Probabilistic Reasoning [Ch. 14]
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Bayes Networks – Part 1
◦ Syntax
◦ Semantics
◦ Parameterized distributions
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Inference – Part2
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Exact inference by enumeration
Exact inference by variable elimination
Approximate inference by stochastic simulation
Approximate inference by Markov chain Monte
Carlo
Bayesian networks
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A simple, graphical notation for conditional
independence assertions and hence for compact
specification of full joint distributions
Syntax:
◦ a set of nodes, one per variable
◦ a directed, acyclic graph (link ≈ “directly influences”)
◦ a conditional distribution for each node given its parents:
P(X i |Parents(X i ))
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In the simplest case, conditional distribution
represented as a conditional probability table (CPT)
giving the distribution over X i for each combination
of parent values
Bayes’ Nets
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A Bayes’ net is an
efficient encoding
of a probabilistic
model of a domain
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Questions we can ask:
◦ Inference: given a fixed BN, what is P(X | e)?
◦ Representation: given a BN graph, what kinds of
distributions can it encode?
◦ Modeling: what BN is most appropriate for a
given domain?
Bayes’ Net Semantics
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Let’s formalize the semantics of a Bayes’ net
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A set of nodes, one per variable X
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A directed, acyclic graph
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A conditional distribution for each node
◦ A collection of distributions over X, one for each
combination of parents’ values
◦ CPT: conditional probability table
◦ Description of a noisy “causal” process
A Bayes net = Topology (graph) + Local Conditional Probabilities
Topology Limits Distributions
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Given some graph topology G,
only certain joint distributions
can be encoded
The graph structure guarantees
certain (conditional)
independences
(There might be more
independence)
Adding arcs increases the set of
distributions, but has several
costs
Full conditioning can encode
any distribution
Independence in a BN
Important question about a BN:
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◦
◦
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Are two nodes independent given certain evidence?
If yes, can prove using algebra (tedious in general)
If no, can prove with a counter example
Example:
X
Y
Z
◦ Question: are X and Z necessarily independent?
• Answer: no. Example: low pressure causes rain, which
causes traffic.
• X can influence Z, Z can influence X (via Y)
• Addendum: they could be independent: how?
Causal Chains
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This configuration is a “causal chain”
X: Low pressure
X
Y
Z
Y: Rain
Z: Traffic
◦ Is X independent of Z given Y?
Yes!
◦ Evidence along the chain “blocks” the influence
Common Cause
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Another basic configuration:
two effects of the same cause
Y
◦ Are X and Z independent?
◦ Are X and Z independent given Y?
X
Z
Y: Project due
X: Newsgroup busy
Yes!
◦ Observing the cause blocks
influence between effects.
Z: Lab full
Common Effect
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Last configuration: two causes of
one effect (v-structures)
◦ Are X and Z independent?
• Yes: the ballgame and the rain cause
traffic, but they are not correlated
• Still need to prove they must be (try it!)
◦ Are X and Z independent given Y?
• No: seeing traffic puts the rain and the
ballgame in competition as explanation?
◦ This is backwards from the other cases
• Observing an effect activates influence
between possible causes.
X
Z
Y
X: Raining
Z: Ballgame
Y: Traffic
The General Case
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Any complex example can be analyzed
using these three canonical cases
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General question: in a given BN, are two
variables independent (given evidence)?
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Solution: analyze the graph
Reachability
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Recipe: shade evidence nodes
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Attempt 1: if two nodes are
connected by an undirected
path not blocked by a shaded
node, they are conditionally
independent
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Almost works, but not quite
◦ Where does it break?
◦ Answer: the v-structure at T
doesn’t count as a link in a path
unless “active”
L
R
D
B
T
Reachability (D-Separation)
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Question: Are X and Y
conditionally independent given
evidence vars {Z}?
◦ Yes, if X and Y “separated” by Z
◦ Look for active paths from X to Y
◦ No active paths = independence!
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A path is active if each triple is
active:
◦ Causal chain X  B  Y where B is
unobserved (either direction)
◦ Common cause X  B  Y where B is
unobserved
◦ Common effect (aka v-structure)
X  B  Y where B or one of its
descendents is observed
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All it takes to block a path is a
single inactive segment
Active
Triples
X
X
B
B
X
Inactive
Triples
Y
Y
X
Y
X
B
X
B
B
Y
Y
Y
B
Y
B
X
Example
Inactive Triples
X
Yes
R
B
X
B
Y
B
Y
X
Y
B
Active Triples
T
X
X
T’
B
B
X
Y
Y
Y
B
Example
Inactive Triples
X
L
X
Yes
R
Yes
B
B
Y
B
Y
X
Y
B
Active Triples
X
D
T
X
B
B
X
Yes
T’
Y
Y
Y
B
Example
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Variables:
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Inactive Triples
X
R: Raining
T: Traffic
D: Roof drips
S: I’m sad
X
B
Y
B
Y
X
Y
B
Questions:
Active Triples
X
Yes
X
B
B
X
Y
Y
Y
B
Causality?
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When Bayes’ nets reflect the true causal patterns:
◦ Often simpler (nodes have fewer parents)
◦ Often easier to think about
◦ Often easier to elicit from experts
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BNs need not actually be causal
◦ Sometimes no causal net exists over the domain
◦ E.g. consider the variables Traffic and Drips
◦ End up with arrows that reflect correlation, not causation
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What do the arrows really mean?
