Logic, Probability and Learning

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1 Logic, Probability and Learning Luc De Raedt

2 Overview Logic Learning Probabilistic Learning Probabilistic Logic Learning

3 Closely following : Russell and Norvig, AI: a modern approach Freiburg AI course slides (Burgard, Nebel, De Raedt) Probabilistic Learning

4 Probabilistic Learning 1. Probability theory 2. Bayesian Networks 3. Probabilistic Learning 4. Alternative models

5 Probabilistic Learning 1. Probability theory 2. Bayesian Networks 3. Probabilistic Learning 4. Alternative models

6 Daughter in Florence

7 Degree of belief and Probability Theory We (and other agents) are convinced by facts and rules only up to a certain degree. One possibility for expressing the degree of belief is to use probabilities. The agent is 90% (or 0.9) convinced by its sensor information = in 9 out of 10 cases, the information is correct (the agent believes). Probabilities sum up the uncertainty that stems from lack of knowledge.

8 Unconditional Probabilities P(A) denotes the unconditional probability or prior probability that A will appear in the absence of any other information, for example: P(Cavity) = 0.1 Cavity is a proposition. We obtain prior probabilities from statistical analysis or general rules.

9 Unconditional Probabilities (2) In general, a random variable can take on true and false values, as well as other values: P(Weather=Sunny) = 0.7 P(Weather=Rain) = 0.2 P(Weather=Cloudy) = 0.08 P(Weather=Snow) = 0.02 P(Headache=TRUE) = 0.1 Propositions can contain equations over random variables. Logical connectors can be used to build propositions, e.g. P(Cavity & Injured) = 0.06.

10 Unconditional Probabilities P(x) is the vector of probabilities for the (ordered) domain of the random variable X: P(Headache) = 0.1, 0.9 P(Weather) = 0.7, 0.2, 0.08, 0.02 define the probability distribution for the random variables Headache and Weather. P(Headache, Weather) is a 4x2 table of probabilities of all combinations of the values of a set of random variables. Headache = TRUE Headache = FALSE Weather = Sunny P(W = Sunny & Headache) P(W = Sunny & Headache) Weather = Rain Weather = Cloudy Weather = Snow

11 Conditional Probabilities (1) New information can change the probability. Example: The probability of a cavity increases if we know the patient has a toothache. If additional information is available, we can no longer use the prior probabilities! P(A B) is the conditional or posterior probability of A given that all we know is B: P(Cavity Toothache) = 0.8 P(X Y) is the table of all conditional probabilities over all values of X and Y.

12 Conditional Probabilities (2) P(Weather Headache) is a 4x2 table of conditional probabilities of all combinations of the values of a set of random variables. Headache = TRUE Headache = FALSE Weather = Sunny P(W = Sunny Headache) P(W = Sunny Headache) Weather = Rain Weather = Cloudy Weather = Snow Conditional probabilities result from unconditional probabilities (if P(B)>0) (by definition) P(A B) = P(A & B) P(B)

13 Conditional Probabilities (3) P(X,Y) = P(X Y) P(Y) corresponds to an equality system: P(W = Sunny & Headache) = P(W = Sunny Headache) P(Headache) P(W = Rain & Headache) = P(W = Rain Headache) P(Headache)... =... P(W = Snow & Headache) = P(W = Snow Headache) P( Headache)

14 Conditional Probabilities (4) P(A&B) P(A B) = P(B) Product rule: P(A&B) = P(A B) P(B) Analog: P(A&B) = P(B A) P(A) A and B are marginally independent if P(A B) = P(A) (equiv. P(B A) = P(B)). Then (and only then) it holds that P(A&B) = P(A) P(B).

15 Joint Probability The agent assigns probabilities to every proposition in the domain. An atomic event is an assignment of values to all random variables X 1,, X n (= complete specification of a state). Example: Let X and Y be boolean variables. Then we have the following 4 atomic events: X. Y, X. Y, X. Y, X. Y. The joint probability distribution P(X 1,, X n ) assigns a probability to every atomic event. Toothache Toothache Cavity Cavity Since all atomic events are disjoint, the sum of all fields is 1 (disjunction of events). The conjunction is necessarily false.

