Demspter-Shafter theory application

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

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

 m2({C})=0.8        {C}  0.48                 {C}   0.32

 m2({@})=0.2       {C}   0.12                {@}   0.08

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

 m3({A})=0.7               NULL  0.644                          {A}   0.056

 m3({@})=0.3                {C}  0.276                             {@} 0.024

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

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

=>    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