Probabilistic Graphical Modelspeople.csail.mit.edu/dsontag/courses/pgm13/slides/lecture1.pdf ·...

Probabilistic Graphical Models

David Sontag

New York University

Lecture 1, January 31, 2013

David Sontag (NYU) Graphical Models Lecture 1, January 31, 2013 1 / 44

One of the most exciting advances in machine learning (AI, signalprocessing, coding, control, . . .) in the last decades


How can we gain global insight based on local observations?


Key idea

1 Represent the world as a collection of random variables X1, . . . ,Xn

with joint distribution p(X1, . . . ,Xn)

2 Learn the distribution from data

3 Perform “inference” (compute conditional distributionsp(Xi | X1 = x1, . . . ,Xm = xm))


Reasoning under uncertainty

As humans, we are continuously making predictions under uncertainty

Classical AI and ML research ignored this phenomena

Many of the most recent advances in technology are possible becauseof this new, probabilistic, approach


Applications: Deep question answering


Applications: Machine translation


Applications: Speech recognition


Applications: Stereo vision

output: disparity!input: two images!


Key challenges

1 Represent the world as a collection of random variables X1, . . . ,Xn

with joint distribution p(X1, . . . ,Xn)

How does one compactly describe this joint distribution?Directed graphical models (Bayesian networks)Undirected graphical models (Markov random fields, factor graphs)

2 Learn the distribution from data

Maximum likelihood estimation. Other estimation methods?How much data do we need?How much computation does it take?

3 Perform “inference” (compute conditional distributionsp(Xi | X1 = x1, . . . ,Xm = xm))


Syllabus overview

We will study Representation, Inference & Learning

First in the simplest case

Only discrete variablesFully observed modelsExact inference & learning

Then generalize

Continuous variablesPartially observed data during learning (hidden variables)Approximate inference & learning

Learn about algorithms, theory & applications


Logistics: class

Class webpage:http://cs.nyu.edu/~dsontag/courses/pgm13/

Sign up for mailing list!Draft slides posted before each lecture

Book: Probabilistic Graphical Models: Principles and Techniques byDaphne Koller and Nir Friedman, MIT Press (2009)

Required readings for each lecture posted to course website.Many additional reference materials available!

Office hours: Wednesday 5-6pm and by appointment. 715Broadway, 12th floor, Room 1204

Teaching Assistant: Li Wan ([email protected])

Li’s Office hours: Monday 5-6pm. 715 Broadway, Room 1231


http://cs.nyu.edu/~dsontag/courses/pgm13/

Logistics: prerequisites & grading

Prerequisites:Previous class on machine learningBasic concepts from probability and statisticsAlgorithms (e.g., dynamic programming, graphs, complexity)Calculus

Grading: problem sets (65%) + in class final exam (30%) +participation (5%)

Class attendance is required.7-8 assignments (every 1–2 weeks). Both theory and programming.First homework out today, due next Thursday (Feb. 7) at 5pmImportant: See collaboration policy on class webpage

Solutions to the theoretical questions require formal proofs.

For the programming assignments, I recommend Python, Java, orMatlab. Do not use C++.


Review of probability: outcomesReference: Chapter 2 and Appendix A

An outcome space specifies the possible outcomes that we wouldlike to reason about, e.g.

IntroducNon&to&probability:&outcomes&

•  An&outcome(space&specifies&the&possible&outcomes&that&we&would&like&to&reason&about,&e.g.&

Ω = Coin toss ,

Ω = Die toss , , , , ,

•  We&specify&a&probability(p(x)&for&each&outcome&x(such&that(X

x2

p(x) = 1p(x) 0,E.g., p( ) = .6

p( ) = .4

We specify a probability p(ω) for each outcome ω such that

p(ω) ≥ 0,∑

ω∈Ω

p(ω) = 1

Review of probability: outcomesReference: Chapter 2 and Appendix A

An outcome space specifies the possible outcomes that we wouldlike to reason about, e.g.



Ω = Coin toss ,

Ω = Die toss , , , , ,


x

p(x) = 1p(x) 0,E.g., p( ) = .6

p( ) = .4

We specify a probability p(x) for each outcome x such that



Ω = Coin toss ,

Ω = Die toss , , , , ,


x

p(x) = 1p(x) 0,E.g., p( ) = .6

p( ) = .4



Review of probability: events

An event is a subset of the outcome space, e.g.

