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Bayesian Networks
• Using random variables to represent
objects and events in the world
– Various instantiations to these variables can
model the current state of the world
• Computing joint probabilities over these variables
• Estimating joint probabilities over all
random variables often leads to a
combinatorial explosion
– Estimating probabilities of every combination
of values for the variables involved
Conditional independence
• Use the chain rule to replace joint
probabilities with conditional ones:
P(A1, A2, A3, A4, A5) = P(A1 | A2, A3, A4,A5) · P(A2
| A3, A4, A5) · P(A3 | A4, A5) · P(A5)
• Bayesian Networks allow to reduce the
terms even further, by taking into account
conditional independence:
– P(A|C1, …, Cn, U) = P(A|C1, …, Cn) for some
set U of random variables
– Given {C1, …, Cn}, A is independent of U
Conditional independence (cont’d)
• The structure of a Bayesian Network
reflects a number of independence
assumptions (d-separation criterion)
• Directed arcs between random variables
represent conditional dependencies
• Combined with the chain rule, conditional
independencies allow us to manipulate
smaller conditional probability tables to
estimate the joint probabilities
Formal notation
• Each random variable A is conditionally
independent of all other random variables
that are not descendants of A, given A’s
n
parents
p( x1 , , xn )   p( xi | x1 ,  xi 1 )
i 1
n
p( x1 ,, xn )   p( xi | pa i )
i 1
Ind( Xi ; {X1,…,Xi-1}\Pai | Pai )
where Pai are parents of Xi
Example: modeling real world
• “Mary walks outside and finds that the street and
lawn are wet. She concludes that it has just
rained recently. Furthermore, she decides that
she doesn’t need to water her roses.”
• Mary’s logic:
– rain or sprinklers  street = wet
– rain or sprinklers  lawn = wet
– lawn = wet  soil = moist
– soil = moist  roses = OK
Mary’s world (cont’d)
• Let’s compute the probability of the
following (intuitively unlikely) world event:
– The roses are OK
– The soil is dry
– The lawn is wet
– The street is wet
– The sprinklers are off
– It’s raining
Where are the missing probabilities,
e.g., P(sprinklers=F) or
P(street=dry|rain=T, sprinklers=T) ?
Computing the probability of
a complex event
• P(sprinklers = F, rain = T, street = wet, lawn =
wet, soil = dry, roses = OK) =
P(roses = OK | soil = dry) *
P(soil = dry | lawn = wet) *
P(lawn = wet | rain = T, sprinklers = F) *
P(street = wet | rain = T, sprinklers = F) *
P(sprinklers = F) * P(rain = T) =
0.2 * 0.1 * 1.0 * 1.0 * 0.6 * 0.7 = 0.0084
 The event is quite unlikely !
Usage scenarios
• There are 2 types of computations
performed with Bayesian Networks
– Belief updating
• Computation of probabilities over random variables
– Belief revision
• Finding maximally probably global assignment
• Given some evidence or observation, our task is to
come up with a set of hypotheses (variable
assignments) that together constitute the most
satisfactory explanation/interpretation of the
evidence at hand.
Belief revision
• Let W be a set of all random variables in
the network
• Let e be the evidence (e is a subset of W)
• Any instantiation of all the variables in W
that is consistent with e is called an
explanation or interpretation of e
• The problem is to find an explanation w*
such that P(w* | e) = max P(w | e)
– w* is called the most probable explanation
Belief updating
• Computing marginal probabilities of a
subset of random variables given the
evidence
• Typically, the task is to determine the best
instantiation of a single random variable
given the evidence
Belief revision - example
• Evidence: e = {roses = OK}
• Goal: determine the assignment to all the
random variables that maximizes P(W|e)
– P(e) is constant, and e is a subset of W
it’s sufficient to maximize P(W)
– Intuitively, non-evidence random variables in
W can be viewed as possible hypotheses for e
• Solution: P(sprinklers=F, rain=T, street=wet,
lawn=wet, soil=wet, roses=OK) = 0.2646
Belief updating - example
• Evidence: e = {roses = OK}
• Goal: determine the probability that the
lawn is either wet or dry given this
observation
– P(lawn=dry | roses = OK) = 0.1190
– P(lawn=wet | roses = OK) = 0.8810