Minimax Rates for Homology Inference
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The document discusses minimax rates for homology inference in topological inference, focusing on estimating the topology of a smooth manifold using points drawn from a distribution derived from it. It explores geometric and statistical assumptions required for well-posed problems, sampling complexities, and the application of le cam's lemma for proving minimax lower bounds. Fundamental open problems in the field are also highlighted, including questions regarding manifold parameters and algorithm efficiency.
Minimax Rates for Homology Inference
- 1. Minimax Rates for Homology Inference Don Sheehy Joint work with Sivaraman Balakrishan, Alessandro Rinaldo, Aarti Singh, and Larry Wasserman
- 2. Something like a joke.
- 3. Something like a joke. What is topological inference?
- 4. Something like a joke. What is topological inference? It’s when you infer the topology of a space given only a finite subset.
- 5. Something like a joke. What is topological inference? It’s when you infer the topology of a space given only a finite subset.
- 6. We add geometric and statistical hypotheses to make the problem well-posed. Geometric Assumption: The underlying space is a smooth manifold M. Statistical Assumption: The points are drawn i.i.d. from a distribution derived from M.
- 7. We add geometric and statistical hypotheses to make the problem well-posed. Geometric Assumption: The underlying space is a smooth manifold M. Statistical Assumption: The points are drawn i.i.d. from a distribution derived from M.
- 8. We add geometric and statistical hypotheses to make the problem well-posed. Geometric Assumption: The underlying space is a smooth manifold M. Statistical Assumption: The points are drawn i.i.d. from a distribution derived from M.
- 9. We add geometric and statistical hypotheses to make the problem well-posed. Geometric Assumption: The underlying space is a smooth manifold M. Statistical Assumption: The points are drawn i.i.d. from a distribution derived from M.
- 11. Input: n points from a d-manifold M in D-dimensions.
- 12. Input: n points from a d-manifold M in D-dimensions. Output: The homology of M.
- 13. Input: n points from a d-manifold M in D-dimensions. Output: The homology of M. Upper bound: What is the worst case complexity?
- 14. Input: n points from a d-manifold M in D-dimensions. Output: The homology of M. Upper bound: What is the worst case complexity? Lower Bound: What is the worst case complexity of the best possible algorithm?
- 15. sam pled i.i.d. Input: n points from a d-manifold M in D-dimensions. Output: The homology of M. Upper bound: What is the worst case complexity? Lower Bound: What is the worst case complexity of the best possible algorithm?
- 16. ion pled i.i.d. distribut d on sam supporte Input: n points from a d-manifold M in D-dimensions. Output: The homology of M. Upper bound: What is the worst case complexity? Lower Bound: What is the worst case complexity of the best possible algorithm?
- 17. ion pled i.i.d. distribut d on sam supporte Input: n points from a d-manifold M in D-dimensions. e with nois Output: The homology of M. Upper bound: What is the worst case complexity? Lower Bound: What is the worst case complexity of the best possible algorithm?
- 18. ion pled i.i.d. distribut d on sam supporte Input: n points from a d-manifold M in D-dimensions. e estimate of with nois an Output: The homology of M. Upper bound: What is the worst case complexity? Lower Bound: What is the worst case complexity of the best possible algorithm?
- 19. ion pled i.i.d. distribut d on sam supporte Input: n points from a d-manifold M in D-dimensions. e estimate of with nois an Output: The homology of M. Upper bound: What is the worst case complexity? ing probabil ity of giv r a wro ng answe Lower Bound: What is the worst case complexity of the best possible algorithm?
- 20. ion pled i.i.d. distribut d on sam supporte Input: n points from a d-manifold M in D-dimensions. e estimate of with nois an Output: The homology of M. Upper bound: What is the worst case complexity? ing probabil ity of giv r a wro ng answe Lower Bound: What is the worst case complexity of the best possible algorithm? The Goal: Matching Bounds (asymptotically)
- 21. Minimax risk is the error probability of the best estimator on the hardest examples.
- 22. Minimax risk is the error probability of the best estimator on the hardest examples. Minimax Risk: Rn = inf sup n ˆ Q (H = H(M )) ˆ H Q∈Q
- 23. Minimax risk is the error probability of the best estimator on the hardest examples. Minimax Risk: Rn = inf sup n ˆ Q (H = H(M )) ˆ H Q∈Q the best r estimato
- 24. Minimax risk is the error probability of the best estimator on the hardest examples. Minimax Risk: Rn = inf sup n ˆ Q (H = H(M )) ˆ H Q∈Q the best r t t he hardes n estimato d istributio
- 25. Minimax risk is the error probability of the best estimator on the hardest examples. Minimax Risk: Rn = inf sup n ˆ Q (H = H(M )) ˆ H Q∈Q product ion the best r t t he hardes n distribut estimato d istributio
- 26. Minimax risk is the error probability of the best estimator on the hardest examples. Minimax Risk: Rn = inf sup n ˆ Q (H = H(M )) ˆ H Q∈Q product ion the true the best r t he hardes n distribut estimato t homology d istributio
- 27. Minimax risk is the error probability of the best estimator on the hardest examples. Minimax Risk: Rn = inf sup n ˆ Q (H = H(M )) ˆ H Q∈Q product ion the true the best r t he hardes n distribut estimato t homology d istributio Sample Complexity: n( ) = min{n : Rn ≤ }
- 28. We assume manifolds without boundary of bounded volume and reach.
