Axioms and AlgorithmsAxioms and Algorithmsfor Inferencesfor Inferences InvolvingInvolving

Probabilistic IndependenceProbabilistic Independence

Dan Geiger, Azaria Paz, and Judea Pearl, Information and Computation 91(1), March 1991,

128-141.

Presentation by Guy Moses & Omer Weissbrod

for the course 236372 - Bayesian NetworksComputer Science Faculty, Technion – winter 2009

partially based on the presentation by Ilan Gronau

2

What’s ahead?

IntroductionIntroduction - some definitions, notations and reminders.

Proof of CompletenessProof of Completeness. - “if it’s true – it can be proved”.

Preparations for the Membership AlgorithmPreparations for the Membership Algorithm – more definitions, and some theoretical groundwork.

The Membership AlgorithmThe Membership Algorithm – description, proof of correctness, complexity analysis.

3

DefinitionsDefinitions• U (Universe) – set of random variables with probability

distribution P.

• X,Y – finite sets of random variables:

X = x1,…,xn, Y = y1,…,ym

• P(X,Y) = P(X)·P(Y) - a short-hand notation for the equality:

Pr{x1=a1,…, xn=an, y1=b1, …, ym=bm} =

Pr{x1=a1,…, xn=an} · Pr{y1=b1, …, ym=bm}

for every choice of a1, …, an, b1, …, bm

• (X,Y) – short-hand for P(X,Y) = P(X)·P(Y)

This is called an independence statement.

*note that X,Y are disjoint sets of variables (XY = ).

4

NotationsNotations - a specific independence statement of the form

(X,Y)

- a set of independence statements of the form

(X,Y): = 1, … , k

• XY - short-hand notation for the union X Y

• P satisfies = (X,Y) means: P(X,Y) = P(X)·P(Y) for that specific P.

5

Soundness and CompletenessSoundness and CompletenessDefinitions:

iff every distribution that satisfies also satisfies .

iff cl(),i.e. there exists a derivation chain 1,…,n=

s.t. for each j, either j or j is derived by an axiom from

the previous statements.

For a set of axioms A:

Soundness: A is sound iff for every and :

Completeness: A is complete iff for every and :

Completeness - Alternative definition:

A is complete iff for every and every cl() there exists a

distribution P that satisfies cl)( and does not satisfy .

6

Independence AxiomsIndependence Axioms

We saw (in 1st lecture) that axioms 1a-1d are sound (always infer correctly).

Today we’ll show they are complete (can derive every true statement).

1a Trivial Independence:

1b Symmetry:

1c Decomposition

( , )

( , ) ( , )

( , ) ( , )

( , ) (

:

1d M , ) ( ,i ing: )x

X

X Y Y X

X YZ X Y

X Y XY Z X YZ

7

What’s ahead?

IntroductionIntroduction - some definitions, notations and reminders.

Proof of CompletenessProof of Completeness. - “if it’s true – it can be proved”.

Preparations for the Membership AlgorithmPreparations for the Membership Algorithm – more definitions, and some theoretical groundwork.

The Membership AlgorithmThe Membership Algorithm – description, proof of correctness, complexity analysis.

8

Minimal StatementMinimal Statement• Definition: =(X,Y) cl() is minimal if for every non-empty X’,Y’

s.t. X’X, Y’ Y, X’Y’XY we have (X’,Y’) cl().

• For every =(X,Y) cl() we can find an appropriate minimal

’=(X’,Y’)cl() through iterative decomposition.

• Observation:

P satisfies P satisfies ’ (decomposition soundness),

Therefore:

P doesn’t satisfy ’ P doesn’t satisfy .

• Our plan: Given an arbitrary cl(), We will find a distribution P that

satisfies cl() but doesn’t satisfy ’. This will prove completeness (using the alternative completeness definition and the observation above).

• To simplify annotation, we will assume WLOG that =(X,Y) is already minimal.

