Artificial Intelligence CS 6364

Transcription

1 Artificial Intelligence CS 6364 rofessor Dan Moldovan Section 12 robabilistic Reasoning

2 Acting under uncertainty Logical agents assume propositions are - True - False - Unknown acting under uncertainty Example: diagnosis for medicine dental diagnosis using 1 st order logic: Wrong: not all patients p with toothaches have cavities! Some have gum disease, an abscess or other problems p Symptom p, Toothache Disease p, Cavity Conclusion: To make the rule true, we have to add an almost unlimited list of possible causes p Symptom p, Toothache Disease p, Cavity Disease p, GumDisease Disease p, Abscess 429

3 Medical Diagnosis Trying to use first-order logic to cope with a domain like medical diagnosis fails because 1. Laziness too much work to list the complete list of rules + too hard to use such rules Example of causal rule: p Disease p, Cavity Symptom p, Toothache wrong, not all cavities cause pain need to augment the antecedent with all conditions that cause toothaches 2. Theoretical ignorance: Medical science has no complete theory for the domain 3. ractical ignorance: Even if we know all the rules, we might be uncertain about a particular patient, because not all necessary tests have been or can be run 430

4 Degree of belief When propositions are not known to be true or false, the agent can at best provide a degree of belief in relevant sentences. The main tool for dealing with degrees of belief is the robability Theory function DT-AGENTpercept returns an action static belief_state, probabilistic beliefs about the current state of the world action, the agent s action update belief_state based on action and percept calculate outcome probabilities for actions, given action descriptions and current belief_state select action with highest expected utility given probabilities of outcomes and utility information return action 431

5 Basic robability Notation Random Variables thought as referring to a part of the world whose status is unknown Random variables have domains values they may take. Depending on the domain, random variables may be classified as - Boolean random variables have the domain <true, false> Example: Cavity = true Cavity = false cavity - Discrete random variables take values from a countable domain - Continuous random variables values from the real numbers Example: the proposition x=4.20 asserts that the random variable x has the exact value We can also have propositions that use inequalities like x

6 Atomic Events An atomic event is a complete specification of the state of the world about which the agent is uncertain. If the world is described by a set of random variables, an atomic event is a particular assignment of values to the random variables Example: 2 random variables: Cavity and Toothache How many atomic events? 4 E E E E cavity cavity cavity cavity false toothache false toothache true toothache true toothache false true false true roperties: a Mutually exclusive: at most one can be true b The set of all possible atomic events is exhaustive 433

7 rior probability rior probability or unconditional probability associated with proposition a is the degree of belief accorded to it in the absence of any other information Example: pcavity = true = 0.1 Important: a can be used only when there is no other information. As soon as some new information is known, we must reason with the conditional probability of a, given that new information Sometimes we are interested in all possible values of a random variable use expressions such as weather which denotes a vector of values for the probabilities of each individual state of the weather Example: weather = sunny = 0.7 weather = rain = 0.2 weather = cloudy = 0.08 weather = snow = 0.02 also written as: weather = <0.7, 0.2, 0.08, 0.02> 434

8 robability distributions Example 1: Hair_color is a discrete random variable. It has the domain <blond, brown, red, black, white, none> Out of a sample of people, we find that 1872 had blond hair, 4325 had brown hair, 2135 had black hair, 652 had red hair, 321 had white hair and the rest were bald The probability distribution is: Hair_color = <0.1872, , , , , > Example 2: UTD_student is a binary random variable the domain is True or False Out of a sample of 1000 youngsters between 19 and 21 in the Dallas area, 321 were students at UTD UTD_student = <0.321, 0.679> 435

9 Full Joint robability Distribution Suppose the world consists only of the variables Cavity Toothache Weather 2 values binary random variable 2 values binary random variable 4 values sunny, rain cloudy, snow Cavity = <0.23,0.77> Toothache = <0.35,0.65> Weather = <0.7,0.2,0.08,0.02> Represent the joint probability distribution: Cavity Cavity Weather Toothache Toothache Toothache Toothache Sunny???? Rain???? Cloudy???? 2x2x4=16 entries for the joint distribution We need to know to compute conditional probabilities Snow???? 436

10 robability Density Functions For continuous variables, it is not possible to write out the entire distribution as a table, because there are infinitely many values. Instead, we define the probability that a random variable takes on some value x as a parameterized function of x. Example: Let the random variable X denote tomorrow s maximum rain fall in Dallas. The sentence X=x = U[1,3]x expresses the belief that X is distributed uniformly between 1in and 3in 437

