Inference III: Approximate Inference

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Inference IIIApproximate Inference

• Slides by Nir Friedman

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Global conditioning
Fixing value of A B
Fixing values in the beginning of the summation
can decrease tables formed by variable
elimination. This way space is traded with time.
Special case choose to fix a set of nodes that
break all loops. This method is called
cutset-conditioning. Alternatively, choose to fix
some variables from the largest cliques in a
clique tree.
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Approximation

• Until now, we examined exact computation
• In many applications, approximation are
  sufficient
• Example P(X xe) 0.3183098861838
• Maybe P(X xe) ? 0.3 is a good enough
  approximation
• e.g., we take action only if P(X xe) gt 0.5
• Can we find good approximation algorithms?

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Types of Approximations

• Absolute error
• An estimate q of P(X x e) has absolute error
  ?, if
• P(X xe) - ? ? q ? P(X xe) ?
• equivalently
• q - ? ? P(X xe) ? q ?
• Absolute error is not always what we want
• If P(X x e) 0.0001, then an absolute error
  of 0.001 is unacceptable
• If P(X x e) 0.3, then an absolute error of
  0.001 is overly precise

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q
2?
0
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Types of Approximations

• Relative error
• An estimate q of P(X x e) has relative error
  ?, if
• P(X xe)(1 - ?) ? q ? P(X xe)(1 ?)
• equivalently
• q/(1 ?) ? P(X xe) ? q/(1 - ?)
• Sensitivity of approximation depends on actual
  value of desired result

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q/(1-?)
q
q/(1?)
0
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Complexity

• Recall, exact inference is NP-hard
• Is approximate inference any easier?
• Construction for exact inference
• Input a 3-SAT problem ?
• Output a BN such that P(Xt) gt 0 iff ? is
  satisfiable

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Complexity Relative Error

• Suppose that q is a relative error estimate
  ofP(X t),
• If ? is not satisfiable, then P(X t)0 . Hence,

0 P(X t)(1 - ?) ? q ? P(X t)(1 ?) 0
namely, q0. Thus, if q gt 0, then ? is
satisfiable
An immediate consequence Thm Given ?, finding
an ?-relative error approximation is NP-hard
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Complexity Absolute error

• We can find absolute error approximations to P(X
  x) with high probability (via sampling).
• We will see such algorithms shortly
• However, once we have evidence, the problem is
  harder
• Thm
• If ? lt 0.5, then finding an estimate of P(Xxe)
  with ? absulote error approximation is NP-Hard

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Proof

• Recall our construction

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Proof (cont.)

• Suppose we can estimate with ? absolute error
• Let p1 ? P(Q1 t X t)
• Assign q1 t if p1 gt 0.5, else q1 f

Let p2 ? P(Q2 t X t, Q1 q1 ) Assign q2
t if p2 gt 0.5, else q2 f
Let pn ? P(Qn t X t, Q1 q1, , Qn-1
qn-1 ) Assign qn t if pn gt 0.5, else qn f
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Proof (cont.)

• Claim if ? is satisfiable, then q1 ,, qn is a
  satisfying assignment
• Suppose ? is satisfiable
• By induction on i there is a satisfying
  assignment with Q1 q1, , Qi qi

Base case If Q1 t in all satisfying
assignments, ? P(Q1 t X t) 1 ? p1 ? 1 - ?
gt 0.5 ? q1 t If Q1 f, in all satisfying
assignments, then q1 f Otherwise, statement
holds for any choice of q1
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Proof (cont.)

• Claim if ? is satisfiable, then q1 ,, qn is a
  satisfying assignment
• Suppose ? is satisfiable
• By induction on i there is a satisfying
  assignment with Q1 q1, , Qi qi

Induction argument If Qi1 t in all satisfying
assignments s.t.Q1 q1, , Qi qi ? P(Qi1 t
X t, Q1 q1, , Qi qi ) 1 ? pi1 ? 1 - ?
gt 0.5 ? qi1 t If Qi1 f in all satisfying
assignments s.t.Q1 q1, , Qi qi then qi1 f
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Proof (cont.)

• We can efficiently check whether q1 ,, qn is a
  satisfying assignment (linear time)
• If it is, then ? is satisfiable
• If it is not, then ? is not satisfiable
• Suppose we have an approximation procedure with ?
  relative error
• ? we can determine 3-SAT with n procedure calls
• ? approximation is NP-hard

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When can we hope to approximate?

• Two situations
• Peaked distributions
• improbable values are ignored
• Highly stochastic distributions
• Far evidence is discarded

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Peaked distributions

• If the distribution is peaked, then most of the
  mass is on few instances
• If we can focus on these instances, we can ignore
  the rest

Instances
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Stochasticity Approximations

• Consider a chain
• P(Xi1 t Xi t) 1- ?P(Xi1 f Xi f)
  1- ?
• Computing the probability of Xn given X1 , we get

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Plot of P(Xn t X1 t)
?
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Stochastic Processes

• This behavior of a chain (a Markov Process) is
  called Mixing. We return to this as a tool in
  approximation
• In general networks there is a similar behavior
• If probabilities are far from 0 1, then effect
  of far evidence vanishes (and so can be
  discarded in approximations).

