Explanation:why do I need a new starter motor? Sensitivity analysis:which probability values are most critical? Conjunctive queries:P(X:XE=e)=P(XE=e)P(XX E=e) Inference tasks *#*9829月38 Outline CHAPTER 14.4-5 INFERENCE IN BAYESIAN NETWORK Q Evaluation tree Enumeration algorithm Recursive depth-first enumeration:O(n)space,O()time Simple query on the burglary network: Inference by enumeration
Inference in Bayesian networks Chapter 14.4–5 Chapter 14.4–5 1 Outline ♦ Exact inference by enumeration ♦ Exact inference by variable elimination ♦ Approximate inference by stochastic simulation ♦ Approximate inference by Markov chain Monte Carlo Chapter 14.4–5 2 Inference tasks Simple queries: compute posterior marginal P(Xi |E = e) e.g., P(NoGas|Gauge = empty, Lights = on, Starts = false) Conjunctive queries: P(Xi , Xj |E = e) = P(Xi |E = e)P(Xj |Xi , E = e) Optimal decisions: decision networks include utility information; probabilistic inference required for P(outcome|action, evidence) Value of information: which evidence to seek next? Sensitivity analysis: which probability values are most critical? Explanation: why do I need a new starter motor? Chapter 14.4–5 3 Inference by enumeration Slightly intelligent way to sum out variables from the joint without actually constructing its explicit representation Simple query on the burglary network: B E J A M P(B|j, m) = P(B, j, m)/P(j, m) = αP(B, j, m) = α Σe Σa P(B, e, a, j, m) Rewrite full joint entries using product of CPT entries: P(B|j, m) = α Σe Σa P(B)P(e)P(a|B, e)P(j|a)P(m|a) = αP(B) Σe P(e) Σa P(a|B, e)P(j|a)P(m|a) Recursive depth-first enumeration: O(n) space, O(d n ) time Chapter 14.4–5 4 Enumeration algorithm function Enumeration-Ask(X, e, bn) returns a distribution over X inputs: X, the query variable e, observed values for variables E bn, a Bayesian network with variables {X} ∪ E ∪ Y Q(X ) ← a distribution over X, initially empty for each value xi of X do extend e with value xi for X Q(xi) ← Enumerate-All(Vars[bn], e) return Normalize(Q(X )) function Enumerate-All(vars, e) returns a real number if Empty?(vars) then return 1.0 Y ← First(vars) if Y has value y in e then return P(y | Pa(Y )) × Enumerate-All(Rest(vars), e) else return P y P(y | Pa(Y )) × Enumerate-All(Rest(vars), ey ) where ey is e extended with Y = y Chapter 14.4–5 5 Evaluation tree .70 P(m|a) .90 P(j|a) .01 P(m| a) .05 P(j| a) .70 P(m|a) .90 P(j|a) .01 P(m| a) .05 P(j| a) .001 P(b) .002 P(e) .998 P( e) .95 P(a|b,e) .06 P( a|b, e) .05 P( a|b,e) .94 P(a|b, e) Enumeration is inefficient: repeated computation e.g., computes P(j|a)P(m|a) for each value of e Chapter 14.4–5 6
ssuming do not depend on P(Bl3.m) Variable elimination algorithm mingout a variable from a product of factors: Variable elimination:Basic operations Inference by variable elimination 1.A vB v C equivalent to counting 3SAT model#P-complete can reduce 3SAT to exact inference NP.hard Complexity of exact inference Thm 2:Y is irrelevant if m-separated from X by E Defn:A from B by Ciff separated by Cin the moral grap Irrelevant variables contd. (Compare this to backward chaining from the query in Hom dlause KBs) Sum over m is ickentically 1:is irrelevant to the query Consider the query P(JohnCalls Bur glary=true) Irrelevant variables Q