11 Conditional robability When new evidence concerning a previously unknown random variable is found, prior probabilities no longer apply. We use conditional probabilities: - If a a random variable - b a random variable - a b denotes the probability of a, given that all we know is b Example: cavity toothache = 0.8 How do we compute a b? a b a b b Same rule: ab = a bb the product rule! 438

12 Conditional Distributions If two random variables X and Y define the world, X Y gives the values for X=x i Y=y i for each possible i and j. Expressed with the product rule, this becomes: X=x 1 Y=y 1 = X=x 1 Y=y 1 Y=y 1 X=x 1 Y=y 2 = X=x 1 Y=y 2 Y=y 2 This can be combined in a single equation: X,Y = X YY This denotes a set of equations relating the corresponding individual entries in the tables, not a matrix multiplication of the tables! 439

13 Axioms of robability 1. 0 a 1 2. true = 1, false = 0 3. ab = a + b ab We can deduce a = 1 a because true = a + a false aa =1 ab = 0 a + a = 1 440

14 Inference Using Full Joint Distribution Example: The domains consist of three boolean variables: Toothache, Cavity and Catch the dentist s nasty steel probe catches in my tooth Toothache Toothache catch catch catch catch Cavity Cavity How many atomic events? 2 3 = 8 as many entries as in the table! e 1 cavity=falsetoothache=falsecatch=false e 1 =0.576 e 2 cavity=falsetoothache=falsecatch=true e 2 =0.144 e 3 cavity=falsetoothache=truecatch=false e 3 =0.064 e 4 cavity=falsetoothache=truecatch=true e 4 =0.016 e 5 cavity=truetoothache=falsecatch=false e 5 =0.008 e 6 cavity=truetoothache=falsecatch=true e 6 =0.072 e 7 cavity=truetoothache=truecatch=false e 7 =0.012 e 8 cavity=truetoothache=truecatch=true e 8 =

15 Inferring robabilities Given any proposition a, we can derive its probability as the sum of the probabilities of the atomic events in which it holds Example 1: a = cavity toothache Toothache Toothache catch catch catch catch Cavity Cavity a e e a i e a = cavity toothache Six events: e 5, e 6, e 7, e 8, e 3, e 4 cavity = =0.2 i e 1 cavity=falsetoothache=falsecatch=false e 1 =0.576 e 2 cavity=falsetoothache=falsecatch=true e 2 =0.144 e 3 cavity=falsetoothache=truecatch=false e 3 =0.064 e 4 cavity=falsetoothache=truecatch=true e 4 =0.016 e 5 cavity=truetoothache=falsecatch=false e 5 =0.008 e 6 cavity=truetoothache=falsecatch=true e 6 =0.072 e 7 cavity=truetoothache=truecatch=false e 7 =0.012 When adding all probabilities in a row we obtain the unconditional or marginal probability e 8 cavity=truetoothache=truecatch=true e 8 =

16 Marginalization and Conditioning Rules Given any two random variables Y and Z, Y Y, Z z marginalization rule A variant involves conditional probabilities, instead of joint probabilities: conditioning rule cavity = 0.2 cavity=0.8 toothache= =0.2 toothache=0.8 catch= =0.34 catch=0.66 Y Y Z Z z 443

17 444 Marginalization and Conditioning Rules If we have three random variables X, Y and Z From the product rule: y z z y X X,, y z z z y z y X X,,,,,, c c b c b a c b c b a bc a,, c c b bc a c b a Why? marginalization rule

18 Conditional robabilities cavity toothache cavity toothache toothache toothache toothache catch catch catch catch cavity cavity Also: cavity toothache cavity toothache toothache Notice: 1toothache remains the same in both calculations it acts like a normalization constant for the distribution Cavity toothache 445

19 Normalization Constants normalization constant Cavity toothache = Cavity,toothache = [Cavity,toothache,catch + Cavity,toothache,catch] = [<0.108, 0.016> + <0.012, 0.064>] = [<0.12, 0.08>] toothache toothache catch catch catch catch cavity cavity Normalization Constants are useful shortcuts in many probability computations! = <0.6, 0.4> = = 1/ = 1/0.2 = =50.12= =50.08=

20 General inference rocedure A notation: - X is a query variable Cavity in the example - E is the set of evidence variables Toothache in the example - e are the observed values for the evidence The query: X e Evaluated as X e X, e X, e, y y 447

21 robabilistic Inference by Enumeration Given a full join distribution to work with, ENUMERATE-JOIN-ASK is a complete algorithm for answering probabilistic queries for discrete variables 448

