首先提出，由Shafer进一步发展起来的一种不精确推理理论，也称为 Dempster-Shafer (DS) 证据理论。证据理论可以在没有先验概率的情况下，灵活并有效地对不确定性建模。
证据理论的基本理论
辨识框架 (Frame of discernment)
辨识框架 Ω \Omega Ω 是一个由问题的所有假设 (hypothesis) 组成穷举集合，所有假设是相互排斥的。设 Ω \Omega Ω 包含 N N N 个元素， Ω \Omega Ω 可以表示为： Ω = { H 1 , H 2 , ⋯ , H N } \Omega=\{H_1,H_2,\cdots, H_N\} Ω={H1,H2,⋯,HN}
Ω \Omega Ω 的子集 A A A 称为命题 (proposition)， Ω \Omega Ω 的幂集 2 Ω 2^\Omega 2Ω 由 Ω \Omega Ω 的所有子集组成，包含 2 N 2^N 2N 个元素， 2 Ω 2^\Omega 2Ω 可以表示为： 2 Ω = { ∅ , { H 1 } , { H 2 } , ⋯ , { H N } , { H 1 , H 2 } , ⋯ , { H 1 , H 2 , ⋯ , H i } , ⋯ , Ω } 2^\Omega=\{\emptyset ,\{H_1\},\{H_2\},\cdots,\{H_N\},\{H_1,H_2\},\cdots,\{H_1,H_2,\cdots, H_i\},\cdots,\Omega\} 2Ω={∅,{H1},{H2},⋯,{HN},{H1,H2},⋯,{H1,H2,⋯,Hi},⋯,Ω}
基本概率分配函数 (Basic probability assignment, bpa) / m 函数 (Mass function)
基本概率分配函数是从 2 Ω 2^\Omega 2Ω 到 [ 0 , 1 ] [0,1] [0,1] 的映射： m : 2 Ω → [ 0 , 1 ] m:2^\Omega\rightarrow[0,1] m:2Ω→[0,1] 它满足以下两个条件： m ( ∅ ) = 0 a n d ∑ A ⊆ Ω m ( A ) = 1 m(\emptyset)=0 \quad and \quad \sum_{A\subseteq\Omega}m(A)=1 m(∅)=0andA⊆Ω∑m(A)=1 m ( A ) m(A) m(A) 的值表示证据对命题 A A A 的支持程度。
对于 A ∈ 2 Ω A\in2^\Omega A∈2Ω，如果 m ( A ) > 0 m(A)>0 m(A)>0，则称 A A A 为一个焦元 (focal element)。
信任函数 (Belief function)
信任函数定义如下： B e l ( A ) = ∑ B ⊆ A m ( B ) Bel(A)=\sum_{B\subseteq A}m(B) Bel(A)=B⊆A∑m(B) B e l ( A ) Bel(A) Bel(A)表示对A的总的信任程度。根据基本概率分配函数的特点，我们可以知道： B e l ( ∅ ) = m ( ∅ ) = 0 Bel(\emptyset)=m(\emptyset)=0 Bel(∅)=m(∅)=0 B e l ( Ω ) = ∑ B ⊆ Ω m ( B ) = 1 Bel(\Omega)=\sum_{B\subseteq\Omega}m(B)=1 Bel(Ω)=B⊆Ω∑m(B)=1
似然函数 (Plausibility function)
似然函数定义如下： P l ( A ) = ∑ B ∩ A ≠ ∅ m ( B ) Pl(A)=\sum_{B\cap A\neq\emptyset}m(B) Pl(A)=B∩A=∅∑m(B)似然函数也可以表示为： P l ( A ) = 1 − B e l ( A ˉ ) Pl(A)=1-Bel(\bar{A}) Pl(A)=1−Bel(Aˉ)其中 A ˉ = Ω − A \bar{A}=\Omega-A Aˉ=Ω−A。似然函数表示不否定A的信任程度。似然函数有如下特点： P l ( ∅ ) = 0 Pl(\emptyset)=0 Pl(∅)=0 P l ( Ω ) = 1 Pl(\Omega)=1 Pl(Ω)=1信任函数与似然函数的关系： P l ( A ) ≥ B e l ( A ) Pl(A)\geq Bel(A) Pl(A)≥Bel(A)
Example:
Ω = { A , B , C } \Omega=\{A,B,C\} Ω={A,B,C} m ( { A } ) = 0.3 , m ( { A , B } ) = 0.2 , m ( Ω ) = 0.2 m ( { B } ) = 0 , m ( { A , C } ) = 0.2 , m ( ∅ ) = 0 m ( { C } ) = 0.1 , m ( { B , C } ) = 0 \begin{aligned} &m(\{A\})=0.3,&\quad m(\{A,B\})=0.2, &\quad m(\Omega)=0.2 \\ &m(\{B\})=0, &\quad m(\{A,C\})=0.2,&\quad m(\emptyset)=0 \\ &m(\{C\})=0.1,&\quad m(\{B,C\})=0 \quad \end{aligned} m({A})=0.3,m({B})=0,m({C})=0.1,m({A,B})=0.2,m({A,C})=0.2,m({B,C})=0m(Ω)=0.2m(∅)=0 B e l ( { A } ) = m ( { A } ) + m ( ∅ ) = 0.3 P l ( { A } ) = m ( { A } ) + m ( { A , B } ) + m ( { A , B } ) + m ( Ω ) = 0.3 + 0.2 + 0.2 + 0.2 = 0.9 \begin{aligned} &Bel(\{A\})=m(\{A\})+m(\emptyset)=0.3 \\ &Pl(\{A\})=m(\{A\})+m(\{A,B\})+m(\{A,B\})+m(\Omega)=0.3+0.2+0.2+0.2=0.9 \end{aligned} Bel({A})=m({A})+m(∅)=0.3Pl({A})=m({A})+m({A,B})+m({A,B})+m(Ω)=0.3+0.2+0.2+0.2=0.9
Dempster-Shafer 合成公式
对于相互独立的不同证据源，有不同的基本概率分配函数。Dempster-Shafer 合成公式采用正交和将不同的基本概率分配函数合成为一个新的基本概率分配函数。公式定义如下： m ( A ) = 1 1 − k ∑ A 1 ∩ A 2 ∩ A 3 ⋯ = A m 1 ( A 1 ) m 2 ( A 2 ) m 3 ( A 3 ) ⋯ m(A)=\frac{1}{1-k}\sum_{A_1\cap A_2\cap A_3\cdots=A}m_1(A_1)m_2(A_2)m_3(A_3)\cdots m(A)=1−k1A1∩A2∩A3⋯=A∑m1(A1)m2(A2)m3(A3)⋯ k = ∑ A 1 ∩ A 2 ∩ A 3 ⋯ = ∅ m 1 ( A 1 ) m 2 ( A 2 ) m 3 ( A 3 ) ⋯ = 1 − ∑ A 1 ∩ A 2 ∩ A 3 ⋯ ≠ ∅ m 1 ( A 1 ) m 2 ( A 2 ) m 3 ( A 3 ) ⋯ 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=1-\sum_{A_1\cap A_2\cap A_3\cdots\neq\emptyset}m_1(A_1)m_2(A_2)m_3(A_3)\cdots k=A1∩A2∩A3⋯=∅∑m1(A1)m2(A2)m3(A3)⋯=1−A1∩A2∩A3⋯=∅∑m1(A1)m2(A2)m3(A3)⋯ 其中 k k k 是冲突系数， k k k 越接近1表示证据源之间冲突越严重， k k k 接近0表示证据源彼此一致。
当 k → 1 k\rightarrow 1 k→1 时，表示证据源高度冲突，这时候采用DS合成公式会得出违反直觉的结果。而且，即使增加彼此一致的信息源的数量，也无法降低冲突系数 k k k。Example 2 和 Example 3 展示了这一问题。我在证据理论入门笔记（2）中总结了其他几种合成公式。
Examples
Example 1
Ω = { A , B , C } m 1 : m 1 ( { A } ) = 0.4 , m 1 ( { A , B } ) = 0.2 , m 1 ( { C } ) = 0.4 m 2 : m 2 ( { A } ) = 0.7 , m 2 ( Ω ) = 0.3 \begin{aligned} &\Omega=\{A,B,C\} \\ &m_1:m_1(\{A\})=0.4,\quad m_1(\{A,B\})=0.2,\quad m_1(\{C\})=0.4 \\ &m_2:m_2(\{A\})=0.7,\quad m_2(\Omega)=0.3 \end{aligned} Ω={A,B,C}m1:m1({A})=0.4,m1({A,B})=0.2,m1({C})=0.4m2:m2({A})=0.7,m2(Ω)=0.3 k = m 1 ( { C } ) m 2 ( { A } ) = 0.28 m ( { A } ) = m 1 ( { A } ) m 2 ( { A } ) + m 1 ( { A } ) m 2 ( Ω ) + m 1 ( { A , B } ) m 2 ( { A } ) 1 − k = 0.4 × 0.7 + 0.4 × 0.3 + 0.2 × 0.7 1 − 0.28 = 0.75 m ( { A , B } ) = m 1 ( { A , B } ) m 2 ( Ω ) 1 − k = 0.2 × 0.3 1 − 0.28 = 0.0833 m ( { C } ) = m 1 ( { C } ) m 2 ( Ω ) 1 − k = 0.4 × 0.3 1 − 0.28 = 0.1667 \begin{aligned} k&=m_1(\{C\})m_2(\{A\})=0.28 \\ m(\{A\})&=\frac{m_1(\{A\})m_2(\{A\})+m_1(\{A\})m_2(\Omega)+m_1(\{A,B\})m_2(\{A\})}{1-k} \\ &=\frac{0.4\times0.7+0.4\times0.3+0.2\times0.7}{1-0.28} \\ &=0.75 \\ m(\{A,B\})&=\frac{m_1(\{A,B\})m_2(\Omega)}{1-k} \\ &=\frac{0.2\times0.3}{1-0.28} \\ &=0.0833 \\ m(\{C\})&=\frac{m_1(\{C\})m_2(\Omega)}{1-k} \\ &=\frac{0.4\times0.3}{1-0.28} \\ &=0.1667 \\ \end{aligned} km({A})m({A,B})m({C})=m1({C})m2({A})=0.28=1−km1({A})m2({A})+m1({A})m2(Ω)+m1({A,B})m2({A})=1−0.280.4×0.7+0.4×0.3+0.2×0.7=0.75=1−km1({A,B})m2(Ω)=1−0.280.2×0.3=0.0833=1−km1({C})m2(Ω)=1−0.280.4×0.3=0.1667
