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Fundamentals of Dempster-Shafer Theory presented by Zbigniew W. Ras University of North Carolina, Charlotte, NC College of Computing and Informatics University.

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Presentation on theme: "Fundamentals of Dempster-Shafer Theory presented by Zbigniew W. Ras University of North Carolina, Charlotte, NC College of Computing and Informatics University."— Presentation transcript:

1 Fundamentals of Dempster-Shafer Theory presented by Zbigniew W. Ras University of North Carolina, Charlotte, NC College of Computing and Informatics University of North Carolina, Charlotte

2 Dempster-Shafer Theory based on the idea of placing a number between zero and one to indicate the degree of belief of evidence for a proposition.

3 Basic Probability Assignment - function m: 2^X  [0,1] such that: (1) m(  )=0, (2)  [m(Y) : Y  X] = 1 /total belief/. m(Y) – basic probability number of Y. Belief function over X - function Bel: 2^X  [0,1] such that: Bel(Y)=  [m(Z): Z  Y]. FACT 1: Function Bel: 2^X  [0,1] is a belief function iff (1)Bel(  )=0, (2) Bel(X)= 1, (3) Bel(  {A(i): i  {1,2,…,n}) =  [(-1)^{|J|+1}  Bel(  {A(i): i  J}) : J  {1,2,…,n}] for every positive integer n and all subsets A(1), A(2), …, A(n) of X FACT 2: Basic probability assignment can be computed from: m(Y) =  [ (-1)^{|Y – Z|  Bel(Z): Z  Y], where Y  X.

4 Example: basic probability assignment Xabcd x1x1 0L x2x2 0SL x3x3 P1L x4x4 3R1L x5x5 22L x6x6 P2L x7x7 3P2H m_a({x1,x2,x3,x6})=[2+2/3]/7=8/21 m_a({x3,x6,x5})=[1+2/3]/7=5/21 m_a({x3,x6,x4,x7})=[2+2/3]/7=8/21 Basic probability assignment (given) m({x1,x2,x3,x6})=8/21 m({x3,x6,x5})=5/21 m({x3,x6,x4,x7})=8/21 defines attribute m_a 1) m_a uniquely defined for x1,x2,x4,x5,x7. 2)m_a undefined for x3,x6. m_a(x1)=m_a(x2) =a1, m_a(x5)=a2,…..

5 Example: basic probability assignment Basic probability assignment – m: X={x1,x2,x3,x4,x5} m(x1,x2,x3)=1/2, m(x1,x2)=1/4, m(x2,x4)=1/4 Belief function: Bel({x1,x2,x3,x5})= ½ + ¼ = ¾, ……….. Focal Element and Core Y  X is called focal element iff m(Y) > 0. Core – the union of all focal elements. Doubt Function - Dou: 2^X  [0,1], Y  X Dou(Y) = Bel(  Y). Plausibility Function – Pl(Y) = 1 – Dou(Y) Pl(Y)=  [m(Z): Z  Y   ]

6 {1,2} 1/4 {1,2} {2,3} 3/4 {1,2} {1,3} 1/2 {1,2} {1} 0 {1,2} {2} 0 {1,2} {3} 1/2 m({3})=1/2, m({2,3})=1/4, m({1,2})=1/4. Pl({1,2}) = m({2,3})+m({1,2}) = ½, Pl({1,3})= m({3})+m({2,3})+m({1,2}) = 1 Core={1,2,3}

7 Properties: - Bel(  ) = Pl(  ) = 0 - Bel(X) = Pl(X) = 1 - Bel(Y)  Pl(Y) - Bel(Y) + Bel(  Y)  1 - Pl(Y) + Pl(  Y)  1 - if Y  Z, then Bel(Y)  Bel(Z) and Pl(Y)  Pl(Z) Bel: 2^X  [0,1] is called a Bayesian Belief Function iff 1)Bel(  ) = 0 2)Bel(X) = 1 3)Bel (Y  Z)= Bel(Y) + Bel(Z), where Y, Z  X, Y  Z =  Fact: Any Bayesian belief function is a belief function.

8 The following conditions are equivalent: 1)Bel is Bayesian 2)All focal elements of Bel are singletons 3)Bel = Pl 4)Bel(Y) + Bel(  Y) = 1 for all Y  X

9 Dempster’s Rule of Combination Bel1, Bel2 – belief functions representing two different pieces of evidence which are independent. Domain = {x1,x2,x3} Bel1  Bel2 – their orthogonal sum /Dempster’s rule of comb./ m1, m2 – basic probability assignments linked with Bel1, Bel2. {x1,x2} 1/4 {x1,x2,x3} 3/2 {x2,x4} 1/4 {x2} 3/8 {x2} 3/32 {x2} 3/16 {x2} 3/32 {x1,x2,x4} 3/8 {x1,x2} 3/32 {x1,x2} 3/16 {x2,x4} 3/32 {x1,x2,x3} 1/4 {x1,x2} 1/16 {x1,x2,x3} 1/8 {x2} 1/16 m1 m2 (m1  m2)({x1,x2})=3/32+3/16+1/16=11/32 (m1  m2)({x1,x2,x3})=1/8 (m1  m2)({x2})=3/32+3/16+3/32+1/16=7/16 (m1  m2)({x2,x4})=3/32

10 Thank You Questions?


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