22 robabilistic Inference by Enumeration roblems: domain described by n Boolean variables Table of size O2 n and O2 n time to process it! Full joint distribution in tabular form is not a practical tool for building reasoning systems. It can be viewed as the theoretical foundation on which more effective approaches may be built. 449

23 Independence toothache toothache catch catch catch catch cavity cavity Let us add a fourth variable Weather The full distribution becomes Toothache, Catch, Cavity, Weather which has 32 entries 8 before 4 values for Weather This table contains four editions of the full table, one for each kind of weather. What relations do these editions have to each other and to the original 3- variable table? How are toothache, catch, cavity, Weather=cloudy and toothache, catch, cavity related? Use the product rule: toothache, catch, cavity, Weather=cloudy = = Weather=cloudy toothache,catch,cavity toothache,catch,cavity But Weather=cloudy toothache,catch,cavity= Weather=cloudy 450

24 Absolute Independence Weather is independent of one s dental problems. Toothache, Catch, Cavity, Weather = Toothache, Catch, Cavity Weather The 32-element table can be constructed from one 8-element table and one 4-element table. 451

25 Independence in Equations If propositions a and b are independent a b = a b a b = a a b a b b Independence between variables X and Y is written: X, Y = X Y X Y = X Y X = Y 452

26 453 Bayes Rule From the product rule: a b = a b b = b a a Then: a b b = b a a For multi-valued variables, Bayes Rule is: a set of equations, each dealing with specific values of the variables. a b b a a b X Y Y X X Y Bayes Rule

27 Examples Cavity Toothache Toothache Cavity Cavity Toothache cavity toothache toothache cavity cavity toothache cavity toothache toothache cavity cavity toothache cavity toothache toothache cavity cavity toothache cavity toothache toothache cavity cavity toothache 454

28 Applying Bayes Rule: The Simple Case Example: Medical diagnosis A doctor knows that meningitis causes a stiff neck 50% of the time: s m=0.5 The doctor also knows some unconditional facts: - the prior probability that the patient has meningitis is 1/50,000 m=1/50,000 - the prior probability that any patient has a stiff neck is 1/20 s=1/20 s m m 0.51/ 50,000 m s s 1/ Only 1 in 5,000 patients with a stiff neck is expected to have meningitis 455

29 456 Another way We can still compute m s without knowing s - instead compute the posterior probability for each value of the query variable here m and m and normalizing the results: s=s mm + s mm Then: Similarly: M s = < s mm, s mm> this can be obtained also from applying Bayes Rule with normalization 20 1/ 1/ 50, s m m s s m small because m=1/50,000 << s=1/ m m s m m s m m s m m s m m s s m 1 1 m m s m m s s m

30 Bayes Rule with Normalization M s = < s mm, s mm> this can be obtained also from applying Bayes Rule with normalization Y X = X YY is a normalization constant needed to make the entries in Y X sum to 1 457

31 Example of Bayes Rule with Normalization We have two discrete random variables: X describing weather conditions, with the domain: X={sunny, rain, cloudy, snow} Y describing clothes: Y={t-shirt, long-sleeves, coat} The distributions of X and Y are: X = <0.331, 0.26, 0.159, 0.25> Y = <0.5, 0.3, 0.2> We also have values of joint probabilities: t-shirt, sunny=0.32 t-shirt, rain=0.08 t-shirt, cloudy=0.09 t-shirt, snow=0.01 long-sleeves, sunny=0.01 long-sleeves, rain=0.15 long-sleeves, cloudy=0.05 long-sleeves, snow=0.09 coat, sunny=0.001 coat, rain=0.03 coat, cloudy=0.019 coat, snow=

32 Example of Bayes Rule with Normalization From these calculate conditional probabilities Y X, that is probability of clothing given weather. t-shirt sunny=t-shirt, sunny/sunny=0.32/ =.967 long-sleeves sunny= long-sleeves, sunny/sunny= 0.01/0.331= coat sunny= coat, sunny/sunny= 0.001/0.331= t-shirt rain= t-shirt, rain/rain= 0.08/0.26=0.307 long-sleeves rain= long-sleeves, rain/rain= 0.15/0.26=0.577 coat rain= coat, rain/rain= 0.03/0.26= t-shirt cloudy= t-shirt, cloudy/cloudy= 0.09/0.159=0.566 long-sleeves cloudy= long-sleeves, cloudy/cloudy= 0.05/0.159=0.314 coat cloudy=coat, cloudy/cloudy = 0.019/0.159= t-shirt snow= t-shirt, snow/snow= 0.01/0.25=0.04 long-sleeves snow= long-sleeves, snow/snow= 0.09/0.25=0.36 coat snow= coat, snow/snow= 0.15/0.25=

