Bayesian Networks Unit 3: Discrete Uncertainty Inference

1

Bayesian Networks
Unit 3
Uncertainty Inference:Discrete
Wang, Yuan-Kai, 王元凱
ykwang@mails.fju.edu.tw
http://www.ykwang.tw

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

2006~2011
Reference this document as:
Wang, Yuan-Kai, “Uncertainty Inference - Discrete,"
Lecture Notes of Wang, Yuan-Kai, Fu Jen University, Taiwan, 2011.
Fu Jen University Department of Electrical Engineering Wang, Yuan-Kai Copyright

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

4

王元凱 Unit - Uncertainty Inference (Discrete) p. 5

Contents
1. Acting Under Uncertainty …………………. 6
2. Basic Probability ..................………..……. 15
3. Marginal Probability ..…….......................... 27
4. Inference Using Full Joint Distribution ... 30
5. Independence ............................................ 43
6. Bayes' Rule and Its Use ............................ 47
7. Summary ……………………………………. 62

5


1. Acting Under Uncertainty
sensors
?
?
environment
agent ?
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.



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



• Where is it
– 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 • Feature of color (Average)
RGB=(255,0,0)
– Red RGB=(220,10,10)
RGB=(223,0,0)
Uncertainty
RGB=(180,20,20)

– Light red RGB=(150,30,30)
RGB=(147,25,25)
...


Example 3-Target Tracking for Robot

• The robot must keep
the target in view
• The target’s trajectory
is not known in
advance target
robot
• The environment may
or may not be known


Inaccuracy & Uncertainty
 Sensor Inaccuracy

• Movement Inaccuracy

 Environmental
Uncertainty


Degree of Belief
• Probability theory
– Assigns a numerical degree of belief
between 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 millimeter
We 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


• 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
Rain Drizzle Drench Cloudy
Wind
Strong 0.15 0.12 0.06
Weak 0.55 0.08 0.04
P(X1=drizzle  X2=strong) = 0.15, ...


• 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)
toothache toothache
cavity 0.04 0.06
cavity 0.01 0.89
P(cavity  toothache)  0.04  0.01  0.06  0.11
P (cavity | toothache)
P (cavity  toothache) 0.04
   0.80
P (toothache) 0.04  0.01


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


CPT v.s. FJD
toothache toothache
cavity 0.04 0.06
cavity 0.01 0.89
Sum of all atomic events = 1

Cavity P(toothache|Cavity) P(toothache|Cavity)
T 0.90 0.1
F 0.05 0.95
Sum of a row = 1


3. Marginal Probability
P( X i )   P(e )
e j E ( X i )
j Marginal probability
• Probability of a random variable is the
sum of the probabilities of the atomic
events containing the random variable
• Marginalization (summing out)
P(toothache)
toothache toothache =P(toothache  cavity)+
cavity 0.04 0.06
P(toothache  cavity)
cavity 0.01 0.89
=0.04 + 0.01 = 0.05


Marginal Probability (1/2)
• Suppose a problem of a world
contains only 3 random variables
{X1, X2, X3}
 P( X 1  X2  X3) 1

P( X 1  X 2 )   P( X
x3 X 3
1  X 2  X 3  x3 )

P( X1  x1  X2  x2 )
 P( X1  x1  X2  x2  X3  x3 )
x3X3


Marginal Probability (2/2)
• Using higher order joint probability
to calculate marginal and other
lower order joint probability
P( X 1  x1 )   P( X
x 2  X 2
1  x1  X 2  x2 )

   P( X
x 2  X 2 x3 X 3
1  x1  X 2  x2  X 3  x3 )



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(toothachecatch cavity)=0.108
– P(toothachecatch cavity)=0.16
– P(toothachecatch cavity)=0.012
– P(toothachecatch cavity)=0.064
– ...


