3. Markov property. Markov property for MRFs. HammersleyClifford theorem. Markov property for Bayesian networks. Imap, Pmap, and chordal graphs


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1 Markov property 33. Markov property Markov property for MRFs HammersleyClifford theorem Markov property for ayesian networks Imap, Pmap, and chordal graphs
2 Markov Chain X Y Z X = Z Y µ(x, Y, Z) = f(x, Y )g(y, Z) Q. What independence does MRF imply? x 0 x x 3 x x x 4 x x 9 x 5 x 7 x8 x 6 Markov property 3
3 Markov property 33 Markov property A let A C be a partition of V Definition: graph separation C separates A from C if any path starting in A and terminating in C has at least one node in Definition: global Markov property distribution µ over X V satisfies the global Markov property on G if for any partition (A,, C) such that separates A from C, µ(x A, x C x ) = µ(x A x ) µ(x C x )
4 Markov property 34 Markov property for undirected graphs We say µ( ) satisfy the global Markov property (G) w.r.t. a graph G if for any partition (A,, C) such that separates A from C, µ(x A, x C x ) = µ(x A x ) µ(x C x ) We say µ( ) satisfy the local Markov property (L) w.r.t. a graph G if for any i V, µ(x i, x rest x i ) = µ(x i x i ) µ(x rest x i ) We say µ( ) satisfy the pairwise Markov property (P) w.r.t. a graph G if for any i, j V that are not connected by an edge obviously: (G) (L) (P) µ(x i, x j x rest ) = µ(x i x rest ) µ(x j x rest )
5 Proof of (L) (P): suppose (L) holds, then for any i and j not connected by an edge. since, for any deterministic function h( ), proofs of () and (): X Y Z = X Y (Z, h(y )) () X Y Z = X h(y ) Z () x i x rest x i () = x i x rest (x V \{i,j} ) () = x i x j (x V \{i,j} ) µ(x, y, h(y ) = h, z) = µ(x, y, z)i(h(y) = h) = f(x, z)g(y, z)i(h(y) = h) this implies both X (Y, h(y )) Z and X Y (Z, h(y )) µ(x, h, z) = y µ(x, y, h, z) = f(x, z) g(y, z)i(h(y) = h) y }{{} g(h,z) this implies X h(y ) Z Markov property 35 = = =
6 what conditions do we need for (P ) (G) to hold? Markov property 36
7 Markov property 37 in general (P ) (G), but (P ) (G) if the following holds for all disjoint subsets A,,C, and D V : intersection lemma if x A x (x C, x D ) and x A x C (x, x D ), then x A (x, x C ) x D (3) for instance, for a strictly positive µ( ), i.e. if µ(x) > 0 for all x X V, then the above holds [Homework.5]
8 Markov property 38 Proof of (P) (G) when (3) holds: [Pearl,Paz 987] by induction over s when s = n, (P) (G) assume (G) for any with s and prove it for = s i by induction assumption by (3), x C xã (x, x i ) x C x i (x, xã) Ã A C x C (xã, x i ) x by induction, we see that (G) holds for all sizes of.