IntroducNon&to&probability:&events&

•  An&event(is&a&subset&of&the&outcome&space,&e.g.&

O = Odd die tosses , ,

E = Even die tosses , ,

•  The&probability&of&an&event&is&given&by&the&sum&of&the&probabiliNes&of&the&outcomes&it&contains,&

p(E) =X

x2E

p(x) E.g., p(E) = p( ) + p( ) + p( )

= 1/2, if fair die

The probability of an event is given by the sum of the probabilities ofthe outcomes it contains,

p(E ) =∑

ω∈Ep(ω)

IntroducNon&to&probability:&events&

•  An&event(is&a&subset&of&the&outcome&space,&e.g.&

O = Odd die tosses , ,

E = Even die tosses , ,

•  The&probability&of&an&event&is&given&by&the&sum&of&the&probabiliNes&of&the&outcomes&it&contains,&

p(E) =X

x2E

p(x) E.g., p(E) = p( ) + p( ) + p( )

= 1/2, if fair die


Independence of events

Two events A and B are independent if

p(A ∩ B) = p(A)p(B)

Are these two events independent?

IntroducNon&to&probability:&independence&

•  Two&events&A&and&B&are&independent(if&p(A \ B) = p(A)p(B)

BAAre these events independent?

No! p(A \ B) = 0

p(A)p(B) =

1

6

2

•  Suppose&our&outcome&space&had&two&different&die:&

Ω = 2 die tosses , , , … , 62 = 36 outcomes

and each die is (defined to be) independent, i.e.

p( ) = p( ) ) p( p( ) = p( ) ) p(

No! p(A ∩ B) = 0, p(A)p(B) =

(1

6

)2

Now suppose our outcome space had two different die:




No! p(A \ B) = 0

p(A)p(B) =

1

6

2




p( ) = p( ) ) p( p( ) = p( ) ) p(

and the probability distribution is such that each die is independent,




No! p(A \ B) = 0

p(A)p(B) =

1

6

2




p( ) = p( ) ) p( p( ) = p( ) ) p(


Independence of events

Two events A and B are independent if

p(A ∩ B) = p(A)p(B)

Are these two events independent?



Are these events independent? B A

p(A) = p( ) p(B) = p( )

p(A)p(B) =

1

6

2

p( ) ) p(

Yes! p(A \ B) = 0p( )

Yes!




p(A) = p( ) p(B) = p( )

p(A)p(B) =

1

6

2

p( ) ) p(

Yes! p(A \ B) = 0p( )




p(A) = p( ) p(B) = p( )

p(A)p(B) =

1

6

2

p( ) ) p(

Yes! p(A \ B) = 0p( )


Conditional probabilityConditional Probabilities •  A simple relation between joint and conditional probabilities

–  In fact, this is taken as the definition of a conditional probability

T W P

hot sun 0.4

hot rain 0.1

cold sun 0.2

cold rain 0.3

A B

A

Conditional probability

Let S1, S2 be events, p(S2) > 0.

p(S1 | S2) =p(S1 \ S2)

p(S2)

Claim 1:P

!2S p(! | S) = 1

Claim 2: If S1 and S2 are independent, then p(S1 | S2) = p(S1)


B

Let A,B be events, p(B) > 0.

p(A | B) =p(A ∩ B)

p(B)

Claim 1:∑

ω∈S p(ω | S) = 1

Claim 2: If A and B are independent, then p(A | B) = p(A)


Two important rules

1 Chain rule Let S1, . . .Sn be events, p(Si ) > 0.

p(S1 ∩ S2 ∩ · · · ∩ Sn) = p(S1)p(S2 | S1) · · · p(Sn | S1, . . . ,Sn−1)

2 Bayes’ rule Let S1,S2 be events, p(S1) > 0 and p(S2) > 0.

p(S1 | S2) =p(S1 ∩ S2)

p(S2)=

p(S2 | S1)p(S1)

p(S2)


Discrete random variables

Often each outcome corresponds to a setting of various attributes(e.g., “age”, “gender”, “hasPneumonia”, “hasDiabetes”)

A random variable X is a mapping X : Ω→ D

D is some set (e.g., the integers)Induces a partition of all outcomes Ω

For some x ∈ D, we say

p(X = x) = p(ω ∈ Ω : X (ω) = x)

“probability that variable X assumes state x”

Notation: Val(X ) = set D of all values assumed by X(will interchangeably call these the “values” or “states” of variable X )

p(X ) is a distribution:∑

x∈Val(X ) p(X = x) = 1


Multivariate distributions

Instead of one random variable, have random vector

X(ω) = [X1(ω), . . . ,Xn(ω)]

Xi = xi is an event. The joint distribution

p(X1 = x1, . . . ,Xn = xn)

is simply defined as p(X1 = x1 ∩ · · · ∩ Xn = xn)

We will often write p(x1, . . . , xn) instead of p(X1 = x1, . . . ,Xn = xn)

Conditioning, chain rule, Bayes’ rule, etc. all apply


Working with random variables

For example, the conditional distribution

p(X1 | X2 = x2) =p(X1,X2 = x2)

p(X2 = x2).

