CS7015 (Deep Learning) : Lecture 18 Markov Networks Mitesh M. - - PowerPoint PPT Presentation

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CS7015 (Deep Learning) : Lecture 18

Markov Networks Mitesh M. Khapra

Department of Computer Science and Engineering Indian Institute of Technology Madras

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Acknowledgments Probabilistic Graphical models: Principles and Techniques, Daphne Koller and Nir Friedman

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Module 18.1: Markov Networks: Motivation

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To motivate undirected graphical models let us consider a new example

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D B A C To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together To motivate undirected graphical models let us consider a new example

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together To motivate undirected graphical models let us consider a new example Now suppose there was some miscon- ception in the lecture due to some er- ror made by the teacher

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together To motivate undirected graphical models let us consider a new example Now suppose there was some miscon- ception in the lecture due to some er- ror made by the teacher Each one of A, B, C, D could have in- dependently cleared this misconcep- tion by thinking about it after the lec- ture

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together To motivate undirected graphical models let us consider a new example Now suppose there was some miscon- ception in the lecture due to some er- ror made by the teacher Each one of A, B, C, D could have in- dependently cleared this misconcep- tion by thinking about it after the lec- ture In subsequent study pairs, each stu- dent could then pass on this inform- ation to their partner

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together We are now interested in knowing whether a student still has the mis- conception or not

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together We are now interested in knowing whether a student still has the mis- conception or not Or we are interested in P(A, B, C, D)

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together We are now interested in knowing whether a student still has the mis- conception or not Or we are interested in P(A, B, C, D) where A, B, C, D can take values 0 (no misconception) or 1 (misconcep- tion)

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together We are now interested in knowing whether a student still has the mis- conception or not Or we are interested in P(A, B, C, D) where A, B, C, D can take values 0 (no misconception) or 1 (misconcep- tion) How do we model this using a Bayesian Network ?

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together First let us examine the conditional independencies in this problem

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together First let us examine the conditional independencies in this problem A ⊥ C|{B, D} (because A & C never interact)

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together First let us examine the conditional independencies in this problem A ⊥ C|{B, D} (because A & C never interact) B ⊥ D|{A, C} (because B & D never interact)

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together First let us examine the conditional independencies in this problem A ⊥ C|{B, D} (because A & C never interact) B ⊥ D|{A, C} (because B & D never interact) There are no other conditional inde- pendencies in the problem

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D B A C A, B, C, D are four students A and B study together sometimes B and C study together sometimes C and D study together sometimes A and D study together sometimes A and C never study together B and D never study together First let us examine the conditional independencies in this problem A ⊥ C|{B, D} (because A & C never interact) B ⊥ D|{A, C} (because B & D never interact) There are no other conditional inde- pendencies in the problem Now let us try to represent this using a Bayesian Network

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D B A C How about this one?

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D B A C How about this one? Indeed, it captures the following in- dependencies relation A ⊥ C|{B, D}

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D B A C How about this one? Indeed, it captures the following in- dependencies relation A ⊥ C|{B, D} But, it also implies that B ⊥ D|{A, C}

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Let us try a different network

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D B C A Let us try a different network

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D B C A Let us try a different network Again A ⊥ C|{B, D}

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D B C A Let us try a different network Again A ⊥ C|{B, D} But B ⊥ D(unconditional)

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D B C A Let us try a different network Again A ⊥ C|{B, D} But B ⊥ D(unconditional) You can try other networks

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D B C A Let us try a different network Again A ⊥ C|{B, D} But B ⊥ D(unconditional) You can try other networks Turns out there is no Bayesian Net- work which can exactly capture inde- pendence relations that we are inter- ested in

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D B C A Perfect Map: A graph G is a Per- fect Map for a distribution P if the in- dependance relations implied by the graph are exactly the same as those implied by the distribution Let us try a different network Again A ⊥ C|{B, D} But B ⊥ D(unconditional) You can try other networks Turns out there is no Bayesian Net- work which can exactly capture inde- pendence relations that we are inter- ested in