9 Markov property 39 HammersleyClifford Theorem (HammersleyClifford) 97 A strictly positive distribution µ(x) (i.e. µ(x) > 0 for all x) satisfies the global Markov property (G) with respect to G(V, E) if and only if it can be factorized according to G: (F) : µ(x) = Z c C(G) ψ c (x c ) i.e. any µ(x) with Markov property can be represented with MRF (F) (G) is easy [Homework.] (F) (G) requires effort
10 Markov property 30 Proof sketch (Grimmett, ull. London Math. Soc. 973) for every S V, define ψ S (x S ) U S µ(x U, 0 V \U ) ( ) S\U need strict positive µ( ) for the division to make sense example: For S = {i} let µ + i ( ) µ(, 0 V \{i}) ψ i (x i ) = µ+ i (x i) µ + i (0) for S = {i, j} (not necessarily an edge) let µ + ij ( ) µ(, 0 V \{i,j}) ψ ij (x i, x j ) = µ+ ij (x i, x j ) µ + ij (0, 0) µ + ij (x i, 0) µ + ij (0, x j)
11 Markov property 3 claim : ψs (x S ) =const. unless S is a clique claim : for any µ(x), µ(x) = µ(0 V ) S V ψ S (x S ) then, it follows from claims and that µ(x) = µ(0 V ) c C(G) ψ c (x c ) this proves that for any positive µ(x) satisfying (G) we can find a factorization as per (F )
12 Markov property 3 Proof of claim : ψ S (x S ) = U S µ + U (x U) ( ) S\U i j T S = U T ( µ + U {i,j} (x U, x i, x j ) µ + U (x ) ( ) T \U U) µ + U {i} (x U, x i ) µ + U {j} (x U, x j ) and unless S is a clique, there exists two nodes i and j in S that are not connected by an edge, then µ + U {i,j} (x U, x i, x j ) µ + U {i} (x U, x i ) (G) = µ(x i x U, 0 rest ) µ(x U, x j, 0 rest ) µ(x i x U, 0 rest ) µ(x U, 0 j, 0 rest ) = µ(0 i x U, 0 rest ) µ(x U, x j, 0 rest ) µ(0 i x U, 0 rest ) µ(x U, 0 j, 0 rest ) = µ(x U, 0 i, x j, 0 rest ) µ(x U, 0 i, 0 j, 0 rest ) = µ + U {j} (x U, x j ) µ + U (x U)
13 proof of claim (where did ( ) S\U come from?): recall, ψ S (x S ) U S µ(x U, 0 V \U ) ( ) S\U Möbius inversion lemma [see also G.C.Rota, Prob. Theor. Rel. Fields, (964) ] let f, g : { subsets of V } R. Then the following are equivalent f(s) = U S g(u), for all S V g(s) = U S( ) S\U f(u), for all S V let, f(s) := log µ(x S, 0 V \S ), then g(s) = U S( ) S\U log µ(x U, 0 V \U ) = log ψ S (x S ) ( hence, exp(f(v )) = µ(x) = exp log ψ ) U (x U ) Markov property U V 33
14 Give an example of a distribution which is not strictly positive, that does not satisfy (G) [Homework.] Markov property 34
15 Markov property 35 Recap consider a distribution µ(x) that factorizes according to an undirected graphical model on G = (V, E), µ(x) = ψ c (x c ) Z where C is the set of all maximal cliques in G (Global Markov property) for any disjoint subsets A,, C V, µ(x) satisfy x A x x C whenever separates A and C c C A C
16 Markov property 36 x x 3 x 0 x x x 4 x x 9 x 5 x 7 x8 example: color the nodes with {R, G, } such that no adjacent node has the same color µ(x) = I(x i x j ) Z (i,j) E for a node i, if we condition on the color of the neighbors of i, color of i is independent of the rest of the graph x 6
17 HammersleyClifford theorem (pairwise) if positive µ(x) satisfies all conditional independences implied by a graph G without any triangles, then we can find a factorization µ(x) = ψ ij (x i, x j ) Z (i,j) E (general) if positive µ(x) satisfies all conditional independences implied by a graph G, then we can find a factorization µ(x) = ψ c (x c ) Z there are conditional independencies that cannot be represented by an undirected graphical model, but is represented by a ayesian network Example: µ(x) = µ(x )µ(x 3 )µ(x x, x 3 ) c C 3 3 x and x 3 are independent no independence Markov property 37
18 Markov property 38 Markov property of directed graphical models Examples 3 µ(x) = µ(x )µ(x x )µ(x 3 x ) 3 µ(x) = µ(x )µ(x x )µ(x 3 x ) 3 µ(x) = µ(x )µ(x 3 )µ(x x, x 3 ) 3 µ(x) = µ(x )µ(x 3 )µ(x x 3 ) since µ(x 3 x ) = µ(x 3 x, x ), we have x x x 3 since µ(x 3 x ) = µ(x 3 x, x ), we have x x x 3 x and x 3 are independent but not x x x 3 x is independent of (x, x 3 )