This notation meansp(X1 = x1 | X2 = x2) = p(X1=x1,X2=x2)

p(X2=x2) ∀x1 ∈ Val(X1)

Two random variables are independent, X1 ⊥ X2, if

p(X1 = x1,X2 = x2) = p(X1 = x1)p(X2 = x2)

for all values x1 ∈ Val(X1) and x2 ∈ Val(X2).


Example

Consider three binary-valued random variables

X1,X2,X3 Val(Xi ) = 0, 1

Let outcome space Ω be the cross-product of their states:

Ω = Val(X1)×Val(X2)×Val(X3)

Xi (ω) is the value for Xi in the assignment ω ∈ ΩSpecify p(ω) for each outcome ω ∈ Ω by a big table:

x1 x2 x3 p(x1, x2, x3)0 0 0 .110 0 1 .02

...1 1 1 .05

How many parameters do we need to specify?

23 − 1David Sontag (NYU) Graphical Models Lecture 1, January 31, 2013 23 / 44

Marginalization

Suppose X and Y are random variables with distribution p(X ,Y )X : Intelligence, Val(X ) = “Very High”, “High”Y : Grade, Val(Y ) = “a”, “b”

Joint distribution specified by:

X

Y

vh h

a 0.7 0.15

b 0.1 0.05

p(Y = a) = ?= 0.85

More generally, suppose we have a joint distribution p(X1, . . . ,Xn).Then,

p(Xi = xi ) =∑

x1

∑

x2

· · ·∑

xi−1

∑

xi+1

· · ·∑

xn

p(x1, . . . , xn)


Conditioning

Suppose X and Y are random variables with distribution p(X ,Y )X : Intelligence, Val(X ) = “Very High”, “High”Y : Grade, Val(Y ) = “a”, “b”

X

Y

vh h

a 0.7 0.15

b 0.1 0.05

Can compute the conditional probability

p(Y = a | X = vh) =p(Y = a,X = vh)

p(X = vh)

=p(Y = a,X = vh)

p(Y = a,X = vh) + p(Y = b,X = vh)

=0.7

0.7 + 0.1= 0.875.


Example: Medical diagnosis

Variable for each symptom (e.g. “fever”, “cough”, “fast breathing”,“shaking”, “nausea”, “vomiting”)

Variable for each disease (e.g. “pneumonia”, “flu”, “common cold”,“bronchitis”, “tuberculosis”)

Diagnosis is performed by inference in the model:

p(pneumonia = 1 | cough = 1, fever = 1, vomiting = 0)

One famous model, Quick Medical Reference (QMR-DT), has 600diseases and 4000 findings


Representing the distribution

Naively, could represent multivariate distributions with table ofprobabilities for each outcome (assignment)

How many outcomes are there in QMR-DT? 24600

Estimation of joint distribution would require a huge amount of data

Inference of conditional probabilities, e.g.

p(pneumonia = 1 | cough = 1, fever = 1, vomiting = 0)

would require summing over exponentially many variables’ values

Moreover, defeats the purpose of probabilistic modeling, which is tomake predictions with previously unseen observations


Structure through independence

If X1, . . . ,Xn are independent, then

p(x1, . . . , xn) = p(x1)p(x2) · · · p(xn)

2n entries can be described by just n numbers (if |Val(Xi )| = 2)!

However, this is not a very useful model – observing a variable Xi

cannot influence our predictions of Xj

If X1, . . . ,Xn are conditionally independent given Y , denoted asXi ⊥ X−i | Y , then

p(y , x1, . . . , xn) = p(y)p(x1 | y)n∏

i=2

p(xi | x1, . . . , xi−1, y)

= p(y)p(x1 | y)n∏

i=2

p(xi | y).

This is a simple, yet powerful, model


Example: naive Bayes for classification

Classify e-mails as spam (Y = 1) or not spam (Y = 0)Let 1 : n index the words in our vocabulary (e.g., English)Xi = 1 if word i appears in an e-mail, and 0 otherwiseE-mails are drawn according to some distribution p(Y ,X1, . . . ,Xn)

Suppose that the words are conditionally independent given Y . Then,

p(y , x1, . . . xn) = p(y)n∏

i=1

p(xi | y)

Estimate the model with maximum likelihood. Predict with:

p(Y = 1 | x1, . . . xn) =p(Y = 1)

∏ni=1 p(xi | Y = 1)∑

y=0,1 p(Y = y)∏n

i=1 p(xi | Y = y)

Are the independence assumptions made here reasonable?

