03 Uncertainty inference(discrete)

Unit 3 Uncertainty Inference:Discrete

Wang, Yuan-Kai, 王元凱[email protected]

http://www.ykwang.tw

Department of Electrical Engineering, Fu Jen Univ.輔仁大學電機工程系

2006~2011

Bayesian Networks

Fu Jen University Department of Electrical Engineering Wang, Yuan-Kai Copyright

Reference this document as: Wang, Yuan-Kai, “Uncertainty Inference - Discrete,"

Lecture Notes of Wang, Yuan-Kai, Fu Jen University, Taiwan, 2011.

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王元凱 Unit - Uncertainty Inference (Discrete) p.

Goal of this Unit• Review advanced concepts of statistics

– Statistical Inference – Pattern recognition

2



Related Units• Previous unit(s)

– Probability Review– Statistics Review

• Next units– Uncertainty Inference (Continuous)

3



Self-Study• Artificial Intelligence: a modern approach

– Russell & Norvig, 2nd, Prentice Hall, 2003.pp.462~474,

– Chapter 13, Sec. 13.1~13.3• 統計學的世界

– 墨爾著，鄭惟厚譯, 天下文化，2002• 深入淺出統計學

– D. Grifiths, 楊仁和譯，2009, O’ Reilly

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5

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Contents1. Acting Under Uncertainty …………………. 62. Basic Probability ..................………..……. 153. Marginal Probability ..…….......................... 274. Inference Using Full Joint Distribution ... 305. Independence ............................................ 436. Bayes' Rule and Its Use ............................ 477. Summary ……………………………………. 62


王元凱 Unit - Uncertainty Inference (Discrete) p. 6

1. Acting Under Uncertainty

environmentagent

?

sensors

actuators

?

?

?

model



Example 1-Localization (1/3)• Where is it

– It is a robot–Sensor: camera, laser range finder, sonar–State: (x, y, orientation), Prob.




– It is a mobile station/robot–Sensor: Wireless LAN–State: (x, y), Prob.




– It is a moving text–Sensor: computer vision techniques–State: (x, y, moving direction), Prob.

t-3 t-2 t-1 t

Output t



Example 2-Correlation of Features and Words of Color

• Word of color

– Red

– Light red

...

• Feature of color (Average)RGB=(255,0,0)

RGB=(220,10,10)

RGB=(180,20,20)

RGB=(150,30,30)

RGB=(223,0,0)

RGB=(147,25,25)

Uncertainty



Example 3-Target Tracking for Robot

targetrobot

• The robot must keepthe target in view

• The target’s trajectoryis not known in advance

• The environment mayor may not be known



Inaccuracy & Uncertainty

• Movement Inaccuracy

Sensor Inaccuracy

EnvironmentalUncertainty



Degree of Belief• Probability theory

–Assigns a numerical degree of beliefbetween 0 and 1 to an evidence

–Provides a way of summarizing the uncertainty



Techniques for Uncertainty

• Bayes rule/Bayesian network with probability theory

• Certainty factor in expert system• Fuzzy theory with possibility theory• Dempster-Shafer theory



2. Basic Probability• Terms

–Random variables–Full joint distribution (FJD)–Conditional probability table (CPT)



Random Variable• Boolean random variable

–Rain : true, false• Discrete random variable

–Rain: cloudy, sunny, drizzle, drench• Continuous random variable

–Rain: rainfall in millimeterWe will focus on Boolean & discrete cases in most examples of this book



For Boolean & Discrete R.V.• P(X) is a vector• Boolean R.V.

–Rain: true, false–P(Rain) = <0.72, 0.28>

• Discrete R.V.–Rain: cloudy, sunny, drizzle, drench–P(Rain) = <0.72, 0.1, 0.08, 0.1>–Normalized, i.e. sums to 1



Full Joint Distribution (1/3)• For a set of random variables

{X1, X2 , , Xn}• X1 X2 Xn are atomic events• P(X1 X2 Xn) is

–A full joint probability distribution–A table of all joint prob. of all atomic

events, if {X1, , Xn} are discrete• All questions about probability of

joint events can be answered by the table



Full Joint Distribution (2/3)• Ex: X1: Rain, X2: Wind

–X1: drizzle, drench, cloudy, –X2: strong, weak–X1 X2 are atomic events–P(X1 X2) is a 3x2 matrix of values

RainWind Drizzle Drench Cloudy

Strong 0.15 0.12 0.06Weak 0.55 0.08 0.04P(X1=drizzle X2=strong) = 0.15, ...