19 Markov property 39 there are simple independencies we can immediately read from the factorization : x i x pr(i)\π(i) x π(i) (pr(i) is the set of predecessors in a topological ordering) = immediate x x 3 x x 3 x, x 4 x x 4 x, x 3, x 5, x 6 x x 5 x, x 4 x, x 3 x 6 x, x, x 3, x 4 x 5 = = = = = 6 less immediate x 3 x 4 x, but x 3 x 4 x, x 6, etc. = however, there are more independencies that follows from the factorization using the idea of dseparation, there is a way (e.g. ayes ball algorithm) to get all the independencies that are implied by all joint distributions that factorize according to a DAG G
20 note that a particular distribution µ( ) that factorize as G might have more independence than specified by the graph (for example, complete independent distribution (µ(x) = i µ(x i)) always can be represented by any DAG), but we are interested in the family of all µ s that factorize as per G, and statements about the independencies satisfied by all of the µ s in that family Markov property 30
21 Markov property 3 formally, let nd(i) be the set of nondescendants of node i, which are the set of nodes that are not reachable from node i via directed paths we say µ(x) has the local Markov property (DL) w.r.t. a directed acyclic graph G, if for all i µ(x i, x nd(i)\π(i) x π(i) ) = µ(x i x π(i) )µ(x nd(i)\π(i) x π(i) ) Definition: topological ordering is an ordering where any node comes after all of its parents (not necessarily unique) let pr(i) be the set of predecessors in a topological ordering we say µ(x) has the ordered Markov property (DO) w.r.t a directed acyclic graph G, if for all i and all topological ordering µ(x i, x pr(i)\π(i) x π(i) ) = µ(x i x π(i) )µ(x pr(i)\π(i) x π(i) ) (DL) (DO), since for each i there exists a topological ordering that places all x nd(i) before x i
22 we say a distribution µ(x) has the global Markov property (DG) w.r.t. an acyclic directed graph G, if for all subsets A,, C such that A and C are dseparated by, two subset A and C are dseparated by if all trails from A to C are by one of the above cases Markov property 3 µ(x A, x C x ) = µ(x A x )µ(x C x ) or equivalently x A = x C x a trail from i to j in a directed acyclic graph is by S if it contains a (shaded) vertex k S such that one of the following happens not not not
23 Markov property x 3 = x 4 x x 3 x 4 x, x 6 A trail from a node x 4 to another node x 6 in a directed graph is any path from x 4 to x 6 that ignores the direction of each of the edges in the path (you can go in the opposite direction of the edge)
24 Markov property 34 a ayesian network on a directed acyclic graph G factorizes as (DF) : µ(x) = i V µ i (x i x π(i) ) (DF) (DG) (DL) (DO) proof of (DG) (DL): the nondescendants of a node are dseparated from the node i conditioned on π(i) (proof by example) 3 by (DG) it follows that x 4 = x, x x proof of (DL) (DF): follows from the definition of conditional independence and ayes rule
25 Markov property 35 proof of (DF) (DG): requires an alternate (actually the original) definition of global Markov property Lemma : If µ(x) factorizes according to a DAG G, then for any ancestral set A, the marginal distribution µ(x A ) factorizes according to G A, where G A is the subgraph defined from G by deleting all nodes in A C Proof: consider the factorization and marginalize out all not in the ancestral set. Lemma : If µ(x) factorizes according to a DAG G, then x A x x C whenever A and are separated by C in the undirected graph moral(g(an(a C))), which is the moral graph of the smallest ancestral set containing A C. Proof: use Lemma. the above definition of independence in the Lemma. is the original definition of global Markov property for DAGs it is equivalent to the condition (DG) (defined in the previous slide), but we will not prove this equivalence in the lecture =
26 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
27 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
28 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
29 Markov property 36 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm:. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not 4 A = {4} 5 7 = {3, 5} C = {6} 3 6