Inference by variable elimination Variable elimination: carry out summations right-to-left, storing intermediate results (factors) to avoid recomputation P(B|j, m) = α P(B) | {z } B Σe P(e) | {z } E Σa P(a|B, e) | {z } A P(j|a) | {z } J P(m|a) | {z } M = αP(B)ΣeP(e)ΣaP(a|B, e)P(j|a)fM(a) = αP(B)ΣeP(e)ΣaP(a|B, e)fJ(a)fM(a) = αP(B)ΣeP(e)ΣafA(a, b, e)fJ(a)fM(a) = αP(B)ΣeP(e)fA¯JM(b, e) (sum out A) = αP(B)fE¯AJ¯ M(b) (sum out E) = αfB(b) × fE¯AJ¯ M(b) Chapter 14.4–5 7 Variable elimination: Basic operations Summing out a variable from a product of factors: move any constant factors outside the summation add up submatrices in pointwise product of remaining factors Σxf1 × · · · × fk = f1 × · · · × fi Σx fi+1 × · · · × fk = f1 × · · · × fi × fX¯ assuming f1, . . . , fi do not depend on X Pointwise product of factors f1 and f2: f1(x1, . . . , xj , y1, . . . , yk) × f2(y1, . . . , yk, z1, . . . , zl) = f(x1, . . . , xj , y1, . . . , yk, z1, . . . , zl) E.g., f1(a, b) × f2(b, c) = f(a, b, c) Chapter 14.4–5 8 Variable elimination algorithm function Elimination-Ask(X, e, bn) returns a distribution over X inputs: X, the query variable e, evidence specified as an event bn, a belief network specifying joint distribution P(X1, . . . , Xn) factors ← [ ]; vars ← Reverse(Vars[bn]) for each var in vars do factors ← [Make-Factor(var , e)|factors] if var is a hidden variable then factors ← Sum-Out(var,factors) return Normalize(Pointwise-Product(factors)) Chapter 14.4–5 9 Irrelevant variables Consider the query P(JohnCalls|Burglary = true) B E J A M P(J|b) = αP(b) e X P(e) a X P(a|b, e)P(J|a) m X P(m|a) Sum over m is identically 1; M is irrelevant to the query Thm 1: Y is irrelevant unless Y ∈ Ancestors({X}∪ E) Here, X = JohnCalls, E = {Burglary}, and Ancestors({X}∪ E) = {Alarm, Earthquake} so MaryCalls is irrelevant (Compare this to backward chaining from the query in Horn clause KBs) Chapter 14.4–5 10 Irrelevant variables contd. Defn: moral graph of Bayes net: marry all parents and drop arrows Defn: A is m-separated from B by C iff separated by C in the moral graph Thm 2: Y is irrelevant if m-separated from X by E B E J A M For P(JohnCalls|Alarm = true), both Burglary and Earthquake are irrelevant Chapter 14.4–5 11 Complexity of exact inference Singly connected networks (or polytrees): – any two nodes are connected by at most one (undirected) path – time and space cost of variable elimination are O(d kn) Multiply connected networks: – can reduce 3SAT to exact inference ⇒ NP-hard – equivalent to counting 3SAT models ⇒ #P-complete A B C D 1 2 3 AND 0.5 0.5 0.5 0.5 L L L L 3. B v C v D 2. C v D v A 1. A v B v C Chapter 14.4–5 12
Basic idea 酮 Sampling from an empty network whose stationary dist ribution is the true posterior -Mark chain Monte Cark(MCMC):samplefrom a stochastic proces Inference by stochastic simulation 图E S R P(WIS.R) Example P(WIS.R Example 可
Inference by stochastic simulation Basic idea: 1) Draw N samples from a sampling distribution S Coin 0.5 2) Compute an approximate posterior probability Pˆ 3) Show this converges to the true probability P Outline: – Sampling from an empty network – Rejection sampling: reject samples disagreeing with evidence – Likelihood weighting: use evidence to weight samples – Markov chain Monte Carlo (MCMC): sample from a stochastic process whose stationary distribution is the true posterior Chapter 14.4–5 13 Sampling from an empty network function Prior-Sample(bn) returns an event sampled from bn inputs: bn, a belief network specifying joint distribution P(X1, . . . , Xn) x ← an event with n elements for i = 1 to n do xi ← a random sample from P(Xi | parents(Xi)) given the values of Parents(Xi) in x return x Chapter 14.4–5 14 Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 15 Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 16 Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 17 Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 18
