Week9: Uncertainty – COMP30024 – Artificial Intelligence Chapter 12
Week9: Uncertainty
COMP30024 – Artificial Intelligence
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Chapter 12
Dr Wafa Johal
University of Melbourne
• Uncertainty
• Probability
• Syntax and Semantics
• Inference
• Independence and Bayes’ Rule
AIMA Slides © and ; Dr Wafa Johal 2
Uncertainty
Let action At = leave for airport t minutes before flight
get me there on time?
1) partial observability (road state, other drivers’ plans, etc.)
2) noisy sensors (radio traffic reports)
3) uncertainty in action outcomes (flat tyre, etc.)
4) immense complexity of modelling and predicting traffic
Hence a purely logical approach either
1) risks falsehood: “A25 will get me there on time”
or 2) leads to conclusions that are too weak for decision making:
“A25 will get me there on time if there’s no accident on the bridge and it
doesn’t rain and my tires remain intact etc etc.”
(A1440 might reasonably be said to get me there on time but I’d have to stay
overnight in the airport . . .)
AIMA Slides © and ; Dr Wafa Johal 3
Methods for handling uncertainty
Nonmonotonic logic:
Assume my car does not have a flat tire
Assume A25 works unless contradicted by evidence
Issues: What assumptions are reasonable? How to handle contradiction?
Rules with confidence factors:
A25 7→0.3 get there on time
However, consider the example:
Sprinkler 7→0.99 WetGrass
WetGrass 7→0.7 Rain
Issues: Problems with combination, e.g., Sprinkler causes Rain??
Probability
Given the available evidence,
A25 will get me there on time with probability 0.04
(Fuzzy logic handles degree of truth NOT uncertainty e.g.,
WetGrass is true to degree 0.2)
AIMA Slides © and ; Dr Wafa Johal 4
Probability
Probabilistic assertions summarize effects of
laziness: failure to enumerate exceptions, qualifications, etc.
ignorance: lack of relevant facts, initial conditions, etc.
Subjective or Bayesian probability:
Probabilities relate propositions to one’s own state of knowledge
e.g., P(A25|no reported accidents) = 0.06
These are not claims of some probabilistic tendency in the current situation
(but might be learned from past experience of similar situations)
Probabilities of propositions change with new evidence:
e.g., P(A25|no reported accidents, 5 a.m.) = 0.15
AIMA Slides © and ; Dr Wafa Johal 5
Making decisions under uncertainty
Suppose I believe the following:
P(A25 gets me there on time| . . .) = 0.04
P(A90 gets me there on time| . . .) = 0.70
P(A120 gets me there on time| . . .) = 0.95
P(A1440 gets me there on time| . . .) = 0.9999
Which action to choose?
Depends on my preferences for missing flight vs. airport cuisine, etc.
Utility theory is used to represent and infer preferences
Decision theory = utility theory + probability theory
AIMA Slides © and ; Dr Wafa Johal 6
Probability basics
Begin with a set Ω—the sample space
e.g., 6 possible rolls of a die.
ω ∈ Ω is a sample point/possible world/atomic event
A probability space or probability model is a sample space with an assignment
P(ω) for every ω ∈ Ω s.t.
0 ≤ P(ω) ≤ 1∑
e.g., P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6.
