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Causal Clustering of Variables with Multiple Latent Causal Clustering of Variables with Multiple Latent

Causal Clustering of Variables with Multiple Latent - PowerPoint Presentation

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Causal Clustering of Variables with Multiple Latent - PPT Presentation

Causes More Theory than Applied Peter Spirtes Erich Kummerfeld Richard Scheines Joe Ramsey 1 An example Person 1 Stress Depression 3 Religious Coping Task learn causal model ID: 328731

x13 variables x12 model variables x13 model x12 x11 x10 pure x14 measurement latent l5x1 constraints good linear variable

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Slide1

Causal Clustering of Variables with Multiple Latent Causes(More Theory than Applied) Peter Spirtes, Erich Kummerfeld, Richard Scheines, Joe Ramsey

1Slide2

An example

Person 1

Stress

Depression

3. Religious Coping

Task: learn causal model

2

Data from

Bongjae

Lee, described in Silva et al. 2006Slide3

These variables cannot be measured directlyThey are estimated by asking people to answer questions, and constructing a model that relates the measured answers to the unobserved variablesProblems:What is the relationship between the measured variables and the latent variables to be estimated?Some questions Might be caused by multiple latent variablesMight be caused by answers to previous questions

Might be caused by latent variables that are not being estimated

Example

3Slide4

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 L2 L4 L6

Example

This edge is not identifiable (unlike single factor case where all of the latent connections are identifiable if the measurement model is simple).

4Slide5

A set of variables V is causally sufficient iff each cause that is a direct cause relative to V of any pair of variables in V, is also in V. It is minimal if the set formed by removing any latent variables is not causally sufficient.Causal Sufficiency

5Slide6

L1 L3 L5 L2 L4 L6

Structural Graph

The

stuctural

graph has all and only the latent variables, and the edges between the latent variables.

6Slide7

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 L2 L4 L6

Measurement Graph

The measurement graph has a minimal causally sufficient set of variables, and all of the edges except the latent-latent edges.

7Slide8

A pure n-factor measurement model for an observed set of variables O is such that:Each observed variable has exactly n latent parents.No observed variable is an ancestor of other observed variable or any latent variable. A set of observed variables O in a pure n-factor measurement model is a pure cluster if each member of the cluster has the same set of n parents.

8

Pure Measurement ModelsSlide9

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 L2 L4 L6

Impure Measurement Model

9

Strategy

: (1) find a subset of variables for which (i) the measurement model is simple, and (ii) it is possible to determine that it is simple, without knowing the true structural model

; (2) then find structural model. Slide10

L1 L3X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 L2 L4

P

ure Measurement

SubModel

10Slide11

Use of Pure Measurement Submodel

L1 L3

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11

L2 L4

Actual Impure Measurement

ModelSlide12

Use of Pure Measurement Submodel

L1 L3

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11

L2 L4

If treat measurement model as pure, no structural model will fit the data well.

But adding an L1 -> L3 edge may improve the fit because it allows for correlations between X1 – X6 and X7 – X11.

Assumed Pure Measurement

ModelSlide13

Causally unconnected variables are independent.No observed variable is a cause of a latent variable.No correlations are close to 0 or to 1 (pre-process)All of the sub covariance matrices are invertibleNo feedback(In practice) There is a one-factor pure measurement submodel

Each variable is a linear function of its parents in the graph + a noise term that is uncorrelated with any of the other noise terms – linear structural equation model.

Silva 06 (and others) Assumptions

13Slide14

Let be the submatrix with rows from A and columns from BFor each quartet of variables there are 3

different

tetrad constraints

: <

1,2;3,4 >

<1,3;2,4> <1,4;2,3>

Only two of the constraints are independent: any two entail the third.Vanishing Tetrad Constraints

14Slide15

For each sextuple of variables there are 10 different sextad constraints: <1,2,3;4,5,6> <1,2,4;3,5,6> <1,2,5;3,4,6> <1,2,6;3,4,5> <1,3,4;2,5,6> <1,3,5;2,4,6> <1,3,6;2,4,5> <1,4,5;2,3,6> <1,4,6;2,3,5> <1,5,6;2,3,4>Vanishing sextad

constraints

15Slide16

An algebraic constraint is linearly entailed by a DAG if it is true of the implied covariance for every value of the free parameters (the linear coefficients and the variances of the noise terms)Entailed Algebraic Constraints

16Slide17

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14

L2 L4 L6

A

trek

in G from i to j is an ordered pair of directed paths

(P1; P2) where P1 has sink i, P

2 has sink j, and both P1 and P2 have the same source k. (L5,X13;L5,X14); (L6,X13;L6,X14); (X13;X13,X14)

Simple Treks

17Slide18

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14

L2 L4 L6

The two paths of a

simple

trek intersect only at the source.