Full Joint Distribution (3/3)• All questions about probability of

joint events can be answered by the table–P(Wind=Strong),

P(Rain=Drizzle Wind=Strong), ...



Posterior v.s. Prior Probabilities

• P(Cavity|Toothache)–Conditional probability–Posterior probability

(after the fact/evidence)• P(Cavity)

–Prior probability (the fact)



An Example (1/2)• For a dental diagnosis

–Let {Cavity,Toothache} be a set of Boolean random variables

• Denotations for Boolean R.V.–P(Cavity=true) = P(cavity)–P(Cavity=false) = P(cavity)



An Example (2/2)• The full joint probability

distribution P(Toothache Cavity)

cavity 0.04 0.06cavity 0.01 0.89

toothache toothache

11.006.001.004.0)( toothachecavityP

80.001.004.0

04.0)(

)()|(

toothacheP

toothachecavityPtoothachecavityP



Conditional Probability (Math)• P(X1=x1i| X2 =x2j) is a conditional

probability• P(X1 | X2) is a conditional

distribution function–All P(X1=x1i| X2 =x2j) for all possible i, j

• For Boolean & discrete R.V.s, conditional distribution function is a table–Conditional distribution of continuous

R.V. will not used in our discussions



Conditional Probability Table (CPT)

• For the dental diagnosis problem,–Toothache & Cavity are Boolean R.V.s–P(Toothache|Cavity) is a CPT

Cavity P(toothache|Cavity)T 0.90F 0.05

Cavity P(toothache|Cavity) P(toothache|Cavity)T 0.90 0.1F 0.05 0.95



CPT v.s. FJD

Cavity P(toothache|Cavity) P(toothache|Cavity)T 0.90 0.1F 0.05 0.95

cavity 0.04 0.06cavity 0.01 0.89

toothache toothache

Sum of all atomic events = 1

Sum of a row = 1



3. Marginal Probability

• Probability of a random variable is the sum of the probabilities of the atomic events containing the random variable

• Marginalization (summing out)

cavity 0.04 0.06cavity 0.01 0.89

toothache toothache

)(

)()(ij XEe

ji ePXP

P(toothache)=P(toothache cavity)+

P(toothache cavity)=0.04 + 0.01 = 0.05

Marginal probability



Marginal Probability (1/2)• Suppose a problem of a world

contains only 3 random variables {X1, X2, X3}

33

)()( 332121Xx

xXXXPXXP

1)( 321 XXXP

33

)()(

332211

2211

XxxXxXxXP

xXxXP



Marginal Probability (2/2)• Using higher order joint probability

to calculate marginal and other lower order joint probability

)( 11 xXP

22

)( 2211Xx

xXxXP

22 33

)( 332211X Xx x

xXxXxXP



4. Inference Using Full Joint Distributions

• Probabilistic inference–Uses the full joint distribution as the

"knowledge base"– Is the computation from observed

evidence of posterior probabilities for query• Compute conditional probability

• The most simple inference method:Inference by enumeration



The Dental Diagnosis Example• The set of random variables:

Toothache, Cavity, and Catch–All are Boolean random variables–Note: P(toothache) P(Toothache=true)

• The full joint distribution is a 2x2x2 table



The Full Joint Distribution

• 8 atomic events (sum=1)– P(toothachecatch cavity)=0.108– P(toothachecatch cavity)=0.16– P(toothachecatch cavity)=0.012– P(toothachecatch cavity)=0.064– ...



