Fuzzy measures

1. • Fuzzy set  Fuzzy measure
To which degree does an What is the degree of evidence,
individual belong to a set that an individual belongs to a
defined in an inexact way? certain (crisp) set?
• Fuzzy measure g: P(X)  [0,1]
If g(A)g(B) BP(X) then A is the "best guess"
 E.g.: diagnose of a patient with given symptoms
• Usual interpretation: subjective probability,
however: properties are less strict!
• Axioms of fuzzy measures:
g1. g()=0 and g(X)=1 (boundary conditions)
g2. " A,BP(X): if A  B then g(A)g(B) (monotonicity)
g3. " sequence (AiP(X)  i, Ai X):
if A1A2... or A1A2... then
(continuity)
i
i
i
i
g A g A
 

lim lim
( ) ( )
4 / 1

2. • More general definition of fuzzy measure g:
g: B  [0,1]
where BP(X) is a family of subsets:
1. B and XB
2. If AB then B
3. If A,BB then ABB
(B is Borel-field or s-field)
 As ABA and ABB, we have
max(g(A),g(B))g(AB)
also ABA and ABB, so
min(g(A),g(B))g(AB)
• Fuzzy measures are "too general". Practical considerations lead to some
restrictions, so we obtain special classes of measures.
 Clearly, probability is one of them!
4 / 2

3. • Belief measure :
Bel: P(X)  [0,1]
Axioms: g1-g3 and
g4Bel.:
for arbitrary AiP(X)
 n=2 Bel(A1A2) Bel(A1)+Bel(A2)-Bel(A1A2)
n=3 Bel(A1A2A3)Bel(A1)+Bel(A2)+Bel(A3)-Bel(A1A2)-Bel(A1A3)-
-Bel(A2A3)+Bel(A1A2A3)
• Interpretation: Bel(A) is the degree of Belief that a particular xX belongs to A
• If (AiAj)= then
Probability is a special Bel measure!
 g4Belg2 (not independent axiom) so {g1, g3, g4Bel } defines Bel measures
Bel A A Bel A
n i
i
n
( ... ) ( )
1
1
  


Bel A A Bel A Bel A A Bel A A A
n i
i
i j
i j
n
n
( ... ) ( ) ( ) ... ( ) ( ... )
1
1
1 2
1
          
 


4 / 3

4. • Plausability measure :
Pl: P(X)  [0,1]
Axioms: g1-g3 and
g4Pl.:
for arbitrary AiP(X), n
 Pl is dual with Bel!
,
(*)
 Connection of g4Bel and g4Pl:
(*) applied to g4Bel we obtain:
I.e. g4Pl for
)
...
(
)
1
(
...
)
(
)
(
)
...
( 2
1
1
1
1 n
n
j
i
j
i
n
i
i
n A
A
A
Pl
A
A
Pl
A
Pl
A
A
Pl 









 




Pl A Bel A
( ) ( )
 
1
Bel A Pl A
( ) ( )
 
1
1 1 1 1 1
1 1
1
1
         
  


   
Pl A Pl A Pl A A Pl A
i
n
i i
i
n
i j
i j
n
i
n
i
( ) ( ( ) ( ( )) ... ( ) ( ( ))
Pl A Pl A Pl A A Pl A
i
n
i i
i
n
i j
i j
n
i
n
i
( ) ( ) ( ) ... ( ) ( )
  


   
     
1 1
1
1
1
Ai
4 / 4

5.  From g4Bel-Pl we obtain:
• Every Bel and its dual Pl can be expressed by the "basic (probabilistic)
assignment" m:
m: P(X)  [0,1],
m()=0 and
m is the degree of evidence that xX belongs to A but not to any particular
subset of A.
 Different from probabilistic p: x  [0,1] !
 m(X)=1 not necessarily!
m(A) m(B) when AB: not necessarily!
m(A) and m( ) have no relation
m is not a fuzzy measure!
• (**)
Bel A Bel A
( ) ( )
 1
Pl A Pl A
( ) ( )
  1
m A
( ) 