c P(SC) c P(SC) 三88 S RPWIS,R) 色888 P(e)drops off exp Hence Then we have ples have Sp with numbe r of evidence Analysis of rejection sampling P(fain|Sprindfer=trwe)=NORMALIZE(()=(02%6.0.70) Rejection sampling Let Ns)be the number of samples generated for evn Sampling from an empty network contd
Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 19 Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 20 Example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 Chapter 14.4–5 21 Sampling from an empty network contd. Probability that PriorSample generates a particular event SPS(x1 . . . xn) = Πi n = 1P(xi |parents(Xi)) = P(x1 . . . xn) i.e., the true prior probability E.g., SPS(t, f,t,t) = 0.5 × 0.9 × 0.8 × 0.9 = 0.324 = P(t, f,t,t) Let NPS(x1 . . . xn) be the number of samples generated for event x1, . . . , xn Then we have N→∞ lim Pˆ(x1, . . . , xn) = N→∞ lim NPS(x1, . . . , xn)/N = SPS(x1, . . . , xn) = P(x1 . . . xn) That is, estimates derived from PriorSample are consistent Shorthand: Pˆ(x1, . . . , xn) ≈ P(x1 . . . xn) Chapter 14.4–5 22 Rejection sampling Pˆ (X|e) estimated from samples agreeing with e function Rejection-Sampling(X, e, bn,N) returns an estimate of P(X |e) local variables: N, a vector of counts over X, initially zero for j = 1 to N do x ← Prior-Sample(bn) if x is consistent with e then N[x] ← N[x]+1 where x is the value of X in x return Normalize(N[X]) E.g., estimate P(Rain|Sprinkler = true) using 100 samples 27 samples have Sprinkler = true Of these, 8 have Rain = true and 19 have Rain = false. Pˆ (Rain|Sprinkler = true) = Normalize(h8, 19i) = h0.296, 0.704i Similar to a basic real-world empirical estimation procedure Chapter 14.4–5 23 Analysis of rejection sampling Pˆ (X|e) = αNPS(X, e) (algorithm defn.) = NPS(X, e)/NPS(e) (normalized by NPS(e)) ≈ P(X, e)/P(e) (property of PriorSample) = P(X|e) (defn. of conditional probability) Hence rejection sampling returns consistent posterior estimates Problem: hopelessly expensive if P(e) is small P(e) drops off exponentially with number of evidence variables! Chapter 14.4–5 24
S RRWISR) Likelihood weighting example Likelihood weighting example Likelihood weighting 888 Likelihood weighting example 可 Likelihood weighting example Likelihood weighting example
Likelihood weighting Idea: fix evidence variables, sample only nonevidence variables, and weight each sample by the likelihood it accords the evidence function Likelihood-Weighting(X, e, bn,N) returns an estimate of P(X |e) local variables: W, a vector of weighted counts over X, initially zero for j = 1 to N do x,w ←Weighted-Sample(bn) W[x ] ←W[x ] + w where x is the value of X in x return Normalize(W[X ]) function Weighted-Sample(bn, e) returns an event and a weight x ← an event with n elements; w ← 1 for i = 1 to n do if Xi has a value xi in e then w ← w × P(Xi = xi | parents(Xi)) else xi ← a random sample from P(Xi | parents(Xi)) return x, w Chapter 14.4–5 25 Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 Chapter 14.4–5 26 Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 Chapter 14.4–5 27 Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 Chapter 14.4–5 28 Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 × 0.1 Chapter 14.4–5 29 Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 × 0.1 Chapter 14.4–5 30
ea few samples have nearly all the total weight "prior and Likelihood weighting analysis Likelihood weighting example 色88 Likelihood weighting example ! Wander about MCMC example contd. fre,there are four state The Markov chain Can aso choose a variable to sample at random each time Approximate inference using MCMC
Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 × 0.1 Chapter 14.4–5 31 Likelihood weighting example Cloudy Rain Sprinkler Grass Wet F T C .20 .80 P(R|C) F T C .50 .10 P(S|C) S R T T T F F T F F .90 .90 .99 P(W|S,R) P(C) .50 .01 w = 1.0 × 0.1 × 0.99 = 0.099 Chapter 14.4–5 32 Likelihood weighting analysis Sampling probability for WeightedSample is SWS(z, e) = Πi l = 1P(zi |parents(Zi)) Note: pays attention to evidence in ancestors only Cloudy Rain Sprinkler Grass Wet ⇒ somewhere “in between” prior and posterior distribution Weight for a given sample z, e is w(z, e) = Πi m = 1P(ei |parents(Ei)) Weighted sampling probability is SWS(z, e)w(z, e) = Πi l = 1P(zi |parents(Zi)) Πi m = 1P(ei |parents(Ei)) = P(z, e) (by standard global semantics of network) Hence likelihood weighting returns consistent estimates but performance still degrades with many evidence variables because a few samples have nearly all the total weight Chapter 14.4–5 33 Approximate inference using MCMC “State” of network = current assignment to all variables. Generate next state by sampling one variable given Markov blanket Sample each variable in turn, keeping evidence fixed function MCMC-Ask(X, e, bn,N) returns an estimate of P(X |e) local variables: N[X ], a vector of counts over X, initially zero Z, the nonevidence variables in bn x, the current state of the network, initially copied from e initialize x with random values for the variables in Y for j = 1 to N do for each Zi in Z do sample the value of Zi in x from P(Zi|mb(Zi)) given the values of MB(Zi) in x N[x ] ← N[x ] + 1 where x is the value of X in x return Normalize(N[X ]) Can also choose a variable to sample at random each time Chapter 14.4–5 34 The Markov chain With Sprinkler = true, WetGrass = true, there are four states: Cloudy Rain Sprinkler Grass Wet Cloudy Rain Sprinkler Grass Wet Cloudy Rain Sprinkler Grass Wet Cloudy Rain Sprinkler Grass Wet Wander about for a while, average what you see Chapter 14.4–5 35 MCMC example contd. Estimate P(Rain|Sprinkler = true, WetGrass = true) Sample Cloudy or Rain given its Markov blanket, repeat. Count number of times Rain is true and false in the samples. E.g., visit 100 states 31 have Rain = true, 69 have Rain = false Pˆ (Rain|Sprinkler = true, WetGrass = true) = Normalize(h31, 69i) = h0.31, 0.69i Theorem: chain approaches stationary distribution: long-run fraction of time spent in each state is exactly proportional to its posterior probability Chapter 14.4–5 36
州 Markov blanket sampling ⊙
Markov blanket sampling Markov blanket of Cloudy is Cloudy Rain Sprinkler Grass Wet Sprinkler and Rain Markov blanket of Rain is Cloudy, Sprinkler, and WetGrass Probability given the Markov blanket is calculated as follows: P(xi 0 |mb(Xi)) = P(xi 0 |parents(Xi))ΠZj∈Children(Xi )P(zj |parents(Zj)) Easily implemented in message-passing parallel systems, brains Main computational problems: 1) Difficult to tell if convergence has been achieved 2) Can be wasteful if Markov blanket is large: P(Xi |mb(Xi)) won’t change much (law of large numbers) Chapter 14.4–5 37 Summary Exact inference by variable elimination: – polytime on polytrees, NP-hard on general graphs – space = time, very sensitive to topology Approximate inference by LW, MCMC: – LW does poorly when there is lots of (downstream) evidence – LW, MCMC generally insensitive to topology – Convergence can be very slow with probabilities close to 1 or 0 – Can handle arbitrary combinations of discrete and continuous variables Chapter 14.4–5 38