Philosophy: Nearly all probabilistic models are “wrong”, but many arenonetheless useful


Bayesian networksReference: Chapter 3

A Bayesian network is specified by a directed acyclic graphG = (V ,E ) with:

1 One node i ∈ V for each random variable Xi

2 One conditional probability distribution (CPD) per node, p(xi | xPa(i)),specifying the variable’s probability conditioned on its parents’ values

Corresponds 1-1 with a particular factorization of the jointdistribution:

p(x1, . . . xn) =∏

i∈Vp(xi | xPa(i))

Powerful framework for designing algorithms to perform probabilitycomputations


Example

Consider the following Bayesian network:

Grade

Letter

SAT

IntelligenceDifficulty

d1d0

0.6 0.4

i1i0

0.7 0.3

i0

i1

s1s0

0.95

0.2

0.05

0.8

g1

g2

g2

l1l 0

0.1

0.4

0.99

0.9

0.6

0.01

i0,d0

i0,d1

i0,d0

i0,d1

g2 g3g1

0.3

0.05

0.9

0.5

0.4

0.25

0.08

0.3

0.3

0.7

0.02

0.2

What is its joint distribution?

p(x1, . . . xn) =∏

i∈Vp(xi | xPa(i))

p(d , i , g , s, l) = p(d)p(i)p(g | i , d)p(s | i)p(l | g)


More examples

p(x1, . . . xn) =∏

i∈Vp(xi | xPa(i))

Will my car start this morning?

Heckerman et al., Decision-Theoretic Troubleshooting, 1995


More examples

p(x1, . . . xn) =∏

i∈Vp(xi | xPa(i))

What is the differential diagnosis?

Beinlich et al., The ALARM Monitoring System, 1989David Sontag (NYU) Graphical Models Lecture 1, January 31, 2013 33 / 44

Bayesian networks are generative models

naive Bayes

Y

X1 X2 X3 Xn. . .

Features

Label

Evidence is denoted by shading in a node

Can interpret Bayesian network as a generative process. Forexample, to generate an e-mail, we

1 Decide whether it is spam or not spam, by samping y ∼ p(Y )2 For each word i = 1 to n, sample xi ∼ p(Xi | Y = y)


Bayesian network structure implies conditionalindependencies!

Grade

Letter

SAT

IntelligenceDifficulty

The joint distribution corresponding to the above BN factors as

p(d , i , g , s, l) = p(d)p(i)p(g | i , d)p(s | i)p(l | g)

However, by the chain rule, any distribution can be written as

p(d , i , g , s, l) = p(d)p(i | d)p(g | i , d)p(s | i , d , g)p(l | g , d , i , g , s)

Thus, we are assuming the following additional independencies:D ⊥ I , S ⊥ D,G | I , L ⊥ I ,D,S | G . What else?


Bayesian network structure implies conditionalindependencies!

Generalizing the above arguments, we obtain that a variable isindependent from its non-descendants given its parents

Common parent – fixing B decouples A and C

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Qualitative Specification

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Cascade – knowing B decouples A and C

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V-structure – Knowing C couples A and B

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This important phenomona is called explaining away and is whatmakes Bayesian networks so powerful


A simple justification (for common parent)

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We’ll show that p(A,C | B) = p(A | B)p(C | B) for any distributionp(A,B,C ) that factors according to this graph structure, i.e.

p(A,B,C ) = p(B)p(A | B)p(C | B)

Proof.

p(A,C | B) =p(A,B,C )

p(B)= p(A | B)p(C | B)


D-separation (“directed separated”) in Bayesian networks

Algorithm to calculate whether X ⊥ Z | Y by looking at graphseparation

Look to see if there is active path between X and Z when variablesY are observed:

(a)

X

Y

Z X

Y

Z

(b)

(a)

X

Y

Z

(b)

X

Y

Z





(a)

X

Y

Z X

Y

Z

(b)

(a)

X

Y

Z

(b)

X

Y

Z





X Y Z X Y Z

(a) (b)

If no such path, then X and Z are d-separated with respect to Y

d-separation reduces statistical independencies (hard) to connectivityin graphs (easy)

Important because it allows us to quickly prune the Bayesian network,finding just the relevant variables for answering a query


D-separation example 1

1X

2X

3X

X 4

X 5

X6

1X

2X

3X

X 4

X 5

X6


D-separation example 2

1X

2X

3X

X 4

X 5

X6

1X

2X

3X

X 4

X 5

X6


2011 Turing Award was for Bayesian networks


Summary

Bayesian networks given by (G ,P) where P is specified as a set oflocal conditional probability distributions associated with G ’s nodes

One interpretation of a BN is as a generative model, where variablesare sampled in topological order

Local and global independence properties identifiable viad-separation criteria

Computing the probability of any assignment is obtained bymultiplying CPDs

Bayes’ rule is used to compute conditional probabilitiesMarginalization or inference is often computationally difficult


Probabilistic Graphical Modelspeople.csail.mit.edu/dsontag/courses/pgm13/slides/lecture1.pdf ·...

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Transcript of Probabilistic Graphical Modelspeople.csail.mit.edu/dsontag/courses/pgm13/slides/lecture1.pdf ·...