证据的组合
(1)定义
对于相同的证据,由于来源不同,可能得不同的基本概率赋值函数,D-S证据理论采用正交和来组合这些函数。
设m1,m2,...,mn为2U上的n个基本概率赋值函数,它们的正交和表示为m=m1⊕m2⊕...⊕mn,且定义为
若K-1=0,则说明mi(Ai)之间是矛盾的。
(2)举例
例:设U={a,b},且从不同的证据源得知概率赋值函数分别为
m1(Φ,{a},{b},{a,b})=(0,0.4,0.5,0.1)
m2(Φ,{a},{b},{a,b})=(0,0.6,0.2,0.2)
求正交和m=m1⊕m2
解:先求K-1
K-1=
=m1({a})*m2({a})+m1({a})*m2({a,b})+m1({b})*m2({b})
+m1({b})*m2({a,b})+m1({a,b})*m2({b})+m1({a,b})*m2({a,b})
=0.4*0.6+0.4*0.2+0.5*0.2+0.5*0.2+0.1*0.6+0.1*0.2+0.1*0.2
=0.62
再求m(A)=K*
m({a})=[1/0.62]*[m1({a})*m2({a})+m1({a,b})*m2({a})+m1({a})*m2({a,b})]
=[1/0.62]*[0.4*0.6+0.1*0.6+0.4*0.2]
=0.61
m({b})=[1/0.62]*[m1({b})*m2({b})+m1({a,b})*m2({b})+m1({b})*m2({a,b})]
=[1/0.62]*[0.5*0.2+0.1*0.2+0.5*0.2]
=0.36
m({a,b})=[1/0.62]*[m1({a,b})*m2({a,b})]
=[1/0.62]*[0.1*0.2]
=0.03
所以有m(Φ,{a},{b},{a,b})=(0,0.61,0.36,0.03)