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D B C A Perfect Map: A graph G is a Per- fect Map for a distribution P if the in- dependance relations implied by the graph are exactly the same as those implied by the distribution Let us try a different network Again A ⊥ C|{B, D} But B ⊥ D(unconditional) You can try other networks Turns out there is no Bayesian Net- work which can exactly capture inde- pendence relations that we are inter- ested in There is no Perfect Map for the dis- tribution

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D B A C The problem is that a directed graph- ical model is not suitable for this ex- ample

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D B A C The problem is that a directed graph- ical model is not suitable for this ex- ample A directed edge between two nodes implies some kind of direction in the interaction

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D B A C The problem is that a directed graph- ical model is not suitable for this ex- ample A directed edge between two nodes implies some kind of direction in the interaction For example A → B could indicate that A influences B but not the other way round

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D B A C The problem is that a directed graph- ical model is not suitable for this ex- ample A directed edge between two nodes implies some kind of direction in the interaction For example A → B could indicate that A influences B but not the other way round But in our example A&B are equal partners (they both contribute to the study discussion)

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D B A C The problem is that a directed graph- ical model is not suitable for this ex- ample A directed edge between two nodes implies some kind of direction in the interaction For example A → B could indicate that A influences B but not the other way round But in our example A&B are equal partners (they both contribute to the study discussion) We want to capture the strength of this interaction (and there is no dir- ection here)

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D B A C We move on from Directed Graph- ical Models to Undirected Graphical Models

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D B A C We move on from Directed Graph- ical Models to Undirected Graphical Models Also known as Markov Network

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D B A C We move on from Directed Graph- ical Models to Undirected Graphical Models Also known as Markov Network The Markov Network on the left ex- actly captures the interactions inher- ent in the problem

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D B A C We move on from Directed Graph- ical Models to Undirected Graphical Models Also known as Markov Network The Markov Network on the left ex- actly captures the interactions inher- ent in the problem But how do we parameterize this graph?

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Module 18.2: Factors in Markov Network

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Grade SAT Intellligence Letter Difficulty

P(G,S, I, L, D) = P(I)P(D)P(G|I, D)P(S|I)P(L|G) Recall that in the directed case the factors were Conditional Probability Distributions (CPDs)

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Grade SAT Intellligence Letter Difficulty

P(G,S, I, L, D) = P(I)P(D)P(G|I, D)P(S|I)P(L|G) Recall that in the directed case the factors were Conditional Probability Distributions (CPDs) Each such factor captured interaction (dependence) between the connected nodes

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Grade SAT Intellligence Letter Difficulty

P(G,S, I, L, D) = P(I)P(D)P(G|I, D)P(S|I)P(L|G) Recall that in the directed case the factors were Conditional Probability Distributions (CPDs) Each such factor captured interaction (dependence) between the connected nodes Can we use CPDs in the undirected case also ?

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Grade SAT Intellligence Letter Difficulty

P(G,S, I, L, D) = P(I)P(D)P(G|I, D)P(S|I)P(L|G) Recall that in the directed case the factors were Conditional Probability Distributions (CPDs) Each such factor captured interaction (dependence) between the connected nodes Can we use CPDs in the undirected case also ? CPDs don’t make sense in the undir- ected case because there is no direc- tion and hence no natural condition- ing (Is A|B or B|A?)

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D B A C So what should be the factors or para- meters in this case

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D B A C So what should be the factors or para- meters in this case Question: What do we want these factors to capture ?

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D B A C So what should be the factors or para- meters in this case Question: What do we want these factors to capture ? Answer: The affinity between con- nected random variables

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D B A C So what should be the factors or para- meters in this case Question: What do we want these factors to capture ? Answer: The affinity between con- nected random variables Just as in the directed case the factors captured the conditional dependence between a set of random variables, here we want them to capture the af- finity between them

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D B A C However we can borrow the intuition from the directed case.

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D B A C However we can borrow the intuition from the directed case. Even in the undirected case, we want each such factor to capture inter- actions (affinity) between connected nodes

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D B A C However we can borrow the intuition from the directed case. Even in the undirected case, we want each such factor to capture inter- actions (affinity) between connected nodes We could have factors φ1(A, B), φ2(B, C), φ3(C, D), φ4(D, A) which capture the affinity between the cor- responding nodes.

