Bayesian Networks. Probability Theory and Probabilistic Reasoning. Emma Rollon and Javier Larrosa Q

Transcription

1 Bayesian Networks Probability Theory and Probabilistic Reasoning Emma Rollon and Javier Larrosa Q Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

2 Degrees of belief Given a formula F, propositional logics classifies sentences into one of three categories: Sentences that are implied by F (i.e., F = α) Sentences whose negation are implied by F (i.e., F = α) And all other sentences (i.e., F = α and F = α) It imposes a binary classification: any assignment is either possible or impossible whether it satisfies or contradicts F. Under some circumstances, a finer classification is required: we can assign a degree of belief or probability in [0, 1] to each assignment (a.k.a, world). Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

3 Degrees of belief Eartquake Burglary Alarm Prossible? Pr( ) true true true true.0190 true true false false.0010 true false true true.0560 true false false false.0240 false true true true.1620 false true false false.0180 false false true true.0072 false false false true.7128 Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

4 Probabilistic model A set Z of mutually exclussive outcomes (Z is usually defined as the crossed product of a set of random variables) A probability measure P : Z R that satisfies the following axioms: 0 P(z) 1 for every z Z z Z P(z) = 1 (This is a normalization convention that allows us to directly compare the degrees of belief held by different assignments). The Table mapping each assignment with its probability is known as joint probability distribution. Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

5 Probability of a sentence A sentence α (a.k.a., event) will refer to the set of assignments z that satisfies α. The probability of a sentence α is: Pr(α) = z =α Pr(z) Recall that this expression is known as prior probability. Given two sentences α and β: α β: assignments satisfying α union assignments satisfying β. α β: assignments satisfying α intersection assignments satisfying β. α: assignments not satisfying α. Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

6 Probability of a sentence: example Eartquake Burglary Alarm Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 What would be the probability of Earthquake (i.e., Pr(Earthquake)), Burglary (Pr(Burglary)), and not Earthquake (Pr( Earthquake))? Observation The joint probability distribution is usually too large to allow a direct representation as a table. Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

7 Properties of Beliefs Prop1: 0 Pr(α) 1 Prop2: Pr(α) = 1, iff α is a tautology Prop3: Pr(α) = 0, iff α is a contradiction Prop4: Since each assignment must either satisfy α or α and Pr(z) = 1, then Pr(α) + Pr( α) = 1 z Prop5: The assignments satisfying α β can be partition into those satisfying α, those satisfying β, and those satisfying α β. Then, Pr(α β) = Pr(α) + Pr(β) Pr(α β) Recall that this property is known as the inclusion-exclusion principle. Observation Is there any case where Pr(α β) = Pr(α) + Pr(β)? Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

8 Properties of Beliefs: example Eartquake Burglary Alarm Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 What would be the probability of Earthquake Burglary? Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

9 Updating beliefs Suppose that we know that the Alarm has triggered (i.e., Alarm = true). This new piece of information is called evidence. We have to update the beliefs in Pr( ) into a new state of beliefs Pr( Alarm = true) (i.e., we condition the old table Pr on evidence): Eartquake Burglary Alarm Pr( ) Pr( Alarm) true true true /.2442 true true false true false true /.2442 true false false false true true /.2442 false true false false false true /.2442 false false false Bayes conditioning: Pr(α β) = Pr(α β) Pr(β), when Pr(β) 0 Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

10 Updating beliefs: example Eartquake Burglary Alarm Pr( ) Pr( Alarm) true true true true true false true false true true false false false true true false true false false false true false false false Some changes in beliefs induced by this new evidence: Pr(Burglary) =.2, Pr(Burglary Alarm) =.741 Pr(Earthquake) =.1, Pr(Earthquake Alarm) =.307 Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

11 Beliefs dynamics Let s see how some beliefs would change upon accepting the evidence: Earthquake: Pr(Burglary) =.2, Pr(Burglary Earthquake) =.2 Pr(Alarm) =.2442, Pr(Alarm Earthquake) =.75 Burglary: Pr(Alarm) =.2442, Pr(Alarm Burglary) =.905 Pr(Earthquake) =.1, Pr(Earthquake Burglary) =.1 Alarm and Earthquake: Pr(Burglary) =.2 Pr(Burglary Alarm) =.741 Pr(Burglary Alarm Earthquake) =.253 Alarm and Earthquake: Pr(Burglary Alarm Earthquake) =.957 Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

12 Event independence An event α is independent of event β (noted α β) iff Pr(α β) = Pr(α) Equivalently, Pr(α β) = Pr(α)Pr(β) Example: Pr(Burglary) =.2, Pr(Burglary Earthquake) =.2 (i.e., Burglary Earthquake) Pr(Earthquake) =.1, Pr(Earthquake Burglary) =.1 An event α is conditional independent of event β given event δ (noted α β δ) iff Pr(α β δ) = Pr(α δ) Example: Pr(Burglary Earthquake Alarm) =.253 (i.e., Burglary Earthquake Alarm) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

13 Event independence: comments Some warnings on conditional independence: α β δ does not imply α β α β does not imply α β δ However, α β iff β α Difference between independence and logical disjointness: Logical disjointness (mutual exclusiveness) iff the models of α and the models of β don t intersect. Independence depends on the actual beliefs (it is a property of the beliefs). Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

14 Variable independence Variables A and B are independent (noted A B) if P(a b) = P(a) for all possible instantiation a, b of A, B. Variables A and B are independent given C (noted A B C) if P(a b, c) = P(a c) for all possible instantiation a, b, c of A, B, C. Observation Variable independence is stronger than event independence. Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

