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Fundamentals of Dempster -Shafer Theory

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

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Fundamentals of Dempster -Shafer Theory

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

  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 • 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 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 • 1) m_a uniquely defined • for x1,x2,x4,x5,x7. • m_a undefined for • x3,x6. defines attribute m_a 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} {1,2} {1,2} {1,2} 1/4 {2,3} 3/4 {1,3} 1/2 m({3})=1/2, m({2,3})=1/4, m({1,2})=1/4. {1,2} {1,2} {1,2} {3} 1/2 {2} 0 {1} 0 Core={1,2,3} Pl({1,2}) = m({2,3})+m({1,2}) = ½, Pl({1,3})= m({3})+m({2,3})+m({1,2}) = 1

  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 • Bel() = 0 • Bel(X) = 1 • 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: • Bel is Bayesian • All focal elements of Bel are singletons • Bel = Pl • 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. 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. Questions? Thank You

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