D-S evidence theory

D-S证据理论的基本概念

定义1： basic possibility assignment (BPA)
Let U be discernment frame, a function $m:2^u\to [0,1]$ follows the two conditions:
(1) $m(\varphi )=0$
(2) ${\sum }_{A\subset U}m(A)=1$
then $m(A)$ denotes 根本赋值 to A, m(A)=0 denotes as the belief degree to A, that is mass function。

定义2 信任函数 （Belief Function）

定义3 似然函数（Plausibility Function）

定义4 信任区间（Belief Interval)
　[Bel(A)，pl(A)]表示命题A的信任区间，Bel(A)表示信任函数为下限，pl(A)表示似真函数为上限。
　举例：如(0.25,0.85),表示A为真有0.25的信任度，A为假有0.15的信任度，A不确定度为0.6。

D-S证据理论的组合规则

BPA from the different sources can be combined with the Dempster orthogonal rule, which incorporates the joint information provided by the sources and can be represented as follows:
$$\begin{array}{l}{m}_{1,2}(\varnothing )=0\\ m_{1,2}(A)=(m_1\oplus m_2)(A)=\frac{1}{1-K}{\sum }_{B\cap C=A\ne \varnothing }{m}_1(B)m_2(C)\end{array}$$
where,
$$K=\sum _{B\cap C=\varnothing }{m}_1(B)m_2(C)$$
$K$ characterizes the a measure of conflict between the different mass sets.

In order to explain the D-S evidence theory, the follow example is provided. For a murder crime, there are three suspects consisting in the frame of discernment {Peter, Paul, Mary}, two witnesses (W1, W2) provide the following BPA to the Police.

空	$m_1()$	$m_2()$	$m_{12}()$
{Peter}	0.98	0	0.49
{Paul}	0.01	0.01	0.015
{Mary}	0	0.98	0.49
{Peter, Paul Mary}	0.01	0.01	0.005

A: Calculate the constant K,
$$\begin{gathered} K=1-{\sum }_{B\cup C=\varnothing }{m}_1(B)\cdot m_2(C) \\ =1-[m_1(Peter)\cdot m_2(Paul)+m_1(Peter)\cdot m_2(Mary)+m_1(Paul)\times m_2(Peter)+m_1(Paul)\cdot m_2(Mary)+m_1(Mary)\times m_2(Peter)+m_1(Mary)\times m_2(Paul)] \\ =1-(0.98\times 0.01+0.98\times 0.98+0.01\times 0+0.01\times 0.98+0\times 0+0\ast 0.98) \\ =0.02 \end{gathered}$$
or
$$\begin{gathered} K={\sum }_{B\cap C\ne \varnothing }{m}_1(B)\cdot m_2(C) \\ =m_1(Peter)\cdot m_2(Peter)+m_1(Peter)\cdot m_2(\Theta ) \\ +m_1(Paul)\cdot m_2(Paul)+m_1(Paul)\cdot m_2(\Theta ) \\ +m_1(Mary)\cdot m_2(Mary)+m_1(Mary)\cdot m_2(\Theta ) \\ +m_1(\Theta )\cdot m_2(Peter)+m_1(\Theta )\cdot m_2(Paul)+m_1(\Theta )\cdot m_2(Mary)+m_2(\Theta )\cdot m_2(Theta) \\ =0.98\times 0.01+0.98\times 0 \\ +0.01\times 0.01+0.01\times 0.01 \\ +0\times 0.98+0\times 0.01 \\ +0.01\times 0+0.01\times 0.01+0.01\times 0.98+0.01\times 0.01 \\ =0.02 \end{gathered}$$

(1) calculate the composite mass function about Peter
m 1 ⊕ m 2 ( { P e t e r } ) = 1 K ∑ B ∩ C = { P e t e r } m 1 ( B ) ⋅ m 2 C = 1 K [ m 1 ( { P e t e r } ) ⋅ m 2 ( { P e t e r } ) + m 1 ( { P e t e r } ) ⋅ m 2 ( Θ ) + m 1 ( Θ ) ⋅ m 2 ( { P e t e r } ) ] = 1 0.02 × ( 0.98 × 0 + 0.98 × 0.01 + 0.01 × 0 ) = 0.49 m_1\oplus m_2(\{Peter\}) = \frac{1}{K}\sum_{B\cap C=\{Peter\}}{m_1(B)\cdot m_2{C}}\\ =\frac{1}{K}[m_1(\{Peter\})\cdot m_2(\{Peter\})+m_1(\{Peter\})\cdot m_2(\Theta)+m_1(\Theta)\cdot m_2(\{Peter\})]\\ =\frac{1}{0.02}\times (0.98\times 0+ 0.98\times 0.01+0.01\times 0)=0.49 m1​⊕m2​({ Peter})=K1​∑B∩C={ Peter}​m1​(B)⋅m2​C=K1​[m1​({ Peter})⋅m2​({ Peter})+m1​({ Peter})⋅m2​(Θ)+m1​(Θ)⋅m2​({ Peter})]=0.021​×(0.98×0+0.98×0.01+0.01×0)=0.49