9

Let =(X,Y) cl() be a minimal statement where:

X={x1,…,xn}, Y={y1,…,ym}, and Z={z1,z2,…,zk} stand for the rest

of the variables in U.

We will construct P as follows: All variables, except x1, are fair coins

(probability for each of their two values)

x1 is defined thus:

Part 1: P does not satisfy

We will inspect the following scenario: x1=1, all other variables are 0.

P(x1, … , xn, y1, … , ym) P(x1, … , xn)·P(y1, … , ym)

Therefore, P does not satisfy , as required.

Completeness ProofCompleteness Proof

12 1

mod2n m

i ji j

x x y

=0 =0.5n =0.5m

10

Completeness Proof – cont’dCompleteness Proof – cont’dPart 2: P satisfies cl()

Let (V,W) cl(). We will show that P(V,W)=P(V)·P(W). This is done by inspecting different scenarios:

Scenario 1: either V or W contains only elements of Z. We will assume WLOG that W contains only elements of Z.

all variables in Z are independent under P and therefore:

WWVV

X

ZY

Y

Y

YY

YX

X

X

Z

Z

Z

Z

Z

Z

Z

Z

ZZ

ZZ

Z

i

iz W

P VW P V P z P V P W

11

Completeness Proof – cont’dCompleteness Proof – cont’dPart 2: P satisfies cl()

Let (V,W) cl(). We will show that P(V,W)=P(V)·P(W). This is done by inspecting different scenarios:

Scenario 2: Both V and W contain elements of X Y,but V W doesn’t contain all elements of X Y.

Without full information about the assignments of the variables in X Y, x1 could turn out to be 0 or 1 with probability , and therefore:

0.5 0.5

0.5 0.5

V W V W

V W

P VW

P V P W

WW

VVX

ZY

Y

Y

YY

YX

X

X

Z

Z

Z

Z

Z

Z

Z

Z

ZZ

ZZ

Z

12

WWX

ZY

Y

Y

YY

YX

X

X

Z

Z

Z

Z

Z

Z

Z

Z

ZZ

ZZ

Z

VV

Completeness Proof – cont’dCompleteness Proof – cont’dPart 2: P satisfies cl() - continued

Scenario 3: Both V and W contain elements of X Y, and (X Y)(V W).

We will show a derivation chain for =(X,Y), contradicting our original assumption that cl():

Mark: (V,W)=(XVYVZV, XWYWZW)cl()

where: Y=YVYW, X=XVXW, ZVZWZ, V=XVYVZV, W=XWYWZW

Remove all z’s by decomposition: (XVYV, XWYW)cl()

Due to minimality of =(X,Y): (XV,YV)cl() and (XW,Y)cl()

(XV,YV)(XVYV, XWYW) (XV,YV XWYW) = (XV,XWY)

(XW,Y) (XWY, XV) (Y, XVXW) = (Y,X) =

mix

mix

13

Completeness Proof – SummaryCompleteness Proof – SummaryReminder: Completeness - Alternative definition:

A is complete iff for every and every cl() there exists a

distribution P that satisfies cl)( and does not satisfy .

We’ve shown: given a minimal cl(), there exists a

distribution P that obeys:

1. P does not satisfy .

2. P satisfies .

Given a non-minimal cl(), we will derive its minimal statement ’, and devise a distribution P’ that satisfies but does not satisfy ’. Due to soundness of decomposition, P’ cannot satisfy as well.

14

Scope of CompletenessScope of CompletenessThe proof uses P- a binary p.d. (probability distribution

function) therefore:all p.d.’s over

Udiscrete p.d.’s

binary p.d.’s

normalp.d.’s

however,

for normal p.d.’s, the axiom set a1-d1 is not complete.

a stronger axiom is required:

replace:

with:

P P P( , ) ( , ) ( ,1d Mixin : )g I X Y I X Y Z I X Y Z

P P P1d' Composition: ( , ) ( , ) ( , )I X Y I X Z I X Y Z

•P

15

What’s ahead?

