modelling uncertainty

Dr hab. inż. Joanna Józefowska, prof. PP

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Modelling uncertainty


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Probability of an event

• Classical method: If an experiment has n possible outcomes assign a probability of 1/n to each experimental outcome.

• Relative frequency method:Probability is the relative frequency of the number of events satisfying the constraints.

• Subjective method:Probability is a number characterising the likelihood of an event – degree of belief


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Axioms of the probability theory

Axiom I The probability value assigned to each experimental outcome must be between 0 and 1.

Axiom II The sum of all the experimental outcome probabilities must be 1.


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Conditional probability

denoted by P(A|B) expresses belief that event A is true assuming that event B is true (events A and B are dependent)

Definition

Let the probability of event B be positive. Conditional probability of event A under condition B is calculated as follows:

0P(B) re whe,P(B)

B)P(A,B)|P(A


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Joint probability

If events A1, A2,... Are mutually exclusive and cover the

sample space , and P(Ai) > 0 for i = 1, 2,... then for any

event B the following equality holds:

i

ii ))P(AA|P(BP(B)


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Bayes’ Theorem

If the events A1, A2,... fulfil the assumptions of the joint

probability theorem, and P(B) > 0, then for i =1, 2,... The

following equality holds:

iii

iii ))P(AA|P(B

))P(AA|P(BB)|P(A

Thomas Bayes (1701-1761)


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Bayes’ Theorem

Let us denote:

H – hipothesis

E – evidence

The Bayes’ rule has the form:

P(E)H)P(H)|P(E

E)|P(H

Prior probabilities

New information

Bayes’ theorem

Posterior probabilities


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Difficulties with joint probability distribution (tabular approach)

• the joint probability distribution has to be defined and stored in memory

• high computational effort required to calculate marginal and conditional probabilities


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B E A J M P(B,E,A,J,M)1 1 1 1 1 0,00011970001 1 1 1 0 0,00001330001 1 1 0 1 0,00005130001 1 1 0 0 0,00000570001 1 0 1 1 0,00000000501 1 0 1 0 0,00000009501 1 0 0 1 0,00000049501 1 0 0 0 0,00000940501 0 1 1 1 0,00580356001 0 1 1 0 0,00064484001 0 1 0 1 0,00248724001 0 1 0 0 0,00027636001 0 0 1 1 0,00000029401 0 0 1 0 0,00000558601 0 0 0 1 0,00002910601 0 0 0 0 0,00055301400 1 1 1 1 0,00361746000 1 1 1 0 0,00040194000 1 1 0 1 0,00155034000 1 1 0 0 0,00017226000 1 0 1 1 0,00000702900 1 0 1 0 0,00013355100 1 0 0 1 0,00069587100 1 0 0 0 0,01322154900 0 1 1 1 0,00061122600 0 1 1 0 0,00006791400 0 1 0 1 0,00026195400 0 1 0 0 0,00002910600 0 0 1 1 0,00048461490 0 0 1 0 0,00920768310 0 0 0 1 0,0479768751

n sample points

2n probabilities

P(B,M)

P(M)

M)P(B,M)|P(B


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Certainty factor

• Buchanan, Shortliffe 1975• Model developed for the rule expert system

MYCIN

If E then H

evidence (observation)

hipothesis


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Belief

othewise

1P(H) if

P(H)max{1,0}

P(H)P(H)}E),|max{P(H

1

E]MB[H,

• MB[H, E] – measure of the increase of belief that H is true based on observation E.


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Disbelief

• MD[H, E] – measure of the increase of disbelief that H is true based on observation E.

otherwise

0P(H) if

P(H)min{1,0}

P(H)P(H)}E),|min{P(H

1

E]MD[H,


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Certainty factor

E]MD[H,E]MB[H,E)CF(H,

CF [–1, 1]


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Interpretation of the certainty factor

Certainty factor is associated with a rule:

If evidence then hipothesis

and denotes the change in belief that H is true after observation E.

E HCF(H, E)


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Uncertainty propagation

E1

H

CF(H, E1)

E2 CF(H, E2)

Parallel rules

E1, E2 HCF(H, E1&E2)

otherwise )]E MB(H,[1-*)E MB(H,)E MB(H,

1)E&EMD(H, if

)E&EMB(H,

121

21

21

0

otherwise )]E MD(H,[1-*)E MD(H,)E MD(H,

1)E&EMB(H, if

)E&EMD(H,

121

21

21

0


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Uncertainty propagation

E1 HCF(E2, E1)

E2

CF(H, E2)

Serial rules

E1 HCF(H, E1)

otherwise )E)CF(H,E,CF(E

0)E,CF(E if )E)CF(H,E,CF(E)ECF(H,

212

122121

If CF(H,E2) is not defined, it is assumed to be 0.


