t02640010220114056bayes theorem & bayes nets_examples

7/31/2019 T02640010220114056Bayes Theorem & Bayes Nets_examples

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Solution Examples2012dks


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2

The notion of conditional probability: P(H\E)

Let:

P(Hi\E) =the probability that hypothesis Hi is true given evidence E

P(E\Hi) =the probability that we will observe evidence E

given that hypothesis i is true

P(Hi) =the a priori probability that hypothesis i is true in theabsence

of specific evidence.

k =the number of possible hypotheses

Bayes' theorem then states that

Further, if we add a new piece of evidence, e, then

P(H\E,e)= P(H\E ). P(e \E, H)

P(e\E)

Bayes Theorem

k

1n)n).P(HnH\P(E

)i

).P(Hi

H\P(EE)\

iP(H


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Conditional Probability

Definition:

Therefore,

can also be obtained as

3

)(

)()|(

Bp

BApBAp

)( BAp

)|()()|()()( ABpApBApBpBAp


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Bayess theorem

is a mechanism for combining new andexisting evidence, usually given asubjective probabilities.

used to revise existing prior probabilitiesbased on new information.

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Bayess theorem

The probability of concluding the hypothesis(hi), given the evidence (e) has the form of aconditional probability:

5

ehp i |


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Examples 1

7

Assume that University ABC has an expert system to

consult the administration officers about studentsadmission.

There are two hypothesis:

h1= accept the student and

h2 = reject the student.

The evidence that can be used in both of thehypothesis is e = the student has good grade.

It was revealed that the system accepts 40% of thestudents who applied to the University.

p(h1) = 0.4

p(h2) = 0.6


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Example 1

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According to the Universitys records, of thosewho have been admitted, 85% of them hasgood grade, and 20% were rejected evenwith good grade.

Consequently, the conditional probabilitiesare:

p(e|h1) = 0.85 p(e|h2) = 0.20


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h1 = accept the student

p(h1 ) = 0.4

h2 = reject the student

p(h2 ) = 0.6

e = the student hasgood grade

p(e/h1 ) = 0.85 p(e/h2 ) = 0.20


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221111

hpe|h+ phpe|hp

hpe|hpp(h1|e) =

0.85 0.4

0.85 0.4 + 0.6 0.2

=

= 0.74


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p(h2|e) =

0.12+0.34

0.12=

0.26

2211

22

hpe|h+ phpe|hp

hpe|hp

=


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Example 2

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h1 = approve the loan

p(h1 ) = 0.70

h2 = reject the load

p(h2 ) = 0.30

e = the applicant hasa steady job

p(e/h1 ) = 0.80 p(e/h2 ) = 0.10


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Example

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2211

11

hpe|h+ phpe|hp

hpe|hpp(h1|e) =

= 0.95


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Limitations of BayesianApproach

14

the number of required conditionalprobabilities increases as more pieces ofevidence are used in the system

even if we can obtain values for theprobabilities estimates, the calculations ofthe posterior probabilities will be too time

consuming


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Kerjakan contoh-contoh tadi tanpamenggunakan rumus, akan tetapi gunakanlah

probability tree!

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Soal 1: Student admission

P(h1) = 0.4 [accepted] P(h2) = 0.6 [rejected]

P(e/h1) = 0.85 e = good grade

P(e/h2) = 0.20

Soal 2: Loan Proposal

P(h1) = 0.70 [approved]

P(h2) = 0.30 [rejected]

P(e/h1) = 0.80 e = steady job

P(e/h2) = 0.10

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S

a

0.4

e

0.85

e

0.15

a

0.6

e

0.20

e

0.80

h1 h2

e e


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1000

10

8 2

990

95 895

(1) Choose a population big enough to guarantee that thenumbers you are dealing with are whole numbers.

(2) Break the population down into a tree diagram given theinformation available.

(3) Calculate the probability in the standard way: favorable

outcomes divided by total outcomes: 8/8+95 = .078


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1% of the population has X disease. A screening test accuratelydetects the disease for 90% of people with it. The test alsoindicates the disease for 15% of the people without it (the falsepositives). Suppose a person screened for the disease testspositive. What is the probability they actually have it?

popul

-ation

0.01

T T

D

T

0.99

D

T

0.850.150.100.90

[0,0571]


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Another example

Lets consider another example of the use of BayesTheorem:

In a certain clinic 0.15 of the patients have got

the HIV virus. Suppose a blood test is carried out on a patient:

If the patient has got the virus the test will turn out

positive with probability 0.95 If the patient does not have the virus the test will turn

out positive with probability 0.02 [false positive]

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If the test is positive what are theprobabilities that the patient

a) has the virus

b) does not have the virus?

If the test is negative what are theprobabilities that the patient

c) has the virusd) does not have the virus?

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B i N t k


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Bayesian Networks


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Here H has two causes S and D. We need to know the probability of H

given each of the four possible combinations of S and D.

Lets suppose a medical survey gives us the following data.


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Given the information we can now answer any

question concerning this network.

Suppose we want to know what the probability of

heart disease is: P(H)

In a Bayesian Network if we wish to know theprobability that nodeNis true, we have to look at

its parent nodes (i.e. its causes).

Listall possible combinationsof the values ofthe parent nodesand then consider the probability

thatNis true for each combination

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In this problem we have to consider all the possible

combinations of the two parent nodes Sand D. There arefour ways he can have heart disease:

He smokes and has bad diet and he has heart disease

He does not smoke and has bad diet and has heartdisease

He smokes and does not have bad diet and has heart

disease

He does not smoke and does not have bad diet and has

heart disease

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We have to work out the probabilities of all four situations and addthem together. Just as in the previous problem we use the SecondAxiom of Probability. For example the probability of each of these

cases is:

The values of all the quantities on the right-hand side are given

above. So we can work out the probabilities of all four cases. P(H) isjust the sum of all four.

P(H) = 0.80.30.4 + 0.50.70.4 + 0.40.30.6 + 0.10.70.6= 0.35


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Suppose we wish to know the probability of heart disease andsmoking [ P(H S) ], i.e. you are told that a patient smokesand has heart disease but you do not know whether he hasbad diet or not. There are two ways in which a patient couldhave heart disease and smoke:

He smokes and has bad diet and he has heart disease He smokes and does not have bad diet and has heart

disease [see previous slide]

From this we can now work out the probability that a patienthas heart disease given that he smokes P(H| S). We use theSecond Axiom


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Tentukan:1. P(H)2. P(HS)3. P(H\S)4. P(E\S)


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t02640010220114056bayes theorem & bayes nets_examples

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