Inference by Enumeration• P(cavity|toothache)

• We can answer the query by–Enumerating P(cavitytoothache)

from the full joint distribution–Enumerating P(toothache) from the full

joint distribution

)()(

toothachePtoothachecavityP



Joint Probability• P(cavity toothache)

= 0.016+0.064 = 0.08

Marginal probability

Order-2 joint probability



Marginal Probability• P(toothache)

= 0.108+0.012+0.016+0.064 = 0.2

Marginal probability of Toothache



Conditional Probability• P(cavity|toothache)

)()(

toothachePtoothachecavityP

4.02.008.0



An Exercise• P(cavity toothache)

P(Cavity=true Toothache=true)= 0.108+0.012+0.072+0.008+0.016+0.064= 0.28



Normalization (1/2)• P(cavity|toothache) and

P(cavity|toothache) have the same denominator P(toothache)

)()()|(

toothachePtoothachecavityPtootheachecavityP

)()()|(

toothachePtoothachecavityPtootheachecavityP

• 1/P(toothache) can be viewed as a normalization constant for probability calculation and derivation



Normalization (2/2)• P(cavity|toothache)

= P(cavitytoothache)• P(cavity|toothache)

= P(cavitytoothache)= [ P(cavitytoothache catch)

+ P(cavitytoothache catch) ]= [ 0.108 + 0.012 ] = 0.12 = 0.6



A General Inference Procedure (1/2)

• Let P(X|E=e) be the query– X be the query variable– E be the set of evidence variables– e be the observed values of E– H be the remaining unobserved variables

(Hidden variables)• Inference of the query P(X|E=e) is

HhhHeEXPeEXPeEXP)()()|(

1

2



A General Inference Procedure (2/2)

• E.x.: Query P(cavity|toothache)–X: Cavity, E/e: Toothache/true,

H: Catch)|( toothachecavityP )|( eEXP

)( toothachecavityP

)]( )([

catchtoothachecavityPcatchtoothachecavityP

)( eEXP

Hh

hHeEXP )(



Complexity of the Enumeration-based Inference

Algorithm• For n Boolean R.V.s

– Space complexity: O(2n)(Store the full joint distribution)

– Time complexity: O(2n) (worst case)• For n discrete R.V.s, all have d discrete

values– Time complexity: O(dn) (worst case)– Space complexity: O(dn)

• It is not a practical algorithm, • More efficient algorithm is needed



5. Independence• A and B are independent iff

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

A B



The Dental Diagnosis Example (1/2)

• Original example: 3 random variables• Add a fourth variable: Weather

– The Weather variable has 4 values– The full joint distribution:

P(Toothache,Catch,Cavity,Weather)has 32 elements (2x2x2x4)

• By commonsense, Weather is independentof the original 3 variables– P(ToothacheCatch Cavity Weather)

= P(Toothache Catch Cavity)P(Weather)



The Dental Diagnosis Example (2/2)

• P(Toothache,Catch,Cavity,Weather)= P(Toothache,Catch,Cavity)P(Weather)– The 32-element table for 4 variables can be

constructed from • One 8-element table, and• One 4-element table

–Reduced from 32 elements to 12 elements



Advantages of Independence• Independence can dramatically

reduce the amount of elements of the full distribution table– i.e., independence can help in reducing

• The size of the domain representation, and • The complexity of the inference problem

• Independence are usually based on knowledge of the domain



6. Bayes' Rule and Its Use• Given two random variables X, Y• By product rule

)()|()( YPYXPYXP

)()()|()|(

XPYPYXPXYP

)()|()( XPXYPYXP

• Bayes' rule underlies all modern AI systems for probabilistic inference



Bayes’ Theorem

)()()/()/(

ePhPhePehP

Posterior

Prior

Probability of Evidence

Likelihood

Probability of an hypothesis, h, can be updated when evidence, e, has been obtained.

Note: it is usually not necessary to calculate P(e) directly as it can be obtained by normalizing the posterior probabilities, P(hi | e).



A Simple ExampleConsider two related variables:1. Drug (D) with values y or n2. Test (T) with values +ve or –ve

And suppose we have the following probabilities:P(D = y) = 0.001P(T = +ve | D = y) = 0.8P(T = +ve | D = n) = 0.01

These probabilities are sufficient to define a joint probability distribution.

