Structured Gaussian Graphical Models Ashish Katiyar, Jessica - - PowerPoint PPT Presentation

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Structured Gaussian Graphical Models Ashish Katiyar, Jessica - - PowerPoint PPT Presentation

Robust Estimation of Tree Structured Gaussian Graphical Models Ashish Katiyar, Jessica Hoffmann, Constantine Caramanis The University of Texas at Austin Undirected Probabilistic Graphical Models Represent conditional independence relations


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Robust Estimation of Tree Structured Gaussian Graphical Models

Ashish Katiyar, Jessica Hoffmann, Constantine Caramanis The University of Texas at Austin

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Undirected Probabilistic Graphical Models

  • Represent conditional

independence relations among random variables in the form of a graph.

  • Any random variable

conditioned on the random variable it has an edge with, is independent of all the remaining random variables.

X₁ X2 X3 X4 X5 X7 X6

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Why Graphical Models?

  • Identify interactions among

variables in large systems. (e.g. Gene Interaction Networks.)

  • Makes inference in large scale

systems tractable.

Sample Data Graphical Model Inference

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Gaussian Graphical Models

  • Jointly

Gaussian random variables with covariance .

  • Support of inverse covariance

gives the graphical model structure.

X X 0 0 X X X 0 0 0 0 0 X X X 0 X 0 0 X X X 0 0 0 X X 0 X X X 0 0 0 X X

=

X₁ X2 X3 X4 X5 X7 X6

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Effect Of Noise

  • Additive Gaussian noise in the

random variables breaks down the conditional independence.

  • Intuitively – Noisy samples do

not convey the whole information.

X₁ X2 X3 X4 X5 X7 X6

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SLIDE 6

Effect Of Noise

  • Additive Gaussian noise in the

random variables breaks down the conditional independence.

  • Intuitively – Noisy samples do

not convey the whole information.

X₁ X2 X3 X4 X5 X7 X6 X3

  • Noisy node
  • Extra Edges
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SLIDE 7

Effect Of Noise

  • Additive Gaussian noise in the

random variables breaks down the conditional independence.

  • Example: if X – Y – Z is a Markov

chain, then X and Z are no longer independent when conditioned

  • n a noisy version of Y.

X₁ X2 X3 X4 X5 X7 X6

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Problem Statement

  • Suppose the graphical model

have a tree structure .

  • We observe .

where is an unknown positive diagonal noise matrix.

  • Goal: recover given .

= +

X₁ X2 X3 X4 X5 X7 X6 X₁ X2 X3 X4 X5 X7 X6

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Bad News! Unidentifiability

  • Even for arbitrarily small noise

the problem is unidentifiable!

  • There are covariance matrices

that differ only on diagonal entries, but their inverses have different sparsity pattern.

1 2 3 1 2 3 1 2 3 1 3 2 2 1 3

1 2 3 True Tree Noisy graph (a) Underlying tree and the noisy graphical model. (b) Graphical models for 3 different decompositions of . 1 2 3

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Good News! Limited Unidentifiability

  • The ambiguity in tree structure

is highly limited.

  • The only ambiguity is between a

leaf node and its immediate neighbor.

10

8 9

11 12 13 14

1 2 3 4 5 6 7

Trees formed by permuting nodes within the dotted regions form an equivalence class .

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Proof - Key Idea

  • Off-diagonal covariance entries

have information about the tree structure.

  • They can be used to categorize

any set of 4 nodes as a star or non-star.

  • Non-star – Exactly 2 nodes lie in
  • ne subtree.

2 6 5 10 11 12 13 1 4 9 14 3 7 8 Node Set Classification {0, 1, 3, 7} Non-Star {7, 8, 9, 10} Non-Star {14, 2, 10, 11} Non-Star {7, 9, 1, 6} Star {1, 14, 10, 6} Star

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Proof – Key Idea

  • Categorization of any set of 4

nodes as star/non-star defines all possible partitions in 2 subtrees with minimum 2 nodes.

  • Thus the off diagonal elements

define the tree upto the equivalence class .

2 6 5 10 11 12 13 1 4 9 14 3 7 8

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Identifiability with Side Information

  • Diagonal Majorization Condition – If the precision matrix is known to

have diagonal entries greater than the absolute off diagonal entries.

  • Minimum Eigenvalue Condition – If a lower bound on the minimum

eigenvalue of the covariance is known and the noise variance is smaller than this lower bound.

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Cluster Tree

2 6 5 10 11 12 13 1 4 9 14 3 7 8 EC 3 EC 4 EC 5 EC 6 EC 2 EC 0 EC 1

EC 0 EC 2 EC 6 EC 5 EC 1 EC 4 EC 3

EC – Equivalence Cluster Original Tree Cluster Tree

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Algorithm - Initialization

  • Split the tree in 2 subtrees.
  • Find the root equivalence

equivalence cluster of each subtree.

2 6 5 10 11 12 13 1 4 9 14 3 7 8 Root Equivalence Clusters

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Algorithm – Recursion step

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Conclusion

  • Unidentifiability of learning tree structured Gaussian Graphical

Models in presence of noise.

  • Identifiability conditions with side information.
  • Algorithm to find the tree structure.