(L5,X13;L5,X14); (L6,X13;L6,X14); (X13;X13,X14) X13 side; X14 sideSimple Treks

18Slide19

Two-Factor ModelA = {1,2,3} B = {4,5,6} CA = {L1} CB = {L2}

A is t-separated from B by <CA,CB> ->

19Slide20

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 L2 L4 L6Let

A

,

B

,

CA, and CB be four subsets of V

(G) whichneed not be disjoint. The pair (CA;CB)

trek separates (or t-separates) A from B if for every trek (P1; P2) from a vertex in A to a vertex in B, either P1

contains a vertex in CA or P2 contains a vertex in CB.

T-separation

20Slide21

The submatrix ΣA,B has rank less than or equal to r for all covariance matrices consistent with the graph G if and only if there exist subsets (CA,CB) included in V(

G

) with #C

A

+ #CB ≤

r such that (CA,CB) t

-separates A from B. Consequently, rk(ΣA,B) ≤ min{#CA + #CB : (CA,CB)

t-separates A from B};and equality holds for covariance matrices consistent with G

(Lebesgue measure 1 over parameters).If rank of submatrix is n, then the determinant of every n+1 x n+1 determinant is zero

Choke Set Theorem

21Slide22

Algebraic Constraint Faithfulness Assumption: If an algebraic constraint holds in the population distribution, then it is linearly entailed to hold by the causal DAG.Partial CorrelationsTetradsSextadsStrong Faithfulness Assumption (for finite sample sizes) A causal DAG does not have parameters such that non-entailed vanishing sextad constraints are very close to zero.

Algebraic Constraint Faithfulness Assumption

22Slide23

Violations of Algebraic Faithfulness Assumption are Lebesgue measure 0.There is a lower dimensional surface in the space of parameters on which faithfulness is violated. Violations of Strong Algebraic Faithfulness Assumption are not Lebesgue measure 0.The surface of parameters on which almost faithfulness is violated is not lower dimensional than the space of parametersAs the number of variables grows, the probability of some violation of faithfulness becomes large.

Algebraic Constraint Faithfulness Assumption

23Slide24

AdvantagesNo need for estimation of model.No iterative algorithmNo local maxima.No problems with identifiability.Fast to compute.DisadvantagesDoes not contain information about inequalities.Power and accuracy of tests?

Difficulty in determining implications among constraints

Advantages and Disadvantages of Algebraic Constraints

24Slide25

Input – Data from observed variable in linear model Output – Set of variables that appear in (almost) pure measurement model, clustered into (almost) pure subsetsWe haven’t defined almost pure (not Silva 06 sense) – there is a list of impurities that can’t be detected by constaint search, but we don’t know whether it is complete. The basic idea with trivial modifications (in theory) can be applied to arbitrary numbers of latent parents, using different constraints.

FindTwoFactorClusters

: Algorithm Sketch (from

Kummerfeld

)

25Slide26

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14

L2 L4 L6

Complete Sextet – All 10

sextads

hold

<1,2,3;4,5,6> <1,2,4;3,5,6> <1,2,5;3,4,6> <1,2,6;3,4,5> <1,3,4;2,5,6> <1,3,5;2,4,6> <1,3,6;2,4,5> <1,4,5;2,3,6> <1,4,6;2,3,5> <1,5,6;2,3,4>

26Slide27

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14

L2 L4 L6

Complete Sextet – All 10

sextads

hold

<1,2,3;4,5,8> <1,2,4;3,5,8> <1,2,5;3,4,8> <1,2,8;3,4,5> <1,3,4;2,5,8> <1,3,5;2,4,8> <1,3,8;2,4,5> <1,4,5;2,3,8> <1,4,8;2,3,5> <1,5,8;2,3,4>

27Slide28

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 L2 L4 L6<X13,X14> not appear in any entailed

sextad

. Remove one of the variables.

Heuristic – remove the variable which appears in the fewest

sextads

that hold.

1. Remove one of pair of variables that appear in no sextads that hold

28Slide29

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 L2 L4 L6<X13,X14> not appear in any entailed

sextad

. Remove one of the variables.

