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Bayesian networks, introduction
Graphical models:
nodes (vertices)
links (edges)
A graph can be
disconnected:
or connected: ; undirected:
or directed:
the edges are
one-directed
arrows
cyclic:
or acyclic:
possible to start in
one node and
“come back”
Examples:
I2A
Transport routes:
F
I1
S
I2B
Acyclic, but not completely directed
Junction trees:
A
B
D
F
G
C
A
E
D
H
F
B
C
E
G
H
BCE
ABD
BDG
DFG
BEG
EGH
From 8 nodes to 6 nodes (Source: Wikipedia)
Markov random field
Given the light blue nodes, the middle blue node is conditionally
independent of all other nodes (the white nodes)
Bayesian (belief) networks
A Bayesian network is a connected directed acyclic graph (DAG) in
which
• the nodes represent random variables
• the links represent direct relevance relationships among variables
Examples:
X
Y
This small network has two nodes representing the random variable X
and Y.
The directed link gives a relevance relationship between the two
variables that means Pr (Y = y | X = x, I )  Pr (Y = y | I )
X
Y
Z
This network has three nodes representing the
random variables X, Y and Z.
The directed links give relevance relationships
that means
Pr ( Y = y | X = x, I )  Pr ( Y = y | I )
Pr ( Z = z | X = x, I )  Pr ( Z = z | I )
but also (as will be seen below)
Pr ( Z = z | Y = y, X = x, I ) = Pr ( Z = z | X = x, I )
Structures in a Bayesian network
There are two classifications for nodes: parent nodes and child nodes
parent node
child node
parent nodes
Thus, a node can be solely a
parent node, solely a child
node or both!
child nodes
Probability “tables”
Each node represents a random variable.
This random variable has either assigned probabilities (nominal scale or
discrete) or an assigned probability density function (continuous scale)
for its states.
For a node that is solely a parent node:
The assigned probabilities or density function are conditional on
background information only (may be expressed as unconditional)
For a node that is a child node (solely or joint parent/child):
The assigned probabilities or density function are conditional on the
states of its parent nodes (and on background information).
Example:
Probability tables
X
Y
X has the
states x1 and x2
X
Probabilities
x1
Pr (X = x1 | I )
x2
Pr (X = x2 | I )
Y has the states
y1 and y2
Probabilities
X:
Y:
x1
x2
y1
Pr (Y = y1 | X = x1, I )
Pr (Y = y1 | X = x2, I )
y2
Pr (Y = y2 | X = x1, I )
Pr (Y = y2 | X = x2, I )
Example Dyes on banknotes (from previous lectures)
Two states:
A?
A?
Probabilities
A : " Dye is present"
A
0.001
A : " Dye is absent"
A
0.999
Probabilities
Two states:
B?
B : " Result is positive"
B : " Result is negative"
A?:
B?:
B
B
A
0.99
A
0.02
0.01
0.98
More about the structure…
Ancestors and descendants:
A node X is an ancestor of a node Y and Y is in turn a descendant of X
if there is a unidirectional path from X to Y
A
D
B
C
E
F
G
H
I
Ancestor
Descendants
A
D, E, G, I
B
E, F, G, H, I
C
F, H, I
E
G, I
F
H, I
G
I
H
I
Different connections:
A
diverging connection
B
A
C
B
A
C
serial connection
B
converging connection
C
Conditional independence and d-separation
1) Diverging connection
A
B
C
There is a path between B and C even if it not unidirectional
 B may be relevant for C (and vice versa)
However, if the state of A is known this relevance is lost.: The
path is blocked
 B and C are conditionally independent given A
Example:
Assume the old rat Willie was caught in a trap.
We have also found a sac with wheat grains with a small hole where
grains have leaked out, and we suspect that Willie made this hole.
Examining the sac and Willie we find
• traces of wheat grain in the jaw of Willie
• traces of saliva at the damage on the sac that matches the DNA of
Willie.
The whole scenario can be described with three random variables :
A with the two states:
A1: “Willie made the whole in the sac”
A2: “Willie has not been near the sac” ,
B with the two states:
B1: “Traces of wheat grain found in Willie’s jaw”
B2: “No traces of wheat grain found in Willie’s
jaw” ,
C with the two states:
C1: “Match between saliva DNA and Willie’s
DNA”
C2: “No match in DNA between saliva and
Willie”
Note that the states of B and C are actually given, but the description gives
a complete model
First we assume none of the states are given:
Is B relevant for C ?
Yes, because if B1 is true, i.e. we have found wheat grains in Willie’s jaw,
the conditional probability of obtaining a match in DNA would be
different from the corresponding conditional probability if B2 was true.
Now assume for A that the state A2 is given, i.e. Willie was never near the
sac.
Under this condition B can no longer be relevant for C as whether we find
a match in DNA between the saliva trace and Willie or not can have
nothing to do with the grains we have found in Willie’s jaw.
Now assume for A that the state A1 is given, i.e. Willie made the hole.
Under this condition it is tempting to think that B is relevant for C, but the
relevance is actually lost. Whether we find a match in DNA or not cannot
have any impact on whether we find grains in the jaw or not once we have
stated that Willie made the hole.
The scenario can be described with the Bayesian network
A
B
C
i.e. a diverging connection
When a state of a node is assumed to be given we say the node is
instantiated
In the example, once A is instantiated the relevance relationship
between B and C is lost.
B and C are thus conditionally independent given a state of A
2) Serial connection
A
B
C
There is a path between A and C (unidirectional from A to C)
 A may be relevant for C (and vice versa)
If the state of B is known this relevance is lost.: The path is blocked
 A and C are conditionally independent given (a state of) B
Example The Willie case with another description
Let
A be a random variable with states A1: “Willie made the hole in the sac”
A2: “Willie did not make the hole” ,
B be a random variable with states B1: “Willie left saliva on the damage”
B2: “Willie left no saliva” ,
C be a random variable with states C1: “There is a match in DNA”
C2: “There is no match”
Assuming no state is given, there is a relevance relationship between A
and C:
PrC1 A1 , I   PrC1 A2 , I 
Now assuming state B1 of B is given, i.e. we assume there was a
contact between Willie’s jaw and the damage.
A can no longer be relevant for C as once we have stated that Willie left
saliva it does not matter for C whether he made the hole or not.
The scenario can be described with the Bayesian network
A
B
C
Once B is instantiated the relevant relationship between A and C is
lost.
A and C are conditionally independent given a state of B
3) Converging connection
A
B
C
There is a path between A and B (not unidirectional)
 A may be relevant for B (and vice versa)
If the state of C is (completely) unknown this relevance does not
exist.
If the state of C is known (exactly or by a modification of the state
probabilities) the path is opened
 A and C are conditionally dependent given information about
the states of C, otherwise they are (conditionally) independent
Example Paternity testing: child, mother and the true father
Let
A be a random variable representing the mother’s genotype in a specific
locus
B be a random variable representing the true father’s genotype in the
same locus
C be a random variable representing the child’s genotype in that locus
A:
A1
A2
B:
B1
B2
C:
C1
C2
If we know nothing about C (C1 and C2 are both unknown) , then
• information about A cannot have any impact on B and vice versa.
If we on the other hand know the genotype of the child (C1 and C2 are
both known or one of them is) then
• knowledge of the genotype of the mother has impact on the
probabilities of the different genotypes that can be possessed by the
true father since the child must have inherited half of the genotype
from the mother and the other half from the father.
Bayesian network:
A
B
C
d-separation
In a directed acyclic graph (DAG) the concept of d-separation is
defined as:
Let SX, SY and SX be three disjoint subsets of variables included in the
DAG
The sets SX and SY are d-separated given SZ if every path between a
variable X in SX and a variable Y in SY contains
either
• a serial connection through a variable Z in SZ or a divergent
connection diverging from a variable Z in SZ
or
• a converging connection converging to a variable W not in SZ and
of which no descendants belong to SX
X1
X2
Z2
Y1
Z1
Y3
Y2
Z3
W1
W2
• No direct link from red area
to blue area or vice versa
• No convergence from blue
area and red area to green area
The Markov property - formal definition of a Bayesian
network
Consider a variable X in a DAG
Let PA(X ) be the set of all parents to X and DE(X ) be the set of all
descendants to X.
Let SY be a set of variables that does not include any variables in DE(X ),
i.e. are not descendants of X
Then, the DAG is a Bayesian network if and only if
PrX PA  X , SY , I   PrX PA  X , I 
i.e. X is conditionally independent of SY given PA(X )
This is also known as the Markov property
Note , by Pr(X | … ) we mean the probability of X having a particular state
Pr X PA  X , SY , I  
Example
 Pr X Y3 ,W1, Y1 , Y2 , Y3 , Y4 , I  
Y1
Y2
 Pr X Y3 ,W1 , Y1 , Y2 , Y4 , I  
Y4

