Outline Chapter13

20070531 Chap13 1

Chapter13

Uncertainty

Outline

• Uncertainty

• Probability

• Syntax and Semantics

• Inference

• Independence and Bayes' Rule

20070531 Chap13 3

Uncertainty

Let action A_t= leave for airport t minutes before flight Will A_tget me there on time?

Problems:

1. partial observability (road state, other drivers' plans, etc .) 2. noisy sensors (traffic reports)

3. uncertainty in action outcomes (flat tire, etc .) 4. immense complexity of modeling and predicting traffic Hence a purely logical approach either

1. risks falsehood: “A₂₅will get me there on time”, or

2. leads to conclusions that are too weak for decision making:

“A₂₅will get me there on time if there's no accident on the bridge and it doesn't rain and my tires remain intact etc.”

(A₁₄₄₀might reasonably be said to get me there on time but I'd have to stay overnight in the airport … )

20070531 Chap13 4

Methods for handling uncertainty

• Defaultor nonmonotoniclogic:

- Assume my car does not have a flat tire

- Assume A₂₅works unless contradicted by evidence

• Issues:

What assumptions are reasonable?

How to handle contradiction?

• Rules with fudge factors: - A₂₅|→_0.3get there on time - Sprinkler |→_0.99WetGrass - WetGrass |→_0.7Rain

• Issues: Problems with combination, e.g., Sprinkler causes Rain??

• Probability

- Model agent's degree of belief - Given the available evidence,

- A₂₅will get me there on time with probability 0.04

20070531 Chap13 5

Probability

Probabilistic assertions summarizeeffects of

- laziness: failure to enumerate exceptions, qualifications, etc.

- ignorance: lack of relevant facts, initial conditions, etc.

Subjectiveprobability:

• Probabilities relate propositions to agent's own state of knowledge

e.g., P(A₂₅| no reported accidents) = 0.06 These are notassertions about the world.

Probabilities of propositions change with new evidence:

e.g., P(A₂₅| no reported accidents, 5 a.m.) = 0.15

Making decisions under uncertainty

Suppose I believe the following:

P(A₂₅gets me there on time | …) = 0.04 P(A₉₀gets me there on time | …) = 0.70 P(A₁₂₀ gets me there on time | …) = 0.95 P(A₁₄₄₀gets me there on time | …) = 0.9999

• Which action to choose?

Depends on my preferencesfor missing flight vs. time spent waiting, etc.

- Utility theoryis used to represent and infer preferences - Decision theory = probability theory + utility theory

• Principle of Maximum Expected Utility(MEU)

An agent is rational iff it chooses the action that yields the highest expected utility, averaged over all the possible outcomes of the actions.

20070531 Chap13 7

Syntax

• Basic element: random variable

• Similar to propositional logic: possible worlds defined by assignment of values to random variables.

• Booleanrandom variables

e.g., cavity(do I have a cavity?)

• Discreterandom variables

e.g., weather is one of <sunny,rainy,cloudy,snow>

• Domain values must be exhaustive and mutually exclusive

• Elementary proposition constructed by assignment of a value to a random variable

e.g., weather = sunny, cavity = false (abbrev. as ¬cavity)

• Complex propositions formed from elementary propositions and standard logical connectives

e.g., weather = sunny ∨ cavity = false

20070531 Chap13 8

Syntax

• Atomic event: A complete specification of the state of the world about which the agent is uncertain.

e.g., if the world consists of only two Boolean variables cavity and toothache, then there are 4 distinct atomic events:

cavity = false ∧toothache = false cavity = false ∧ toothache = true cavity = true ∧ toothache = false cavity = true ∧toothache = true

• Atomic events are mutually exclusive and

exhaustive.

20070531 Chap13 9

Axioms of probability

• For any propositions A, B - 0 ≤ P(A) ≤ 1

- P(true) = 1 and P(false) = 0

- P(A ∨ B) = P(A) + P(B) - P(A ∧ B )

Prior probability

• Prioror unconditional probabilitiesof propositions correspond to belief prior to arrival of any (new) evidence e.g., P(cavity = true) = 0.1 and P(weather = sunny) = 0.72

• Probability distribution

gives values for all possible assignments:

e.g., P(weather) = <0.72, 0.1, 0.08, 0.1> (normalized, i.e., sums to 1)

• Joint probability distributionfor a set of random variables gives the probability of every atomic event

e.g., P(weather,cavity)= a 4 × 2 matrix ofvalues:

weather = sunny rainy cloudy snow

cavity = true 0.144 0.02 0.016 0.02 cavity = false 0.576 0.08 0.064 0.08

• Every question about a domain can be answered by the joint distribution.

