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。
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
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\Omega=\{H_1,H_2,\cdots, H_N\}
Ω={H1,H2,⋯,HN}
Ω
\Omega
Ω 的子集
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A 称为命题 (proposition)，
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Ω 的幂集
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2^\Omega
2Ω 由
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\Omega
Ω 的所有子集组成，包含
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2^N
2N 个元素，
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2Ω 可以表示为：
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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}
基本概率分配函数是从
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2^\Omega
2Ω 到
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[0,1]
[0,1] 的映射：
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m:2^\Omega\rightarrow[0,1]
m:2Ω→[0,1] 它满足以下两个条件：
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m(\emptyset)=0 \quad and \quad \sum_{A\subseteq\Omega}m(A)=1
m(∅)=0andA⊆Ω∑m(A)=1
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m(A)
m(A) 的值表示证据对命题
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对于
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A\in2^\Omega
A∈2Ω，如果
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m(A)>0
m(A)>0，则称
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A
A 为一个焦元 (focal element)。
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
似然函数定义如下：
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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：
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Pl(A)=1-Bel(\bar{A})
Pl(A)=1−Bel(Aˉ)其中
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\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：
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Pl(\emptyset)=0
Pl(∅)=0
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Pl(\Omega)=1
Pl(Ω)=1信任函数与似然函数的关系：
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≥
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Pl(A)\geq Bel(A)
Pl(A)≥Bel(A)
Example:
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\Omega=\{A,B,C\}
Ω={A,B,C}
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\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
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\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
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：
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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)⋯
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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，
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k 接近0表示证据源彼此一致。
当
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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。
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
（证据源高度冲突的情况）
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\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
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\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
（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）
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\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
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\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