Y3
W1
X
W2
Serial links Y1  Y3  X ,
Y4  (W2 )W1  X
 Pr X Y3 ,W1 , I 
Pr  X PA  X , SW , I  
 Pr  X Y3 , W1, W1 , W2 , I  
D1
D1
 Pr  X Y3 , W1 , W2 , I  
 Serial link W2  W1  X 
D3
 Pr  X Y3 , W1 , I 

Software
GeNIe (Graphical network Interface)
• Software free-of-charge
• Powerful for building complex network and running with moderately
large probability tables
• Download from http://genie.sis.pitt.edu/
HUGIN
• Commercial software
• Probably today’s most powerful software for Bayesian networks
• A demo version (less powerful than GeNIe) can be downloaded from
www.hugin.com
Example
A burglary was done in a shop. On the shop floor the police have secured
a shoeprint. In the home of a suspect a shoe is found with a sole pattern
that matches that of the shoeprint.
In a compiled database of shoeprints it is found that the particular pattern
is prevalent on 3 out of 657 prints.
Hypotheses (usually called propositions in forensic literature):
Hp : “The shoeprint was made by the found shoe”
Hd : “The shoeprint was made by some other shoe”
“p” in Hp stands for “Prosecutor” (incriminating proposition)
“d” in Hd stands for “Defence” (alternative to the incriminating)
Evidence:
E : “There is a match in pattern between shoeprint and the found
shoe”
Default settings
Table
automatically set
from table of H
Setting the probability table for node E
If proposition Hp (Shoeprint was made by found shoe) is true:

