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Discovering Cyclic Causal Models by Independent Components Analysis Gustavo Lacerda Peter Spirtes Joseph Ramsey Patrik O. Hoyer

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Structural Equation Models (SEMs) Graphical models that represent causal relationships. Manipulating x3 to a fixed value… x1x2 x3 x4 x1x2 x3 x4 f3(x1, x2)x3 = x4 = f4(x3) k M: M(do (x3 = k)):

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Structural Equation Models (SEMs) Can be acyclic …or cyclic The data produced by cyclic models can be interpreted as equilibrium points of dynamical systems x1x2 x3 x4

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Linear Structural Equation Models (SEMs) (deterministic example) The structural equations are linear e.g.: x3 = 1.2 x1 + 0.9 x2 - 3 x4 = -5 x3 + 1 Each edge weight tells us the corresponding coefficient x1x2 x3 x4 1.2 0.9 -5

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Linear Structural Equation Models (SEMs) (with randomness) Now, each variable has an additive noise term with non- zero variance. x1 = e1 x2 = e2 x3 = 1.2 x1 + 0.9 x2 – 3 + e3 x4 = -5 x3 + 1 + e4 x = B x + e x1x2 x3 x4 1.2 0.9 -5 e1 e2 e3 e4

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Linear Structural Equation Models (SEMs) (with randomness) x = B x + e Solving for x, we get: x = (I – B) -1 e Let A = (I – B) -1 then x = A e A is called the “mixing matrix”. x1x2 x3 x4 1.2 0.9 -5 e1 e2 e3 e4

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Linear Structural Equation Models (SEMs) (with randomness) The “mixing matrix” shows how the noise propagates: x1x2 x3 x4 1.2 0.9 -5 e1 e2 e3 e4

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Linear Structural Equation Models (SEMs) (with randomness) The “mixing matrix” shows how the noise propagates: Done. x1x2x3x4 e1e2e3e4 x1x2 x3 x4 1.2 0.9 -5 e1 e2 e3 e4 Let’s make it: 1 11 1 1.2 -6 0.9 -5 -4.5

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What can we learn from observational data alone? Until recently, the best we could do was identify the d-separation equivalence class We couldn’t tell the difference between: x1 x2 x1 x2 M1:M2:

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Why not? Because it was assumed that the error terms are Gaussian …and when they are Gaussian, these two graphs are distribution-equivalent x1 x2 x1 x2 M1:M2:

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Independent Components Analysis (ICA) Cocktail party problem You want to get back the original signals, but all you have are the mixtures. What can you do? x = A e x1x2 e2e1

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Independent Components Analysis (ICA) Cocktail party problem This equation has infinitely many solutions! For any invertible A, there is a solution! But if you assume that the signals are independent, it is possible to estimate A and e from just x. How? x = A e x1x2 e2e1

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Independent Components Analysis (ICA) Cocktail party problem Any choice of A implies a list of samples of e Each list of implied samples of e has a degree of independence We want the A for which the implied e’s are maximally independent e’s maximally independent ↔ e’s maximally non-Gaussian Intuition: Central Limit Theorem x = A e x1x2 e2e1

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Independent Components Analysis (ICA) We don’t know which source signal is which, i.e. which is Alex and which is Bob Scaling: when used with SEMs, the variance of each error term is confounded with its coefficients on each x.

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The LiNGAM approach (Shimizu et al, 2006) What happens if we generate data from this linear SEM … and then run ICA? x1 x2x3 x4 e1 e2 e3 e4 1.5 -2 1.1

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The LiNGAM approach We would expect to see: Except that ICA doesn’t know the scaling x1x2x3x4 e1e2e3e4 1.5-3 1 1.1 -2 111

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The LiNGAM approach So we should expect to see something like: …and we’d need to normalize by dividing all children of e1 by 2 x1x2x3x4 e1e2e3e4 3-6 2 2.2 -2

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The LiNGAM approach getting us: Except that ICA doesn’t know the order of the e’s, i.e. which e’s go with which x’s… x1x2x3x4 e1e2e3e4 1.5-3 1 1.1 -2

