# Dempster-Shafer Theory

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

Examples – Datasets (Information Systems)

Information System How to interpret term “blond”?
Color_of_Hair Nationality X1 Blond X2 Spanish X3 Black X4 German How to interpret term “blond”? How to interpret “blond + Spanish”, “Blond  Spanish”? I(+) =  , I() = . We have several options: 1. I(Blond) = {X1,X4} - pessimistic 2. I(Blond) = {X1,X4,X2} – optimistic 3. I(Blond) = [{X1,X4}, {X2}] – rough 4. I(Blond) = {(X1,1), (X4,1), (X2,1/2)} assuming that X2 is either Blond or Black and the chances are equal for both colors. Option 1. 0 = I(Black  Blond) = I(Black)  I(Blond) = {X3}  {X1,X4} = 0 X = I(Black + Blond) = I(Black)  I(Blond) = {X3}  {X1,X4} = {X1,X3,X4} Option 2. 0 = I(Black  Blond) = I(Black)  I(Blond) = {X2, X3}  {X1,X4,X2} = {X2} X = I(Black + Blond) = I(Black)  I(Blond) = {X2, X3}  {X1,X4,X2} = X

Query Languages (Syntax) – built from
values of attributes (if they are not local, we call them foreign), functors +, ,  predicates = , <, , ….. Queries form a smallest set T such that: 1. If w  V then w  T. 2. If t1 , t2  T then (t1+t2), (t1t2), (t1)  T Extended Queries form a smallest set F such that: 1. If t1 , t2  T then (t1 = t2), (t1 < t2), (t1  t2)  F 2. If 1 , 2  F then ~1, ( 1  2), ( 1  2)  F Query Languages (Local Semantics) – domain of interpretation has to be established Example: Query – “blond  Spanish”, Extended query - “blond  Spanish = black”

CERTAIN RULES: (Headache, Yes)  (Temp, High  Very High)  (Flu, Yes) (Temp, Normal)  (Flue, No) POSSIBLE RULES: (Headache, No)  (Temp, High)  (Flue, Yes), Conf= ½ (Headache, No)  (Temp, High)  (Flue, No), Conf= ½ ……………………………………………

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. 19

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. 20

Example: basic probability assignment
d x1 L x2 S x3 P 1 x4 3 R x5 2 x6 x7 H 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,….. 21

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  ]

{1,2} {1,2} {1,2} 1/4 {2,3} 3/4 {1,2} {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 {1} {2} 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

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.

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

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} {x2} 3/8 3/32 3/16 {x1,x2,x4} 1/16 1/8 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

Failing Query Problem

Questions? Thank You

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