cs 4100 artificial intelligence prof. c. hafner class notes feb 28 and march 13-15, 2012

CS 4100 Artificial Intelligence

Prof. C. HafnerClass Notes Feb 28 and March 13-15, 2012

Uncertain knowledge/beliefs arise from

• Chance (probability)• Incomplete knowledge

– Defeasible (“default”) reasoning– Reasoning from ignorance

• Ambiguous or conflicting evidence– Evidential reasoning

• Vagueness– Fuzzy / open-textured concepts

Examples• Chance

– Will your first child be a boy or girl (before conceived)– Will the coin come up heads or tails

• Defeasible reasoning:– If I believe Tweety is a bird, then I think he can fly– If I learn Tweety is a penguin then I think he can’t fly

• Assuming the normal case unless/until you learn otherwise

• Reasoning from ignorance: – I believe the President of the US is alive

Examples (cont.)• “Evidential reasoning” - how strongly do I believe P

based on the evidence? (Confidence levels)– Quantitative [ 0 , 1 ], [ -1 , 1 ]– Qualitative {definite; very likely, likely, neutral, unlikely, very unlikely,

definitely not}

• Fuzzy concepts: – John is tall– My friend promises to return a book “soon”

• Add “degree” to fuzzy assertions (between 0 and 1)

Some problems

Using probability, if we know P(A) and P(B) we have methods for calculating P(~A), P(A and B), P(A or B),

P(A | B)

Using evidential or fuzzy models, these cause problems of consistency:

If “John is tall” .8 and “John is smart” .6 what can be said about “John is both tall and smart” ?

We will mainly focus on probabilistic reasoning models

Syntax for probabilistic reasoning• Basic element: random variable

• Similar to propositional logic: possible worlds defined by assignment of values to random variables.

• Boolean random variablese.g., Cavity (do I have a cavity?)

• Discrete random variablese.g., Weather is one of <sunny,rainy,cloudy,snow>

• Domain values must be exhaustive and mutually exclusive

• Elementary proposition constructed by assignment of a value to a single random variable: e.g., Weather = sunny, Cavity = false (sometime abbreviated as cavity)

• Complex propositions formed from elementary propositions and standard logical connectives e.g., Weather = sunny Cavity = false

–

Syntax• Atomic event: A complete specification of the state of the world

about which the agent is uncertain

Cavity = false Toothache = falseCavity = false Toothache = trueCavity = true Toothache = falseCavity = true Toothache = true

• Atomic events are mutually exclusive and exhaustive (often called “Outcomes”)

• Events in general are sets of atomic events, such as “Cavity = true”

• E.g., if the world consists of only two Boolean variables Cavity and Toothache, then there are 4 distinct atomic events:

Axioms of probability

• For any propositions A, B

– 0 ≤ P(A) ≤ 1– P(true) = 1 and P(false) = 0– P(A B) = P(A) + P(B) - P(A B)

Prior probability• Prior or unconditional probabilities of propositions

e.g., P(Cavity = true) = 0.1 and P(Weather = sunny) = 0.72 correspond to belief prior to arrival of any (new) evidence

• Probability distribution gives values for all possible assignments:P(Weather) = <0.72,0.1,0.08,0.1> (sums to 1)

• Joint probability distribution for a set of random variables gives the probability of every atomic event on those random variables: P(Weather,Cavity) = a 4 × 2 matrix of values:

Weather = sunny rainy cloudy snow Cavity = true 0.144 0.02 0.016 0.02Cavity = false 0.576 0.08 0.064 0.08

• Every probability question about a domain can be answered by the joint distribution

Conditional probability• Conditional or posterior probabilities

e.g., P(cavity | toothache) = 0.8i.e., given that toothache is all I know

• (Notation for conditional distributions (use boldface):P(Cavity | Toothache) = 2-element vector of 2-element vectors)

• If we know more, e.g., cavity is also given, then we haveP(cavity | toothache,cavity) = 1