 1
A P(X)
A



A
B
B
m
A
Bel )
(
)
(
Pl A m B
B A
( ) ( )

 

4 / 5

6. Interpretation:
m(A): degrees of evidence that xX belongs to A alone
Bel(A): degrees of evidence that xX belongs to some subsets of A
Pl(A): degrees of evidence that xX belongs to A or any subsets of X which
overlaps with A
 Pl(A)Bel(A) " AP(X)
• If m(A)>0 : A is a focal element (of m)
• Body of evidence: <F, m>
F: set of focal elements
m: basic assignment
 F={X}, m(X)=1 : total ignorance
 Bel(X)=1, Bel(A)=0 " AX
 Pl()=0, Pl(A)=1 " A
• Simple support function (focused at A):
m(A)=s, m(X)=1-s,
and m(B)=0 " BA,X
4 / 6

7.  Basic assignment derived from Bel:
 Example X={a,b,c}
P(X)={,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}
m: {a}  0.1
{b}  0.2
{a,b}  0.3 F={{a},{b},{a,b},{b,c},{a,b,c}}
{b,c}  0.3
{a,b,c}  0.1
Bel({a,b})=m()+m{a}+m{b}+m{a,b}=0+0.1+0.2+0.3=0.6
Pl({a,b})=m{a,b,c}+m{a,b},+m{a,c}+m{b,c}+m{a}+m{b}=
=0.1+0.3+0+0.3+0.1+0.2=1.0
Bel({c})=m()+m{c}=0+0=0
Pl({c})=m{a,b,c}+m{a,c}+m{b,c}+m{c}=0.1+0+0.3+0=0.4
Bel()=0 Bel({a})=0.1 Bel({b})=0.2
Bel({a,c}=0.1 Bel({b,c})=0.5 Bel({a,b,c})=1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
m()=0 m({a})=Bel({a})=0.1
m({b})=Bel({b})=0.2 m({c})=0
m({a,b})=Bel({a,b})-Bel({a})-Bel({b})=0.6-0.1-0.2=0.3
Etc.
m A Bel B
A B
B A
( ) ( ) ( )
  

 1
4 / 7

8. • Evidence from two independent sources: m1,m2
Joint basic assignment m1,2:
A (***)
m1,2()=0 (Dempster's rule of combination)
The sum of products m1(B)m2(C) for all focal elements of B (m1) and C (m2) so that
BC, equals (in the denominator); this is why the sum of
products is normalized
 Example (from Klir-Folger textbook)
Assume that an old painting was discovered that strongly resembles paintings by Raphael. Such a discovery
is likely to generate various questions regarding the status of the painting. Assume the following three
questions:
1. Is the discovered painting a genuine painting by Raphael?
2. Is the discovered painting a product of one of Raphael's many disciples?
3. Is the discovered painting a counterfeit?
Let R, D and C denote subsets of our universal set X - the set of paintings - that contain the set of all
paintings by Raphael, the set of all paintings by disciples of Raphael, and the set of all counterfeits of
Raphael's paintings, respectively.
Assume now, that two experts performed careful examinations of the painting and subsequently provided us
with basic assignments m1 and m2 specified in the table below. These are the degrees of evidence that each
expert obtained by the examination and that support the various claims that the painting belongs to one of the
sets of our concern. For example, m1(RD)=0.15 is the degree of evidence obtained by the first expert that
the painting was done by Raphael himself or that the painting was done by one of his disciples.
m A
m B m C)
m B m C)
B C A
B C
1 2
1 2
1 2
1
, ( )
( ) (
( ) (


 
 