IntroductionIntroduction - some definitions, notations and reminders.

Proof of CompletenessProof of Completeness. - “if it’s true – it can be proved”.

Preparations for the Membership AlgorithmPreparations for the Membership Algorithm – more definitions, and some theoretical groundwork.

The Membership AlgorithmThe Membership Algorithm – description, proof of correctness, complexity analysis.

16

Some more Definitions and ToolsSome more Definitions and Tools

Definition: Spanspan(): the set of elements represented in statement .

Example: span(x1x2,x3,x4) = {x1,x2,x3,x4}

span(): the set of elements represented in all statements of .

Example: span({(x1,x2),(x1,x3)}) = {x1,x2,x3}

17

Some more Definitions and ToolsSome more Definitions and Tools

Definition: ProjectionThe projection of on X, denoted (X), is the statement

derived from by removing all elements not in X from .

Example:

if =(x1x2x3, x4x5) and X={x2,x3,x4} then

(X)=(x2x3, x4).

The projection of on X, denoted (X), is {(X) | }.

18

Some more Definitions and ToolsSome more Definitions and Tools

Projection Lemma: iff ‘ , where ’ = (span())

) if ' then clearly because all the statements in ‘ can be derived from the statements in by decomposition.

19

Some more Definitions and ToolsSome more Definitions and Tools

Projection Lemma: iff ’ , where ’ = (span()), s = span()

) if then there is a derivation chain for : 1, 2, … , k.

For each j:

if k j, k<j, (by symmetry or decomposition)

then k(s) j(s) by symmetry or decomposition respectively.

Similarly,

if j is derived from k and l by mixing,

then j(s) is derived from k(s), l(s) by mixing.

20

Some more Definitions and ToolsSome more Definitions and Tools

Projection Lemma: iff ’ , where ’ = (span()), s = span()

Observations from projection lemma:• Variables not in are unnecessary for determining whether

.

• The problem of verifying whether can be simplified to the problem of verifying whether ' , where '= (span()).

• This problem can be solved with a possibly reduced time and space complexity.

21

Conditions for Conditions for Inference of IndependenceInference of Independence

Maim claim: for a given , we have ’ iff:

1. is trivial: =(X,) (up to symmetry)

OR

2. is in ’: ’ (up to symmetry)

OR

3. is derivable from ’:

there exists ’’ s.t. span() = span(’)

and for ’=(AP,BQ) =(AQ,BP) (A,B,Q,P may be empty)

’ (A,P), ’ (B,Q) (up to symmetry)

22

Proof of Main ClaimProof of Main ClaimMaim claim: for a given , we have ’ iff:

1. is trivial*: =(X,) *up to symmetry

2. is in* ’: ’

3. is derivable* from ’: ’’ s.t. span() = span(’)

and for ’=(AP,BQ) =(AQ,BP) : ’ (A,P), ’ (B,Q)

) if 1. is trivial*

OR 2. is in* ’. than the proof is immediate.

otherwise,

3. there exists ’’ s.t. span() = span(’)

and for ’=(AP,BQ) =(AQ,BP) : ’ (A,P), ’ (B,Q)

we will show a constructive proof under these conditions

23

Proof of Main ClaimProof of Main ClaimMaim claim: for a given , we have ’ iff:

1. is trivial*: =(X,) *up to symmetry

2. is in* ’: ’

3. is derivable* from ’: ’’ s.t. span() = span(’)

and for ’=(AP,BQ) =(AQ,BP) : ’ (A,P), ’ (B,Q)

) (contd.) given that ’ (AP,BQ), ’ (A,P), ’ (B,Q).

1. (A,P)(AP,BQ) (A,PBQ)

2. (B,Q)(AP,BQ) (APB,Q) (PB,Q)

3. (PB,Q)(A,PBQ) (AQ,PB) = (AQ, BP) =

We’ve proven this direction.

mix

mix

mix

dec.