Suppose an athlete tests positive. What is the probability that he has taken the drug?

074.09990010001080

001080)()|()()|(

)()|(

......

nDPnDveTPyDPyDveTPyDPyDveTPve)y|TP(D



A More Complex CaseSuppose now that there is a similar link between Lung Cancer (L) and a chest X-ray (X) and that we also have the following relationships:

History of smoking (S) has a direct influence on bronchitis (B) and lung cancer (L);

L and B have a direct influence on fatigue (F).

What is the probability that someone has bronchitis given that they smoke, have fatigue and have received a positive X-ray result?

lb

l

lxfsbP

lxfsbP

xfsPxfsbPxfsbP

,),,,,(

),,,(

),,(),,,(),,|(

111

1111

111

11111111

where, for example, the variable B takes on values b1 (has bronchitis) and b2(does not have bronchitis).R.E. Neapolitan, Learning Bayesian Networks (2004)



An Example• Prior knowledge

– The probability of a patient has meningitis is 1/50,000: P(m)=1/50000

– The probability of a patient has stiff neck is 1/20: P(s)=1/20

– The meningitis causes the patient to have a stiff neck 50% of the time: P(s|m)=0.5

• The probability of a stiff-neck patient has meningitis

0002.020/150000/15.0

)()()|()|(

sPmPmsPsmP



Generalization of Bayes’ Rule• Conditionalized on more evidences, say

e

)|()|(),|(),|(

eXPeYPeYXPeXYP

• P(Cavity|toothachecatch)= (P(toothachecatch|Cavity)P(Cavity))

/ P(toothachecatch)=> P(toothache catch Cavity)

/ P(toothachecatch) Better way?



Use Independence in Conditional (1/2)

• Probe Catch and Toothache are independent, given the presence or the absence of a Cavity– Both Catch and Toothache are caused by

Cavity, but neither has a direct effect on the other

• P(toothachecatch|Cavity)= P(toothache|Cavity) P(catch|Cavity)

• Conditional independence



Conditional Independency for the Dentist Diagnosis Problem (1/2)

• The conditional independence, like independence, can also decompose the full joint distribution into smaller pieces

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

(by product rule)• = P(Toothache|Cavity) P(Catch|Cavity)

P(Cavity)(by definition of conditional independence)



Conditional Independency for the Dentist Diagnosis Problem (2/2)

• P(Toothache Catch Cavity) has 23-1 elements

• P(Toothache|Cavity)P(Catch|Cavity)P(Cavity) has 2+2+1=5 elements

cavity 0.9 0.1cavity 0.05 0.95

P(toothache|Cavity) P(toothache|Cavity)

Conditional Probability Table (CPT)• Conditional independence can reduce the

amount of probabilities



Probabilistic Modeling of Problems (1/2)

• Usually random variables have two semantics–Cause–Effect

• Conditional probability P(Y|X) can be rewritten as P(Cause|Effect) or P(Effect|Cause)



Probabilistic Modeling of Problems (2/2)

• We usually assume some effects are conditionally independent given a cause–P(effect1 effect2 | cause)

= P(effect1 | cause) P(effect2 | cause)The conditional independence can be drawn as a graphic model



Advantages of Conditional Independence

• The use of conditional independence reduces the size of the joint distribution from O(2n) to O(n)

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



Pairwise Independence• A generalization of pairwise independence

– We say that the variables X1, …, Xn are mutually conditionally independent, given a set X

n

iin XXpXXXXp

121 )|()|,,(

When X is empty, we have p(X1,X2, …,Xn)=p(X1)p(X2)…p(Xn)

Why?),,|(),,,( 11

121 XXXPXXXP ii

n

in

By chain rule

),,,|()|,,,( 111

21 XXXXPXXXXP ii

n

in

Since Xi is conditional independent of the others given X

n

iin XXpXXXXp

121 )|()|,,(



7. Summary• The full joint distribution can

answer any query of the domain–However, it is intractable

• Independence and conditional independence is important for the reduction of the full joint distribution

• We can now move on to Bayesian networks

03 Uncertainty inference(discrete)

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