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Certainty measure

E HCF(H, E)

C(E) C(H)

0H)(EC(H)CF' if |}H)(ECF'|,|C(H)min{|1

H)(ECF'C(H)

0H)(ECF'C(H), if H)(EC(H))CF'(1C(H)

0H)(ECF'C(H), if H)(EC(H))CF'(1C(H)

(H)C'

C(E)}max{0,H)CF(EH)(ECF'

Grzymała-Busse 1991


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Example 1

C(s1 s2) = min(0,2; – 0,1) = – 0,1

CF’(h, s1 s2) = 0,4 * 0 = 0

s1h

CF(h, s1 s2) = 0,4

s2

C(s1) = 0,2

C(s2) = – 0,1C(h) = 0,3

C’(h) = 0,3 + (1– 0,3) * 0 = 0,3 + 0 = 0,3

0H)(ECF'C(H), if H)(EC(H))CF'(1C(H)(H)C'



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Example 2

C(s1 s2) = min(0,2; 0,8) = 0,2

CF’(h, s1 s2) = 0,4 * 0,2 = 0,08

s1h

CF(h, s1 s2) = 0,4

s2

C(s1) = 0,2

C(s2) = 0,8C(h) = 0,3

C’(h) = 0,3 + (1– 0,3) * 0,08 = 0,3 + 0,7 * 0,08 = 0,356

0H)(ECF'C(H), if H)(EC(H))CF'(1C(H)(H)C'



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Dempster-Shafer theory

Each hipothesis is characterised by two values: balief and plausibility.

It models not only belief, but also the amount of acquired information.


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Density probability function

ΘA

Θ

1m(A)

0]m[

0,12 :m


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Belief

Belief Bel [0,1] measures the value of acquired information supporting the belief that the considered set hipothesis is true.

ABm(B)Bel(A)


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Plausibility

Plausibility Pl [0,1] measures how much the belief that A is true is limited by evidence supporting A.

A)Bel(1Pl(A)


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Combining various sources of evidence

Assume two sources of evidence: X and Y represented by

respective subsets of : X1,...,Xm and Y1,...,Yn. Probability

density functions m1 and m2 are defined on X and Y

respectively. Combining observations from two sources a

new value m3(Z) is calculated for each subset of as

follows:

jYiX j2i1

ZjYiX j2i1

3 )(Y)m(Xm1

)(Y)m(Xm

(Z)m


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Example

={A, F, C, P} m1() = 1

Observation 1 m2({A, F, C}) = 0,6

m2() = 0,4

m1() = 1

m2({A, F, C}) = 0,6

m3({A, F, C}) = 0,6

m2() = 0,4

m3() = 0,4

A – allergyF – fluC – coldP - pneumonia


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Example

Observation 2 m4({F,C,P}) = 0,8

m4() = 0,2

m4({F,C,P}) = 0,8

m5({F,C}) = 0,48

m4() = 0,2

m5({A,F,C}) = 0,12

m3({A, F, C}) = 0,6 m3() = 0,4

m3({A,F,C}) = 0,6

m3() = 0,4 m5({F,C,P}) = 0,32 m5() = 0,08


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Example

Observation 3 m6({A}) = 0,75

m6() = 0,25

m6({A}) = 0,75

m7() = 0,36

m6() = 0,25

m7({F,C}) = 0,12

m7({A}) = 0,09 m7({A,F,C}) = 0,03

m5({F,C}) = 0,48 m5({A,F,C}) = 0,12

m5({F,C,P}) = 0,32 m5() = 0,08

m5({F,C}) = 0,48

m5({A,F,C}) = 0,12

m5({F,C,P}) = 0,32

m5() = 0,08

m7() = 0,24 m7({F,C,P}) = 0,08

m7({A}) = 0,06 m7() = 0,02

m7({A}) = 0,15


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Example

Observation 3 m6({A}) = 0,75

m6() = 0,25

m6({A}) = 0,75

m7() = 0,36

m6() = 0,25

m7({F,C}) = 0,12

m7({A}) = 0,09 m7({A,F,C}) = 0,03

m5({F,C}) = 0,48 m5({A,F,C}) = 0,12

m5({F,C,P}) = 0,32 m5() = 0,08

m5({F,C}) = 0,48

m5({A,F,C}) = 0,12

m5({F,C,P}) = 0,32

m5() = 0,08

m7() = 0,24 m7({F,C,P}) = 0,08

m7({A}) = 0,06 m7() = 0,02

m7() = 0,6


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Example

m7({F,C}) = 0,12

m7({A}) = 0,15

m7({A,F,C}) = 0,03

m7({F,C,P}) = 0,08

m7() = 0,02

m7({F,C}) = 0,3

m7({A}) = 0,375

m7({A,F,C}) = 0,075

m7({F,C,P}) = 0,2

m7() = 0,05

{A}: [0,375, 0,500]

{F}: [0, 0,625]

{C}: [0, 0,625]

{P}: [0, 0,250]

1 – 0,3 – 0,2

1 – 0,375

1 – 0,375 – 0,3 – 0,075


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Fuzzy sets (Zadeh)

Rough sets (Pawlak)


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Probabilistic reasoning

alarm

earthquakeburglary

John Mary


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B – burglary

E – earthquake

A – alarm

J – John calls

M – Mary calls

Joint probability distribution – P(B,E,A,J,M)

?


