Independence-6

Independence-6 - 1 The Tragedy of Conditional Probability...

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Unformatted text preview: 1 The Tragedy of Conditional Probability Thanks xkcd! http://xkcd.com/795/ Not Everything is Equally Likely • Say n balls are placed in m urns Each ball is equally likely to be placed in any urn • Counts of balls in urns are not equally likely! Example: two balls (A and B) placed with equal likelihood in two urns (Urn 1 and Urn 2) Possibilities: Counts: Urn 1 Urn 2 A, B- A B B A- A, B Urn 1 Urn 2 Prob 2 1/4 1 1 2/4 2 1/4 A Few Useful Formulas • For any events A and B: P(A B) = P(B A) (Commutativity) P(A B) = P(A | B) P(B) (Chain rule) = P(B | A) P(A) P(A B c ) = P(A) – P(AB) (Intersection) P(A B) ≥ P(A) + P(B) – 1 (Bonferroni) Generality of Conditional Probability • For any events A, B, and E, you can condition consistently on E, and these formulas still hold: P(A B | E) = P(B A | E) P(A B | E) = P(A | B E) P(B | E) P(A | B E) = (Bayes Thm.) • Can think of E as “everything you already know” • Formally, P( | E) satisfies 3 axioms of probability P(B | A E) P(A | E) P(B | E) Dissecting Bayes Theorem • Recall Bayes Theorem (common form): • Odds(H | E): • How odds of H change when evidence E observed Note that P(E) cancels out in odds formulation • This is a form of probabilistic inference P(E | H) P(H) P(E) P(H | E) = P(E | H) P(H) P(E | H c ) P(H c ) P(H | E) P(H c | E) = “Prior” “Likelihood” “Posterior” It Always Comes Back to Dice • Roll two 6-sided dice, yielding values D 1 and D 2 Let E be event: D 1 = 1 Let F be event: D 2 = 1 • What is P(E), P(F), and P(EF)? P(E) = 1/6, P(F) = 1/6, P(EF) = 1/36 P(EF) = P(E) P(F) E and F independent • Let G be event: D 1 + D 2 = 5 {(1, 4), (2, 3), (3, 2), (4, 1)} • What is P(E), P(G), and P(EG)?...
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This document was uploaded on 12/24/2011.

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Independence-6 - 1 The Tragedy of Conditional Probability...

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