1 1 2
 
 
m B m C)
B C
( ) (
4 / 8

9. U
singEq. (**) wecan easilycalculatethetotal evidence, Bel1 and Bel2, in each set, as shown in thetable.
Applying Dempster's rule (Eq. (***)) to m1 and m2, we obtain the joint basic assignment m1,2, which is also
shown in thetable. To determinethevalues of m1,2, wecalculatethenormalization factor 1-Kfirst:

K m B m C
B C
 
 
1 2
( ) ( )
K=m1(R)m2(D)+m1(R)m2(C)+m1(R)m2(DC)+m1(D)m2(R)+m1(D)m2(C)+m1(D)m2(RC)+m1(C)m2(R)+
 +m1(C)m2(D)+m1(C)m2(RD)+m1(RD)m2(C)+m1(RC)m2(D)+m1(DC)m2(R)=0.03
Thenormalization factor is then 1-K=0.97. Values of m1,2 arecalculated byEq. (***). For example:
m1,2(R)=[m1(R)m2(R)+m1(R)m2(RD)+m1(R)m2(RC)+m1(R)m2(RDC)+m1(RD)m2(R)+
 +m1(RD)m2(RC)+m1(RC)m2(R)+m1(RC)m2(RD)+m1(RDC)m2(R)]/0.97=0.21
m1,2(D)=[m1(D)m2(D)+m1(D)m2(RD)+m1(D)m2(DC)+m1(D)m2(RDC)+m1(RD)m2(D)+
 +m1(RD)m2(DC)+m1(DC)m2(D)+m1(DC)m2(RD)+m1(RDC)m2(D)]/0.97=0.01
m1,2(RC)=[m1(RC)m2(RC)+m1(RC)m2(RDC)+m1(RDC)m2(RC)]/0.97=0.2
m1,2(RDC)=[m1(RDC)m2(RDC)]/0.97=0.31
and similarly for the remaining
focal elements C, RD and DC.
The joint basic assignment can now
be used to calculate the joint Belief
Bel1,2 (see table) and joint
plausibilityPl1,2.

Expert 1 Expert 2 Combined evidence
Focal
elements m1 Bel1 m2 Bel2 m1,2 Bel1,2
R 0.05 0.05 0.15 0.15 0.21 0.21
D 0 0 0 0 0.01 0.01
C 0.05 0.05 0.05 0.05 0.09 0.09
RD 0.15 0.2 0.05 0.2 0.12 0.34
RC 0.1 0.2 0.2 0.4 0.2 0.5
DC 0.05 0.1 0.05 0.1 0.06 0.16
RDC 0.6 1 0.5 1 0.31 1
Combination of degrees of evidence fromtwo independent sources
4 / 9

10. • Marginal basic assignment:
Given m: P(X×Y)  [0,1],
Focal elements are binary relations R on X×Y
If RX is RX and RY is RY, the projections of m on X and Y, resp. are the
marginal basic assignments:
" AP(X)
" AP(X) (****)
• Marginal bodies of evidence:
<FX,mX>, <FY,mY>
• Noninteractive marginal basic assignment:
m(A×B)=mX(A)mY(B) " AFX, " BFY
and m(R)=0 for RA×B
m A m R
X
R A RX
( ) ( )
:
 

m B m R
Y
R B RY
( ) ( )
:
 