An event A is any subset of Ω
E.g., P(die roll < 4) = 1/6 + 1/6 + 1/6 = 1/2 AIMA Slides © and ; Dr Wafa Johal 7 Random variables A random variable is a function from sample points to some range, e.g., the reals or Booleans e.g., Odd(1) = true. P induces a probability distribution for any r.v. X: P(X = xi) = {ω:X(ω)=xi} e.g., P(Odd = true) = 1/6 + 1/6 + 1/6 = 1/2 AIMA Slides © and ; Dr Wafa Johal 8 Propositions Think of a proposition as the event (set of sample points) where the proposition is true Given Boolean random variables A and B: event a = set of sample points where A(ω) = true event ¬a = set of sample points where A(ω) = false event a ∧ b = points where A(ω) = true and B(ω) = true Often in AI applications, the sample points are defined by the values of a set of random variables, i.e., the sample space is the Cartesian product of the ranges of the variables With Boolean variables, sample point = propositional logic model e.g., A = true, B = false, or a ∧ ¬b. Proposition = disjunction of atomic events in which it is true e.g., (a ∨ b) = uiv(¬a ∧ b) ∨ (a ∧ ¬b) ∨ (a ∧ b) =⇒ P(a ∨ b) = P(¬a ∧ b) + P(a ∧ ¬b) + P(a ∧ b) AIMA Slides © and ; Dr Wafa Johal 9 Why use probability? The definitions imply that certain logically related events must have related probabilities e.g., P(a ∨ b) = P(a) + P(b)− P(a ∧ b) de Finetti (1931): an agent who bets according to probabilities that violate these axioms can be forced to bet so as to lose money regardless of outcome. AIMA Slides © and ; Dr Wafa Johal 10 Syntax for propositions Propositional or Boolean random variables e.g., Cavity (do I have a cavity?) Discrete random variables (finite or infinite) e.g., Weather is one of 〈sunny, rain, cloudy, snow〉 Weather = rain is a proposition Values must be exhaustive and mutually exclusive Continuous random variables (bounded or unbounded) e.g., Temp = 21.6; also allow, e.g., Temp < 22.0. Arbitrary Boolean combinations of basic propositions AIMA Slides © and ; Dr Wafa Johal 11 Prior probability Prior or unconditional probabilities of propositions e.g., P(Cavity = true) = 0.1 and P(Weather = sunny) = 0.72 correspond to belief prior to arrival of any (new) evidence Probability distribution gives values for all possible assignments: P(Weather) = 〈0.72, 0.1, 0.08, 0.1〉 (normalized, i.e., sums to 1) Joint probability distribution for a set of r.v.s gives the probability of every atomic event on those r.v.s (i.e., every sample point) P(Weather,Cavity) = a 4× 2 matrix of values: Weather = sunny rain 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 because every event is a sum of sample points AIMA Slides © and ; Dr Wafa Johal 12 Probability for continuous variables Express distribution as a parameterized function of value: P(X = x) = U[18, 26](x) = uniform density between 18 and 26 Here P is a density; integrates to 1. P(X = 20.5) = 0.125 really means P(20.5 ≤ X ≤ 20.5 + dx)/dx = 0.125 AIMA Slides © and ; Dr Wafa Johal 13 Gaussian density AIMA Slides © and ; Dr Wafa Johal 14 Conditional probability Conditional or posterior probabilities e.g., P(cavity|toothache) = 0.8 i.e., given that toothache is all I know NOT “if toothache then 80% chance of cavity” (Notation for conditional distributions: P(Cavity|Toothache) = 2-element vector of 2-element vectors) If we know more, e.g., cavity is also given, then we have P(cavity|toothache, cavity) = 1 Note: the less specific belief remains valid after more evidence arrives, but is not always useful New evidence may be irrelevant, allowing simplification, e.g., P(cavity|toothache, carltonWins) = P(cavity|toothache) = 0.8 This kind of inference, sanctioned by domain knowledge, is crucial AIMA Slides © and ; Dr Wafa Johal 15 Conditional probability Definition of conditional probability: P(a|b) = P(a ∧ b) if P(b) 6= 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 4× 2 set of equations, not matrix multiplication) Chain rule is derived by successive application of product rule: P(X1, . . . ,Xn) = P(X1, . . . ,Xn−1) P(Xn|X1, . . . ,Xn−1) = P(X1, . . . ,Xn−2) P(Xn−1|X1, . . . ,Xn−2) P(Xn|X1, . . . ,Xn−1) i=1 P(Xi|X1, . . . ,Xi−1) AIMA Slides © and ; Dr Wafa Johal 16 Inference by enumeration Start with the joint distribution: catch catch catch catch For any proposition ϕ, sum the atomic events where it is true: ω:ω|=ϕ P(ω) AIMA Slides © and ; Dr Wafa Johal 17 Inference by enumeration Start with the joint distribution: catch catch catch catch For any proposition ϕ, sum the atomic events where it is true: ω:ω|=ϕ P(ω) P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2 AIMA Slides © and ; Dr Wafa Johal 18 Inference by enumeration Start with the joint distribution: catch catch catch catch For any proposition ϕ, sum the atomic events where it is true: ω:ω|=ϕ P(ω) P(cavity ∨ toothache) = 