• New evidence may be irrelevant, allowing simplification, e.g., cavity does not depend on weather:P(cavity | toothache, sunny) = P(cavity | toothache) = 0.8

• This kind of inference, sanctioned by domain knowledge, is crucial for probabilistic reasoning in AI. WHY?–

Conditional probability• Definition of conditional probability:

P(a | b) = P(a b) / P(b) if P(b) > 0

• Product rule gives an alternative formulation:P(a b) = P(a | b) P(b) = P(b | a) P(a)

• A general version holds for whole distributions, e.g.,P(Weather,Cavity) = P(Weather | Cavity) P(Cavity)

• (View as a set of 4 × 2 equations, not matrix mult.)

• Chain rule is derived by successive application of product rule:P(X1, …,Xn) = P(X1,...,Xn-1) P(Xn | X1,...,Xn-1) = P(X1,...,Xn-2) P(Xn-1 | X1,...,Xn-2) P(Xn | X1,...,Xn-1) = …

= P(X1) P(X2 | X1) P(X3 | X1, X2) . . . P(Xn | X1, . . ., Xn-1)

OR: πi= 1 to n P(Xi | X1, … ,Xi-1)–

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is true: P(φ) = Σω:ω φ╞ P(ω)

•

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is true: P(φ) = Σω:ω φ╞ P(ω)

• P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2

•

Inference by enumeration

• Start with the joint probability distribution:

• Can also compute conditional probabilities:

P(cavity | toothache) = P(cavity toothache)P(toothache)

= 0.016+0.064 0.108 + 0.012 + 0.016 + 0.064

= 0.4

In class exercise• Given the joint distribution shown on slide and the

definition P(a | b) = P(a b) / P(b): – What is P(Cavity = True) ?– What is P(Weather = Sunny) ?– What is P(Cavity = True | Weather = Sunny)

• Given the meta-equation:– P(Weather,Cavity) = P(Weather | Cavity) P(Cavity)

What are the 8 equations represented here?

Weather = sunny rainy cloudy snow Cavity = true 0.144 0.02 0.016 0.02Cavity = false 0.576 0.08 0.064 0.08

Independence• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)

P(Toothache, Catch, Cavity, Weather)= P(Toothache, Catch, Cavity) P(Weather)

• 32 entries reduced to 12; for n independent biased coins, O(2n) →O(n)

• Absolute independence powerful but rare

• Dentistry is a large field with hundreds of variables, none of which are independent. What to do?

•

Bayes' Rule• Product rule P(ab) = P(a | b) P(b) = P(b | a) P(a)

Bayes' rule: P(a | b) = P(b | a) P(a) / P(b)• or in distribution form

P(Y|X) = P(X|Y) P(Y) / P(X) = αP(X|Y) P(Y)

• Useful for assessing diagnostic probability from causal probability:– P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect)

– E.g., let M be meningitis, S be stiff neck:P(m|s) = P(s|m) P(m) / P(s) = 0.8 × 0.0001 / 0.1 = 0.0008

– Note: posterior probability of meningitis still very small!•

•

•

Example: Expert Systems for Medical Diagnosis

• 100 diseases• 20 symptoms

• # of parameters needed to calculate P(Di) when a patient provides his/her symptoms

• Strategy to reduce the size: assume independence of symptoms

• Recalculate number of parameters needed

Bayes' Rule and conditional independence

P(Cavity | toothache catch) = αP(toothache catch | Cavity) P(Cavity) = αP(toothache | Cavity) P(catch | Cavity) P(Cavity)

• This is an example of a naïve Bayes model:

P(Cause,Effect1, … ,Effectn) = P(Cause) πiP(Effecti|Cause)

• Total number of parameters is linear in n

––

Conditional independence• P(Toothache, Cavity, Catch) has 23 – 1 = 7 independent entries

• If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:(1) P(catch | toothache, cavity) = P(catch | cavity)

• The same independence holds if I haven't got a cavity:(2) 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)

––

»

Conditional independence contd.• Write out full joint distribution using chain rule:

P(Toothache, Catch, Cavity)= P(Toothache | Catch, Cavity) P(Catch, Cavity)= P(Toothache | Catch, Cavity) P(Catch | Cavity) P(Cavity)= P(Toothache | Cavity) P(Catch | Cavity) P(Cavity)

I.e., 2 + 2 + 1 = 5 independent numbers

• In most cases, the use of conditional independence reduces the size of the representation of the joint distribution from exponential in n to linear in n.