30 Markov property 36 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm:. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not 4 A = {4} 5 7 = {3, 5} C = {6} 3 6
31 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
32 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
33 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
34 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not A = {4} = {3, 5} C = {6} 3 6
35 for any distribution µ(x) that factorizes accroding to a directed acyclic graph (DAG) G, and for any disjoint subsets A,, C V, we can test whether x A x x C using ayes ball algorithm: Markov property 36. shade all nodes in. place a ball at each node in A 3. let balls propagate, duplicating itself, following the rules shown on the right (Remark: balls do not interact) 4. if no ball can reach C, then x A x x C not not not balls are destroyed Q.when do we stop? A = {4} = {3, 5} C = {6} implies x 4 {x 3, x 5 } x 6
36 x {x, x 4 } x 3, x {x, x 3 } x 4 Markov property 37 if a distribution µ(x) factorizes according to a DAG G, i.e. µ(x) = i V µ i (x i x π(i) ) then µ(x) satisfies all the conditional independencies obtainable by ayes ball if a distribution µ(x) satisfies all the conditional independencies obtainable by ayes ball on a DAG G, then we can find a factorization of µ(x) on G in the worstcase ayes ball has runtime O( E ) [Shachter 998] there are conditional independencies that cannot be represented by a ayesian network, but is represented by a MRF, Example: µ(x) = ψ (x, x )ψ 3 (x, x 3 )ψ 34 (x 3, x 4 )ψ 4 (x 4, x ) 4 3
37 Markov property 38 Markov property of factor graphs Factor graphs are more fine grained than undirected graphical models ψ(x, x, x 3 ) ψ (x, x )ψ 3 (x, x 3 )ψ 3 (x 3, x ) ψ 3 (x, x, x 3 ) all three encodes same independencies, but different factorizations (in particular 3 X vs. X 3 ) set of independencies represented by MRF is the same as FG but FG can represent a larger set of factorizations for a factor graph G = (V, F, E), for any disjoint subsets A,, C V, µ(x) satisfy x A x x C whenever separates A and C
38 Markov property 39 Independence maps (Imaps) which graph (graphical model) is good? let I(G) denote all conditional independencies implied by a graph G let I(µ) denote all conditional independencies of a distribution µ( ) G is an Imap of µ if µ satisfy all the independencies of the graph G, i.e. I(G) I(µ) given µ(x), can we construct a G that captures as many independencies as possible? G is a minimal Imap for µ(x) if G is an Imap for µ(x), and removing a single edge from G causes the graph to no longer be an Imap
39 Markov property 330 Constructing minimal Imaps Constructing minimal Imaps for ayesian network. choose an (arbitrary) ordering of x, x,..., x n. consider a directed graph for ayes rule µ(x) = µ(x )µ(x x ) µ(x n x,..., x n ) 3. for each i, select its parents π(i) to be the minimal subset of {x,..., x i } such that x i x π(i) {x..., x i } \ x π(i) minimal Imap for N is not unique there are n! choices of the ordering, and a priori we cannot tell which ordering is best even for the same ordering, the Imap might not be unique e.g. X = X and X 3 = X + Z, then x 3 x x and x 3 x x, but x 3 (x, x ) so there are two Imaps for the ordering (,,3) = =
40 Markov property 33 Constructing minimal Imaps for Markov random fields. consider a complete undirected graph. for each edge (i, j), remove the edge if i and j are conditionally independent given all the rest of nodes if µ(x) > 0, then the resulting graph is the unique minimal Imap of µ can you prove this? idea: when µ > 0, graph obtained from the pairwise independencies is exactly the same as the graph obtained from the global independencies (from HammersleyClifford) and hence there is no loss in only considering the pairwise conditional independencies when constructing the graph when µ(x) = 0 for some x, then Imap might not be unique [Homework.]