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Bounded conditioning
Fixing value of A B
By examining only the probable assignment of A
B, we perform several simple computations instead
of a complex one
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Bounded conditioning

• Choose A and B so that P(Y,e a,b) can be
  computed easily. E.g., a cycle cutset.
• Search for highly probable assignments to A,B.
• Option 1--- select a,b with high P(a,b).
• Option 2--- select a,b with high P(a,b e).
• We need to search for such high mass values and
  that can be hard.

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Bounded Conditioning

• Advantages
• Combines exact inference within approximation
• Continuous more time can be used to examine more
  cases
• Bounds unexamined mass used to compute
  error-bars
• Possible problems
• P(a,b) is prior mass not the posterior.
• If posterior is significantly different P(a,b
  e), Computation can be wasted on irrelevant
  assignments

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Network Simplifications

• In these approaches, we try to replace the
  original network with a simpler one
• the resulting network allows fast exact methods

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Network Simplifications

• Typical simplifications
• Remove parts of the network
• Remove edges
• Reduce the number of values (value abstraction)
• Replace a sub-network with a simpler one(model
  abstraction)
• These simplifications are often w.r.t. to the
  particular evidence and query

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Stochastic Simulation

• Suppose we can sample instances ltx1,,xngt
  according to P(X1,,Xn)
• What is the probability that a random sample
  ltx1,,xngt satisfies e?
• This is exactly P(e)
• We can view each sample as tossing a biased coin
  with probability P(e) of Heads

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Stochastic Sampling

• Intuition given a sufficient number of samples
  x1,,xN, we can estimate
• Law of large number implies that as N grows, our
  estimate will converge to p with high probability

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Sampling a Bayesian Network

• If P(X1,,Xn) is represented by a Bayesian
  network, can we efficiently sample from it?
• Idea sample according to structure of the
  network
• Write distribution using the chain rule, and then
  sample each variable given its parents

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Logic sampling
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
e
P(r)
0.3
0.001
a
P(c)
0.8
0.05
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Logic sampling
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
e
P(r)
0.3
0.001
a
P(c)
0.8
0.05
e
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Logic sampling
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
e
P(r)
0.3
0.001
a
P(c)
0.8
0.05
e
a
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Logic sampling
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
e
P(r)
0.3
0.001
a
P(c)
0.8
0.05
e
a
c
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Logic sampling
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
e
P(r)
0.3
0.001
a
P(c)
0.8
0.05
e
a
c
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Logic Sampling

• Let X1, , Xn be order of variables consistent
  with arc direction
• for i 1, , n do
• sample xi from P(Xi pai )
• (Note since Pai ? X1,,Xi-1, we already
  assigned values to them)
• return x1, ,xn

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Logic Sampling

• Sampling a complete instance is linear in number
  of variables
• Regardless of structure of the network
• However, if P(e) is small, we need many samples
  to get a decent estimate

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Can we sample from P(X1,,Xn e)?

• If evidence is in roots of network, easily
• If evidence is in leaves of network, we have a
  problem
• Our sampling method proceeds according to order
  of nodes in graph
• Note, we can use arc-reversal to make evidence
  nodes root.
• In some networks, however, this will create
  exponentially large tables...

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Likelihood Weighting

• Can we ensure that all of our sample satisfy e?
• One simple solution
• When we need to sample a variable that is
  assigned value by e, use the specified value
• For example we know Y 1
• Sample X from P(X)
• Then take Y 1
• Is this a sample from P(X,Y Y 1) ?

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Likelihood Weighting

• Problem these samples of X are from P(X)
• Solution
• Penalize samples in which P(Y1X) is small
• We now sample as follows
• Let xi be a sample from P(X)
• Let wi be P(Y 1X x i)

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Likelihood Weighting

• Why does this make sense?
• When N is large, we expect to sample NP(X x)
  samples with xi x
• Thus,

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Likelihood Weighting
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
r
r
P(r)
0.3
0.001
a
a
P(c)
B E A C R
0.8
0.05
Samples
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Likelihood Weighting
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
r
r
P(r)
0.3
0.001
a
P(c)
0.8
0.05
e
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Likelihood Weighting
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
r
r
P(r)
0.3
0.001
a
P(c)
0.8
0.05
0.6
e
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Likelihood Weighting
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
r
r
P(r)
0.3
0.001
a
P(c)
0.8
0.05
0.6
e
c
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Likelihood Weighting
P(b)
0.03
P(e)
0.001
b e
P(a)
0.4
0.01
0.98
0.7
r
r
P(r)
0.3
0.001
a
P(c)
B E A C R
0.8
0.05
0.6
0.3
e
c
r
Samples
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Likelihood Weighting

• Let X1, , Xn be order of variables consistent
  with arc direction
• w 1
• for i 1, , n do
• if Xi xi has been observed
• w ?w ? P(Xi xi pai )
• else
• sample xi from P(Xi pai )
• return x1, ,xn, and w

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Likelihood Weighting

• What can we say about the quality of answer?
• Intuitively, the weights of sample reflects their
  probability given the evidence. We need collect a
  certain mass.
• Another factor is the extremeness of CPDs.
• Thm
• If P(Xi Pai) ?l,u for all CPDs, and
• then with probability 1-?, the estimate is ?
  relative error approximation

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