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11.4 Dempster-Shafer Theory
                                                       1.0                                            393

                                                           don’t know  Empty  = Empty  don’t know   U
                                                                       don’t know
                                                                      U
                                                                                  don’t know
                                                                                  = don’t know
                                                  Bel            0.6 X 0. 6 = 0.36  0.6 X 0. 4 = 0.24
                                                     1
                                                      0.4
                                                                                 don’t know   U
                                                           occupied  Occupied  = φ U  Empty   = Occupied
                                                                                  Occupied

                                                                 0.4 X 0. 6 = 0.24  0.4 X 0. 4 = 0.16
                                                       0.0
                                                                     empty         don’t know
                                                            0.0               0.6          1.0
                                                                           Bel
                                                                              2

                                                   Figure 11.7  Example of the need for normalization.




                                      belief. For the previous example the total area of valid belief was 1.0, so the
                                      belief for a set was computed as:

                                              P
                                                 (m(C k )
                                      m(C)      k           =
                                                 1:0
                                        If a portion of that area belongs to ; (or there are p areas for ;), then the
                                      total area is 1:0  m(;):

                                               P  (m(C k )
                                      m(C)       k P        =
                                              1     m(; p )
                                                   p
                                        Normalizing leads to:

                                       m(O        )  =    0:16  =  0:21
                                          ccupied
                                                         1  0:24
                                         m(E      )  =    0:36  =  0:47
                                            mpty
                                                         1  0:24
                                       m(dontknow  )  =   0:24  =  0:32
                                                         1  0:24
                                        This idea of summing up the areas for a particular proposition and nor-
                                      malizing can be captured with a mathematical formula known as Dempster’s
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