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CS2800-Probability_part_b_v.2

# CS2800-Probability_part_b_v.2 - Discrete Math CS 2800 Prof...

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1 Discrete Math CS 2800 Prof. Bart Selman [email protected] Module Probability --- Part b) Bayes’ Rule Random Variables

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2 Bayes’ Theorem How to assess the probability that a particular event will occur on the basis of partial evidence? Examples: What is the likelihood that people who test positive to a particular disease (e.g., HIV), actually have the disease? What is the probability that an e-mail message is spam? Key idea: one should factor in additional information regarding occurrence of events.
Assume that with respect to events F and E (“E” for “Evidence”): We know P(F) probability that event F occurs (e.g. probability that email message is spam; this is given by what fraction of email is spam) We also know event E has occurred. (e.g., email message contains words “sale” and “bargain”) Therefore the probability conditional probability that F occurs given that E occurs, P(F|E), is a more realistic estimate that F occurs than P(F). How do we compute P(F|E)? E.g., based on P(F), P(E|F), and P(E| ¬F) Note: ¬F is also referred to as complement of F (F C or F).

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Bayesian Bayesian Inference Inference Evidence Evidence Original Belief Original Belief (Prior Probability) (Prior Probability) Modified Modified Belief Belief Hypothesis Hypothesis Theory Theory P(F) E P(F|E)
Experiment: Pick one box at random (p = 0.5) and than a ball at random from that box. Assume you picked a red ball. What’s the probability that it came form the left box? > 0.5 ? Define: E – you choose a red ball. (therefore ¬ E – you choose the green ball) F – you choose the left box. (therefore ¬ F– you choose the right box) We want to know P(F|E) Box A Box B Why?

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What we know: P(E|F) = and P(E|¬F) = Given that the boxes are selected at random: P(F) = P(¬F)=1/2 P(F|E) = P(E∩F)/P(E) so we need to compute P(E∩F) and P(E) . P(F|E)? We know P(E|F) = P(E∩F) /P(F). So, P(E∩F) = P(E|F) P(F) = 7/9 * 1/2 = 7/18. What about P(E) ? Note that P(E) = P(E F) +P (E∩ ¬F) . Why? Note also that P (E ∩ ¬F) = P(¬F) P(E|¬F) = 1/2 * 3/7 = 3/14 So, P(E) = P(E F) +P (E∩ ¬F) = 7/18 + 3/14 = 38/63 And therefore P(F|E) = P(E∩F)/P(E) = (7/18) / (38/63) = 49/76 0.645 7/9 3/7 E – red color F – left box Indeed > 0.5 phew!
Bayesian Bayesian Inference Inference Concrete (new) Evidence Concrete (new) Evidence Red ball picked (E) Red ball picked (E) Original Belief Original Belief there is a 0.5 that you will pick left box there is a 0.5 that you will pick left box (P(F)). (P(F)). Modified Belief Modified Belief Increased to 0.65 Probability Increased to 0.65 Probability (P(F|E)) (P(F|E))

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( | ) ( ) ( | ) ( | ) ( ) ( | ) ( ) C C P E F P F P F E P E F P F P E F P F = + Theorem: Bayes’ Theorem Suppose that E and F are events from a sample space S such that P(E) ≠0 and P(F)≠0. Then Proof: , ( ) ( | ) ( ) exp ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( | ) ( ) ( | ) ( ( | ) ( ) / ( ) ( | ) ( | ) ( ) ( | ) ( ) ( | ) ) / ( ) ) ( | C C C C C So P E F P E F P F therefore We only need an ression for P E E E S E F F E F E F So P E P E F P F E P P P E F P F P E F E F P E P E F P F E P P E F P F P E F P F So P E P F E F P E E F = = = = = + = + = = = = ( | ) ( ) ( | ) ( ) ( | ) ) ( ) ( ) ( ) C C P E F P F P E F F P F P P E F P F P E F + = ( | ) ( ) ( | ) ( ) P E F P F P F E P E =
Example --- A Classic!

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