 

 Pr No match H , I   Pr E H , I   0
Pr Match H p , I  Pr E H p , I  1
p
p
If proposition Hd (Shoeprint was made by another shoe) is true:
PrMatch H d , I   PrE H d , I   
where  is the proportion shoes in the (relevant) population of shoes
having the observed pattern


 PrNo match H d , I   Pr E H d , I  1  
The proportion  is unknown, but an estimate from the database can be
used
ˆ 
3
 0.0046
657
Run the network
Instantiate the match
B  LR 
Pr H p E , I  Pr H d E , I 
Pr H p I  Pr H d I 

0.995421 0.004579

 0.995421 0.004579  217
0.5 0.5
On the other we could directly have computed
1
LR 
 219
3 657
which is more accurate
Alternative network
Propositions (as before):
Hp : “The shoeprint was made by the found shoe”
Hd : “The shoeprint was made by some other shoe”
Evidence:
X : Sole pattern of the found shoe
States: q (the observed pattern)
non-q
Y : Pattern of the shoe print
States: q
non-q
LR 






 
Pr X  q, Y  q H p , I
Pr  X  q, Y  q H d , I 
Pr Y  q X  q, H p , I  Pr X  q H p , I



Pr Y  q X  q, H d , I  Pr  X  q H d , I 


Pr X  q H p , I 

 Pr X  q H d , I



Pr Y  q X  q, H p , I
Pr Y  q X  q, H d , I 
Pr Y  q X  q, H p , I
Pr Y  q H d , I 
1

3 657

1

Pr Y  q H d , I 
in a network…
Probability table for X
H
X
X
Probability
q

non-q
1–
Y
Probability table for Y
H
Hp
X
Y
Hd
q
non-q
q
non-q
q
1
0
3/657
3/657
non-q
0
1
654/657
654/657
Note!
We need to give a probability table for X to make the software work.
However, we do not know  but it does not matter what value we set
here.
Instantiate nodes X and Y both to q
Example: (more complex) In the head of the experienced examiner
Assume there is a question whether an individual has a specific
disease A or another disease B.
What is observed is
The individual has an increased level of substance 1
The individual has recurrent fever attacks
The individual has light recurrent pain in the stomach
The experience of the examining physician says
1. If disease A is present it is quite common to have an increased level
of substance 1.
2. If disease B is present it is less common to have an increased level
of substance 1.
3. If disease A is present it is not generally common to have recurrent
fever attacks, but if there is also an increased level of substance 1
such events are very common
4. Recurrent fever attacks are quite common when disease B is
present regardless of the level of substance 1
5. Recurrent pain in the stomach are generally more common when
disease B is present than when disease A is present, and regardless
of the level of substance 1 and whether fever attacks are present or
not
6. If a patient has disease A, increased levels of substance 1 and
recurrent fever attacks he/she would almost certainly have
recurrent pain in the stomach. Otherwise, if disease A is present
recurrent pain in the stomach is equally common.
Can we put this up in a network?
Let the “disease node” be H, with states A and B
Let the “evidence” nodes be
X with states
x1 : “The individual has an increased level of substance 1”
x2 : “The individual has a normal level of substance 1”
Y with states
y1 : “The individual has recurrent fever attacks”
y2 : “The individual has no fever attacks”
Z with states
z1 : “The individual has light recurrent pain in the stomach”
z2 : “The individual has no pain in the stomach”
Probability table for X
H
H:
X
X
A
B
x1


x2
1–
1–
Probability table for Y
Y
H:
A
X:
Z
Y
y1
B
x1
x2
x1
x2




1–
1–
1– 1–
y2
Probability table for Z
H:
A
X:
x1
Y:
Z
B
x2
x1
x2
y1
y2
y1
y2
y1
y2
y1
y2
z1
1







z1
0
1–
1–
1–
1– 1– 1– 1–
The probabilities set out in the tables take into account some of the
experience listed (e.g. that some probabilities are equal)
However, we need to estimate numbers for , , , , ,  and 
Experience 1 & 2   >> 
Assume   0.8 and   0.2
Experience 3   high and  < 0.5 Assume   0.9,   0.3
Experience 4 Assume   0.8
Experience 5 & 6   >  Assume   0.6 and   0.4
Run network
Instantiate the nodes X, Y and Z
The likelihood ratio of the evidence becomes
88.325
LR 
 7.5
11.764
Thus the three observations are combined 7.5 times more probable if
disease A is present than if disease B is present.