Upload
ptkngoan
View
222
Download
0
Embed Size (px)
Citation preview
7/31/2019 Presentace Karny
1/27
Some applications of Bayesian networks
Ji r Vomlel
Institute of Information Theory and AutomationAcademy of Sciences of the Czech Republic
This presentation is available athttp://www.utia.cas.cz/vomlel/
1
7/31/2019 Presentace Karny
2/27
Contents Brief introduction to Bayesian networks
Typical tasks that can be solved using Bayesian networks
1: Medical diagnosis (a very simple example)
2: Decision making maximizing expected utility (another simpleexample)
3: Adaptive testing (a case study)
4: Decision-theoretic troubleshooting (a commercial product)
2
7/31/2019 Presentace Karny
3/27
Bayesian network a directed acyclic graph G = ( V , E)
each node i V corresponds to a random variable Xi with anite set X i of mutually exclusive states
pa(i) denotes the set of parents of node i in graph G
to each node i V corresponds a conditional probability tableP(Xi | (X j) j pa(i) )
the DAG implies conditional independence relations between(Xi) i V
d-separation (Pearl, 1986) can be used to read the CI relationsfrom the DAG
3
7/31/2019 Presentace Karny
4/27
Using the chain rule we have that:
P(( Xi) i V ) = i V
P(Xi | Xi 1, . . . , X1)
Assume an ordering of Xi, i V such that if j pa(i) then j < i.From the DAG we can read conditional independence relations
Xi Xk | (X j) j pa(i) for i V and k < i and k pa(i)
Using the conditional independence relations from the DAG we get
P(( Xi) i V ) = i V
P(Xi | (X j) j pa(i) ) .
It is the joint probability distribution represented by the Bayesiannetwork .
4
7/31/2019 Presentace Karny
5/27
Example:X1 X2P(X1) P(X2)
P(X3 | X1)P(X4 | X2)
P(X6 | X3 , X4)
P(X9 | X6)P(X8 | X7 , X6)
P(X5 | X1)
P(X7 | X5)
X5
X7
X3
X8
X6
X9
X4
P(X1, . . . , X9) == P(X9 |X8, . . . , X1) P(X8 |X7, . . . , X1) . . . P(X2 |X1) P(X1)
= P(X9 |X6) P(X8 |X7, X6) P(X7 |X5) P(X6 |X4, X3)P(X5 |X1) P(X4 |X2) P(X3 |X1) P(X2) P(X1)
5
7/31/2019 Presentace Karny
6/27
7/31/2019 Presentace Karny
7/27
Simplied diagnostic exampleWe have a patient.Possible diagnoses: tuberculosis, lung cancer, bronchitis.
7
7/31/2019 Presentace Karny
8/27
We dont know anything about the pa-tient
Patient is a smoker.
8
7/31/2019 Presentace Karny
9/27
Patient is a smoker. ... and he complains about dyspnoea
9
7/31/2019 Presentace Karny
10/27
Patient is a smoker and complainsabout dyspnoea
... and his X-ray is positive
10
7/31/2019 Presentace Karny
11/27
7/31/2019 Presentace Karny
12/27
Application 2 :Decision making
The goal: maximize expected utility
Hugin example: mildew4.net
12
7/31/2019 Presentace Karny
13/27
Fixed and Adaptive Test Strategies
wrong
correct wrong
wrong correct
wrong
wrong
correct
correctwrong wrongcorrectcorrectcorrect
Q5
Q4
Q3
Q2
Q1
Q2
Q1
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q6 Q8
Q7
Q8
Q4 Q7 Q10
Q6
Q7
Q9
13
7/31/2019 Presentace Karny
14/27
X3
X1
X3
X3
X2
X3
X2
X1
X2
X1
X2
X2
X3
X1
X1
For all nodes n of a strategy s wehave dened:
evidence e n , i.e. outcomes ofsteps performed to get to noden,
probability P(e n ) of getting to
node n, and utility f (e n ) being a real num-
ber.
Let L( s ) be the set of terminalnodes of strategy s .Expected utility of strategy isE f (s ) = L (s ) P(e ) f (e ).
14
7/31/2019 Presentace Karny
15/27
X3
X1
X3
X3
X2
X3
X2
X1
X2
X1
X2
X2
X3
X1
X1
Strategy s is optimal iff it maxi-mizes its expected utility.
Strategy s is myopically optimal iffeach step of strategy s is selectedso that it maximizes expected utilityafter the selected step is performed(one step look ahead ).
15
7/31/2019 Presentace Karny
16/27
Application 3 : Adaptive test of basicoperations with fractions
Examples of tasks:
T 1:34
56
18 =
1524
18 =
58
18 =
48 =
12
T 2: 16 +1
12 =2
12 +1
12 =3
12 =14
T 3: 14 112 =
14
32 =
38
T 4: 12 12 13 + 13 = 14 23 = 212 = 16 .