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Joint probability distribution

B E A J M P(B,E,A,J,M)1 1 1 1 1 0,00011970001 1 1 1 0 0,00001330001 1 1 0 1 0,00005130001 1 1 0 0 0,00000570001 1 0 1 1 0,00000000501 1 0 1 0 0,00000009501 1 0 0 1 0,00000049501 1 0 0 0 0,00000940501 0 1 1 1 0,00580356001 0 1 1 0 0,00064484001 0 1 0 1 0,00248724001 0 1 0 0 0,00027636001 0 0 1 1 0,00000029401 0 0 1 0 0,00000558601 0 0 0 1 0,00002910601 0 0 0 0 0,00055301400 1 1 1 1 0,00361746000 1 1 1 0 0,00040194000 1 1 0 1 0,00155034000 1 1 0 0 0,00017226000 1 0 1 1 0,00000702900 1 0 1 0 0,00013355100 1 0 0 1 0,00069587100 1 0 0 0 0,01322154900 0 1 1 1 0,00061122600 0 1 1 0 0,00006791400 0 1 0 1 0,00026195400 0 1 0 0 0,00002910600 0 0 1 1 0,00048461490 0 0 1 0 0,00920768310 0 0 0 1 0,0479768751


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What is the probability of a burglary if Mary called? P(B=y|M=y) ?

JE,A,

M)J,A,E,P(B,M)P(B,

Marginal probability:

Conditional probability:

0.1333130.055205370.0084917

0.0084917P(M)

B)P(M,M)|P(B

B M P(B,E,A,J,M)1 1 0,00849171 0 0,00150830 1 0,055205370 0 0,93479463


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Advantages of probabilistic reasoning

• Sound mathematical theory • On the basis of the joint probability distribution one can

reason about:– the reasons on the basis of the observed

consequences,– consequences on the basis of given evidence,– Any combination of the above ones.

• Clear semantics based on the interpretation of probability.

• Model can be taught with statistical data.


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Complexity of probabilistic reasoning

• in the „alarm” example– (25 – 1) = 31 values,– direct acces to unimportant information, e.g.

P(B=1,E=1,A=1,J=1,M=1) – calculating any practical value, e.g. P(B=1|M=1)

requires 29 elementary operations.

• in general– P(X1, ..., Xn) requires storing 2n-1 values

– difficult knowledge acquisition (not natural)– exponential complexity


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Bayes’ theorem

P(E)H)P(H)|P(E

E)|P(H


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Bayes’ theorem

BA

B depends on A

P(B)A)P(A)|P(B

B)|P(A

P(B|A)

A

A)P(A)|P(BP(B)


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The chain rule

P(X1,X2) = P(X1)P(X2|X1)

P(X1,X2,X3) = P(X1)P(X2|X1)P(X3|X1,X2)

................................................................

P(X1,X2,...,Xn) = P(X1)P(X2|X1)...P(Xn|X1,...,Xn-1)


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Conditional independence of variables in a domain

In any domain one can define a set of variables pa(Xi){X1, ..., Xi–1} such that Xi is independent of variables from the set {X1, ..., Xi–1} \ pa(Xi).

Thus

P(Xi|X1, ..., Xi – 1) = P(Xi|pa(Xi))

and

P(X1, ..., Xn) = P(Xi|pa(Xi))i=1

n


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Bayesian network

B1

A

B2 Bn

C1

.....

Bi directly influences A

Cm.....

P(A|B1, ..., Bn)


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Example

alarm

Mary callsJohn calls

earthquakeburglary

burglary earthquake P(alarm|burglary, earthquake)

true falsetrue true 0.950 0.050true false 0.940 0.060 false true 0.290 0.710false false 0.001 0.999


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Example

A

MJ

EB

P(B)

0.001P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05


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Complexity of the representation

• Instead of 31 values it is enough to store 10.

• Easy construction of the model– Less parameters.– More intuitive parameters.

• Easy reasoning.


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Bayesian networks

Bayesian network is an acyclic directed graph which

• nodes represent formulas or variables in the considered domain,

• arcs represent dependence relation of variables, with related probability distributions.