33 Computing the Normalization Constant Bayes s Rule: trying to guess the weather from the clothes people wear: X Y = Y XX = X,Y X Y add up to 1 for each value of Y 1. sunny t-shirt= 1 t-shirt, sunny= = = rain t-shirt= 1 t-shirt, rain= =2 0.08= cloudy t-shirt= 1 t-shirt, cloudy= =2 0.09= snow t-shirt= 1 t-shirt, snow= =2 0.01= = 1 1 =1/0.5 = 2 1. sunny long-sleeves= 2 long-sleeves, sunny= = = rain long-sleeves= 2 long-sleeves, rain= = = cloudy long-sleeves= 2 long-sleeves, cloudy= = = snow long-sleeves= 2 long-sleeves, snow= = = = 1 2 =1/0.3 = 3, sunny coat= 3 coat, sunny= = = rain coat= 3 coat, rain= =5 0.03= cloudy coat= 3 coat, cloudy= = = snow coat= 3 coat, snow= =5 0.15= = 1 3 =1/0.2 = 5 460

34 Example - Summary X sunny rain cloudy snow t-shirt Joint robabilities X, Y Y long-sleeve coat Y weather X t-shirt long-sleeve coat sunny X rain Y X clothes Y cloudy snow Conditional robabilities X clothes Y sunny rain cloudy snow t-shirt X Y weather X Y long-sleeve coat Conditional robabilities 461

35 More on Bayes rule What happens if one has the conditional probability in one direction but not the other? - Example: meningitis domain The doctor knows that a stiff neck implies meningitis in 1 of 5000 cases the doctor has quantitative information in the diagnostic direction from symptoms to causes. Lucky case the doctor has no need to use Bayes s rule 462

36 Causal Knowledge Note: Unfortunately, diagnostic knowledge is often more fragile than causal knowledge. Why? If there is a sudden epidemic of meningitis, the prior probability of meningitis m will go up. Because of this, the doctor who designed the diagnostic probability m S directly from statistical information, will not know how to update m s. But m s should go up s m m m s proportionally with m s Important: s m is unaffected by epidemic because it reflects how meningitis works. Conclusions: Using Causal or model-based knowledge provides robustness feasible probabilistic reasoning 463

37 Combining Evidence Until now, we considered probabilistic information available in the form effect cause single evidence What happens when there are multiple pieces of evidence? Example: dentist domain Cavity toothache catch = <0.108,0.016> <0.871, 0.129> This will not scale up to larger numbers of variables Cavity toothache catch = toothache catch CavityCavity we need to know the values of the conditional probabilities of the conjunction toothache catch for all values of Cavity If we have n possible evidence variables X rays, diet, oral hygiene, these are 2 n possible combinations of observed values, and for each we need to know the conditional probabilities. 464

38 Solution Consider the notion of independence Three variables: Cavity, Toothache, Catch which are independent? - Cavity, Catch? if the probe catches in the tooth, it probably has a cavity and that probably causes a toothache - Toothache, Catch? if there is a cavity, there will be a toothache, regardless of the probable catching the tooth - Toothache,Cavity? if there is a cavity, it might cause a toothache, but toothaches are not only caused by cavities 465

39 Conditional Independence toothach e catch Cavity toothache Cavity catch Cavity Conditional independence of toothache and catch given Cavity We also know: Cavity toothache catch toothache catch Cavity Cavity then interpret: Cavity toothache catch toothache Cavity catch Cavity cavity Cavity toothache catch cause effect1 effect2 toothache Cavity catch Cavity Cavity effect1 given cause effect2 given cause cause 466

40 Computational Complexity General definition of conditional independence of two variables X and Y given a third variable Z X, Y Z X Z Y Z Example: Toothach e, Catch Cavity Toothache Cavity Catch Cavity We can now derive the decomposition: Toothache, Catch, Cavity Toothache, Catch Cavity Cavity Toothache Cavity Catch Cavity Cavity Table 1 Table 2 Table 3 product rule Initial table-size = = 7 values 2 3 all, but since they sum to 1, we do not need the last one Table1-size = 2 values ; each row sums to 1 Table2-size = 2 values Table3-size = 1 value Thus total number of values = 2+2+1=5 instead of 7. In general the size of the representation grows as On instead of O2 n 467