20070531 Chap13 11

Conditional probability

• Conditional or posterior probabilities

e.g., P(cavity | toothache) = 0.8

i.e., the prob. of having a cavity will be 0.8 given that all we know is toothache

• If we know more, e.g., cavity is also given, then we have

• New evidence may be irrelevant, allowing simplification,

e.g., P(cavity | toothache, sunny) = P(cavity | toothache) = 0.8

• This kind of inference, sanctioned by domain knowledge, is crucial.

20070531 Chap13 12

Conditional probability

• Definition of conditional probability:

P(a | b) = P(a∧ b) / P(b) if P(b) > 0

• Product rule gives an alternative formulation:

P(a∧ b) = P(a | b) * P(b) = P(b | a) * P(a )

• A general version holds for whole distributions,

e.g., P(weather, cavity) = P(weather | cavity) * P(cavity) (View as a set of 4 × 2 equations, notmatrix mult .)

• Chain rule is derived by successive application of product rule:

P(X₁, …, X_n) = P(X₁, ..., X_n-1) * P(X_n| X₁, ..., X_n-1)

= P(X₁, ..., X_n-2) *P(X_n-1| X₁, ..., X_n-2) *P(X_n| X₁, ..., X_n-1)

= …

= P(X₁) *P(X₂| X₁) *P(X₃ | X₁, X₂ ) * … * P(X_n| X₁, ..., X_n-1)

20070531 Chap13 13

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is true: P(φ) = Σ

P(ω )

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is true: P(φ) = Σ

P(ω )

P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2

20070531 Chap13 15

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is true: P(φ) = Σ

P(ω )

P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2 P(cavity∨toothache) = 0.108 + 0.012 + 0.016 + 0.064

+ 0.072 + 0.008 = 0.28

20070531 Chap13 16

Inference by enumeration

• Start with the joint probability distribution:

• Can also compute conditional probabilities:

P(¬cavity | toothache)

= P(¬cavity∧ toothache) P(toothache)

= 0.016+0.064 = 0.08 0.108 + 0.012 + 0.016 + 0.064 0.2

= 0.4

20070531 Chap13 17

Normalization

• Denominator can be viewed as a normalization constantα

P(cavity | toothache) = α P(cavity, toothache)

= α [P(cavity, toothache, catch) + P(cavity, toothache, ¬ catch)]

= α [<0.108, 0.016> + <0.012, 0.064>]

= α<0.12, 0.08> = <0.6, 0.4>

• General idea: compute distribution on query variable by fixing evidence variablesand summing over hidden variables

P(X | e) = αP(X, e) = αΣ_yP(X, e, y)

Inference by enumeration

Typically, we are interested in

the posterior joint distribution of the query variablesX given specific values efor the evidence variablesE

Let the hidden variables(remaining unobserved variable) be Y Then the required summation of joint entries is done by summing

out the hidden variables:

P(X | E = e) = αP(X, E = e) = αΣ_yP(X,E= e, Y = y)

• The terms in the summation are joint entries because X, E and Y together exhaust the set of random variables

• Obvious problems:

1. Worst-case time complexity O(dⁿ) where d is the largest arity 2. Space complexity O(dⁿ)to store the joint distribution

3. How to find the numbers for O(dⁿ) entries?

20070531 Chap13 19

Independence

• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) * P(B )

P(toothache, catch, cavity, weather)

= P(toothache, catch, cavity) * P(weather)

• 32 entries reduced to 8 + 4;

for n independent biased coins, O(2ⁿ) →O(n)

• Absolute independence is powerful but rare.

• Dentistry is a large field with hundreds of variables, none of which are independent. What to do?