15 Summary of properties of beliefs Prior probability: Pr(α) = z =α Pr(z) Inclusion-exclusion principle: Bayes conditioning: Pr(α β) = Pr(α) + Pr(β) Pr(α β) Pr(α β) = Pr(α β) Pr(β) Product rule: Pr(α β)pr(β) = Pr(α β) Chain rule: Pr(α β 1... β n ) = Pr(α β 1... β n )Pr(β 1 β 2... β n )... Pr(β n ) Bayes rule: Pr(α β)pr(β) = Pr(β α)pr(α) Low of total probability: Given that β 1,..., β n are mutually exclusive: n n Pr(α) = Pr(α β i ) = Pr(α β i )Pr(β i ) i=1 Emma Rollon and Javier Larrosa Bayesian Networks Q / 26 i=1

16 Example: Diagnosis We want to diagnose whether a patient has a certain disease (e.g., cancer) or not. To do that, a test on that disease is available. Let the proposition variable C stand for disease (i.e., C = true (noted c) means that the patience has the disease, and C = false (noted c) means the patience is healthy), and the proposition variable T stand for test (t when the test comes out positive, t otherwise). If we know that: P(c) =.01, P(t c) =.9, P(t c) =.2, P( c) = P( t c) = P( t c) = Observe that... P(C T ) = P(T C)P(C) P(C T ) = P(C) P(c t) =.04 P(c t) =.001 P(t) =.207 P(T C) P(T ) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

17 Example: Diagnosis with two independent tests Now, we have available two tests (T 1 and T 2 ). We know that: P(c) =.01, P(t i c) =.9, P(t i c) =.2, P( c) = P( t i c) = P( t i c) = and that T 1 and T 2 are conditionally independent given C (i.e., T 1 T 2 C). Observe that... P(C, T 1, T 2 ) = P(T 1 C, T 2 )P(T 2 C)P(C) = P(T 1 C)P(T 2 C)P(C) P(c t 1 ) =.04 P(c t 1, t 2 ) =.16 P(t 1 ) =.207 P(t 1 t 2 ) =.23 Observation Note that T 1 and T 2 are NOT independent. Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

18 Example: Conditional independence We have the weather forecast of three consecutive days (we know whether those days were sunny or not). Let the propositional variables S 1, S 2, and S 3 refer to each of those days, and let S i = s when that day was sunny and S i = s otherwise. We know that: P(s 1 ) =.9, P(s i s i 1 ) =.95, P(s i s i 1 ) =.6 P( s 1 ) =.1, P( s i s i 1 ) =.005, P( s i s i 1 ) =.4 and that S 1 S 3 S 2. Observe that... P(S 1, S 2, S 3 ) = P(S 3 S 2 )P(S 2 S 1 )P(S 1 ) P(s 1 ) =.90, P(s 2 ) =.91, P(s 3 ) =.92 P(s 1 s 3 ) =.91, P(s 3 s 1 ) =.93 Observation Note that S1 and S 3 are NOT independent. Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

19 Example: Explaining away I am happy (H) when it is sunny (S) or when I get a salary raise (R). We know that: P(s) =.7 P(r) =.01 P(h s, r) = 1 P(h s, r) =.9 P(h s, r) =.7 P(h s, r) =.1 and we assume S R. P(H, S, R) = P(H S, R)P(S, R) = P(H S, R)P(S)P(R) P(r h) =.018 P(r h, s) =.014 Observation Note that R and S are NOT independent given H Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

20 Most usual queries Let X = E Q Z, where E are the evidence variables, e are the observed values, Q are the variables of interest, Z are the rest of the variables. Probability of Evidence (Q = ): Prior Marginals (E = ): Posterior Marginals (E ): Pr(e) = Z Pr(e, Z) Pr(Q) = Z Pr(Q, Z) Pr(Q e) = Z Pr(Q, Z e) Most Probable Explanation (MPE) (E, Z = ): q = arg max Q Pr(Q e) Maximum a Posteriori Probability (MAP): q = arg max Q Pr(Q e) = arg max Q Z Pr(Q, Z e) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

21 Most usual queries: Example Recall our Earthquake, Burglary and Alarm example, with joint probability: Eartquake (T) Burglary (B) Alarm (A) Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

22 Most usual queries: Example (Probability of Evidence) Eartquake (T) Burglary (B) Alarm (A) Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 Pr(A = true) = T,B Pr(T, B, A = true) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

23 Most usual queries: Example (Prior Marginals) Eartquake (T) Burglary (B) Alarm (A) Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 Pr(T ) = A,B Pr(T, B, A) Eartquake (T) true false Pr( ) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

24 Most usual queries: Example (Posterior Marginals) Eartquake (T) Burglary (B) Alarm (A) Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 Pr(T A = true) = B Pr(T, B A = true) = B Pr(T,B,A=true) Pr(A=true) Eartquake (T) true false Pr( A = true) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

25 Most usual queries: Example (Most Probable Explanation) Eartquake (T) Burglary (B) Alarm (A) Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 arg max T,B Pr(T, B A = true) = arg max T,B Pr(T,B,A=true) Pr(A=true) = arg max T,B Pr(T, B, A = true) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26

26 Most usual queries: Example (Maximum a Posteriori Probability) Eartquake (T) Burglary (B) Alarm (A) Pr( ) true true true.0190 true true false.0010 true false true.0560 true false false.0240 false true true.1620 false true false.0180 false false true.0072 false false false.7128 arg max T Pr(T A = true) = arg max T B Pr(T, B A = true) = Pr(T,B,A=true) arg max T B Pr(A=true) = arg max T B Pr(T, B, A = true) Emma Rollon and Javier Larrosa Bayesian Networks Q / 26