The document discusses Bayesian belief networks (BBNs), which represent probabilistic relationships between variables. BBNs consist of a directed acyclic graph showing the dependencies between nodes/variables, and conditional probability tables quantifying the effects. They allow representing conditional independence between non-descendant variables given parents. The document provides an example BBN modeling a home alarm system and neighbors calling police. It then shows calculations to find the probability of a burglary given one neighbor called police using the network. Advantages are handling incomplete data, learning causation, and using prior knowledge, while a disadvantage is more complex graph construction.

1. BAYESIAN BELIEF NETWORKS
Submitted by
KODAM SAI KUMAR, 213CS2157, M.Tech – IS,
Department of Computer Science and Engineering
NATIONAL INSTITUTE OF TECHNOLOGY, ROURKELA

2. Contents:
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•
•
•
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Why BAYESIAN BELIEF NETWORKS
Definition
Incremental Network Construction
Conditional Independence
Example
Advantages and Disadvantages

3. Why BAYESIAN BELIEF NETWORKS ?
To represent the probabilistic relationships
between different classes.
To avoid dependences between value of
attributes by joint conditional probability
distribution.
In Naive Bayes Classifier,
conditionally independent
attributes are

4. Bayesian Belief Network (BN) Definition:
BN are also known as Bayesian Networks, Belief Networks,
and Probabilistic Networks.
A BN is defined is defined by two parts, a directed acyclic
graph (DAG) and a set of conditional probability tables (CPT).
Nodes
Links
Variables
Dependency

5. A Formal Definition
A BN is a graph with the following properties
Nodes: Set of random variables.
Directed Links: The real meaning of a link from
node X to node Y is that X has a direct influence
on Y.
Each node has a CPT that quantifies the effects
that the parent have on the node.
The Graph has no directed cycles.

6. If an arc drawn from Y to Z, then Y is a parent
or immediate predecessor of Z, and Z is a
descendant of Y.
Y
Z
Fig: 1
Each variable is conditionally independent of
its nondescendants in the graph, given its
parents.

7. Incremental Network Construction:
1. Choose the set of relevant variables Xi that
describes the domain
2. Choose an ordering for the variables (very
important step)
3. While there are variable left
a) Pick a variable X and add a node for it
b) Set parent (X) to some minimal set of existing nodes
such that the conditional independence property is
satisfied.
c) Define the CPT for X

8. Conditional Independence:
P(X1,X2,…..Xn) = P(Xn/X n-1,X n-2,...,X1)P(Xn-1,X n-2,...,X1)
=P(Xn/Xn-1,Xn-2,...,X1)P(Xn-1/Xn-2,..,X1)….P(X2/X1)P(X1)
=
A BN represents Conditional Independence
P(Xn/X n-1,X n-2,...,X1) = P(Xn/Parents(X n))

9. Example:
Burglar Alarm at home
• Fairy reliable at detecting a Burglary.
• Responds at times of Earthquakes.
Two neighbors on hearing Alarm, calls police
• John always calls when he hears the alarm, but
sometimes confuses the telephone ringing with the
alarm and calls then too.
Mary likes loud music and sometimes misses the
alarm altogether.

10. P(B)=0.001
A simple BN:
P(E)=0.002
Barglary
Earthquake
B
A
T
0.95
F
0.95
F
T
0.29
F
F
0.001
0.90
F
T
T
P(J)
P(A)
T
Alarm
E
0.05
A
T
John Calls
P(M)
0.90
F
Fig: 2
0.05
Mary Calls

11. How to find P(Burglary / JohnCalls)...?
P(B / J) = P(J B) / P(J) -----------------------------------------------------------Eqn (1)
1. How to find P(J B) ?
P(J B) = P(J A B) + P(J A’ B)
= P(J / A B) P(A B) + P(J / A’ B) P(A’ B)
= P(J / A) P(A B) + P(J / A’) P(A’ B)
=(0.9)*P(A B) + (0.05)*P(A’ B) --------------------------------Eqn (2)
P(A B) = P(A B E) + P(A B E’)
=P(A / B E) P(B E) + P(A / B E’) P(B E’)
=P(A / B) P(B) P(E) + P(A / B) P(B) P(E’)
=(0.95) (0.001) (0.002) + (0.95) (0.001) (0.998)
=0.00095
P(A’ B) = P(A’ B E) + P(A’ B E’)
=P(A’ / B E) P(B E) + P(A’ / B E’) P(B E’)
=P(A’ / B) P(B) P(E) + P(A’ / B) P(B) P(E’)
=(0.05) (0.001) (0.002) + (0.05) (0.001) (0.998)
=0.00005
Substitute P(A B) & P(A’ B) values in Eqn (2)
P(J B)=(0.9)*(0.00095) + (0.05)*(0.00005)
P(J B)= 0.00086

12. 2. How to find P(J) ?
P(J) = P(J A) + P(J A’)
= P(J / A) P(A) + P(J / A’) P(A’)
=(0.9) P(A) + P(0.05) P(A’) --------------------------------Eqn (3)
P(A)= P(A B E) + P(A B E’) + P(A B’ E) + P(A B’ E’)
= P(A / B E) P(B E) + P(A / B E’) P(B E’) + P(A/ B’ E) P(B’ E) + P(A / B’ E’) P(B’ E’)
= (0.95) P(B) P(E) + (0.95) P(B) P(E’) + (0.29) P(B’) P(E) + (0.001) P(B’) P(E’)
=(0.95)(0.001)(0.002)+(0.95)(0.001)(0.998)+(0.29)(0.999)(0.002)+(0.001)(0.998) (0.999)
= 0.0025
P(A) = 0.0025 P(A’) = 0.9975 Substitute these values in Eqn (3)
P(J) = (0.9) P(A) + P(0.05) P(A’)
= (0.9) (0.0025) + (0.05) (0.9975)
= 0.052125
P(J)= 0.052125
Substitute P(J B) and P(J) values in Eqn (1)
P(B / J) = P(J B) / P(J)
= (0.00086) / (0.052125)
P(B / J) = 0.016

13. Advantages and Disadvantages:
• It can readily handle incomplete data sets.
• It allow one to learn about causal relationships.
• It readily facilitate use of prior knowledge.
• It is more complex to construct the graph

14. CONCLUSION:
Bayesian networks provide a natural
representation for conditional independence.

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16. THANK YOU…!!!