证据理论(1)—— DS证据理论基本理论
證據理論
??證據理論 (Theory of Evidence) 是由 Dempster 首先提出,由Shafer進一步發展起來的一種不精確推理理論,也稱為 Dempster-Shafer (DS) 證據理論。證據理論可以在沒有先驗概率的情況下,靈活并有效地對不確定性建模。
證據理論的基本理論
辨識框架 (Frame of discernment)
??辨識框架 Ω\OmegaΩ 是一個由問題的所有假設 (hypothesis) 組成窮舉集合,所有假設是相互排斥的。設 Ω\OmegaΩ 包含 NNN 個元素,Ω\OmegaΩ 可以表示為:Ω={H1,H2,?,HN}\Omega=\{H_1,H_2,\cdots, H_N\}Ω={H1?,H2?,?,HN?}
??Ω\OmegaΩ 的子集 AAA 稱為命題 (proposition),Ω\OmegaΩ 的冪集 2Ω2^\Omega2Ω 由 Ω\OmegaΩ 的所有子集組成,包含 2N2^N2N 個元素,2Ω2^\Omega2Ω 可以表示為:2Ω={?,{H1},{H2},?,{HN},{H1,H2},?,{H1,H2,?,Hi},?,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}
基本概率分配函數 (Basic probability assignment, bpa) / m 函數 (Mass function)
??基本概率分配函數是從 2Ω2^\Omega2Ω 到 [0,1][0,1][0,1] 的映射:m:2Ω→[0,1]m:2^\Omega\rightarrow[0,1] m:2Ω→[0,1] 它滿足以下兩個條件:m(?)=0and∑A?Ωm(A)=1m(\emptyset)=0 \quad and \quad \sum_{A\subseteq\Omega}m(A)=1m(?)=0andA?Ω∑?m(A)=1 m(A)m(A)m(A) 的值表示證據對命題 AAA 的支持程度。
??對于 A∈2ΩA\in2^\OmegaA∈2Ω,如果 m(A)>0m(A)>0m(A)>0,則稱 AAA 為一個焦元 (focal element)。
信任函數 (Belief function)
??信任函數定義如下:Bel(A)=∑B?Am(B)Bel(A)=\sum_{B\subseteq A}m(B)Bel(A)=B?A∑?m(B) Bel(A)Bel(A)Bel(A)表示對A的總的信任程度。根據基本概率分配函數的特點,我們可以知道:Bel(?)=m(?)=0Bel(\emptyset)=m(\emptyset)=0Bel(?)=m(?)=0 Bel(Ω)=∑B?Ωm(B)=1Bel(\Omega)=\sum_{B\subseteq\Omega}m(B)=1Bel(Ω)=B?Ω∑?m(B)=1
似然函數 (Plausibility function)
??似然函數定義如下:Pl(A)=∑B∩A≠?m(B)Pl(A)=\sum_{B\cap A\neq\emptyset}m(B)Pl(A)=B∩A=?∑?m(B)似然函數也可以表示為:Pl(A)=1?Bel(Aˉ)Pl(A)=1-Bel(\bar{A})Pl(A)=1?Bel(Aˉ)其中 Aˉ=Ω?A\bar{A}=\Omega-AAˉ=Ω?A。似然函數表示不否定A的信任程度。似然函數有如下特點:Pl(?)=0Pl(\emptyset)=0Pl(?)=0 Pl(Ω)=1Pl(\Omega)=1Pl(Ω)=1信任函數與似然函數的關系:Pl(A)≥Bel(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)=0.2m({B})=0,m({A,C})=0.2,m(?)=0m({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})=0?m(Ω)=0.2m(?)=0? 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\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
??對于相互獨立的不同證據源,有不同的基本概率分配函數。Dempster-Shafer 合成公式采用正交和將不同的基本概率分配函數合成為一個新的基本概率分配函數。公式定義如下:m(A)=11?k∑A1∩A2∩A3?=Am1(A1)m2(A2)m3(A3)?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)\cdotsm(A)=1?k1?A1?∩A2?∩A3??=A∑?m1?(A1?)m2?(A2?)m3?(A3?)? k=∑A1∩A2∩A3?=?m1(A1)m2(A2)m3(A3)?=1?∑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)\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)\cdotsk=A1?∩A2?∩A3??=?∑?m1?(A1?)m2?(A2?)m3?(A3?)?=1?A1?∩A2?∩A3??=?∑?m1?(A1?)m2?(A2?)m3?(A3?)? 其中 kkk 是沖突系數,kkk 越接近1表示證據源之間沖突越嚴重,kkk 接近0表示證據源彼此一致。
??當 k→1k\rightarrow 1k→1 時,表示證據源高度沖突,這時候采用DS合成公式會得出違反直覺的結果。而且,即使增加彼此一致的信息源的數量,也無法降低沖突系數 kkk。Example 2 和 Example 3 展示了這一問題。我在證據理論入門筆記(2)中總結了其他幾種合成公式。
Examples
Example 1
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\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=m1({C})m2({A})=0.28m({A})=m1({A})m2({A})+m1({A})m2(Ω)+m1({A,B})m2({A})1?k=0.4×0.7+0.4×0.3+0.2×0.71?0.28=0.75m({A,B})=m1({A,B})m2(Ω)1?k=0.2×0.31?0.28=0.0833m({C})=m1({C})m2(Ω)1?k=0.4×0.31?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
(證據源高度沖突的情況)
Ω={A,B,C}m1:m1({A})=0.99,m1({B})=0.01,m1({C})=0m2:m2({A})=0,m2({B})=0.01,m2({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=m1({A})m2({B})+m1({A})m2({C})+m1({B})m2({C})=0.99×0.01+0.99×0.99+0.01×0.99=0.9999m({A})=m1({A})m2({A})1?k=0.99×01?0.9999=0m({B})=m1({B})m2({B})1?k=0.01×0.011?0.9999=1m({C})=m1({C})m2({C})1?k=0×0.991?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
(三個證據源,其中證據源1和證據源2高度沖突,證據源1和證據源3一致)
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})1?k=0.98×0×0.91?0.99901=0m({B})=0m({C})=m1({C})m2({C})m3({C})1?k=0.01×0.99×0.11?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?
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