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 a0 b0 a0 b0 a0 b0 a0 b1 a0 b1 a0 b1 a0 b1 a1 b0 a1 b0 a1 b0 a1 b0 a1 b1 a1 b1 a1 b1 a1 b1

Intuitively, it makes sense to have these factors associated with each pair of connected random variables.

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

But who will give us these values ?

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

But who will give us these values ? Well now you need to learn them from data (same as in the directed case)

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

But who will give us these values ? Well now you need to learn them from data (same as in the directed case) If you have access to a lot of past interac- tions between A&B then you could learn these values(more on this later)

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

But who will give us these values ? Well now you need to learn them from data (same as in the directed case) If you have access to a lot of past interac- tions between A&B then you could learn these values(more on this later)

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors Roughly speaking φ1(A, B) asserts that it is more likely for A and B to agree [∵ weights for a0b0, a1b1 > a0b1, a1b0]

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

But who will give us these values ? Well now you need to learn them from data (same as in the directed case) If you have access to a lot of past interac- tions between A&B then you could learn these values(more on this later)

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors Roughly speaking φ1(A, B) asserts that it is more likely for A and B to agree [∵ weights for a0b0, a1b1 > a0b1, a1b0] φ1(A, B) also assigns more weight to the case when both do not have a mis- conception as compared to the case when both have the misconception a0b0 > a1b1

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 b1 10 a1 b1 100 a1 b1 1 a1 b1 100

But who will give us these values ? Well now you need to learn them from data (same as in the directed case) If you have access to a lot of past interac- tions between A&B then you could learn these values(more on this later)

Intuitively, it makes sense to have these factors associated with each pair of connected random variables. We could now assign some values of these factors Roughly speaking φ1(A, B) asserts that it is more likely for A and B to agree [∵ weights for a0b0, a1b1 > a0b1, a1b0] φ1(A, B) also assigns more weight to the case when both do not have a mis- conception as compared to the case when both have the misconception a0b0 > a1b1 We could have similar assignments for the other factors

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

Notice a few things

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

Notice a few things These tables do not represent prob- ability distributions

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

Notice a few things These tables do not represent prob- ability distributions They are just weights which can be interpreted as the relative likelihood

• f an event

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

Notice a few things These tables do not represent prob- ability distributions They are just weights which can be interpreted as the relative likelihood

• f an event

For example, a = 0, b = 0 is more likely than a = 1, b = 1

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

But eventually we are interested in probability distributions

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

But eventually we are interested in probability distributions In the directed case going from factors to a joint probability dis- tribution was easy as the factors were themselves conditional probab- ility distributions

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

But eventually we are interested in probability distributions In the directed case going from factors to a joint probability dis- tribution was easy as the factors were themselves conditional probab- ility distributions We could just write the joint probab- ility distribution as the product of the factors (without violating the axioms

• f probability)

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D B A C

φ1(A, B) φ2(B, C) φ3(C, D) φ4(D, A) a0 b0 30 a0 b0 100 a0 b0 1 a0 b0 100 a0 b1 5 a0 b1 1 a0 b0 100 a0 b1 1 a1 b0 1 a1 b0 1 a1 b1 100 a1 b0 1 a1 a1 10 a1 b1 100 a1 b1 1 a1 b1 100

But eventually we are interested in probability distributions In the directed case going from factors to a joint probability dis- tribution was easy as the factors were themselves conditional probab- ility distributions We could just write the joint probab- ility distribution as the product of the factors (without violating the axioms

• f probability)

What do we do in this case when the factors are not probability distribu- tions

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Assignment Unnormalized Normalized a0 b0 c0 d0 300,000 4.17E-02 a0 b0 c0 d1 300,000 4.17E-02 a0 b0 c1 d0 300,000 4.17E-02 a0 b0 c1 d1 30 4.17E-06 a0 b1 c0 d0 500 6.94E-05 a0 b1 c0 d1 500 6.94E-05 a0 b1 c1 d0 5,000,000 6.94E-01 a0 b1 c1 d1 500 6.94E-05 a1 b0 c0 d0 100 1.39E-05 a1 b0 c0 d1 1,000,000 1.39E-01 a1 b0 c1 d0 100 1.39E-05 a1 b0 c1 d1 100 1.39E-05 a1 b1 c0 d0 10 1.39E-06 a1 b1 c0 d1 100,000 1.39E-02 a1 b1 c1 d0 100,000 1.39E-02 a1 b1 c1 d1 100,000 1.39E-02