• Conditional independence is our most basic and robust form of knowledge about uncertain environments.

–

Bayesian networks

• A simple, graphical notation for conditional independence assertions and hence for compact specification of full joint distributions

• Syntax:– a set of nodes, one per variable– a directed, acyclic graph (link ≈ "directly influences")– a conditional distribution for each node given its parents:

P (Xi | Parents (Xi))

• In the simplest case, conditional distribution represented as a conditional probability table (CPT) giving the distribution over Xi for each combination of parent values

Probabilities/Bayesian Inference Nets

• Thanks to Andrew Moore from CMU for some great slides. (used with permission)

Extend to P(A ^ B ^ C ^ …) = ?

Review: Conditional probabilities and JPD (joint distribution)

Chain rule follows from this definition

• Product ruleP(a b) = P(a | b) P(b) = P(b | a) P(a)

• Chain rule is derived by successive application of product rule:P(X1, …,Xn) can also be written P(X1 ^ ... ^ Xn) = P([Xn ^ [X1 ,. . . Xn-1]) = P(X1,...Xn-1) P(Xn | X1,...,Xn-1) = P(X1,...,Xn-2) P(Xn-1 | X1,...,Xn-2) P(Xn | X1,...,Xn-1) = …

= P(X1) P(X2 | X1) P(X3 | X1, X2) . . . P(Xn | X1, . . ., Xn-1)

–

Conditional Prob. example

Example

Likes Football Dislikes Neutral

Male .25 .1 .15

Female .1 .3 .1

In-class exercise:Calculate:

P(Likes Football | Male )P( ~ Likes Football | Female)

Review the Joint Distribution (JPD)

What assumption can we make ?

Test your understanding: Fill in the table

Structure for CP-based AI Models Given a set of RV’s X, typically, we are interested in

the posterior joint distribution of the query variables Y given specific values e for the evidence variables E

Let the hidden variables be H = X - Y – E

Then the required calculation of P(Y | E) is done by summing out the hidden variables:

Note: what is α ?

Given the definition: P(a | b) = P(a b) / P(b)

α is the denominator 1/P(E=e). P(E=e) can be calculatedfrom the joint distribution as: ΣhP(E= e ^ H = h)P( Y | E = e) = αP(Y ^ E = e) or αΣhP(Y ^ E= e ^ H

= h)

Example (medical diagnosis)Causal model: D I S (Y H E)

Cancer anemia fatigueKidney disease anemia fatigue

P(Y=cancer | E=fatigue) = α [ P(Y=cancer ^ E=fatigue ^ anemia) + P(Y=cancer ^ E=fatigue ^ ~anemia) ]

α = 1/P(E = fatigue) or 1/[P(E=fatigue ^ anemia) + P(E=fatigue ^ ~anemia) ]

Analysis

• The terms in the summation are joint entries because Y, E and H together exhaust the set of random variables

• Obvious problems:1. Time and space complexity O(dn) where d is the largest arity2. How to find the numbers to solve real problems?

(A solution to 1. : assume independence !!)