4 / 10

11.  Example: body of evidence (only focal elements are given)
This is also a case of noninteractive marginal basic assignments
E.g.: m(R1)=0.0625 m(R1)=mX({2,3})*mY({b,c})=0.25*0.25
{2,3}x{b,c}={2b,2c,3b,3c}=R1
Or m(R7)=0.0375 m(R7)=mX({1,3})*mY({a})=0.15*0.25
XxY
1a 1b 1c 2a 2b 2c 3a 3b 3c m(Ri)
R1 = 0 0 0 0 1 1 0 1 1 0.0625
R2 = 0 0 0 1 0 0 1 0 0 0.0625
R3 = 0 0 0 1 1 1 1 1 1 0.125
R4 = 0 1 1 0 0 0 0 1 1 0.0375
R5 = 0 1 1 0 1 1 0 0 0 0.075
R6 = 0 1 1 0 1 1 0 1 1 0.075
R7 = 1 0 0 0 0 0 1 0 0 0.0375
R8 = 1 0 0 1 0 0 0 0 0 0.075
R9 = 1 0 0 1 0 0 1 0 0 0.075
R10 = 1 1 1 0 0 0 1 1 1 0.075
R11 = 1 1 1 1 1 1 0 0 0 0.15
R12 = 1 1 1 1 1 1 1 1 1 0.15
m: P(XxY)  [0,1]
By (****) we obtain:
X
1 2 3 mX(A
)
A = 0 1 1 0.25
1 0 1 0.15
1 1 0 0.3
1 1 1 0.3
mx: P(X)  [0,1]
Y
a b c mY(B)
B = 0 1 1 0.25
1 0 0 0.25
1 1 1 0.5
mY
: P(Y)  [0,1]
4 / 11

12. • Probability measures
If g4Bel is replaced by g4P:
Bel(AB)=Bel(A)+Bel(B), if AB= (additivity)
we obtain a special class of belief measures: probability measures (Bayesian
belief measures)
 Bel is P iff the basic assignment:
m({x})=Bel({x}) and m(A)=0 if A is not singleton
 Probability measures are fully represented by
p: x  [0,1] probability distribution function (as focal elements are only
singletons)
 " AP(X)
i.e.
• Total ignorance:
• Joint and marginal probability distributions - similar way
 Very Broad field
Bel A Pl A m x
x A
( ) ( ) ({ })
  

Bel A Pl A p x
x A
( ) ( ) ( )
  

 
p x
X
m x
( ) ({ })
 
1
4 / 12

13. • Nested subsets:
• Consonant Bel / Pl
The family of focal elements is nested (consonant body of evidence <F, m> )
 Given a consonant <F, m> then
" A,B P(X)
• Necessity measure = consonant Bel
Possibility measure = consonant Pl
(very typical – ‘axiomatic’ – prop’s)
 Every possibility measure can be uniquely determined by a possibility distribution f’n
So that
4 / 13
X
A
A
A
A i
n 


 
2
1
))
(
),
(
max(
)
(
))
(
),
(
min(
)
(
B
Pl
A
Pl
B
A
Pl
B
Bel
A
Bel
B
A
Bel




))
(
),
(
max(
)
(
))
(
),
(
min(
)
(
B
A
B
A
B
A
B
A










)
(
1
)
( A
A 
 

]
1
,
0
[
: 
X
r
)
(
max
π(A) x
r
A
x


14. • Let
where (possibility distribution) length of r: n
• Let when i < j
is the set of all possibility distributions of length n
• Given two possibility distributions (length n)
 is a lattice: lattice of possibility distributions of length n
n = 3
4 / 14
}
,
,
,
{ 2
1 n
x
x
x
X 

)
,
,
,
( 2
1 n
r 

 

)
( i
i x
r


j
i 
 
R
n
R
R
N
n
n



))
,
min(
,
),
,
(min(
))
,
max(
,
),
,
(max(
)
,
,
,
(
)
,
,
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1
1
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2
1
1
1
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n
j
n
i
j
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j
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n
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j
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n
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n
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r
r
r
r
N
i
IFF
r
r
R
r
R
r

























"







,
R
n



 3
3
3 r
r
r i
)
1
,
1
,
1
(
3 

r
)
,
,
( 3
2
1
3 

 i
i
i
i
r 
)
0
,
0
,
1
(
3 

r
Total ignorance
Perfect evidence

15. • ,  is defined on P(X) in terms of its m. So all
focal elements are nested subsets:
4 / 15
}
,
,
,
{ 2
1 n
x
x
x
X 

