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Bayesian Networks • Using random variables to represent objects and events in the world – Various instantiations to these variables can model the current state of the world • Computing joint probabilities over these variables • Estimating joint probabilities over all random variables often leads to a combinatorial explosion – Estimating probabilities of every combination of values for the variables involved Conditional independence • Use the chain rule to replace joint probabilities with conditional ones: P(A1, A2, A3, A4, A5) = P(A1 | A2, A3, A4,A5) · P(A2 | A3, A4, A5) · P(A3 | A4, A5) · P(A5) • Bayesian Networks allow to reduce the terms even further, by taking into account conditional independence: – P(A|C1, …, Cn, U) = P(A|C1, …, Cn) for some set U of random variables – Given {C1, …, Cn}, A is independent of U Conditional independence (cont’d) • The structure of a Bayesian Network reflects a number of independence assumptions (d-separation criterion) • Directed arcs between random variables represent conditional dependencies • Combined with the chain rule, conditional independencies allow us to manipulate smaller conditional probability tables to estimate the joint probabilities Formal notation • Each random variable A is conditionally independent of all other random variables that are not descendants of A, given A’s n parents p( x1 , , xn ) p( xi | x1 , xi 1 ) i 1 n p( x1 ,, xn ) p( xi | pa i ) i 1 Ind( Xi ; {X1,…,Xi-1}\Pai | Pai ) where Pai are parents of Xi Example: modeling real world • “Mary walks outside and finds that the street and lawn are wet. She concludes that it has just rained recently. Furthermore, she decides that she doesn’t need to water her roses.” • Mary’s logic: – rain or sprinklers street = wet – rain or sprinklers lawn = wet – lawn = wet soil = moist – soil = moist roses = OK Mary’s world (cont’d) • Let’s compute the probability of the following (intuitively unlikely) world event: – The roses are OK – The soil is dry – The lawn is wet – The street is wet – The sprinklers are off – It’s raining Where are the missing probabilities, e.g., P(sprinklers=F) or P(street=dry|rain=T, sprinklers=T) ? Computing the probability of a complex event • P(sprinklers = F, rain = T, street = wet, lawn = wet, soil = dry, roses = OK) = P(roses = OK | soil = dry) * P(soil = dry | lawn = wet) * P(lawn = wet | rain = T, sprinklers = F) * P(street = wet | rain = T, sprinklers = F) * P(sprinklers = F) * P(rain = T) = 0.2 * 0.1 * 1.0 * 1.0 * 0.6 * 0.7 = 0.0084 The event is quite unlikely ! Usage scenarios • There are 2 types of computations performed with Bayesian Networks – Belief updating • Computation of probabilities over random variables – Belief revision • Finding maximally probably global assignment • Given some evidence or observation, our task is to come up with a set of hypotheses (variable assignments) that together constitute the most satisfactory explanation/interpretation of the evidence at hand. Belief revision • Let W be a set of all random variables in the network • Let e be the evidence (e is a subset of W) • Any instantiation of all the variables in W that is consistent with e is called an explanation or interpretation of e • The problem is to find an explanation w* such that P(w* | e) = max P(w | e) – w* is called the most probable explanation Belief updating • Computing marginal probabilities of a subset of random variables given the evidence • Typically, the task is to determine the best instantiation of a single random variable given the evidence Belief revision - example • Evidence: e = {roses = OK} • Goal: determine the assignment to all the random variables that maximizes P(W|e) – P(e) is constant, and e is a subset of W it’s sufficient to maximize P(W) – Intuitively, non-evidence random variables in W can be viewed as possible hypotheses for e • Solution: P(sprinklers=F, rain=T, street=wet, lawn=wet, soil=wet, roses=OK) = 0.2646 Belief updating - example • Evidence: e = {roses = OK} • Goal: determine the probability that the lawn is either wet or dry given this observation – P(lawn=dry | roses = OK) = 0.1190 – P(lawn=wet | roses = OK) = 0.8810