Inference by Enumeration
• P(cavity|toothache)
P(cavity  toothache)

P(toothache)
• We can answer the query by
–Enumerating P(cavitytoothache)
from the full joint distribution
–Enumerating P(toothache) from the full
joint distribution


Joint Probability
• P(cavity  toothache) Order-2 joint
= 0.016+0.064 = 0.08 probability

Marginal probability


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

Marginal probability of Toothache


Conditional Probability
P (cavity  toothache) 0.08
   0 .4
P (toothache) 0.2



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)
P(cavity  toothache)
P(cavity | tootheache) 
P(toothache)
P(cavity  toothache)
P(cavity | tootheache) 
P(toothache)
• 1/P(toothache) can be viewed as a
normalization constant for probability
calculation and derivation


Normalization (2/2)
=  P(cavitytoothache)
• P(cavity|toothache)
=  P(cavitytoothache)
=  [ P(cavitytoothache  catch)
+ P(cavitytoothache  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
P ( X | E  e)  P ( X  E  e )
1

   P ( X  E  e  H  h)
2 hH


A General Inference
Procedure (2/2)
• E.x.: Query P(cavity|toothache)
– X: Cavity, E/e: Toothache/true,
H: Catch
P (cavity | toothache ) P ( X | E  e)
 P(cavity  toothache)  P ( X  E  e)
  [ P(cavity  toothache  catch )    P( X  E  e  H  h)
 P (cavity  toothache  catch )] hH



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(AB) = P(A)P(B)

A B 



(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 independent
of the original 3 variables
– P(ToothacheCatch  Cavity  Weather)
= P(Toothache  Catch  Cavity)P(Weather)


(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
P( X  Y )  P( X | Y )P(Y )
P( X  Y )  P(Y | X )P( X )
P( X | Y ) P(Y )
 P(Y | X ) 
P( X )
• Bayes' rule underlies all modern AI
systems for probabilistic inference


Bayes’ Theorem
Likelihood Prior

P (e / h ) P ( h )
P (h / e) 
P (e)

Posterior Probability of
Evidence
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 Example
Consider two related variables:
1. Drug (D) with values y or n
2. Test (T) with values +ve or –ve
And suppose we have the following probabilities:
P(D = y) = 0.001
P(T = +ve | D = y) = 0.8
P(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?
P (T   ve | D  y ) P ( D  y )
P(D  y|T   ve) 
P (T   ve | D  y ) P ( D  y )  P (T   ve | D  n ) P ( D  n )
0.8  0.001

0.8  0.001  0.01  0.999
 0.074


A More Complex Case
Suppose 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

 P(b , s f , x , l )
fatigue and have received a positive X-ray result?
1 1 1 1
P(b , s , f , x )
P(b | s , f , x )  1
 1 1 1 l

 P(b, s , f , x , l )
1 1 1 1
P( s , f , x ) 1 1 1 1 1 1
b ,l
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
P(s | m)P(m) 0.51/ 50000
P(m | s)    0.0002
P(s) 1/ 20


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(toothachecatch|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
P(toothache|Cavity) P(toothache|Cavity)
cavity 0.9 0.1
cavity 0.05 0.95
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
n
set X p( X 1 , X 2 , X n | X )   p( X i | X )
Why? i 1
n
By chain rule P( X 1 , X 2 ,, X n )   P( X i | X i 1 ,, X 1 )
n i 1
 P( X 1 , X 2 ,  , X n | X )   P( X i | X i 1 ,  , X 1 , X )
i 1
Since Xi is conditional independent of the others given X
n
 p( X 1 , X 2 , X n | X )   p( X i | X )
i 1
When X is empty, we have p(X1,X2, …,Xn)=p(X1)p(X2)…p(Xn)


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

Bayesian Networks Unit 3: Discrete Uncertainty Inference

Recommended

Recommended

More Related Content

Viewers also liked

Viewers also liked (6)

Similar to Bayesian Networks Unit 3: Discrete Uncertainty Inference

Similar to Bayesian Networks Unit 3: Discrete Uncertainty Inference (20)

Recently uploaded

Recently uploaded (20)

Bayesian Networks Unit 3: Discrete Uncertainty Inference