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PROBABILITY. Uncertainty  Let action A t = leave for airport t minutes before flight from Logan Airport  Will A t get me there on time ? Problems :

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Presentation on theme: "PROBABILITY. Uncertainty  Let action A t = leave for airport t minutes before flight from Logan Airport  Will A t get me there on time ? Problems :"— Presentation transcript:

1 PROBABILITY

2 Uncertainty  Let action A t = leave for airport t minutes before flight from Logan Airport  Will A t get me there on time ? Problems : 1. Partial observability ( road state, other drivers ' plans, etc.). 2. Noisy sensors ( traffic reports ). 3. Uncertainty in action outcomes ( flat tire, etc.). 4. Immense complexity of modeling and predicting traffic.

3 Uncertainty  Let action A t = leave for airport t minutes before flight from Logan Airport  Will A t get me there on time ? A purely logical approach either : 1) risks falsehood : “ A 120 will get me there on time,” or 2) leads to conclusions that are too weak for decision making : “ A 120 will get me there on time if there ' s no accident on I -90 and it doesn ' t rain and my tires remain intact, etc., etc.” ( A 1440 might reasonably be said to get me there on time but I ' d have to stay overnight in the airport … )

4 Methods for handling uncertainty  Default logic :  Assume my car does not have a flat tire  Assume A 120 works unless contradicted by evidence  Issues : What assumptions are reasonable ? How to handle contradiction ?  Probability  Model agent ' s degree of belief  Given the available evidence, A 120 will get me there on time with probability 0.4

5 Probability  Probabilistic assertions summarize effects of  laziness : failure to enumerate exceptions, qualifications, etc.  ignorance : lack of relevant facts, initial conditions, etc.  Subjective probability :  Probabilities relate propositions to agent ' s own state of knowledge e. g., P ( A 120 | no reported accidents ) = 0.6  These are not assertions about the world, but represent belief about the whether the assertion is true.  Probabilities of propositions change with new evidence :  e. g., P ( A 120 | no reported accidents, 5 a. m.) = 0.15

6 Making decisions under uncertainty  Suppose I believe the following : P ( A 135 gets me there on time |...) = 0.04 P ( A 180 gets me there on time |...) = 0.70 P ( A 240 gets me there on time |...) = 0.95 P ( A 1440 gets me there on time |...) =  Which action to choose ?  Depends on my preferences for missing flight vs. airport cuisine, etc.  Utility theory is used to represent and infer preferences  Decision theory = utility theory + probability theory

7 Quick Question  You go to the doctor and are tested for a disease. The test is 98% accurate if you have the disease. 3.6% of the population has the disease while 4% of the population tests positive.  How likely is it you have the disease ?

8 Syntax  Basic element : random variable  Boolean random variables  e. g., Cavity ( do I have a cavity ?)  Discrete random variables  e. g., Weather is one of  Continuous random variables  represented by probability density functions ( PDFs ), valued [0,1]

9 Syntax  Elementary propositions constructed by assignment of a value to a random variable  e. g., Weather = sunny, Cavity = false  ( abbreviated as sunny or  cavity )  Complex propositions formed from elementary propositions and standard logical connectives  e. g., ( Weather = sunny  Cavity = false )  e. g., ( sunny   cavity )

10 Syntax  Atomic event :  A complete specification of the state of the world.  A complete assignment of values to variables.  E. g., if the world consists of only two Boolean variables Cavity and Toothache, then there are 4 distinct atomic events : Cavity = false  Toothache = false Cavity = false  Toothache = true Cavity = true  Toothache = false Cavity = true  Toothache = true  Atomic events are mutually exclusive and exhaustive

11 Axioms of probability ( Kolmogorov ’ s Axioms )  For any propositions A, B 1. 0 ≤ P ( A ) ≤ 1 2. P ( true ) = 1 and P ( false ) = 0 3. P ( A  B ) = P ( A ) + P ( B ) - P ( A  B )

12

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14 Priors  Prior or unconditional probabilities of propositions correspond to belief prior to arrival of any ( new ) evidence. e. g., P ( cavity ) = P ( Cavity = true )= 0.1 P ( Weather = sunny ) = 0.72 P ( cavity  Weather = sunny ) = 0.072

15 Distributions  Probability distribution gives probabilities of all possible values of the random variable.  Weather is one of  P ( Weather ) =  ( Normalized, i. e., sums to 1. Also note the bold font..)

16 Joint Probability The entire table sums to 1.

17 Inference by Enumeration

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19 Conditional Probability

20 Inference by Enumeration

21 Conditional Probability

22 Absolute Independence

23  Absolute independence powerful but rare  For n independent biased coins, O (2 n ) →O ( n ) 2*2*2*4=32 entries 2*2*2+4=12 entries

24 Conditional independence  If I have a cavity, the probability that the probe catches in it doesn ' t depend on whether I have a toothache :  P ( catch | toothache, cavity ) = P ( catch | cavity )  The same independence holds if I haven ' t got a cavity :  P ( catch | toothache,  cavity ) = P ( catch |  cavity )  Catch is conditionally independent of Toothache given Cavity :  P ( Catch | Toothache, Cavity ) = P ( Catch | Cavity )  Equivalent statements : P ( Toothache | Catch, Cavity ) = P ( Toothache | Cavity ) P ( Toothache, Catch | Cavity ) = P ( Toothache | Cavity ) P ( Catch | Cavity )

25 Bayes ' Rule

26 Much easier to calculate/estimate than the part on the left!

27 Bayes Rule : Example Note: probability of meningitis still very small!

28 Bayes ' Rule and conditional independence

29 Naïve Bayes Model

30 Summary  Probability is a formalism for uncertain knowledge  Basic probability statements include prior probabilities and conditional probabilities  The full joint probability distribution specifies probability of each complete assignment of values to variables.  Usually too large to create and use in its explicit form.  Absolute independence between subsets of random variables allows the full joint distribution to be factored into smaller joint distributions.  Absolute independent rarely occurs in practice.  Bayes ’ rule allows unknown probabilities to be computed from known conditional probabilities.  Applying Bayes ’ rule with many variables runs into the same scaling problem as above.  Conditional independence, brought about by direct causal relationships in the domain, may allow the full joint distribution to be factored into smaller, conditional distributions.  Naïve Bayes model assumes conditional independence of all effect variables.


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