Dempster-Shafer theory in decision making
Let us define
A= “not crowed at all”
B= “normal”
C=“ very crowed”
Then whole Environment E={A,B,C}.
Vehicle Message content Message reputation score
Vehicle Message content
V1 C 0.6
V2 C 0.8
V3 A 0.7
(1) considering V1and V2(@ here means Unknown)
m1({C})=0.6 m1({@})=0.4
Thus m1+m2({C})=0.48+0.32+0.12=0.92 m1+m2({@})=0.08
(2) take V3 into consideration
m1+m2 ({C})=0.92 m1+m2 ({@})=0.08
Which means m1+m2+m3({NULL})=0.644, m1+m2+m3({A})=0.056
m1+m2+m3({C})=0.276, m1+m2+m3({@})=0.024
(3) standardization because of Empty set
(3) standardization because of Empty set
Because the m1+m2+m3({NULL})=0.644 should be 0, so after standard process:
m1+m2+m3({A})=0.056/(1-0.644)=0.157
m1+m2+m3({C})=0.276/(1-0.644)=0.775
m1+m2+m3({@})= 0.024/(1-0.644)=0.068
So, confidence interval :
A=[0.157, 0.157+0.068]=[0.157,0.225]
C=[0.775, 0.775+0.068]=[0.775,0.843]
Bad case for Dempster-Shafer theory
two doctor and one patient
1st doctor diagnoses the patient was attacted by tumor with 0.9 propability, and cancer with 0.1;
2nd doctor diagnoses the patient was attacted by concussion with 0.9 propability, and cancer with 0.1;
thus
m1({tumor})=0.9 m1({cancer})=0.1
m2({concussion})=0.9 {NULL} 0.81 {NULL} 0.09
m2({cancer})=0.1 {NULL} 0.09 {cancer} 0.01
Bad case for Dempster-Shafer theory
two doctor and one patient
1st doctor diagnoses the patient was attacted by tumor with 0.9 propability, and cancer with 0.1;
2nd doctor diagnoses the patient was attacted by concussion with 0.9 propability, and cancer with 0.1;
thus
m1({tumor})=0.9 m1({cancer})=0.1
=> m1+m2({NULL})=0.81+0.09+0.09=0.99 m1+m2({tumor})=0.01
K= m1+m2({NULL})=0.99
after standardization:
m1+m2({tumor})=0.01/(1-K)=1
Which would be greatly contrary with our intuition.
So, when K is equal or very close to 1, we should just cancle the standardization




