人工智能的课件CH13-Uncertainty.ppt

1、智能计算研究中心XIII. UncertaintyAutumn 2012Instructor: Wang XiaolongHarbin Institute of Technology, Shenzhen Graduate SchoolIntelligent Computation Research Center(HITSGS ICRC)2Outline Uncertainty Probability Syntax and Semantics Inference Independence and Bayes Rule3UncertaintyLet action At = leave for ai

2、rport t minutes before flightWill At get 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 trafficHence a purely logical appro

3、ach either1. risks falsehood: “A25 will get me there on time”, or2. leads to conclusions that are too weak for decision making:“A25 will get me there on time if theres no accident on the bridge and it doesnt rain and my tires remain intact etc.”(A1440 might reasonably be said to get me there on time

4、 but Id have to stay overnight in the airport )4Methods for handling uncertaintyDefault or nonmonotonic logic: Assume my car does not have a flat tire Assume A25 works unless contradicted by evidenceIssues: What assumptions are reasonable? How to handle contradiction?Rules with fudge factors: A25 |0

5、.3 get there on time Sprinkler | 0.99 WetGrass WetGrass | 0.7 RainIssues: Problems with combination, e.g., Sprinkler causes Rain?Probability Model agents degree of belief Given the available evidence, A25 will get me there on time with probability 0.045ProbabilityProbabilistic assertions summarize e

6、ffects of laziness: failure to enumerate exceptions, qualifications, etc. ignorance: lack of relevant facts, initial conditions, etc.Subjective probability:Probabilities relate propositions to agents own state of knowledgee.g., P(A25 | no reported accidents) = 0.06These are not assertions about the

7、worldProbabilities of propositions change with new evidence:e.g., P(A25 | no reported accidents, 5 a.m.) = 0.156Making decisions under uncertaintySuppose 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(A1

8、440 gets me there on time | ) = 0.9999 Which action to choose?Depends on my preferences for missing flight vs. time spent waiting, etc. Utility theory is used to represent and infer preferences Decision theory = probability theory + utility theory7SyntaxBasic element: random variableSimilar to propo

9、sitional logic: possible worlds defined by assignment of values to random variables.Boolean random variablese.g., Cavity (do I have a cavity?)Discrete random variablese.g., Weather is one of Domain values must be exhaustive and mutually exclusiveContinuous random variablesElementary proposition cons

10、tructed by assignment of a value to a random variable: e.g., Weather = sunny, Cavity = false (abbreviated as cavity)Complex propositions formed from elementary propositions and standard logical connectives e.g., Weather = sunny Cavity = false8Syntax Atomic event: A complete specification of the stat

11、e of the world about which the agent is uncertainE.g., if the world consists of only two Boolean variables Cavity and Toothache, then there are 4 distinct atomic events:Cavity = false Toothache = falseCavity = false Toothache = trueCavity = true Toothache = falseCavity = true Toothache = true Atomic

12、 events are mutually exclusive and exhaustive9Axioms of probabilityFor 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)de Finetti (1931):if an agent has some degree of belief in a proposition a, then the agent should be able to state odds at which it is indif

13、ferent to bet for or against a.stronger: if agent 1 express a set of degrees of belief that violate the axioms of probability theory then there is a combination of bets by agent 2 that guarantees that agent 1 will lose money every time.10Prior probabilityPrior or unconditional probabilities of propo

14、sitionse.g., P(Cavity = true) = 0.1 and P(Weather = sunny) = 0.72 correspond to belief prior to arrival of any (new) evidenceProbability distribution gives values for all possible assignments:P(Weather) = (normalized, i.e., sums to 1)Joint probability distribution for a set of random variables gives

15、 the probability of every atomic event on those random variablesP(Weather,Cavity) = a 4 2 matrix of values:Weather =sunnyrainycloudysnow Cavity = true 0.1440.02 0.016 0.02Cavity = false 0.5760.08 0.064 0.08Every question about a domain can be answered by the joint distribution11Conditional probabili

