1. Bayesian Networks (Directed Acyclic Graphical Models)
Coin1
Coin2
Bell
The situation of a bell that rings whenever the outcome of two coins are equal can not be well represented by undirected graphical models.
A clique will be formed because of induced dependency of the two coins given the bell.

2. Bayesian Networks (Directed Graphical Models)

4. Example I X1 X2 X3 X4
ID( X3 ; X1 | X2)
ID( X4 ; {X1, X2}| X3)

5. Example II In the order V,S,T,L,B,A,X,D, we have: ID( S; V )
ID( T; S | V )
ID( l; {T, V} | S ) …
ID( X; {V,S,T,L,B,D} | A)
Does ID( {X, D} ; V | A ) also hold ?
To answer this question one needs to analyze the types of paths that connect {X, D} and V.

6. Paths Intuition: dependency must “flow” along paths in the graph
A path is a sequence of neighboring variables
Examples:
X  A  D  B
A  L  S  B V S L T A B X D

7. Path blockage Every path is classified given the evidence:
active -- creates a dependency between the end nodes
blocked – does not create a dependency between the end nodes
Evidence means the assignment of a value to a subset of nodes.

8. Path Blockage Three cases: Common cause Blocked Blocked Active S L B S

9. Path Blockage Three cases: Common cause Intermediate cause Blocked
Active
Blocked S A L

10. Path Blockage Three cases: Common cause Intermediate cause
Common Effect
Blocked
Active
Blocked T L X A T L X A

11. Definition of Path Blockage
Definition: A path is active, given evidence Z, if
Whenever we have the configuration then either A or one of its descendents is in Z
No other nodes in the path are in Z.
Definition: A path is blocked, given evidence Z, if it is not active. T L A
Definition: X is d-separated from Y, given Z, if all paths from a node in X and a node in Y are blocked, given Z.

12. Example
ID(T,S|) = yes V S L T A B X D

13. Example
ID (T,S |) = yes
ID(T,S|D) = no V S L T A B X D

14. Example ID (T,S |) = yes ID(T,S|D) = no ID(T,S|{D,L,B}) = yes V S L T

15. d-separation The definition of ID (X; Y | Z) is such that:
Soundness [Theorem 9]: ID (X; Y | Z) = yes implies IP(X;Y|Z) follows from Basis(G)
Completeness [Theorem 10]: ID (X; Y | Z) = no implies IP(X;Y|Z) does not follow from Basis(G)

16. Revisiting Example II So does IP( {X, D} ; V | A ) hold ? V S L T A B

20. Extension of the Markov Chain Property

23. How Expressive are Bayesian Networks

26. Quantifying the links of Bayesian Networks
p(v)
p(s) V S L T A B X D
p(t|v)
p(l|s)
p(b|s)
p(a|t,l)
p(d|a,b)
p(x|a)
Bayesian network = Directed Acyclic Graph (DAG), annotated with conditional probability distributions.

27. Bayesian Network (cont.)
Each Directed Acyclic Graph defines a factorization of the form:
p(t|v) V S L T A B X D
p(x|a)
p(d|a,b)
p(a|t,l)
p(b|s)
p(l|s)
p(s)
p(v)

28. Independence in Bayesian networks
IP( Xi ; {X1,…,Xi-1}\Pai | Pai )
This set of independence assertions is denoted Basis(G) .
All other independence assertions that are entailed by (*) are derivable using the semi-graphoid axioms.

29. Local distributions- Asymmetric independence
Lung Cancer
(Yes/No)
Tuberculosis
Abnormality
in Chest
(Yes/no)
p(A|T,L)
Table:
p(A=y|L=n, T=n) = 0.02
p(A=y|L=n, T=y) = 0.60
p(A=y|L=y, T=n) = 0.99
p(A=y|L=y, T=y) = 0.99