0.108 + 0.012 + 0.072 + 0.008 + 0.016 + 0.064 = 0.28 AIMA Slides © and ; Dr Wafa Johal 19 Inference by enumeration Start with the joint distribution: catch catch catch catch Can also compute conditional probabilities: P(¬cavity|toothache) = P(¬cavity ∧ toothache) P(toothache) 0.016 + 0.064 0.108 + 0.012 + 0.016 + 0.064 AIMA Slides © and ; Dr Wafa Johal 20 Normalization catch catch catch catch Denominator can be viewed as a normalization constant α 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 variables and summing over hidden variables AIMA Slides © and ; Dr Wafa Johal 21 Inference by enumeration, contd. Typically, we are interested in the posterior joint distribution of the query variables Y given specific values e for the evidence variables E Let the hidden variables be H = X − Y − E Then the required summation of joint entries is done by summing out the hidden variables: P(Y|E = e) = αP(Y,E = e) = α P(Y,E = e,H = h) The terms in the summation are joint entries because Y, E, and H 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??? AIMA Slides © and ; Dr Wafa Johal 22 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) Toothache Catch decomposes into Toothache Catch P(Toothache,Catch,Cavity,Weather) = P(Toothache,Catch,Cavity)P(Weather) 32 entries reduced to 12; for n independent biased coins, 2n → n Absolute independence powerful but rare Dentistry is a large field with hundreds of variables, none of which are independent. What to do? AIMA Slides © and ; Dr Wafa Johal 23 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: (1) P(catch|toothache, cavity) = P(catch|cavity) The same independence holds if I haven’t got a cavity: (2) P(catch|toothache,¬cavity) = P(catch|¬cavity) Catch is conditionally independent of 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) AIMA Slides © and ; Dr Wafa Johal 24 Conditional independence contd. 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) I.e., 2 + 2 + 1 = 5 independent numbers (equations 1 and 2 remove 2) 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. AIMA Slides © and ; Dr Wafa Johal 25 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) or in distribution form P(Y|X) = P(X|Y)P(Y) = αP(X|Y)P(Y) Useful for assessing diagnostic probability from causal probability: P(Cause|Effect) = P(Effect|Cause)P(Cause) E.g., let M be meningitis, S be stiff neck: P(m|s) = P(s|m)P(m) 0.8× 0.0001 Note: posterior probability of meningitis still very small! AIMA Slides © and ; Dr Wafa Johal 26 Bayes’ Rule and conditional independence P(Cavity|toothache ∧ catch) = αP(toothache ∧ catch|Cavity)P(Cavity) = αP(toothache|Cavity)P(catch|Cavity)P(Cavity) This is an example of a naive Bayes model: P(Cause,Effect1, . . . ,Effectn) = P(Cause) P(Effecti|Cause) Effect1 Effectn Total number of parameters is linear in n AIMA Slides © and ; Dr Wafa Johal 27 Case Study: Searching for the wreckage of Air France AF 447 References: based on L. Stone, C. Keller, T. Kratzke, J. Strumpfer “Search for the Wreckage of Air France Flight AF 447”, to appear in Statistical Science 1 and L. Stone, C. Keller, T. Kratzke, J. Strumpfer “Search Analysis for the Underwater Wreckage of Air France Flight 447” Fusion 2011 - July 7, Chicago, USA 2 AIMA Slides © and ; Dr Wafa Johal 28 http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.370.2913&rep=rep1&type=pdf http://www.sarapp.com/docs/AF447 Slides for INFORMS Jun 2011.pdf Air France Flight AF 447 1 June 2009 – AF 447 disappeared in the Atlantic Ocean with the loss of 228 passengers and crew 2 June 2009 – Wreckage sighted by search aircraft May 2010 – Black boxes still not found July 2010 – U.S. company Metron engaged to use Bayesian analysis of evidence to redirect search 20 January 2011 – Metron deliver their report on probability map to guide search Late March 2011 – Search resumed based on prob. map 3 April 2011 – Wreckage found on ocean floor AIMA Slides © and ; Dr Wafa Johal 29 How can we use Bayesian analysis? We’ll use their work to motivate a very simplified example of this type of analysis At the time of writing, a similar search is underway for Malaysian Airlines Flight MH 370 Based on the locations of where “pings” were heard from the black box flight recorders,a grid search is being undertaken in southern Indian Ocean by Australia, China, Japan, Malaysia, , South Korea, United Kingdom and United States AIMA Slides © and ; Dr Wafa Johal 30 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 Examples of skills expected: • Calculate conditional probabilities using inference by enumeration • Use conditional independence to simplify probability calculations • Use Bayes’ rule for solving diagnostic problems AIMA Slides © and ; Dr Wafa Johal 31 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com