(2) Calculate the composite mass function about Paul
m 1 ⊕ m 2 ( { P a u l } ) = 1 K ∑ B ∩ C = { P a u l } m 1 ( B ) ⋅ m 2 ( C ) = 1 K [ m 1 ( { P a u l } ) ⋅ m 2 ( { P a u l } ) + m 1 ( { P a u l } ) ⋅ m 2 ( Θ ) + m 1 ( T h e t a ) ⋅ m 2 ( { P a u l } ) ] = 1 0.02 × ( 0.01 × 0.01 + 0.01 × 0.01 + 0.01 × 0.01 ) = 0.015 m_1\oplus m_2(\{Paul\})=\frac{1}{K}\sum_{B\cap C=\{Paul\}}m_1(B)\cdot m_2(C)\\ =\frac{1}{K}[m_1(\{Paul\})\cdot m_2(\{Paul\})+m_1(\{Paul\})\cdot m_2(\Theta)+m_1(Theta)\cdot m_2(\{Paul\})]\\ =\frac{1}{0.02}\times (0.01\times 0.01+0.01\times 0.01+0.01\times 0.01)=0.015 m1​⊕m2​({ Paul})=K1​∑B∩C={ Paul}​m1​(B)⋅m2​(C)=K1​[m1​({ Paul})⋅m2​({ Paul})+m1​({ Paul})⋅m2​(Θ)+m1​(Theta)⋅m2​({ Paul})]=0.021​×(0.01×0.01+0.01×0.01+0.01×0.01)=0.015

(3) Calculate the composite mass function about Mary
m 1 ⊕ m 2 ( { M a r y } ) = 1 K ∑ B ∩ C = { M a r y } m 1 ( B ) ⋅ m 2 ( C ) = 1 K [ m 1 ( { M a r y } ) ⋅ m 2 ( { M a r y } ) + m 1 ( { M a r y } ) ⋅ m 2 ( Θ ) + m 1 ( Θ ) ⋅ m 2 ( { M a r y } ) ] = 1 0.02 × ( 0 × 0.98 + 0 × 0.01 + 0.01 × 0.98 ) = 0.49 m_1\oplus m_2(\{Mary\})=\frac{1}{K}\sum_{B\cap C=\{Mary\}}m_1(B)\cdot m_2(C)\\ =\frac{1}{K}[m_1(\{Mary\})\cdot m_2(\{Mary\})+m_1(\{Mary\})\cdot m_2(\Theta)+m_1(\Theta)\cdot m_2(\{Mary\})]\\ =\frac{1}{0.02}\times (0\times 0.98+0\times 0.01+0.01\times 0.98)=0.49 m1​⊕m2​({ Mary})=K1​∑B∩C={ Mary}​m1​(B)⋅m2​(C)=K1​[m1​({ Mary})⋅m2​({ Mary})+m1​({ Mary})⋅m2​(Θ)+m1​(Θ)⋅m2​({ Mary})]=0.021​×(0×0.98+0×0.01+0.01×0.98)=0.49

(4) Calculate the composite mass function about Θ = { P e t e r , P a u l , M a r y } \Theta=\{Peter, Paul, Mary\} Θ={ Peter,Paul,Mary}
$$\begin{gathered} m_1\oplus m_2(\Theta )=\frac{1}{K}{\sum }_{B\cap C=\Theta }{m}_1(B)\cdot m_2(C) \\ =\frac{1}{K}{m}_1(\Theta )\times m_2(\Theta ) \\ =\frac{1}{0.02}0.01\times 0.01=0.005 \end{gathered}$$
Further, according to the formulas of belief function and flausible function, we obtain:
Bel ( { P e t e r } = 0.49 \text{Bel}(\{Peter\}=0.49 Bel({ Peter}=0.49; Pl ( { P e t e r } ) = 0.49 + 0.005 = 0.495 \text{Pl}(\{Peter\})=0.49+0.005=0.495 Pl({ Peter})=0.49+0.005=0.495
Bel ( { P a u l } = 0.015 \text{Bel}(\{Paul\}=0.015 Bel({ Paul}=0.015; Pl ( { P a u l } ) = 0.015 + 0.005 = 0.020 \text{Pl}(\{Paul\})=0.015+0.005=0.020 Pl({ Paul})=0.015+0.005=0.020
Bel ( { M a r y } = 0.49 \text{Bel}(\{Mary\}=0.49 Bel({ Mary}=0.49; Pl ( { M a r y } ) = 0.49 + 0.005 = 0.495 \text{Pl}(\{Mary\})=0.49+0.005=0.495 Pl({ Mary})=0.49+0.005=0.495
$\text{Bel}(\Theta )=\text{Pl}(\Theta )=0.49+0.015+0.49+0.005=1$