)
(
2
1 X
A
A
A n 


 
i
n
i
i
A
A
if
A
m
N
i
x
x
A




0
)
(
}
,
,
{ 1 
x1 x2 x3 x4 xn
m(A1)
m(A2) m(A3) m(A4)
m(An)
r(x1)=1
r(x2)=2
r(x3)=3 r(x4)=4
r(xn)=n
…
…
…
Complete sequence of nested subsets of X

16. • Example
4 / 16
x1 x2 x3 x4
m(A2) = 0.3
m(A3) = 0.4 m(A6) = 0.1
r(x1)=1=1
r(x2)=2=1
r(x3)=3=0.7 r(x4)=4=0.3 r(x6)=6=0.3
x5 x6 x7
m(A7) = 0.2
r(x5)=5=0.3
r(x7)=7=0.2
i m(Ai)
1 0
2 0.3
3 0.4
4 0
5 0
6 0.1
7 0.2
A possibility measure defined on X

17. • Basic distribution m
• The set of all basic distributions
 m and r represent each other unambiguously
• (one-to-one correspondance)
4 / 17
1
,
)
(
)
,
,
,
(
1
2
1





n
i
i
i
i
n
A
m
m




 
)
0
(
(*)
:
.
.
)
(
})
({
1
1
2
2
2
1
1
convention
by
i
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i
A
m
x
Pl
n
i
i
i
n
n
n
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n
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i
k
k
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i
k
k
i
i

"










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"
















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









M
M
M
N
n
n
n



(length n)
M
R
t 
:
)
(
)
(
(*)
)
(
2
1
1
1
2
1
m
t
m
t
IFF
m
m
satisfied
is
IFF
m
r
t





Partial ordering defined on according to on
M
n
 R
n

18. • Let’s calculate r from m in the example on 4/16
• Possibility measure:
• Marginal possibility distributions rX and rY of joint possibility distribution r
• Noninteractive sets (possibilistic sense) if
4 / 18




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19. • Example for noninteractive consonant bodies of evidence 4 / 19
y1 y2
x1 x2 x3 x4 y3
1,1 1,2 2,1 2,2 3,1 3,2 4,1 4,2 1,3 2,3 3,3 4,3
X Y
X×Y
1=0.4
3=0.1 4=0.5
12=0.4
32=0.1 42=0.2
43=0.3
2´=0.7
3´=0.3
1=1
2=0.6 3=0.6
4=0.5
11=1
12=1
21=0.6 22=0.6 31=0.6 32=0.6
41=0.5 42=0.5
13=0.3
23=0.3
33=0.3
43=0.3
1´=1
2´=1
3´=0.3
Marginal consonant bodies of evidence
Joint consonant body of evidence
)
,
min( '
j
i
ij 

 
This definition of
noninteraction does not
conform with
Dempster’s rule (4/8)

20. • Example: two consonant bodies of evidence 4 / 20
1 2
.x1 .x2
.y1 .y2
.x1y1
.x2y1
.x1y2
.x2y2
.x1y2
.x1y1
.x2y1
.x2y2
1 0.8
X Y
X×Y
1 0.6
1´ 2´
0.2 0.8
1 2
0.4
0.6
1´
2´
0.2
0.2
0.6
1
0.8
0.6
0.6
0.12
0.08
0.32
0.48
Marginal
possibility
distributions
Marginal
basic
assignment
))
(
),
(
min(
)
,
( y
r
x
r
y
x
r Y
X
 K
C
m
B
m
y
x
m
Y
A
C
B
X

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)
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Joint basic
assignment
Joint possibility
distribution
Joint basic
assigmnet
m r m’
Largest joint poss. distr. that
satisfies the marginal distr’s
Not consonant!
Not a subject of
possibility theory
X×Y

21. • Classification of fuzzy measures
4 / 21
Fuzzy measures
Plausibility m’s
Belief m’s
Nec. m’s
crisp
Poss. m’s
crisp
Probability m’s (additivity)
Nested focal
elements
Focal element is
singleton
Mathematical theory of evidence