Well we could still write it as a product

• f these factors and normalize it appro-

priately

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Assignment Unnormalized Normalized a0 b0 c0 d0 300,000 4.17E-02 a0 b0 c0 d1 300,000 4.17E-02 a0 b0 c1 d0 300,000 4.17E-02 a0 b0 c1 d1 30 4.17E-06 a0 b1 c0 d0 500 6.94E-05 a0 b1 c0 d1 500 6.94E-05 a0 b1 c1 d0 5,000,000 6.94E-01 a0 b1 c1 d1 500 6.94E-05 a1 b0 c0 d0 100 1.39E-05 a1 b0 c0 d1 1,000,000 1.39E-01 a1 b0 c1 d0 100 1.39E-05 a1 b0 c1 d1 100 1.39E-05 a1 b1 c0 d0 10 1.39E-06 a1 b1 c0 d1 100,000 1.39E-02 a1 b1 c1 d0 100,000 1.39E-02 a1 b1 c1 d1 100,000 1.39E-02

Well we could still write it as a product

• f these factors and normalize it appro-

priately P(a, b, c, d) = 1 Z φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a)

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Assignment Unnormalized Normalized a0 b0 c0 d0 300,000 4.17E-02 a0 b0 c0 d1 300,000 4.17E-02 a0 b0 c1 d0 300,000 4.17E-02 a0 b0 c1 d1 30 4.17E-06 a0 b1 c0 d0 500 6.94E-05 a0 b1 c0 d1 500 6.94E-05 a0 b1 c1 d0 5,000,000 6.94E-01 a0 b1 c1 d1 500 6.94E-05 a1 b0 c0 d0 100 1.39E-05 a1 b0 c0 d1 1,000,000 1.39E-01 a1 b0 c1 d0 100 1.39E-05 a1 b0 c1 d1 100 1.39E-05 a1 b1 c0 d0 10 1.39E-06 a1 b1 c0 d1 100,000 1.39E-02 a1 b1 c1 d0 100,000 1.39E-02 a1 b1 c1 d1 100,000 1.39E-02

Well we could still write it as a product

• f these factors and normalize it appro-

priately P(a, b, c, d) = 1 Z φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a) where Z =

• a,b,c,d

φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a)

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Assignment Unnormalized Normalized a0 b0 c0 d0 300,000 4.17E-02 a0 b0 c0 d1 300,000 4.17E-02 a0 b0 c1 d0 300,000 4.17E-02 a0 b0 c1 d1 30 4.17E-06 a0 b1 c0 d0 500 6.94E-05 a0 b1 c0 d1 500 6.94E-05 a0 b1 c1 d0 5,000,000 6.94E-01 a0 b1 c1 d1 500 6.94E-05 a1 b0 c0 d0 100 1.39E-05 a1 b0 c0 d1 1,000,000 1.39E-01 a1 b0 c1 d0 100 1.39E-05 a1 b0 c1 d1 100 1.39E-05 a1 b1 c0 d0 10 1.39E-06 a1 b1 c0 d1 100,000 1.39E-02 a1 b1 c1 d0 100,000 1.39E-02 a1 b1 c1 d1 100,000 1.39E-02

Well we could still write it as a product

• f these factors and normalize it appro-

priately P(a, b, c, d) = 1 Z φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a) where Z =

• a,b,c,d

φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a) Based on the values that we had assigned to the factors we can now compute the full joint probability distribution