16 Working with Joint Probability All relevant probabilities can be computed using the joint probability by expressing them as a disjunction of atomic events. Examples: P(Cavity V Toothache) = P(Cavity & Toothache) + P( Cavity & Toothache) + P(Cavity & Toothache) We obtain unconditional probabilities by adding across a row or column (marginalization): P(Cavity) = P(Cavity & Toothache) + P(Cavity & Toothache) P(Cavity Toothache) = P(Cavity & Toothache) 0.04 = P(Toothache) = 0.80

17 Problems with Joint Distribution We can easily obtain all probabilities from the joint probability. The joint probability, however, involves k n values, if there are n random variables with k values. Difficult to represent Difficult to assess Questions: 1. Is there a more compact way of representing joint probabilities? 2. Is there an efficient method to work with this representation? Not in general, but it can work in many cases. Modern systems work directly with conditional probabilities and make assumptions on the independence of variables in order to simplify calculations.

18 Bayes Rule We know (product rule): P(A.B) = P(A B) P(B) and P(A.B) = P(B A) P(A) By equating the right-hand sides, we get P(A B) P(B) = P(B A) P(A) P(B A) P(A) P(A B) = P(B) For multi-valued variables (set of equalities): P(X Y) P(Y) P(Y X) = P(X) Generalization (conditioning on background evidence E): P(Y X,E) = P(X Y,E) P(Y E) P(X E)

19 Applying Bayes Rule P(Toothache Cavity) = 0.4 P(Cavity) = 0.1 P(Toothache) = 0.05 P(Cavity Toothache) = 0.4 x 0.1 = Why don t we try to assess P(Cavity Toothache) directly? P(Toothache Cavity) (causal) is more robust than P(Cavity Toothache) (diagnostic): P(Toothache Cavity) is independent from the prior probabilities P (Toothache) and P(Cavity). If there is a cavity epidemic and P(Cavity) increases, P(Toothache Cavity) does not change, but P(Toothache) and P(Cavity Toothache) will change proportionally.

20 Multiple Evidence (1) A dentist s probe catches in the aching tooth of a patient. Using Bayes rule, we can calculate: P(Cavity Catch) = 0.95 But how does the combined evidence help? Using Bayes rule, the dentist could establish: P(Cav Tooth & Catch) = P(Tooth & Catch Cav) x P(Cav) P(Tooth & Catch) P(Cav Tooth & Catch) = α P(Tooth & Catch Cav) x P(Cav)

21 Multiple Evidence (2) Problem: The dentist needs P(Tooth & Catch Cav), i.e. diagnostic knowledge of all combinations of symptoms in the general case. It would be nice if Tooth and Catch were independent but they are not: if a probe catches in the tooth, it probably has cavity which probably causes toothache. They are independent given we know whether the tooth has cavity: P(Tooth & Catch Cav) = P(Tooth Cav) P(Catch Cav) Each is directly caused by the cavity but neither has a direct effect on the other.

22 Conditional Independence The general definition of conditional independence of two variables X and Y given a third variable Z is: P(X,Y Z) = P(X Z) P(Y Z) Thus our diagnostic problem turns into: P(Cav Tooth & Catch) = α P(Tooth Cav) P(Catch Cav) P(Cav)

23 Probabilistic Learning 1. Probability theory 2. Bayesian Networks 3. Probabilistic Learning 4. Alternative models

24 Different Types of Models Graphical Models Directed - Bayesian Networks - HMMs (special case) Undirected - Markov Networks - Conditional Random Fields

25 Bayesian Networks (also belief networks, probabilistic networks, causal networks) 1. The random variables are the nodes. 2. Directed edges between nodes represent direct influence. 3. A table of conditional probabilities (CPT) is associated with every node, in which the effect of the parent nodes is quantified. 4. The graph is acyclic (a DAG). Remark: Burglary and Earthquake are denoted as the parents of Alarm

26 Bayesian Networks

27 The Meaning of Bayesian Networks Alarm depends on Burglary and Earthquake. MaryCalls only depends on Alarm. P(MaryCalls Alarm, Burglary) = P(MaryCalls Alarm) A Bayesian Network can be considered as a set of independence assumptions.

28 Conditional Independence Relations in Bayesian Networks (1) A node is conditionally independent of its nondescendants given its parents.

29 Example JohnCalls is cond. independent of Burglary and Earthquake given the value of Alarm.

30 Bayesian Networks and the Joint Probability Bayesian networks can be seen as a more comprehensive representation of joint probabilities. Let all nodes X 1,, X n be ordered topologically according to the the arrows in the network. Let x 1,, x n be the values of the variables. Then P(x 1,, x n ) = P(x n x n-1,, x 1 ) P(x 2 x 1 ) P(x 1 ) = n i=1 P(x i x i-1,, x 1 ) From the independence assumption, this is equivalent to P(x 1,, x n ) = n i=1 P(x i parents(x i )) We can calculate the joint probability from the network topology and the CPTs!

