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. PGM 2002/3 – Tirgul6 Approximate Inference: Sampling
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Approximation u Until now, we examined exact computation u In many applications, approximation are sufficient Example: P(X = x|e) = 0.3183098861838 Maybe P(X = x|e) 0.3 is a good enough approximation e.g., we take action only if P(X = x|e) > 0.5 u 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 = x|e) - q P(X = x|e) + equivalently q - P(X = x|e) q + u 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 0 1 q 22
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Types of Approximations Relative error An estimate q of P(X = x | e) has relative error , if P(X = x|e)(1 - ) q P(X = x|e)(1 + ) equivalently q/(1 + ) P(X = x|e) q/(1 - ) Sensitivity of approximation depends on actual value of desired result 0 1 q q/(1+ ) q/(1- )
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Complexity u Recall, exact inference is NP-hard u Is approximate inference any easier? u Construction for exact inference: Input: a 3-SAT problem Output: a BN such that P(X=t) > 0 iff is satisfiable
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Complexity: Relative Error Suppose that q is an relative error estimate of P(X = t), If is not satisfiable, then P(X = t)(1 - ) q P(X = t)(1 + ) 0 = P(X = t)(1 - ) q P(X = t)(1 + ) = 0 Thus, if q > 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) l We will see such algorithms shortly u However, once we have evidence, the problem is harder Thm If < 0.5, then finding an estimate of P(X=x|e) with absulote error approximation is NP-Hard
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Proof u Recall our construction... 11 Q1Q1 Q3Q3 Q2Q2 Q4Q4 QnQn 22 33 kk A1A1 k-1 A2A2 A k/2 X...
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Proof (cont.) Suppose we can estimate with absolute error Let p 1 P(Q 1 = t | X = t) Assign q 1 = t if p 1 > 0.5, else q 1 = f Let p 2 P(Q 2 = t | X = t, Q 1 = q 1 ) Assign q 2 = t if p 2 > 0.5, else q 2 = f … Let p n P(Q n = t | X = t, Q 1 = q 1, …, Q n-1 = q n-1 ) Assign q n = t if p n > 0.5, else q n = f
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Proof (cont.) Claim: if is satisfiable, then q 1,…, q n is a satisfying assignment Suppose is satisfiable By induction on i there is a satisfying assignment with Q 1 = q 1, …, Q i = q i Base case: If Q 1 = t in all satisfying assignments, P(Q 1 = t | X = t) = 1 p 1 1 - > 0.5 q 1 = t If Q 1 = f, in all satisfying assignments, then q 1 = f Otherwise, statement holds for any choice of q 1
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Induction argument: If Q i+1 = t in all satisfying assignments s.t. Q 1 = q 1, …, Q i = q i P(Q i+1 = t | X = t, Q 1 = q 1, …, Q i = q i ) = 1 p i+1 1 - > 0.5 q i+1 = t If Q i+1 = f in all satisfying assignments s.t. Q 1 = q 1, …, Q i = q i then q i+1 = f Proof (cont.) Claim: if is satisfiable, then q 1,…, q n is a satisfying assignment Suppose is satisfiable By induction on i there is a satisfying assignment with Q 1 = q 1, …, Q i = q i
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Proof (cont.) We can efficiently check whether q 1,…, q n 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 u approximation is NP-hard
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Search Algorithms Idea: search for high probability instances Suppose x[1], …, x[N] are instances with high mass u We can approximate: If x[i] is a complete instantiation, then P(e|x[i]) is 0 or 1
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Search Algorithms (cont) Instances that do not satisfy e, do not play a role in approximation We need to focus the search to find instances that do satisfy e u Clearly, in some cases this is hard (e.g., the construction from our NP-hardness result
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Stochastic Simulation Suppose we can sample instances according to P (X 1,…,X n ) What is the probability that a random sample 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 x[1],…,x[N], we can estimate Law of large number implies that as N grows, our estimate will converge to p with high probability u How many samples do we need to get a reliable estimation? Use Chernof’s bound for binomial distributions
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Sampling a Bayesian Network If P (X 1,…,X n ) is represented by a Bayesian network, can we efficiently sample from it? u Idea: sample according to structure of the network l Write distribution using the chain rule, and then sample each variable given its parents
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Samples: B E A C R Logic sampling P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 b Earthquake Radio Burglary Alarm Call 0.03
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Samples: B E A C R Logic sampling P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 eb Earthquake Radio Burglary Alarm Call 0.001
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Samples: B E A C R Logic sampling P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 eab 0.4 Earthquake Radio Burglary Alarm Call