20070531 Chap13 20

Conditional Independence

• P(toothache, cavity, catch)has 2³–1 = 7 independent entries

• If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:

P(catch | toothache, cavity) = P(catch | cavity)

• The same independence holds if I haven't got a cavity:

P(catch | toothache, ¬cavity) = P(catch | ¬cavity)

• Catch is conditionally independentof toothache given cavity : P(catch | toothache, cavity) = P(catch | cavity)

• Equivalent statements:

P(toothache | catch, cavity) = P(toothache | cavity )

P(toothache, catch | cavity) = P(toothache | cavity) P(catch | cavity)

20070531 Chap13 21

Conditional independence

• Write out full joint distribution using chain rule:

P(toothache, catch, cavity)

= P(toothache | catch, cavity) P(catch, cavity )

= P(toothache | catch, cavity) P(catch | cavity) P(cavity )

= P(toothache | cavity) P(catch | cavity) P(cavity)

• In most cases, the use of conditional independence reduces the size of the representation of the joint distribution from exponential in n to linear in n .

• Conditional independence is our most basic and robust form of knowledge about uncertain

environments.

Bayes' Rule

• Product rule P(a∧b) = P(a|b) * P(b) = P(b|a) * P(a )

⇒ Bayes' rule: P(a|b) = P(b|a) * P(a) / P(b )

• or in distribution form

P(Y|X) = P(X|Y) * P(Y) / P(X) = αP(X|Y) P(Y)

• Useful for assessing diagnostic probability from causal probability:

- P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect ) e.g., let M be meningitis(腦膜炎), S be stiff neck(頸僵硬):

P(m|s) = P(s|m)*P(m) / P(s) = 0.8 * 0.0001 / 0.1 = 0.0008 Note: posterior probability of meningitis still very small!

20070531 Chap13 23

Bayes' Rule and conditional independence

P(Cavity | toothache ∧ catch)

= αP(Cavity) * P(toothache∧ catch | Cavity)

= αP(Cavity) * P(toothache | Cavity) * P(catch | Cavity) (Catch is conditionally independent of toothache given cavity)

• This is an example of a naïve Bayesmodel:

P(Cause, Effect₁, … , Effect_n)

= P(Cause) π_iP(Effect_i|Cause )

• Total number of parameters is linearin n

20070531 Chap13 24

Wumpus World

• P

=true iff [i, j] contains a pit

• B

=true iff [i, j] is breezy

• Include only B

,B

,B

in the probability model

20070531 Chap13 25

Specifying the probability model

• The full joint distribution is P(P_1,1, . . . , P_4,4, B_1,1,B_1,2,B_2,1)

• Apply product rule:

P(B_1,1,B_1,2,B_2,1| P_1,1, . . . , P_4,4) * P(P_1,1, . . . , P_4,4 ) (Do it this way to get P(Effect|Cause).)

• First term: 1 if pits are adjacent to breezes, 0 otherwise

• Second term: pits are placed randomly, probability 0.2 per square:

P(P_1,1, . . . , P_4,4 ) = for n pits.

n n

j

i ₌ Pij = ⁻

∏

Observations and query

• We know the following facts:

b = ¬b

∧ b

∧ b

known = ¬p

∧ ¬p

∧ ¬p

• Query is P(P

|known, b)

• Define Unknown = P

s other than P

and Known

• For inference by enumeration, we have

P(P_1,3|known, b) = αΣ_unknownP(P_1,3, unknown, known, b)

• Grows exponentially with number of squares!

20070531 Chap13 27

Using conditional independence

P(P_1,3 | known, b) =

α’<0.2(0.04 + 0.16 + 0.16), 0.8(0.04 + 0.16)>

≈ <0.31, 0.69>

P(P_2,2 | known, b) ≈ <0.86, 0.14>

[1,3] or [3,1] contains a pit with roughly 31% probability.

[2,2] contains a pit with roughly 86% probability.

The wumpus agent should definitely avoid [2,2]!

20070531 Chap13 28

Summary

• Probability is a rigorous formalism for uncertain knowledge.

• Joint probability distribution specifies probability of every atomic event.

• Queries can be answered by summing over atomic events.

• For nontrivial domains, we must find a way to reduce the joint size.

• Independence and conditional independence

provide the tools.