41 Separation Cause, Effect1,..., Effect n Cause Effect i i Cause Toothache, Catch, Cavity Toothpaste Cavity Catch Cavity Cavity here Cavity separates Toothache and Catch because it is a Cause to both of them! Naïve Bayes model It is called Naïve because it works surprisingly well even when the effect variables are not conditionally independent 468

42 The Wumpus World Revisited After finding a breeze in both [1,2] and [2,1], the agent is stuck because there is no safe place to explore 1,4 2,4 3,4 4,4 1,3 2,3 3,3 4,3 1,2 1,1 B OK OK 2,2 3,2 4,2 2,1 B OK 3,1 4,1 Goal: compute the probability that the 3 neighboring squares contain a pit pit[1,3], pit[2,2], pit[3,1] Information: - a pit causes a breeze in all neighboring squares - each square other than [1,1] contains a pit with probability 0.2 Step 1: Identify random variables i,j =1 if square [i,j] contains a pit [i,j] = only for observed squares, e.g [1,1],[1,2],[2,1] B ij = 1 if square [i,j] is breezy ij, B ij are boolean variables 469

43 robabilistic Reasoning for the Wumpus Next step: specify the full distribution: 11 41, 12, 42 B,, 11 13, B, 43, 12 14, B, 44, B 21 21, 11, B 11 22, 12 23, B,...,, , 31 11, 32,...,, 44 33, 34, The prior probability of a pit configuration: 4,4 11,..., 44 ij i, j1,1 If there are n pits,,..., n 16n

44 Combining Evidence The evidence: - observed breeze in each square visited + each square visited contains no pit b b 11 b 12 known p 11 Query : 13 b 21 p 12 p known, b 21 How likely is that [1,3] contains a pit, given the observation so far 1,4 2,4 3,4 4,4 1,3 2,3 3,3 4,3 1,2 1,1 B OK OK 2,2 3,2 4,2 2,1 B OK 3,1 4,1 1,4 2,4 3,4 4,4 1,3 2,3 3,3OTHER4,3 QUERY 1,2 1,1 B OK KNOWN OK 2,2 3,2 4,2 2,1 FRINGE B OK 3,1 4,1 471

45 Answering the Query To answer 13 known,b we sum over the entries from the full distribution. Let unknown be a composite variable consisting of the ij variables for squares other than known and the query square [1,3] known, b, unknown, known, b unknown How many squares? =12 the summation contains 2 12 = 4096 terms too many! known query 472

46 Careful Computation Why would the contents of [4,4] affect whether [1,3] has a pit? Let us consider the set Fringe made of variables of the visited squares. Fringe={[2,2],[3,1]} Let us also consider the set Other containing variables of the unknown squares 10 of them The observed breezes are conditionally independent 13 known, b b known,, fringe, other, known, fringe, other fringe other fringe other b unknown b known, , unknown, known, fringe 13 13, known, 13, known, unknown fringe, other conditional independent 473

47 Continue Computation 13 known, b b known, 13, fringe 13, known, fringe other fringe, other The first term in this expression does not depend on the other variables we can move the summation inwards fringe b known, 13, fringe 13 other, known, fringe, other The prior term can be factored 13 known, b b known, 13, fringe 13 known fringe other fringe known 13 b known, 13 fringe other b known,, fringe fringe ' fringe, fringe fringe other other 474

48 Finishing 13 known, b ' 13 b known, 13, fringe {[2,2],[3,1]} fringe fringe fringe How do we build models for the Fringe? Since B 12 and B 21 we may have a pit in 13, 22 or 31 a b Model Model Model Model Model5 475

49 Likelihood of it at [1,3] 13 known, b ' 13 b known, 13, fringe fringe fringe Use Model 1,2&3 for 13 = 1 Model 4,5 for 13 = 0 13 known,b = < , > 13 =1 Model1 Model2 Model3 Model4 Model5 = < , > = <0.072, 0.016> <0.3103, > Because = 1/0.232 =

50 Interpretation From 13 known,b = <0.3103, > we know that [1,3] and [3,1] by symmetry contains a pit roughly 31% probability. Similarly, 22 known,b contains a pit roughly 86% probability The wumpus should avoid [2,2]! Lessons: seemingly complicated problems can be formulated precisely in probability theory and solved using simple algorithms Efficient solutions are obtained when independence and conditional independence relationships are used to simplify the summations Independence corresponds to our natural understanding of how the problem should be decomposed 477