CS 415 – A.I. Slide Set 12. Chapter 5 – Stochastic Learning Heuristic – apply to problems who either don’t have an exact solution, or whose state spaces.

CS 415 – A.I.

Slide Set 12

Chapter 5 – Stochastic Learning

Heuristic – apply to problems who either don’t have an exact solution, or whose state spaces are prohibitively large

Stochastic Methodology – also good for these situations

Based on counting the elements of an application domain

Addition and Multiplication Rule

Set A |A| - cardinality of A (number of elts)

A could be: empty, finite, countably infinite, or uncountably infinite

U – Universe (a.k.a. Domain) The set of ALL elements that could be in A

A’ - Compliment Example

U – people in a room A – males from U A’ – females in the room

Other Notations

Subset, Union, Intersection

Permutations and Combinations Permutation – an arranged sequence

of elements of a set (each used only once)

Question: how many unique permutations are there of a set of size n?

n * n-1 * n-2 * n-3 * … * 1 Question: how many ways can we

arrange a set of 10 books on a shelf where only 6 books can fit?

nPr

Combination – Any subset of the elements that can be formed

Question: How many combinations given a set of items?

1 combination for n elements Order DOES NOT MATTER

Question: How many combinations taken r at a time (How many ways can I form a four person committee from 10 people?)

Elements of Probability Theory

Examples

What is the probability of rolling a 7 or 11 from two fair die?

Sample Space Size? 36

Event Size? 8

For 6: 1,6; 2,5; 3,4; 4,3; 5,2; 6,1 For 11: 10,1; 1,10

Probability 8/36 = 2/9 Add them together because they are “independent”

How many four-of-a-kind hands can be dealt in all possible five card hands?

Sample Space? 52 cards taken 5 at a time

Event Space? Multiply number of combinations of 13

cards 1 at a time * Combination of 4 taken 4 at a time * 48

(number of different kinds of cards) * (number of ways to pick all four cards of same kind) * (times the number of ways the fifth card can be chosen)

See top of pg 172 Approx. 0.00024

The probability of any event E from the sample space S is:

The sum of the probabilities of all possible outcomes is 1

The probability of the compliment of an event is

The probability of the contradictory or false outcome of an event

Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 8


Are Two Events Independent?

Random bit strings of length four 2 Events

1. String has even number of ones2. Bit string ends with a zero

A total of 24=16 bit strings – 8 strings end with zero– 8 strings have even number of ones

Are independent events

All of Probability Theory

In a Nut Shell

Probabilistic Inference: Example 3 boolean random variables

All either true or false S – traffic is slowing down A – probability of an accident C – probability of road construction

Given state traffic data in Table 5.1 Next slide Note: all possibilities sum to 1

Can use these numbers to calculate Probability traffic slowdown Probability of construction without

slowdown, etc.

Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005

Table 5.1 The joint probability distribution for the traffic slowdown, S, accident, A, and construction, C, variable of the example of Section 5.3.2

Fig 5.1 A Venn diagram representation of the probability distributions of Table 5.1; S is traffic slowdown, A is accident, C is construction.

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Random Variables Individual Probability Computation

1. Combinatorical Methods (Analytical) Ex: Probability of rolling a 5 on a 6-sided

die

2. Sampling Events (Empirical) For times when it isn’t that simple to

analyze Assumptions

Not all events are equally likely (Easier if they are)

Probability of event lies between 0 and 1 Probabilities of union of sets still holds

Use “Random Variables”


Random Variable Example

Random variable – Season Domain – {spring, summer, fall, winter}

Discrete Random Variable p(Season=spring) = .75

Boolean Random Variable p(Season=spring) = true

Expectation

Expectation – the notion of expected payoff or cost of an outcome

Example A fair roulette wheel

Integers 0 – 36 equally spaced Each player places $1 on any number Wheel stops on that number, wins $35

Else – loses the $1

Reward of win $35 Probability 1/37

Cost of loss $1 Probability 36/37

Ex(E) = 35(1/37) + (-1)(36/37) = -0.027

Conditional Probability

2 kinds of Probablities1. Prior Probabilities

What’s the probability of getting a 2 or a 3 on a fair die?

2. Conditional (Posterior) Probabilities If a patient has system X, Y and Z then

what is the probability that he has the flu?



Fig 5.2 A Venn diagram illustrating the calculations of P(d|s) as a function of p(s|d).

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The chain rule for two sets:

The generalization of the chain rule to multiple sets


We make an inductive argument to prove the chain rule, consider the nth case:

We apply the intersection of two sets of rules to get:

And then reduce again, considering that:

Until is reached, the base case, which we have already demonstrated.

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You say [to ow m ey tow] and I say [t ow m aa t ow]…

- Ira Gershwin, Lets Call The Whole Thing Off


Fig 5.3 A probabilistic finite state acceptor for the pronunciation of “tomato”, adapted from Jurafsky and Martin (2000).

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Phoneme Recognition Problem

Use the Tomato style stochastic finite state acceptor

Interpret ambiguous collections of phonemes See how well the phonemes match the path

through the state machine for this and other words

phoneme is the smallest structural unit that distinguishes meaning. Phonemes are not the physical segments themselves, but, in theoretical terms, cognitive abstractions or categorizations of them.

the /t/ sound in the words tip, stand, water, and cat

Phoneme Recognition Problem Suppose an algorithm has identified the phoneme

/ni/ Occurs just after other recognized speech, /l/

Need to associate phoneme with either a word or the first part of a word

How? Brown corpus

1 Million word collection of sentences from 500 texts

Switchboard corpus 1.4 Million word collection of phone

conversations Together: ~2.5 Million words that let us

sample written and spoken words

How to proceed?

Can determine which word with the phoneme is most likely used

See Table 5.2 Most likely, “the”


Table 5.2 The ni words with their frequencies and probabilities from the Brown and Switchboard corpora of 2.5M words, adapted from Jurafsky and Martin (2000).

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Use Bayes’ theorem p(word | [ni]) = p([ni]|word) x p(word) See Table 5.3 Most likely, “new”

“I new” doesn’t make sense “I need”, does


Table 5.3 The ni phone/word probabilities from the Brown and Switchboard corpora (Jurafsky and Martin, 2000).

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Bayes’ Theorem

Review: one disease and one symptom

Individual hypotheses, hi Each is disjoint

Set of hypotheses, H Set of evidence, E

P(hi|E) = (p(E|hi) x p(hi))/p(E) Can use this to determine which

hypothesis is strongest given E Drop the denominator Arg max (maximum likelihood)

hypothesis

Finding p(E)

Given: entire space is partitioned by the set of hypotheses hi

Partition of a set = split of set into disjoint subsets

p(E) = Σip(E|hi)p(hi)

Bayes’ Theorem : General Form

The general form of Bayes’ theorem where we assume the set of hypotheses H partition the evidence set E:

Example

Suppose you want to buy a car Prob go to dealer 1, d1 Prob purchasing a car at d1, a1

Necessary for using Bayes’ All probabilities with various hi must be

known All relationships between evidence and

hypothesis {p(E|hi)} must be known

The application of Bayes’ rule to the car purchase problem:


Naïve Bayes, or the Bayes classifier, that uses the partition assumption, even when it is not justified:


CS 415 – A.I. Slide Set 12. Chapter 5 – Stochastic Learning Heuristic – apply to problems who either don’t have an exact solution, or whose state spaces.

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