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The LiNGAM approach really, ICA gives us something like: So first we need to find the right permutation of the e’s And then do the scaling Note that, since the model is a DAG, there is exactly one valid way to permute the error terms. x1x2x3x4 e… 3-6 2 2.2 -2 x1x2x3x4 e1e2e3e4 3-6 2 2.2 -2 x1x2x3x4 e1e2e3e4 1.5-3 1 1.1 -2

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The LiNGAM approach After some matrix magic, we get back: x1 x2x3 x4 e1 e2 e3 e4 1.5 -2 1.1 B = I – A -1

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The LiNGAM approach Discovers the full structure of the DAG … by assuming causal sufficiency (i.e. independence of the error terms) “causal sufficiency”: no latent variable is a cause of more than one observed variable linear case, causal sufficiency ↔ independence of the error terms In particular, now M1 and M2 can be distinguished! x1 x2 x1 x2 M1:M2:

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The LiNGAM approach Gaussian Uniform Images by Patrik Hoyer et al, used with permission from “Estimation of causal effects using linear non-Gaussian causal models with hidden variables”

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The LiNGAM approach Note that, once the valid permutation was found, there were no left- pointing arrows. This is because: the generating model was a DAG. we wrote down the x’s in an order compatible with it But it is possible for ICA to return a matrix that does not satisfy the acyclicity assumption LiNGAM will pretend the red edge is not there x1x2x3x4 e1e2e3e4

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The LiNGAM approach LiNGAM cannot discover cyclic models… because: since it assumes the data was generated by a DAG, it searches for a single valid permutation If we search for any number of valid permutations… then we can discover cyclic models too. That’s exactly what we did!

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The LiNG-DG approach When the data looks acyclic, it works just like LiNGAM, and returns a single model. When the data looks cyclic, more than one permutation is considered valid. Thus, it returns a distribution-equivalent set containing more than one model. “distribution-equivalent” means you can’t do better, at least without experimental data or further assumptions.

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The LiNG-DG approach Let’s simulate using this model: Error terms are generated by sampling from a Gaussian and squaring 15000 data points We test which ICA coefficients are zero by using bootstrap sampling followed by a quantile test Ready? x5 e1 x4 x1 x2 x3 e2 e5 e4 e3 3 1.2 -0.3 2

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The LiNG-DG approach LiNG-DG returns a set with 2 models: #1#2

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LiNG-DG + the stability assumption Note that only one of these models is stable. If our data is a set of equilibria, then the true model must be stable. Under what conditions are we guaranteed to have a unique stable model?

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LiNG-DG + the stability assumption Theorem: if the true model’s cycles don’t intersect, then only one model is stable. For simple cycle models, cycle-products are inverted: c1 = 1/c2. So at least one cycle will be > 1 (in modulus) and thus unstable. each cycle works independently, and any valid permutation* will invert at least one cycle, creating an unstable model. *except for the identity permutation

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very large class: not even covariance equivalent d-separation equivalence class What should one use? non-GaussianGaussian DAG DG Constraint-based methods e.g. PC, CPC, SGS (or Geiger and Heckerman 1994 for a Bayesian alternative) LiNGAM unique model Richardson’s CCD LiNG-DG 2 cases acyclicunique model cyclic distribution- equivalence class unknown or both or too little data Check out Hoyer, Hyvärinen, Glymour, Spirtes, Scheines,Ramsey, Lacerda, Shimizu (submitted) ?

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UAI is due soon! Please send me your comments: gusl@cs.cmu.edu

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Appendix 1: self-loops Equilibrium equations usually correspond with the dynamical equations. EXCEPT if a self-loop has coefficient 1, we will get the wrong structure, and the predicted results of intervention will be wrong! self-loop coefficients are underdetermined. Our stability results only hold if we assume no self-loops.

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Appendix 2: search and pruning Testing zeros: local vs non-local methods To estimate the variance of the estimated coefficients, we use bootstrap sampling, carefully. How to find row-permutations of W that have a zeroless diagonal: Acyclic: Hungarian algorithm General: k-best linear assignments, or constrained n- Rooks (put rooks on the non-zero entries)

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