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Bayesian networks

variable A with parent nodes pa(A) = {B1,...,Bn}

conditional probablity table P(A|B1,...,Bn) or P(A|pa(A))

if pa(A) = a priori probability equals P(A)


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Bayesian networks

B1

A

B2 BnB3 .....

pa(A)

P(A|B1, B2, ..., Bn)

Event Bi has no

predecesors (pa(Bi) = ) a priori probability P(Bi)

B1 ... Bn P(A|B1, Bn)

T T 0.18 T F 0.12 ................................. F F 0.28


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Local semantics of Bayesian network

• Only direct dependence relations between variables.

• Local conditional probability distribution.

• Assumption about conditional independence of variables not bounded in the graph.


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Global semantics of bayesian network

Joint probability distribution given implicite.

It can be calculated using the following rule:

i

n1iin1 )A,...,A|P(A)A,...,P(A

))P(AA|)...P(AA,...,A,A|P(A nn1n-n321


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Global semantics of bayesian network

Node numbering: node index is smaller than indices of its predecessors.

Finally:

Bayesian network is a complete probabilistic model.

))pa(A|P(A)A,...,A|P(A iin1ii

i

iin1 ))pa(A|P(A)A,...,P(A


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A2

Global probability distribution

B1

A1

B2 BnB3 .....

P(A1|B1, ...Bn) P(A2|B3, ...Bn)

pa(A1)

pa(A2)

B1 ... Bn A1 A2 P(A1,A2,B1, ...Bn)

T ... T T T T ... T T F ...................................................... F ... F F F


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Global probability distribution

A1

B2 BnB3 .....

A2

pa(A1)

pa(A2)

B1

P(A1|B1, ...Bn)

B1 ... Bn P(A1)

T ... T 0.25 T ... F .......................... F ... F

B1

B1 ... Bn A1 A2 P(A1,A2,B1, ...Bn)

T ... T T T 0.075 T ... T T F ...................................................... F ... F F F

B1

P(A2|B3, ...Bn)

B3 ... Bn P(A2)

T ... T 0.30 T ... F .......................... F ... F


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Reasoning in Bayesian networks

Updating evidence that a hipothesis H is true given some ecidence E, i.e. defining conditional probability distribution P(H|E).

Two types of reasoning:

•probability of a single hipothesis

•probability of all hipothesis.


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Example

A

MJ

EB

P(B)

0.001P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

John calls (J) and Mary calls (M). What is the probability that neither

burglary nor earthquake occurred if the alarm

rang?


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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

E)BAMP(J

E)B)P(E)P(B|A)P(A|A)P(M|P(J

i

iin1 ))pa(A|P(A)A,...,P(A

?E)BAMP(J


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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

?E)BAMP(J

E)BAMP(J E)B)P(E)P(B|A)P(A|A)P(M|P(J


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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

?E)BAMP(J



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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

?E)BAMP(J



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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

?E)BAMP(J



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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

?E)BAMP(J



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Example

A

MJ

EB

P(B)

0.001

P(E)

0.002

B E P(A)

T T 0.950T F 0.940F T 0.290F F 0.001

A P(M)

T 0.70F 0.01

A P(J)

T 0.90F 0.05

?E)BAMP(J

E)BAMP(J

0.00062 0.998*0.999*0.001*0.7*0.9 E)B)P(E)P(B|A)P(A|A)P(M|P(J


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Types of reasoning in Bayesian networks

B

J

Evidence B occurs and we qould like to update probability of hipothesis J.

Interpretation.

There was a burglary, what is the probability that John will call?

A

P(J|B) = P(J|A)P(A|B) = 0.9 * 0.95 = 0.86

P(B) = 0.001

A P(J)

T 0.90F 0.05

B P(A)

T 0.95F 0.01


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Wnioskowanie diagnostyczne

We observe J – what is the probability that B is true?

Diagnosis.

John calls. What is the probability of a burglary?

B

J

A

P(B|J) = P(J|B)*P(B)/P(J) =

(0,95*0,9*0,001)/(0,9+0,05) = 0,0009

P(B) = 0.001

A P(J)

T 0.90F 0.05

B P(A)

T 0.95F 0.01

diagnostic


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B E

We observe E. What is the probability that B is true?

Alarm rang, so P(B|A) = 0.376, but if earthuake is observed as well then P(B|A,E) = 0.03

A


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E

A

J

mixed

We observe E and J What is the probability of A.

John calls and we know that there was an earthquake. What is the probability that alarm rang?

P(A|J,E) = 0.03


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Multiply connected Bayesian network

B1

A2

B2 Bn

C1

.....

Cm.....

...A1 Ak


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Summary

• Models of uncertainty:

• Certainty factor, certainty measure

• Dempster-Shafer theory

• Bayesian networks

• Fuzzy sets

• Raough sets


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Summary

• Bayesian networks represent joint probability distribution.

• Reasoning in multiply connected BN is NP-hard.

• Exponential complexity may be avoided by:

• Constructing the net as a polytree

• Transforming a network to a polytree

• Approximate reasoning

modelling uncertainty

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