Heuristic – remove the variable which appears in the fewest

sextads

that hold.

1. Remove one of pair of variables that appear in no sextads that hold

29Slide30

A subset of 5 variables is a good pentuple iff when add any sixth variable to the pentuple, the resulting sextuple is completeGood Pentuple

30Slide31

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 L2 L4 L6

2. Find all good

pentuples

<1,2,3,4,5,

6

> <1,2,3,4,5,

7

>

<

1,2,3,4,5,

8

> <

1,2,3,4,5,

9

> <1,2,3,4,5,

10

>

<1,2,3,4,5,

11

>

<1,2,3,4,5,

12

> <1,2,3,4,5,

13

>

Any subset of X1-X6 with 5 variables is a good

pentuple

31Slide32

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 L2 L4 L6

<1,2,3,4,7> is not a good

pentuple

<1,2,3,4,7,

6

> <1,2,3,4,7,

5

>

<

1,2,3,4,7,

8

> <

1,2,3,4,7,

9

> <1,2,3,4,7,

10

>

<1,2,3,4,7,

11

>

<1,2,3,4,7,

12

> <1,2,3,4,7,

13

>

32Slide33

<7,8,9,10,12,

1

> <7,8,9,10,12,

2

> <7,8,9,10,12,

3

> <7,8,9,10,12,

4

> <7,8,9,12,11,

5

> <7,8,9,12,11,

6

> <7,8,9,10,12,

11

> <7,8,9,10,12,

13

>

L1 L3 L5

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13

L2 L4 L6

<7,8,9,10,12> is not a good

pentuple

33Slide34

For a given set of variables, if all subsets of 5 are good

pentuples

, merge them.

All subsets of size 5 of X1-X6 are good

pentuples

, so merge.

L1 L3 L5

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13

L2 L4 L6

3. Merge Good

Pentuples

34Slide35

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 L2 L4 L6

<7,8,9,10,11> is a good

pentuple

<7,8,9,10,11,

1

> <7,8,9,10,11,

2

> <7,8,9,10,11,

3

> <7,8,9,10,11,

4

> <7,8,9,10,11,

5

> <7,8,9,10,11,

6

> <7,8,9,10,11,

12

> <7,8,9,10,11,

13

>

35Slide36

X12 and X13 do not appear in any good

pentuples

. If X13 is removed, all subsets of size 5 of X7-X12 become good

pentuples

, so they are merged. (Similarly for X12.)

L1 L3 L5

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12

L2 L4 L6

4. Check whether leftover variables should be removed, and repeat previous

36Slide37

We can (conceptually) remove L5 because it is not needed to make a causally sufficient set. However, L6 has to remain, and X7-X12 is not pure by our definition because X12 has 3 latent parents.

L1 L3

X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12

L2 L4 L6

4. Check whether leftover variables should be removed, and repeat previous

37Slide38

Collider Model – Impure Cluster, but Complete SextetChoke sets <{L1},{L7}> where L7 on the X6 side

38Slide39

Spider Model – Impure Cluster, but Complete SextetChoke sets <{L1},{L1}>

39Slide40

However, the spider model and the collider model do not receive the same chi-squared score when estimated, so in principle they can be distinguished from a 2-factor model. ExpensiveRequires multiple restartsNeed to test only pure clustersIf non-Gaussian, may be able to detect additional impurities. Checking with Estimated Model

40Slide41

For sextads, the first step is to check 10 * n choose 6 sextads.However, a large proportion of social science contexts, there are at most 100 observed variables, and 15 or 16 latents. If based on questionairres, generally can’t get people to answer more questions than that. Simulation studies by

Kummerfeld

indicate that given the vanishing

sextads

, the rest of the algorithm is

subexponential in the number of clusters, but exponential in the size of the clusters.

Complexity

41Slide42

Problems in Testing Constraints

Tests require (algebraic) independence among constraints.

Additional complication – when some correlations or partial correlations are non-zero, additional dependencies among constraints arise

Some models entail that neither of a pair of

sextad

constraints vanish, but that they are equal to each other

42Slide43

For single factor submodels, the algorithm can be applied to more than a hundred measured variables, with comparable accuracy to Silva 06 algorithm.43Preliminary ResultsSlide44

3 latents, 6 measures, 1 crossconstruct impurity, 2 direct edge impurities, 20 trials# 2 cluster – 15/20# 1 cluster – 5/20# 0 clusters – 2/20Average misassigned: 1Average left out if 2 cluster: 1

Average impurities left in: .1

44

Sanity Check Simulation for 2-FactorSlide45

L1 L3 L5X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 L2 L4 L6

Extension to Non-linearity

Theory: As long as

parts

(choke

sets to observed

) of the graph are

linear with additive noise,

t-separation theorem still holds.