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Assignment Unnormalized Normalized a0 b0 c0 d0 300,000 4.17E-02 a0 b0 c0 d1 300,000 4.17E-02 a0 b0 c1 d0 300,000 4.17E-02 a0 b0 c1 d1 30 4.17E-06 a0 b1 c0 d0 500 6.94E-05 a0 b1 c0 d1 500 6.94E-05 a0 b1 c1 d0 5,000,000 6.94E-01 a0 b1 c1 d1 500 6.94E-05 a1 b0 c0 d0 100 1.39E-05 a1 b0 c0 d1 1,000,000 1.39E-01 a1 b0 c1 d0 100 1.39E-05 a1 b0 c1 d1 100 1.39E-05 a1 b1 c0 d0 10 1.39E-06 a1 b1 c0 d1 100,000 1.39E-02 a1 b1 c1 d0 100,000 1.39E-02 a1 b1 c1 d1 100,000 1.39E-02

Well we could still write it as a product

• f these factors and normalize it appro-

priately P(a, b, c, d) = 1 Z φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a) where Z =

• a,b,c,d

φ1(a, b)φ2(b, c)φ3(c, d)φ4(d, a) Based on the values that we had assigned to the factors we can now compute the full joint probability distribution Z is called the partition function.

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Let us build on the original example by adding some more students

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C E F

Let us build on the original example by adding some more students

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C E F

Let us build on the original example by adding some more students Once again there is an edge between two students if they study together

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C E F

Let us build on the original example by adding some more students Once again there is an edge between two students if they study together One way of interpreting these new connections is that {A, D, E} from a study group or a clique

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C E F

Let us build on the original example by adding some more students Once again there is an edge between two students if they study together One way of interpreting these new connections is that {A, D, E} from a study group or a clique Similarly {A, F, B} form a study group and {C, D} form a study group and {B, C} form a study group

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D B A C E F

Now, what should the factors be?

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C E F

Now, what should the factors be? We could still have factors which cap- ture pairwise interactions

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D B A C E F

φ1(A, E)φ2(A, F)φ3(B, F)φ4(A, B) φ5(A, D)φ6(D, E)φ7(B, C)φ8(C, D) Now, what should the factors be? We could still have factors which cap- ture pairwise interactions

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D B A C E F

φ1(A, E)φ2(A, F)φ3(B, F)φ4(A, B) φ5(A, D)φ6(D, E)φ7(B, C)φ8(C, D) Now, what should the factors be? We could still have factors which cap- ture pairwise interactions But could we do something smarter (and more efficient)

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D B A C E F

φ1(A, E)φ2(A, F)φ3(B, F)φ4(A, B) φ5(A, D)φ6(D, E)φ7(B, C)φ8(C, D) Now, what should the factors be? We could still have factors which cap- ture pairwise interactions But could we do something smarter (and more efficient) Instead of having a factor for each pair of nodes why not have it for each maximal clique?

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D B A C E F

φ1(A, E)φ2(A, F)φ3(B, F)φ4(A, B) φ5(A, D)φ6(D, E)φ7(B, C)φ8(C, D) φ1(A, E, D)φ2(A, F, B)φ3(B, C)φ4(C, D) Now, what should the factors be? We could still have factors which cap- ture pairwise interactions But could we do something smarter (and more efficient) Instead of having a factor for each pair of nodes why not have it for each maximal clique?

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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What if we add one more student?

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C F E G

What if we add one more student?

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D B A C F E G

What if we add one more student? What will be the factors in this case?

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C F E G

What if we add one more student? What will be the factors in this case? Remember, we are interested in max- imal cliques

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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D B A C F E G

What if we add one more student? What will be the factors in this case? Remember, we are interested in max- imal cliques So instead of having factors φ(EAG) φ(GAD) φ(EGD) we will have a single factor φ(AEGD) correspond- ing to the maximal clique

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Grade SAT Intellligence Letter Difficulty

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Grade SAT Intellligence Letter Difficulty

A distribution P factorizes over a Bayesian Network G if P can be expressed as P(X1, . . . , Xn) =

n

• i=1

P(Xi|PaXi )

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Grade SAT Intellligence Letter Difficulty

A distribution P factorizes over a Bayesian Network G if P can be expressed as P(X1, . . . , Xn) =

n

• i=1

P(Xi|PaXi )

B C A D E F Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Grade SAT Intellligence Letter Difficulty