16、tyConditional or posterior probabilitiese.g., P(cavity | toothache) = 0.8i.e., given that toothache is all I knowIf we know more, e.g., cavity is also given, then we haveP(cavity | toothache,cavity) = 1New evidence may be irrelevant, allowing simplification, e.g.,P(cavity | toothache, sunny) = P(cav

17、ity | toothache) = 0.8This kind of inference, sanctioned by domain knowledge, is crucial12Conditional 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

18、whole distributions, e.g.,P(Weather,Cavity) = P(Weather | Cavity) P(Cavity) (View as a set of 4 2 equations, not matrix mult.) 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= 1n P(

19、Xi | X1, ,Xi-1)13Inference by enumeration Start with the joint probability distribution: For any proposition , sum the atomic events where it is true: P() = : P()14Inference by enumeration Start with the joint probability distribution: For any proposition , sum the atomic events where it is true: P(

20、) = : P() P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.215Inference by enumeration Start with the joint probability distribution: For any proposition , sum the atomic events where it is true: P() = : P()16Inference by enumeration Start with the joint probability distribution: Can also compute co

21、nditional probabilities:P(cavity | toothache) = P(cavity toothache)P(toothache)= 0.016+0.064 0.108 + 0.012 + 0.016 + 0.064= 0.417Normalization Denominator can be viewed as a normalization constant P(Cavity | toothache) = P(Cavity,toothache) = P(Cavity,toothache,catch) + P(Cavity,toothache, catch)= +

22、 = = General idea: compute distribution on query variable by fixing evidence variables and summing over hidden variables18Inference by enumeration, contd.Let X be all the variables. Typically, we wantthe posterior joint distribution of the query variables Y given specific values e for the evidence v

23、ariables ELet the hidden variables be H = X - Y - EThen the required summation of joint entries is done by summing out the hidden variables:P(Y | E = e) = P(Y,E = e) = hP(Y,E= e, H = h)The terms in the summation are joint entries because Y, E and H together exhaust the set of random variablesObvious

24、 problems:1. Worst-case time complexity O(dn) where d is the largest arity (目)2. Space complexity O(dn) to store the joint distribution3. How to find the numbers for O(dn) entries?19IndependenceA and B are independent iffP(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)P(Toothache, Catch, Cavity

25、, Weather)= P(Toothache, Catch, Cavity) P(Weather)32 entries reduced to 12; for n independent biased coins, O(2n) O(n)Absolute independence powerful but rareDentistry is a large field with hundreds of variables, none of which are independent. What to do?20Conditional independenceP(Toothache, Cavity,

26、 Catch) has 23 1 = 7 independent entriesIf I have a cavity, the probability that the probe catches in it doesnt depend on whether I have a toothache:(1) P(catch | toothache, cavity) = P(catch | cavity)The same independence holds if I havent got a cavity:(2) P(catch | toothache,cavity) = P(catch | ca

27、vity)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)21Conditional independence contd.Write out

28、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 numbersIn most cases, the use of conditional

29、 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.22Bayes RuleProduct rule P(ab) = P(a | b) P(b) = P(b | a) P(a) Bayes rule: P(a | b)

30、 = P(b | a) P(a) / P(b)or in distribution form P(Y|X) = P(X|Y) P(Y) / P(X) = P(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, S be stiff neck:P(m|s) = P(s|m) P(m) / P(s) = 0.8 0.0001 / 0.

31、1 = 0.0008 Note: posterior probability of meningitis still very small!23Bayes Rule and conditional independenceP(Cavity | toothache catch) = P(toothache catch | Cavity) P(Cavity) = P(toothache | Cavity) P(catch | Cavity) P(Cavity) This is an example of a naive Bayes model:P(Cause,Effect1, ,Effectn)

32、= P(Cause) iP(Effecti|Cause) Total number of parameters is linear in n24Summary 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 tools25Assignments Ex 13.15