Practice: The algorithm can be applied (with same caveats) even if the structural model is non-linear or has feedback.

45Slide46

Described algorithm that relies on weakened assumptionsWeakened linearity assumption to linearity below the latentsWeakened assumption of existence of pure submodels to existence of n-pure submodelsConjecture correct if add assumptions of no star or collider models, and faithfulness of constraints

Is there reason to believe in faithfulness of constraints when non-linear relationships among the

latents

?

46

Summary Slide47

Give complete list of assumptions for output of algorithm to be pure.Speed up the algorithm.Modify algorithm to deal with almost unfaithful constraints as much as possible.Add structure learning component to output of algorithm. Silva – Gaussian process model among latents, linearity below latentsIdentifiability questions for

stuctural

models with pure measurement models.

Open Problems

47Slide48

Silva, R. (2010). Gaussian Process Structure Models with Latent Variables. Proceedings from Twenty-Sixth Conference Annual Conference on Uncertainty in Artificial Intelligence (UAI-10). Silva, R., Scheines, R., Glymour, C., & Spirtes, P. (2006a). Learning the structure of linear latent variable models. J Mach Learn Res, 7, 191-246. Sullivant, S., Talaska, K., & Draisma

, J. (2010). Trek Separation for Gaussian Graphical Models.

Ann Stat

,

38

(3), 1665-1685.

References48Slide49

3 latents, 6 measures, 1 crossconstruct impurity, 2 direct edge impurities, 10 trials49

Sanity Check Simulation

Cluster 1

Cluster 2

Cluster 3

Impurities

5/6

4/64/5

23/54/6

4/51

3/5

4/6

4/5

2

5/6

4/6

4/5

2

6/6

6/6

-

3

3/6

3/5

-13/5

3/6-2

---3

5/6--3

3/6--

3Slide50

3 latents, 6 measures, 10 trials50Sanity Check Simulation

Clusters

+

Clusters -

Unassigned

Misassigned

0

042

1110

20

0

4

2

0

0

4

3

0

1

10

2

0

0

4

400

441

0310

041

0042Slide51

Main Example51Sanity Check Simulation

Clusters

+

Clusters -

Unassigned

Misassigned

Impure

0011

011

102

0

0

4

2

0

0

4

3

0

1

10

2

0

0

4

400

441

031

004

100

42Slide52

3 latents, 6 measures, 1 crossconstruct impurity, 2 direct edge impurities, 10 trials52

Sanity Check Simulation for 2-Factor

Unassigned

Misassigned

Impurities

Missed

6

10

100

600

1

0

0

2

1

0

1

2

0

10

0

0

10

0

00

0071

0Slide53

Suppose A = {X2,X3}, B = {X4,X5

},

C

A

= {

L1}, C

B = X2 = 3 X1

+ f2(e2,X6) X4 = 0.6 L1 + f

4(e4)X1 = 2 L

1 + f1(e

1

)

X

5

= 0.9

L

1

+

f

5

(

e5)X3

= 0.8 L1 + f3(e3)

D(CA,A) = {X1,X

2,X3} D(CB,B

) = Illustration of

Linearity Below Choke Set

53Slide54

Theorem: Suppose G is a directed graph containing CA , A, CB , and B, <CA

,

C

B

> t-separates

A and B, and A

and B are linear below their choke sets CA and C

B . Then rank(cov(A,B)) ≤ #CA + #CB .Theorem 2: Suppose G

is a directed graph containing CA , A, CB

, and B, and A and B are linear below

C

A

,

C

B

but <

C

A

,

C

B

> does not t-separate A and B

. Then there is a covariance matrix compatible with the graph in which rank(cov(A,B)) > #CA + #

CB .Proof: This follows from Sullivant et al. for linear models.Question: Is there a natural sense in which the set of parameters for which the rank

(cov(A,B)) ≤ #CA + #CB

is of measure 0 if it is not entailed by t-separation, even for the non-linear case?Extension of Choke Point Theorem

54