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CSCI 121 Special Topics: Bayesian Networks Lecture #2: Bayes Nets

CSCI 121 Special Topics: Bayesian Networks Lecture #2: Bayes Nets. P(B=F) P(B=T) .999 .001 . P(E=F) P(E=T) .998 .002 . Earthquake. Burglary. B E P(A=F) P(A=T) F F .999 .001 T F .06 .94 F T .71 .29 T T .05 .95. Alarm.

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CSCI 121 Special Topics: Bayesian Networks Lecture #2: Bayes Nets

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  1. CSCI 121 Special Topics: Bayesian NetworksLecture #2: Bayes Nets

  2. P(B=F)P(B=T) .999 .001 P(E=F)P(E=T) .998 .002 Earthquake Burglary BEP(A=F)P(A=T) F F .999 .001 T F .06 .94 F T .71 .29 T T .05 .95 Alarm JohnCalls MaryCalls AP(J=F)P(J=T) F .95 .05 T .10 .90 A P(M=F)P(M=T) F .99 .01 T .30 .70

  3. Why Use Bayes Nets? • Why not just use joint prob. (can be used to answer any query)? BEAJMP F F F F F P1 ... T T T T T Pm • Table size is exponential (m=2n entries for n vars.) • Network (graph) is sparse; only encodes local dependencies. • n2k entries, where k = avg. # inputs to node.

  4. Types of Inferences / Queries • Diagnostic: P(B | J) = 0.02 • Causal: P(J | B) = 0.85 • Intercausal: P(B | J &M & ~E) = 0.34 • Explaining Away: A university accepts a student if s/he is a good scholar or good athlete. If an accepted student is a good scholar, is s/he a good athlete? (Berkson's Paradox / selection bias)

  5. Answering Queries • E.g., compute P(B|A) • Product Rule tells us that P(B|A) = P(B&A) / P(A) • Problem: the only values we can obtain directly from the probability tables are the priors and some joint conditionals. For P(B&A) and P(A), the values are confounded with other variables (E). • Solution: we can marginalize(sum out) the other variables to focus on the ones we care about • First, we must convert the tables to joint probabilities of all relevant variables:

  6. P(B) .001 P(E) .002 B E P(A) T T .95 T F .94 F T .29 F F .001 Answering Queries: Marginalization BEAProb T T T .001*.002*.95 = .000001900 T T F .001*.002*.05 = .000000100 T F T .001*.998*.94 = .000938120 T F F .001*.998*.06 = .000059880 F T T .999*.002*.29 = .000579420 F T F .999*.002*.71 = .001418580 F F T .999*.998*.001 = .000997002 F F F .999*.998*.999 = .996004998

  7. Answering Queries: Marginalization Now we get P(A) by summing over rows where A = True: BEAProb T T T .000001900 T T F .000000100 T F T .000938120 T F F .000059880 F T T .000579420 F T F .001418580 F F T .000997002 F F F .996004998 P(A) = .000001900 + .000938120 + .000579420 + .000997002 .002516442

  8. BEAProb T T T .000001900 T T F .000000100 T F T .000938120 T F F .000059880 F T T .000579420 F T F .001418580 F F T .000997002 F F F .996004998 Answering Queries: Marginalization • Next, we need to compute P(B&A) • Again, we marginalize: • So P(B|A) = .000940020 / .002516442= .373551228 P(B&A) = . 000001900 + .000938120 .000940020

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