Example 2
（证据源高度冲突的情况）
Ω = { A , B , C } m 1 : m 1 ( { A } ) = 0.99 , m 1 ( { B } ) = 0.01 , m 1 ( { C } ) = 0 m 2 : m 2 ( { A } ) = 0 , m 2 ( { B } ) = 0.01 , m 2 ( { C } ) = 0.99 \begin{aligned} &\Omega=\{A,B,C\} &\quad &\quad \\ &m_1:m_1(\{A\})=0.99,& m_1(\{B\})=0.01,&\quad m_1(\{C\})=0 \\ &m_2:m_2(\{A\})=0, & m_2(\{B\})=0.01,&\quad m_2(\{C\})=0.99 \\ \end{aligned} Ω={A,B,C}m1:m1({A})=0.99,m2:m2({A})=0,m1({B})=0.01,m2({B})=0.01,m1({C})=0m2({C})=0.99 k = m 1 ( { A } ) m 2 ( { B } ) + m 1 ( { A } ) m 2 ( { C } ) + m 1 ( { B } ) m 2 ( { C } ) = 0.99 × 0.01 + 0.99 × 0.99 + 0.01 × 0.99 = 0.9999 m ( { A } ) = m 1 ( { A } ) m 2 ( { A } ) 1 − k = 0.99 × 0 1 − 0.9999 = 0 m ( { B } ) = m 1 ( { B } ) m 2 ( { B } ) 1 − k = 0.01 × 0.01 1 − 0.9999 = 1 m ( { C } ) = m 1 ( { C } ) m 2 ( { C } ) 1 − k = 0 × 0.99 1 − 0.0.9999 = 0 \begin{aligned} k&=m_1(\{A\})m_2(\{B\})+m_1(\{A\})m_2(\{C\})+m_1(\{B\})m_2(\{C\}) \\ &= 0.99\times0.01+0.99\times0.99+0.01\times0.99 \\ &=0.9999 \\ m(\{A\})&=\frac{m_1(\{A\})m_2(\{A\})}{1-k} \\ &=\frac{0.99\times0}{1-0.9999} \\ &=0 \\ m(\{B\})&=\frac{m_1(\{B\})m_2(\{B\})}{1-k} \\ &=\frac{0.01\times0.01}{1-0.9999} \\ &=1 \\ m(\{C\})&=\frac{m_1(\{C\})m_2(\{C\})}{1-k} \\ &=\frac{0\times0.99}{1-0.0.9999} \\ &=0 \\ \end{aligned} km({A})m({B})m({C})=m1({A})m2({B})+m1({A})m2({C})+m1({B})m2({C})=0.99×0.01+0.99×0.99+0.01×0.99=0.9999=1−km1({A})m2({A})=1−0.99990.99×0=0=1−km1({B})m2({B})=1−0.99990.01×0.01=1=1−km1({C})m2({C})=1−0.0.99990×0.99=0
Example 3
（三个证据源，其中证据源1和证据源2高度冲突，证据源1和证据源3一致）
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 } ) 1 − k = 0.98 × 0 × 0.9 1 − 0.99901 = 0 m ( { B } ) = 0 m ( { C } ) = m 1 ( { C } ) m 2 ( { C } ) m 3 ( { C } ) 1 − k = 0.01 × 0.99 × 0.1 1 − 0.99901 = 1 \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\})&=\frac{m_1(\{A\})m_2(\{A\})m_3(\{A\})}{1-k} \\ &=\frac{0.98\times0\times0.9}{1-0.99901} \\ &=0 \\ m(\{B\})&=0 \\ m(\{C\})&=\frac{m_1(\{C\})m_2(\{C\})m_3(\{C\})}{1-k} \\ &=\frac{0.01\times0.99\times0.1}{1-0.99901} \\ &=1 \\ \end{aligned} km({A})m({B})m({C})=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=1−km1({A})m2({A})m3({A})=1−0.999010.98×0×0.9=0=0=1−km1({C})m2({C})m3({C})=1−0.999010.01×0.99×0.1=1