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Reasoning Patterns

Date post: 31-Dec-2015
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Representation. Probabilistic Graphical Models. Bayesian Networks. Reasoning Patterns. d 0. d 1. i 0. i 1. 0.6. 0.4. 0.7. 0.3. g 1. g 2. g 3. i 0 ,d 0. 0.3. 0.4. 0.3. i 0 ,d 1. 0.05. 0.25. 0.7. s 0. s 1. i 1 ,d 0. 0.9. 0.08. 0.02. i 0. 0.95. 0.05. i 1 ,d 1. 0.5. - PowerPoint PPT Presentation
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Daphne Koller Bayesian Networks Reasoning Patterns Probabilistic Graphical Models Representation
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Daphne Koller

Bayesian Networks

Reasoning Patterns

ProbabilisticGraphicalModels

Representation

Daphne Koller

The Student Network

IntelligenceDifficulty

Grade

Letter

SAT

0.30.080.250.4g2

0.020.9i1,d0

0.70.05i0,d1

0.5

0.3g1 g3

0.2i1,d1

0.3i0,d0

l1l0

0.99

0.40.1 0.9g1

0.01g3

0.6g2

0.20.95

s0 s1

0.8i10.05i0

0.40.6

d1d0

0.30.7i1i0

Daphne Koller

IntelligenceDifficulty

Grade

Letter

SAT

Causal Reasoning

P(l1) ≈ 0.5

P(l1 | i0 ) ≈

P(l1 | i0 , d0) ≈

Difficulty Intelligence

0.39

0.51

Daphne Koller

IntelligenceDifficulty

Grade

Letter

SAT

Evidential ReasoningP(d1) = 0.4 P(i1) = 0.3P(d1 | g3) ≈ P(i1 | g3) ≈

Student gets a C

0.30.080.250.4g2

0.020.9i1,d0

0.70.05i0,d1

0.5

0.3g1 g3

0.2i1,d1

0.3i0,d0

0.63 0.08

Daphne Koller

IntelligenceDifficulty

Grade

Letter

SAT

Intercausal Reasoning

Class is hard!

Student gets a C

P(i1 | g3, d1) ≈ 0.11

P(d1) = 0.4 P(i1) = 0.3P(d1 | g3) ≈ 0.63 P(i1 | g3) ≈ 0.08

Daphne Koller

Intercausal Reasoning Explained

X2X1

Y OR

X1 X2Y Prob

0 0 0 0.25

0 1 1 0.25

1 0 1 0.25

1 1 1 0.25

Daphne Koller

Difficulty IntelligenceDifficulty

Grade

Letter

SAT

Intercausal Reasoning II

Class is hard!

Student gets a B

P(i1) = 0.3

P(i1 | g2, d1) ≈ 0.34P(i1 | g2) ≈ 0.175

g2

Daphne Koller

Student Aces the SAT• What happens to the posterior

probability that the class is hard?Intelligence

Grade

Letter

SAT

Student gets a C

Difficulty

Student aces the SAT

Daphne Koller

Intelligence

Grade

Letter

SAT

Student Aces the SAT

P(i1 | g3, s1) ≈ 0.58P(d1 | g3, s1) ≈ 0.76

Difficulty

Student aces the SAT

P(d1) = 0.4 P(i1) = 0.3P(d1 | g3) ≈ 0.63 P(i1 | g3) ≈ 0.08

Student gets a C


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