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Proof of Main ClaimProof of Main ClaimMaim claim: for a given , we have ’ iff:

1. is trivial*: =(X,) *up to symmetry

2. is in* ’: ’

3. is derivable* from ’: ’’ s.t. span() = span(’)

and for ’=(AP,BQ) =(AQ,BP) : ’ (A,P), ’ (B,Q)

) Given ’ , if 1. is trivial* OR 2. is in* ’,

than the proof is immediate.

Otherwise, since no axiom can add new variables to a statement, there must exist ’’ s.t. span() =

span(’) in the derivation chain of .

also: = (AQ,BP) (A,P)

= (AQ,BP) (Q,B)

dec.

dec.

25

Conclusions from ClaimConclusions from Claim• We’ve seen that, after discarding unneeded

variables,it is possible to tell whether ’ (when it’s not immediately obvious) by:

a. Finding another statement ’’ for whichspan() = span(’),

b. Verifying that ’ (A,P), ’ (B,Q)when ’=(AP,BQ) =(AQ,BP).

• This suggests using a recursive “divide and

conquer” approach.

26

What’s ahead?

IntroductionIntroduction - some definitions, notations and reminders.

Proof of CompletenessProof of Completeness. - “if it’s true – it can be proved”.

Preparations for the Membership AlgorithmPreparations for the Membership Algorithm – more definitions, and some theoretical groundwork.

The Membership AlgorithmThe Membership Algorithm – description, proof of correctness, complexity analysis.

27

The Membership Algorithm The Membership Algorithm Procedure Find(,):

1. set ’ := (span()).

2. if is trivial, or ’ (up to symmetry)then Find(,) := TRUE.

3. else if for all non-trivial ’’: span() span(’),

then Find(,) := FALSE.

4. else there exists ’’: span() = span(’),

and ’=(AP,BQ) =(AQ,BP),

set 1:= (A,P), 2:= (B,Q).

Find(,) := (Find(’,1) Find(’,2))

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Algorithm Correctness ProofAlgorithm Correctness ProofWe will prove that Find(,) := TRUE cl() by

induction on k=.

Induction base: if k=1 then is trivial, therefore the algorithm will return TRUE in step 2 and cl().

29

Algorithm Correctness ProofAlgorithm Correctness ProofInduction assumption: Find(,) := TRUE cl()

for each ’<k.

Induction step: Find(,) := TRUE iff either:

1. Step 2 returns TRUE is trivial or ’ cl().

2. Step 4 returns TRUE

iff

Find(’,1) := TRUE Find(’,2) := TRUE

iff

1cl(’) 2cl(’)

iff

cl(’)

(according to algorithm’s definition)

(according to induction assumption)

(according to main claim)

(according to projection lemma)iff cl()

30

Complexity AnalysisComplexity AnalysisDefinitions:

n = the number of distinct variables in {}.

k = the number of distinct variables in {}.

• First projection cost: O(||·n) – happens only once.

• Recursive step: T)k) ||·k + T(k1) + T(k2)

where k1+k2=k, k1=|1|, k2=|2|

• Can be shown by induction: T)k) ||·k·(depth of recursion)

• Worst case analysis: T)k) ||·k·k= ||·k2

• Total run time is bounded by: O(||·n + ||·k2)which is also: O(||·n2) since k n.

31

Improvements and Variations Improvements and Variations • Instead of arbitrarily choosing ’, find one whose

sub-statements {A,B,P,Q} have balanced size (can improve run-time complexity).

• Using the derivation chain presented in the constructive proof, the algorithm can also return a derivation chain for with a length of O(k).

32

Variations (contd.) Variations (contd.) The algorithm can be expanded into a polynomial

algorithm for the following problems:

• Given two sets and , is cl() cl() ?is cl() = cl() ?

• Minimize the size of while preserving cl():

Start with a maximal-size statement and remove from all statements derivable from it.Repeat with the next largest statement etc.

Thank you!Thank you!