A distribution P factorizes over a Bayesian Network G if P can be expressed as P(X1, . . . , Xn) =

n

• i=1

P(Xi|PaXi )

B C A D E F

A distribution factorizes over a Markov Network H if P can be expressed as P(X1, . . . , Xn) = 1 Z

m

• i=1

φ(Di) where each Di is a complete sub-graph (maximal clique) in H

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Grade SAT Intellligence Letter Difficulty

A distribution P factorizes over a Bayesian Network G if P can be expressed as P(X1, . . . , Xn) =

n

• i=1

P(Xi|PaXi )

B C A D E F

A distribution factorizes over a Markov Network H if P can be expressed as P(X1, . . . , Xn) = 1 Z

m

• i=1

φ(Di) where each Di is a complete sub-graph (maximal clique) in H A distribution is a Gibbs distribution parametrized by a set of factors Φ = {φ1(D1), . . . , φm(Dm)} if it is defined as P(X1, . . . , Xn) = 1 Z

m

• i=1

φi(Di)

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Module 18.3: Local Independencies in a Markov Network

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Let U be the set of all random vari- ables in our joint distribution

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Let U be the set of all random vari- ables in our joint distribution Let X, Y, Z be some distinct subsets

• f U

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Let U be the set of all random vari- ables in our joint distribution Let X, Y, Z be some distinct subsets

• f U

A distribution P

• ver these RVs

would imply X⊥Y |Z if and only if we can write P(X) = φ1(X, Z)φ2(Y, Z)

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18

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Let U be the set of all random vari- ables in our joint distribution Let X, Y, Z be some distinct subsets

• f U

A distribution P

• ver these RVs

would imply X⊥Y |Z if and only if we can write P(X) = φ1(X, Z)φ2(Y, Z) Let us see this in the context of our

• riginal example

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D B A C In this example P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)φ3(C, D)φ4(D, A)]

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D B A C In this example P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)φ3(C, D)φ4(D, A)] We can rewrite this as P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)]

• φ5(B,{A,C})

[φ3(C, D)φ4(D, A)]

• φ6(D,{A,C})

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D B A C In this example P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)φ3(C, D)φ4(D, A)] We can rewrite this as P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)]

• φ5(B,{A,C})

[φ3(C, D)φ4(D, A)]

• φ6(D,{A,C})

We can say that B⊥D|{A, C} which is indeed true

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D B A C In this example P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)φ3(C, D)φ4(D, A)]

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D B A C In this example P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)φ3(C, D)φ4(D, A)] Alternatively we can rewrite this as P(A, B, C, D) = 1 Z [φ1(A, B)φ2(D, A)]

• φ5(A,{B,D})

[φ3(C, D)φ4(B, C)]

• φ6(C,{B,D})

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D B A C In this example P(A, B, C, D) = 1 Z [φ1(A, B)φ2(B, C)φ3(C, D)φ4(D, A)] Alternatively we can rewrite this as P(A, B, C, D) = 1 Z [φ1(A, B)φ2(D, A)]

• φ5(A,{B,D})

[φ3(C, D)φ4(B, C)]

• φ6(C,{B,D})

We can say that A⊥C|{B, D} which is indeed true

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For a given Markov network H we define Markov Blanket of a RV X to be the neighbors of X in H

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For a given Markov network H we define Markov Blanket of a RV X to be the neighbors of X in H Analogous to the case of Bayesian Networks we can define the local in- dependences associated with H to be X⊥(U − {X} − MBH)|MBH(X)

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Bayesian network

Grade SAT Intellligence Letter Difficulty

Local Independencies Xi⊥NonDescendentsXi|ParentG

Xi

Markov network

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Bayesian network

Grade SAT Intellligence Letter Difficulty

Local Independencies Xi⊥NonDescendentsXi|ParentG

Xi

Markov network

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Bayesian network

Grade SAT Intellligence Letter Difficulty

Local Independencies Xi⊥NonDescendentsXi|ParentG

Xi

Markov network Local Independencies Xi⊥NonNeighborsXi|NeighborsG

Xi

Mitesh M. Khapra CS7015 (Deep Learning) : Lecture 18