• P(Y | E = e) = αP(Y ^ E = e) = αΣhP(Y ^ E= e ^ H = h) [repeated]

What is Independence ??• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)

P(Toothache, Catch, Cavity, Weather) JD entries are 2x2x2x4= P(Toothache, Catch, Cavity) P(Weather) entries are 2x2x2 + 4

• 32 entries reduced to 12• In general, total independence assumption reduces

exponential to linear complexity

•

What is Independence ??• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)• Toss 10 coins, different OUTCOMES are 2^10 = 2048• Biased coins whose behavior is independent of each other:

O(2n) →O(n) = can compute P(all outcomes) with 10 values• All coins have the same bias (includes the case of fair coins) ????

How many values are needed ?

Test your understanding:• Consider a “3 sided coin” (or die). How many entries needed to

show the probabilities of all outcomes?• If you toss 10 of those and:• All have the same bias?• Bias unknown, but independence is assumed?• Bias unknown, no independence assumed?

•

Example: Expert Systems for Medical Diagnosis• 10 diseases• 20 symptoms

• # of parameters needed to calculate P(D | S) for all combinations using a JPD

• Strategy to reduce the size of the model: assume mutual independence of symptoms and diseases - Recalculate number of parameters needed

• Absolute independence powerful but rare• Medicine is a large field with hundreds of variables,

many of which are not independent. What to do?

Problem 2: We still need to find the numbers

Assuming independence, doctors may be able to estimate:P(symptom | disease) for each S/D pair (causal reasoning)

While what we need to know s/he may not be able to estimate as easily:

P(disease | symptom)

Thus, the importance of Bayes rule in probabilistic AI

Bayes' Rule• Product rule P(ab) = P(a | b) P(b) = P(b | a) P(a)

Bayes' rule: P(a | b) = P(b | a) P(a) / P(b)• or in distribution form

P(Y|X) = P(X|Y) P(Y) / P(X) = αP(X|Y) P(Y)

• Useful for assessing diagnostic probability from causal probability:

P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect) P(Disease|Symptom) = P(Symptom|Diease) P(Symptom) / (Disease)

– E.g., let M be meningitis, S be stiff neck:P(m|s) = P(s|m) P(m) / P(s) = 0.8 × 0.0001 / 0.1 = 0.0008

– Note: posterior probability of meningitis still very small!•

•

Bayes' Rule and conditional independenceP(Cavity | toothache catch)

= αP(toothache catch | Cavity) P(Cavity) = αP(toothache | Cavity) P(catch | Cavity) P(Cavity)

• We say: “toothache and catch are independent, given cavity”. This is an example of a naïve Bayes model. We will study this later as our simplest machine learning application

P(Cause,Effect1, … ,Effectn) = P(Cause) πiP(Effecti|Cause)

• Total number of parameters is linear in n (number of symptoms). This is our first Bayesian inference net.

––

Conditional independence• P(Toothache, Cavity, Catch) has 23 – 1 = 7 independent entries

• If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache:(1) P(catch | toothache, cavity) = P(catch | cavity)

• The same independence holds if I haven't got a cavity:(2) P(catch | toothache,cavity) = P(catch | cavity)

• Catch is conditionally independent of Toothache given Cavity:P(Catch | Toothache,Cavity) = P(Catch | Cavity)

• Equivalent statements (from original definitions of independence):P(Toothache | Catch, Cavity) = P(Toothache | Cavity)P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)

––

»

Conditional independence contd.• Write out full joint distribution using chain rule:

P(Toothache, Catch, Cavity)= P(Toothache | Catch, Cavity) P(Catch, Cavity)= P(Toothache | Catch, Cavity) P(Catch | Cavity) P(Cavity)= P(Toothache | Cavity) P(Catch | Cavity) P(Cavity)

I.e., 2 + 2 + 1 = 5 independent numbers

• In most cases, the use of conditional independence reduces the size of the representation of the joint distribution from exponential in n to linear in n.

• Conditional independence is our most basic and robust form of knowledge about uncertain environments.

–

Remember this examples

Example of conditional independence

Test your understanding of the Chain Rule

This is our second Bayesian inference net

How to construct a Bayes Net

Test your understanding: design a Bayes net with plausible numbers

Calculating using Bayes’ Nets