31 Example Only the probabilities for positive events are given. The negative probabilities can be found using P( X) = 1 P(X). P(J, M, A, B, E) = = P(J A) P(M A) P(A B, E) P( B) P( E) = 0.9 x 0.7 x x x =

32 Compactness of Bayesian Networks For the explicit representation of Bayesian networks, we need a table of size 2 n where n is the number of variables. In the case that every node in a network has at most k parents, we only need n tables of size 2 k (assuming boolean variables). Example: n = 20 and k = = and 20 x 2 5 = 640 different explicitly-represented probabilities! In the worst case, a Bayesian network can become exponentially large, for example if every variable is directly influenced by all the others. The size depends on the application domain (local vs. global interaction) and the skill of the designer.

33 Inference in Bayesian Networks Instantiating evidence variables and sending queries to nodes. What is P(Burglary JohnCalls) or P(Burglary JohnCalls, MaryCalls)?

34 Conditional Independence Relations in Bayesian Networks (1) A node is conditionally independent of its nondescendants given its parents.

35 Example JohnCalls is independent of Burglary and Earthquake given the value of Alarm.

36 Conditional Independence Relations in Bayesian Networks (2) A node is conditionally independent of all other nodes in the network given the Markov blanket, i.e. its parents, children and children s parents.

37 Example Burglary is independent of JohnCalls and MaryCalls, given the values of Alarm and Earthquake.

38 Various problems for BNs Find P(state) Find P(X e) -- typical query Find most likely state arg max X P(X=x e) Learning

39 Exact Inference in Bayesian Networks Compute the posterior probability distribution for a set of query variables X given an observation, i.e. the values of a set of evidence variables E. Complete set of variables is X E Y Y are called the hidden variables Typical query P(X e) where e are the observed values of E. In the remainder: X is a singleton Example: P(Burglary JohnCalls = true, MaryCalls=true) = (0.284,0.716)

40 Inference by Enumeration P(X e) = α P(X,e) = α P(X,e,y) The network gives a complete representation of the full joint distribution. A query can be answered using a Bayesian network by computing sums of products of conditional probabilities from the network.

41 Example Consider P(Burglary JohnCalls = true, MaryCalls=true) The hidden variables are Earthquake and Alarm. We have: P(B j, m) = α P(B, j, m) = α P(B, j, m, e, a) If we consider the independence of variables, we obtain for B=b P(b j, m) = α P(j a) P(m a) P(a e,b) P(e) P(b)

42 Complexity of Exact Inference If the network is singly connected or a polytree, the time and space complexity of exact inference is linear in the size of the network. The burglary example is a typical singly connected network. For multiply connected networks inference in Bayesian Networks is NP-hard. There are approximate inference methods for multiply connected networks such as sampling techniques or Markov chain Monte Carlo.

43 Probabilistic Learning 1. Probability theory 2. Bayesian Networks 3. Probabilistic Learning 4. Alternative models

44 Probabilistic Learning Two problems parameter estimation structure learning

45 Parameter Estimation Given a set of examples E, a probabilistic model M = (S, λ) with structure S and parameters λ, a probabilistic coverage relation P (e M) that computes the probabilty of observering the example e given the model M, a scoring function score(e, M) that employs the probabilistic coverage relation P (e M) Find the parameters λ that maximize score(e, M), i.e., λ = arg max λ score(e, (S, λ)) (1)

46 Scoring functions maximally a posteriori hypothesis H MAP H MAP = arg max H P (H E) = arg max H P (E H) P (H) P (E) (1) Simplified into maximum likelihood hypothesis H ML equally likely all hypotheses a priori H ML = arg max H P (E H) (2) Assuming independently and identically distributed examples: H ML = arg max P (e i H) (3) H e i E Often easier to maximized the log likelihood H LL = arg max log P (e i H) = arg max H H e i E log P (e i H) (4) e i E

47 Toy Example

48 Parameter Estimation A1 A2 A3 A4 A5 A6 true true false true false false false true true true false false true false false false true true complete data set simply counting X1 X2... XM

49 Parameter Estimation hidden/ latent A1 A2 A3 A4 A5 A6 true true? true false false? true?? false false true false? false true? incomplete data set Real-world data: states of some random variables are missing E.g. medical diagnose: not all patient are subjects to all test Parameter reduction, e.g. clustering,...? missing value