41 Markov property 33 Moralization moralization converts a N D to a MRF G such that I(G) I(D) resulting G is an minimal (undirected) Imap of I(D). retain all edges in D and make them undirected. connect every pair of parents with an edge 3 3 when do we lose nothing in converting a directed graph to an undirected one? (when moralization adds no more edges than already present in D) there exists a G such that I(D) = I(G) if and only if moralization of D does not add any edges proof of : in the example above, the independence is not preserved, and this argument can be made general proof of : we will not prove this in the lecture
42 Perfect maps (Pmaps) G is a perfect map (Pmap) for µ(x) if I(G) = I(µ) set of all distributions on V Markov chains trees perfect undirected chordal undirected perfect directed examples: trees have efficient inference algorithms, which plays a crucial role in developing efficient algorithms for other graphs as well Markov property 333
43 Markov property 334 Trees undirected tree is a connected undirected graph with no cycle directed tree is a connected directed graph where each node has at most one parent conversion from MRF on tree to N on tree: take any node as root and start directing edges away from the root nonunique all conversions result in the same set of independencies example: Markov chain, hidden Markov models exact inference is extremely efficient on trees
44 Markov property 335 Chordal graphs consider an undirected graph G(V, E) we say (i,..., i k ) form a trail in the graph G = (V, E) if for every j {,..., k }, we have (i j, i j+ ) E a loop is a trail (i,..., i k ) where i k = i an undirected graph is chordal if the longest minimal loop is a triangle given G, if there exists a DAG D such that if I(G) = I(D) then G is a chordal graph proof idea: any loop of size > 3 with no chord (a chord is an edge connecting two vertices that are not consecutive in the loop), have independency that cannot be encoded by a DAG
45 the following are equivalent for MRFs () G is chordal () there exists an orientation of the edges in G that gives a DAG D whose moral graph is G (3) there exists a directed graphical model with conditional independencies identical to those implied by G, i.e. I(G) = I(D) proof of () (3): follows from the moralization slide (we did not provide a proof) proof of (3) (): proved in the previous slide proof of () (): Lemma 3. a chordal G is recursively simplicial. given a recursively simplicial graph G, we can construct a DAG D as follows: start with empty D 0 and fix a simplicial ordering (,..., n) start from node x, sequentially adding one node at a time, adding x t to D t and add edges from x t towards x t (unless there is already an edge in the opposite direction) it is clear for the construction that D is acyclic the moral graph of D has no additional edges, since the parents of any node form a clique in the original graph Markov property 336
46 Markov property 337 use triangulation to construct such a directed graph D from an undirected chordal graph G, which we will not cover in this course a node in G is simplicial if the subgraph defined by its neighbors form a complete subgraph a graph G is recursively simplicial if it contains a simplicial node x i and when x i is removed from the (sub)graph, what remains is recursively simplicial proof of Lemma 3. is omitted here, but refer to e.g. bartlett/courses/009fallcs8a/graphnotes.pdf
47 Markov property 338 Roadmap Cond. Indep. Factorization Graphical Graph Cond. Indep. µ(x) µ(x) Model G implied by G x {x, x 3 } x 4 ; Z ψa (x a ) FG Factor Markov x 4 {} x 7 ; Z ψc (x C ) MRF Undirected Markov. ψi (x i x π(i) ) N Directed Markov Undirected graphical models: (F) (G) (L) (P) For any positive µ(x) that satisfy (G), we can find a factorization (F) List of all positive distributions {µ(x)} that factorize according to (F) is equivalent as the list of all distributions that satisfy (G) Directed graphical models: (F) (G) (L) (O)
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