◦ Topology may happen to encode causal structure
◦ Topology only guaranteed to encode conditional
independence
Changing Bayes’ Net Structure
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The same joint distribution can be encoded
in many different Bayes’ nets
◦ Causal structure tends to be the simplest
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Analysis question: given some edges, what
other edges do you need to add?
◦ One answer: fully connect the graph
◦ Better answer: don’t make any false conditional
independence assumptions
Example
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Topology of network encodes conditional
independence assertions:
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Weather is independent of the other variables
Toothache and Catch are conditionally
independent, given Cavity
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Example
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I'm at work, neighbor John calls to say my alarm is ringing,
but neighbor Mary doesn't call. Sometimes it's set off by
minor earthquakes. Is there a burglar?
Variables: Burglar, E arthquake, Alarm, J ohnCalls, M aryCalls
Network topology reflects “causal” knowledge:
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A burglar can set the alarm off
An earthquake can set the alarm off
The alarm can cause Mary to call
The alarm can cause John to call
Example contd.
Probabilities derived
from prior observations
Compactness
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A CPT for Boolean X i with k Boolean parents
has 2k rows for the combinations of parent values
Each row requires one number p for X i =true
(the number for X i =f alse is just 1 - p)
If each variable has no more than k parents,
the complete network requires O(n · 2k) numbers
I.e., grows linearly with n, vs. O(2n) for the full joint distribution
For burglary net, 1 + 1 + 4 + 2 + 2=10 numbers (vs. 25-1 = 31)
Global semantics
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“Global” semantics defines the full joint
distribution as the product of the local
conditional distributions:
P(x1, ... , xn) = Π ni=1
i P(x i|parents(X i))
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e.g., P(j ∧ m ∧ a ∧ ¬b ∧ ¬e)
= P(j|a)P(m|a)P(a|¬b,¬e)P(¬b)P(¬e)
= 0.9×0.7×0.001×0.999×0.998
≈ 0.00063
Local semantics
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Local semantics: each node is conditionally
independent of its nondescendants given its parents
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Theorem: Local semantics ⇔ global semantics
Markov blanket
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Each node is conditionally independent of all others given its
Markov blanket: parents + children + children's parents
Constructing Bayesian networks
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Need a method such that a series of locally testable
assertions of conditional independence guarantees the
required global semantics
1. Choose an ordering of variables X , . . . ,X n
2. For i = 1 to n
add X i to the network
select parents from X , . . . ,X i- such that
P(X i |Parents(X i )) = P(X i | X , . . . ,X i- )
This choice of parents guarantees the global semantics:
P(X , . . . , X n)
= Π ni=1
P(X i| X , . . . ,X i- ) (chain rule)
i
= Π ni=1
P(X i|Parents(X i)) (by construction)
i
Example: Problem formulation
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I'm at work, neighbor John calls to say my alarm is ringing,
but neighbor Mary doesn't call. Sometimes it's set off by
minor earthquakes. Is there a burglar?
Variables: Burglar, E arthquake, Alarm, J ohnCalls, M aryCalls
Network topology reflects “causal” knowledge:
◦
◦
◦
◦
A burglar can set the alarm off
An earthquake can set the alarm off
The alarm can cause Mary to call
The alarm can cause John to call
Example
Suppose we choose the ordering M , J, A, B, E
MaryCalls
JohnCalls
Alarm
Burglary
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P(J |M ) = P(J )? No
P(A|J,M ) = P(A|J )? P(A|J,M ) = P(A)?
P(B|A,J,M ) = P(B|A)? Yes
P(B|A,J,M ) = P(B)? No
P(E |B,A,J,M ) = P(E |A)? No
P(E |B,A,J,M ) = P(E |A,B)? Yes
No
Earthquake
Example contd.
MaryCalls
JohnCalls
Alarm
Burglary
Earthquake
Deciding conditional independence is hard in non-causal directions
(Causal models and conditional independence seem hardwired for
humans!)
Assessing conditional probabilities is hard in non-causal directions
Network is less compact: 1 + 2 + 4 + 2 + 4 = 13 numbers needed
Example: Car diagnosis
Initial evidence: car won't start
Testable variables (green), “broken, so fix it” variables (orange)
Hidden variables (gray) ensure sparse structure, reduce parameters
Example: Car insurance
Compact conditional distributions
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CPT grows exponentially with number of parents
CPT becomes infinite with continuous-valued parent or child
Solution: canonical distributions that are defined compactly
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Deterministic nodes are the simplest case:
X = f (Parents(X )) for some function f
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E.g., Boolean functions
NorthAmerican ⇔ Canadian ∨ US ∨ M exican
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E.g., numerical relationships among continuous variables
Level
 inflow  precipitation - outflow - evaporation
t
Compact conditional distributions
contd.
Noisy-OR distributions model multiple noninteracting causes
1) Parents U1 . . .Uk include all causes (can add leak node)
2) Independent failure probability qi for each cause alone
⇒ P(X |U1 . . .Uj , ¬Uj +1 . . . ¬Uk) = 1 - Π j i =1qi
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Cold
Flu
M alaria
F
F
F
0.0
1.0
F
F
T
0.9
0.1
F
T
F
0.8
0.2
F
T
T
0.98
0.02 = 0.2 x 0.1
T
F
F
0.4
0.6
T
F
T
0.94
0.06 = 0.6 x 0.1
T
T
F
0.88
0.12 = 0.6 x 0.2
T
T
T
0.988
0.012 = 0.6 x 0.2 x 0.1
P(¬Fever)
P(Fever)
Number of parameters linear in number of parents
Know
Infer