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Samples: B E A C R Logic sampling P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 eacb Earthquake Radio Burglary Alarm Call 0.8
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Samples: B E A C R Logic sampling P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 eacb r 0.3 Earthquake Radio Burglary Alarm Call
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Logic Sampling Let X 1, …, X n be order of variables consistent with arc direction for i = 1, …, n do sample x i from P(X i | pa i ) (Note: since Pa i {X 1,…,X i-1 }, we already assigned values to them) return x 1, …,x n
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Logic Sampling u Sampling a complete instance is linear in number of variables l 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( X 1,…,X n |e)? u If evidence is in roots of network, easily u If evidence is in leaves of network, we have a problem l Our sampling method proceeds according to order of nodes in graph u Note, we can use arc-reversal to make evidence nodes root. l 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? u One simple solution: l 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) ? X Y
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Likelihood Weighting Problem: these samples of X are from P(X) u Solution: Penalize samples in which P(Y=1|X) is small u We now sample as follows: Let x[i] be a sample from P(X) Let w[i] be P(Y = 1|X = x [i]) X Y
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Likelihood Weighting u Why does this make sense? When N is large, we expect to sample NP(X = x) samples with x[i] = x u Thus, u When we normalize, we get approximation of the conditional probability
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Samples: B E A C R Likelihood Weighting P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a 0.8 0.05 P(r) e e 0.30.001 b Earthquake Radio Burglary Alarm Call 0.03 Weight = r a = a
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Samples: B E A C R Likelihood Weighting P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 eb Earthquake Radio Burglary Alarm Call 0.001 Weight = r = a
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Samples: B E A C R Likelihood Weighting P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 eb 0.4 Earthquake Radio Burglary Alarm Call Weight = r = a 0.6 a
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Samples: B E A C R Likelihood Weighting P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 ecb Earthquake Radio Burglary Alarm Call 0.05 Weight = r = a a 0.6
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Samples: B E A C R Likelihood Weighting P(b) 0.03 P(e) 0.001 P(a) b e 0.98 0.4 0.7 0.01 P(c) a a 0.8 0.05 P(r) e e 0.30.001 ecb r 0.3 Earthquake Radio Burglary Alarm Call Weight = r = a a 0.6 *0.3
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Likelihood Weighting Let X 1, …, X n be order of variables consistent with arc direction u w = 1 for i = 1, …, n do if X i = x i has been observed w w* P(X i = x i | pa i ) l else sample x i from P(X i | pa i ) return x 1, …,x n, and w
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Importance Sampling A general method for evaluating P(X) when we cannot sample from P(X). Idea: Choose an approximating distribution Q(X) and sample from it Using this we can now sample from Q and then W(X) If we could generate samples from P(X) Now that we generate the sample from Q(X)
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(Unnormalized) Importance Sampling 1. For m=1:M Sample X[m] from Q(X) Calculate W(m) = P(X)/Q(X) 2. Estimate the expectation of f(X) using Requirements: P(X)>0 Q(X)>0 (do not ignore possible scenarios) It is possible to calculate P(X),Q(X) for a specific X=x It is possible to sample from Q(X)
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Normalized Importance Sampling Assume that we cannot now even evalute P(X=x) but can evaluate P’(X=x) = P(X=x) (for example we can evaluate P(X) but not P(X|e) in a Bayesian network) We define w’(X) = P’(X)/Q(X). We can then evaluate : and then: where in the last step we simply replace with the above equation
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Normalized Importance Sampling We can now estimate the expectation of f(X) similarly to unnormalized importance sampling by sampling from Q(X) and then (hence the name “normalized”)
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Importance Sampling to LW We want to compute P(Y=y|e)? (X is the set of random variables in the network and Y is some subset we are interested in) 1) Define a mutilated Bayesian network B Z=z to be a network where: all variables in Z are disconnected from their parents and are deterministically set to z all other variables remain unchanged 2) Choose Q to be B E=e convince yourself that P’(X)/Q(X) is exactly P(Y=y|X) 3) Choose f(x) to be 1(Y[m]=y)/M 4) Plug into the formula and you get exactly Likelihood Weighting Likelihood weighting is correct!!!
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