50 On the Alarm Net

51 Derivation (fully obs) Generalize : P (a, m) = λ 0 (1 λ 2 ) P (E M) = λ a 0.(1 λ 0) a.λ a m 1 (1 λ 1 ) a m.λ a m 2.(1 λ 2 ) a m Use log likelihood: L = log P (E M) = a. log λ 0 + a. log(1 λ 0 ) + a m. log λ 1 + a m. log(1 λ 1 ) + a m. log λ 2 + a m. log(1 λ 2 )

52 Derivation (fully obs) (2) Derivatives are : Yielding as solutions L = a a λ 0 λ 0 1 λ 0 L a m a m = λ 1 λ 1 1 λ 1 L a m a m = λ 2 λ 2 1 λ 2 λ 0 = λ 1 = λ 2 = a a + a a m a m + a m a m a m + a m (1) (2) (3) So, simply counting!

53 Parameter Estimation (part. obs) Idea Expectation Maximization Incomplete data? 1. Complete data (Imputation) most probable?, average?,... value 2. Count 3. Iterate

54 EM Idea: Complete the incomplete data complete A M true true true? false true true false false? P (m a) = 0.6 P (m a) = 0.2 complete data expected counts A true M true N 1.6 iterate true false false false true false λ 1 = P (m a) = λ 2 = P (m a) = = = 0.6

55 EM - formulae Q(λ), the likelihood is a function of λ: Q(λ) = E[L(λ)] = E[log P (E M(λ)] = E ( log P (e i M(λ)) ) e i E = E ( log P (x i, y i M(λ))] ) e i E = P (y i x i, M(λ j )) log P (x i, y i M(λ)) e i E

56 EM

57 Parameter Tying P (E M) = λ a 0.(1 λ 0) a.λ a m 1 (1 λ 1 ) a m.λ a m 2.(1 λ 2 ) a m. λ a j 3.(1 λ 3 ) a j.λ a j 4 (1 λ 4 ) a j Assume λ 1 = λ 3 and λ 2 = λ 4. The likelihood simplifies to P (E M) = λ a 0.(1 λ 0) a.λ a m + a j 1 (1 λ 1 ) a m + a j. λ a m + a j 2.(1 λ 2 ) a m + a m

58 Structure Learning Given a set of examples E a language L M of possible models of the form M = (S, λ) with structure S and parameters λ a probabilistic coverage relation P (e M) that computes the probabilty of observering the example e given the model M, a scoring function score(e, M) that employs the probabilistic coverage relation P (e M) Find the model M = (S, λ) that maximizes score(e, M), i.e., M = arg max M score(e, M) (1)

59 Structure learning H MAP = arg max H = arg max H = arg max H = arg max H P (H E) (1) P (E H)P (H) P (E) (2) P (E H)P (H) (3) log P (E H) + log P (H) (4) Often : heuristic search through space of models Bayesian Nets : add, delete or reverse arcs a kind of refinement operator

60 Probabilistic Learning 1. Probability theory 2. Bayesian Networks 3. Probabilistic Learning 4. Alternative models

61 Markov Models Define probability distribution over sequences. Assume that i, j 1 : P(X i X i 1 ) = P(X j X j 1 ) (1) i, j 1 : P(X i X i 1, X i 2,, X 0 ) = P(X i X i 1 ) (2) The Markov model factorizes as P(X 0,, X n ) = P(X 0 ) P(X i X i 1 ) i>0 (3)

62 Hidden Markov Model Assume also that i, j 0 : P(Y i X i ) = P(Y j X j ) (1) The hidden Markov model factorizes as n P(Y 0,, Y n, X 0,, X n ) = P(X 0 )P(Y 0 X 0 ) P(Y i X i )P(X i X i 1 ) (2) i>0

63 Example Possible states {rain, sun} Observations {umbrella, no umbrella } Questions Given sequence of umbrella, no umbrella -- what was the weather outside? Is there now rain (given a sequence of observations)? Will there be sun tomorrow, given a sequence of observations? Next week?

64 Typical Problems Decoding: compute the probability P(Y 0 = y 0,, Y n = y n ) of a particular observation sequence. Most likely state sequence: compute the most likely hidden state sequence corresponding to a particular observation sequence, that is arg max P (X 0 = x 0,, X n = x n Y 0 = y o,, Y n = y n ) (1) x 0,,x n State prediction:compute the most likely state state based on a particular observation sequence: arg max x t P (X t = x t,, X n = x n Y 0 = y o,, Y n = y n ) (2) If t < n, this is called smoothing; for t = n, filtering, and if t > n it is prediction Parameter estimation For Markov Models : EFFICIENT

65 Conclusions Principles of Probability Theory & Graphical Models A wide variety of such models exist They can be learned from data

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