l08_prob_axioms

l08_prob_axioms - Probability II: Axioms & Properties...

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02/16/2009 CS206 - Intro. to Discrete Structures II 1 Probability II: Reading: Wednesday, February 18, 2009
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02/16/2009 CS206 - Intro. to Discrete Structures II 2 Recall: Probability of Events • If the sample space of an experiment Ω is finite and if all atomic events are equally likely, then the probability of an event E is P(E) = Pr(E) = |E| / | Ω | = #outcomes in E / #outcomes in Ω • I.e., fraction of all outcomes that make E happen. Wednesday, February 18, 2009
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02/16/2009 CS206 - Intro. to Discrete Structures II 3 Axioms of Probability Andrei N. Kolmogorov (1903-1987) A function P that maps events on the set of real numbers, P: E is a probability measure on the sample space Ω if it satisfies the following conditions: 1. 0 P(E) 1 2. P( Ω ) = 1 3. P(A [ B) = P(A) + P(B), A B = (P, Ω ) is called the probability space . Wednesday, February 18, 2009
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02/16/2009 CS206 - Intro. to Discrete Structures II 4 Axiomatic probabilities Consequence : 1. To define a probability measure it is sufficient to assign probabilities to atomic events, 0 P( ω ) 1, ω Ω 2. Probabilities of atomic events have to sum up to 1, ω Ω P( ω ) = 1. 3. Probability of an event is P(E) = ω E P( ω ) Wednesday, February 18, 2009
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02/16/2009 CS206 - Intro. to Discrete Structures II 5 Example: Axiomatic dice Ω = {1,2,3,4,5,6} • Define: P(1)=P(2)=…=P(6)=1/6 • E = {3,4} • P(E) = |E|/| Ω | = |{3,4}|/6 = 2/6 = 1/3. • Axiomatic: P(E) = P({3} {4}) = P({3}) + P({4}) = 1/6 + 1/6 = 1/3. {3} and {4} are mutually exclusive (Axiom 3) Wednesday, February 18, 2009
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02/16/2009 CS206 - Intro. to Discrete Structures II 6 Example: Axiomatic dice (cont’d) • E = { ω : ω is even} • Axiomatic: P(E) = P({2,4,6}) = P({2})+P({4})+P({6}) = 1/6+1/6+1/6 = 1/2 • E = { ω : ω≤ 4} • Axiomatic: P(E) = P({1,2,3,4}) = P({1})+P({2})+P({3})+P({4}) = 4 1/6 = 2/3 Wednesday, February 18, 2009
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CS206 - Intro. to Discrete Structures II 7 Example: “Skewed” axiomatic dice Let’s make this dice “skewed” or unfair: we won’t let it ever show sides 5 & 6. Moreover, we’ll make 1 twice as likely as other outcomes. Define: P(1)=2/5, P(2)=P(3)=P(4)=1/5, P(5)=P(6)=0 E 1 = {3,4} P(E 1 ) = |E 1 |/| Ω | = |{3,4}|/6 = 2/6 = 1/3. WRONG! • Actually, Ω ={1,2,3,4} P(E 1 ) = |E 1 |/| Ω | = |{3,4}|/|{1,2,3,4}|=2/4=1/2. WRONG! • Axiomatic:
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This note was uploaded on 03/24/2011 for the course CS 206 taught by Professor Fredman during the Spring '08 term at Rutgers.

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l08_prob_axioms - Probability II: Axioms & Properties...

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