16
7/31/2019 Presentace Karny
17/27
Elementary and operational skillsCP Comparison (common nu-
merator or denominator)
12
> 13 , 23 > 13
AD Addition (comm. denom.) 17 +27 =
1+ 27 =
37
SB Subtract. (comm. denom.) 25 15 =
2 15 =
15
MT Multiplication 12 35 = 310
CD Common denominator 12 ,23 =
36 ,
46
CL Cancelling out 46 =2223 =
23
CIM Conv. to mixed numbers 72 =32+ 1
2 = 312
CMI Conv. to improp. fractions 3 12 =32+ 1
2 =72
17
7/31/2019 Presentace Karny
18/27
MisconceptionsLabel Description Occurrence
MAD ab +cd =
a+ cb+ d 14.8%
MSB ab cd =
a cb d 9.4%
MMT1 ab cb = acb 14.1%MMT2 ab
cb =
a+ cbb 8.1%
MMT3 ab cd =
adbc 15.4%
MMT4 ab cd =
acb+ d 8.1%
MC abc =
abc 4.0%
18
7/31/2019 Presentace Karny
19/27
Student model
CP
ACD
HV2 HV1
AD SB CMI CIM CL CD MT
MMT1 MMT2 MMT3 MMT4MCMAD MSB
ACMI ACIM ACL
19
7/31/2019 Presentace Karny
20/27
Evidence model for task T 134
56
18 =
1524
18 =
58
18 =
48 =
12
T 1 MT & CL & ACL & SB & MMT 3 & MMT 4 & MSB
CL
MMT4
MSB
SB
MMT3
ACL MT
T1
X1
P(X1 | T1 )
Hugin: model-hv-2.net
20
7/31/2019 Presentace Karny
21/27
Using information gain as the utility function
The lower the entropy of a probability distribution the more we know.
H (P(X )) = x
P(X = x ) log P(X = x )
0
0.5
1
0 0.5 1
e n t r o p
y
robabilit
Information gain in a node n of a strategy
IG(e n) = H (P(S )) H (P(S | e n ))
21
7/31/2019 Presentace Karny
22/27
Skill Prediction Quality
74
76
78
80
82
84
86
88
90
92
0 2 4 6 8 10 12 14 16 18 20
Q u a
l i t y o
f s k
i l l p r e
d i c t i o n s
Number of answered uestions
adaptiveaveragedescending
ascending
22
7/31/2019 Presentace Karny
23/27
Application 4: Troubleshooting
Dezide Advisor customized to a specic portal, seen from the usersperspective through a web browser.
23
7/31/2019 Presentace Karny
24/27
Application 2: Troubleshooting - Light print problem
FF3
F2
F1
F4
FaultsActions
A3
A2
A1
Q1
Problem
Questions
Problems: F1 Distribution problem , F2 Defective toner , F3Corrupted dataow , and F4 Wrong driver setting .
Actions:A1
Remove, shake and reseat toner ,A2
Try another toner , and A3 Cycle power .
Questions: Q1 Is the conguration page printed light?
24
7/31/2019 Presentace Karny
25/27
Troubleshooting strategy
A1 = no
A2 = yes
Q1 = no
A1 = yes A2 = yes
Q1 = yes
A1 = yes
A2 = no
A1 = no A2 = no
A2
Q1
A1
A2 A1
The task is to nd a strategy s S minimising expected cost of repair
ECR(s ) = L (s )
P(e ) ( t(e ) + c(e ) ) .
25
7/31/2019 Presentace Karny
26/27
Expected cost of repair for a given strategy
A1 = no
A2 = yes
Q1 = no
A1 = yes A2 = yes
Q1 = yes
A1 = yes
A2 = no
A1 = no A2 = no
A2
Q1
A1
A2 A1
ECR(s ) =
P(Q1 = no, A1 = yes) cQ1 + c A1+ P(Q1 = no, A1 = no, A2 = yes) cQ1 + c A1 + c A2
+ P(Q1 = no, A1 = no, A2 = no) cQ1 + c A1 + c A2 + cCS+ P(Q1 = yes, A2 = yes) cQ1 + c A2+ P(Q1 = yes, A2 = no, A1 = yes) cQ1 + c A2 + c A1+ P(Q1 = yes, A2 = no, A1 = no) cQ1 + c A2 + c A1 + cCS
Demo: www.dezide.com Products/Demo/Try out expert mode
26
7/31/2019 Presentace Karny
27/27
Commercial applications of Bayesian networks ineducational testing and troubleshooting
Hugin Expert A/S.software product: Hugin - a Bayesian network tool.http://www.hugin.com/
Educational Testing Service (ETS)the worlds largest private educational testing organizationResearch unit doing research on adaptive tests using Bayesiannetworks: http://www.ets.org/research/
SACSO ProjectSystems for Automatic Customer Support Operations
- research project of Hewlett Packard and Aalborg University.The troubleshooter offered as DezisionWorks by Dezide Ltd.http://www.dezide.com/
27