- 29. We assume manifolds without boundary of bounded volume and reach. Let M be the set of compact d-dimensional Riemannian manifolds without boundary such that
- 30. We assume manifolds without boundary of bounded volume and reach. Let M be the set of compact d-dimensional Riemannian manifolds without boundary such that 1 M ⊂ ballD (0, 1)
- 31. We assume manifolds without boundary of bounded volume and reach. Let M be the set of compact d-dimensional Riemannian manifolds without boundary such that 1 M ⊂ ballD (0, 1) 2 vol(M ) ≤ cd
- 32. We assume manifolds without boundary of bounded volume and reach. Let M be the set of compact d-dimensional Riemannian manifolds without boundary such that 1 M ⊂ ballD (0, 1) 2 vol(M ) ≤ cd 3 The reach of M is at most τ .
- 33. We assume manifolds without boundary of bounded volume and reach. Let M be the set of compact d-dimensional Riemannian manifolds without boundary such that 1 M ⊂ ballD (0, 1) 2 vol(M ) ≤ cd 3 The reach of M is at most τ . Let P be the set of probability distributions supported over M ∈ M with densities bounded from below by a constant a.
- 34. We consider 4 different noise models. Noiseless Clutter Tubular Additive
- 35. We consider 4 different noise models. Noiseless Clutter Q=P Tubular Additive
- 36. We consider 4 different noise models. Noiseless Clutter Q = (1 − γ)U + γP Q=P P ∈P U is uniform on ball(0, 1) Tubular Additive
- 37. We consider 4 different noise models. Noiseless Clutter Q = (1 − γ)U + γP Q=P P ∈P U is uniform on ball(0, 1) Tubular Additive Let QM,σ be uniform on M σ . Q = {QM,σ : M ∈ M}
- 38. We consider 4 different noise models. Noiseless Clutter Q = (1 − γ)U + γP Q=P P ∈P U is uniform on ball(0, 1) Tubular Additive Let QM,σ be uniform on M σ . Q = {P Φ : P ∈ P} Φ is Gaussian Q = {QM,σ : M ∈ M} with σ τ or Φ has Fourier transform bounded away from 0 and τ is fixed.
- 39. Le Cam’s Lemma is a powerful tool for proving minimax lower bounds.
- 40. Le Cam’s Lemma is a powerful tool for proving minimax lower bounds. Lemma. Let Q be a set of distributions. Let θ(Q) take values in a metric space (X, ρ) for Q ∈ Q. For any Q1 , Q2 ∈ Q, inf sup EQn ˆ θ(Q)) ≥ 1 ρ(θ(Q1 ), θ(Q2 ))(1−TV(Q1 , Q2 ))2n ρ(θ, ˆ θ Q∈Q 8
- 41. Le Cam’s Lemma is a powerful tool for proving minimax lower bounds. Lemma. Let Q be a set of distributions. Let θ(Q) take values in a metric space (X, ρ) for Q ∈ Q. For any Q1 , Q2 ∈ Q, inf sup EQn ˆ θ(Q)) ≥ 1 ρ(θ(Q1 ), θ(Q2 ))(1−TV(Q1 , Q2 ))2n ρ(θ, ˆ θ Q∈Q 8 0 if x = y For homology, use the trivial metric. ρ(x, y) = 1 if x = y
- 42. Le Cam’s Lemma is a powerful tool for proving minimax lower bounds. Lemma. Let Q be a set of distributions. Let θ(Q) take values in a metric space (X, ρ) for Q ∈ Q. For any Q1 , Q2 ∈ Q, inf sup EQn ˆ θ(Q)) ≥ 1 ρ(θ(Q1 ), θ(Q2 ))(1−TV(Q1 , Q2 ))2n ρ(θ, ˆ θ Q∈Q 8 0 if x = y For homology, use the trivial metric. ρ(x, y) = 1 if x = y n ˆ = H(M )) ≥ 1 (1 − TV(Q1 , Q2 ))2n inf sup Q (H ˆ H Q∈Q 8
- 43. Le Cam’s Lemma is a powerful tool for proving minimax lower bounds. Lemma. Let Q be a set of distributions. Let θ(Q) take values in a metric space (X, ρ) for Q ∈ Q. For any Q1 , Q2 ∈ Q, inf sup EQn ˆ θ(Q)) ≥ 1 ρ(θ(Q1 ), θ(Q2 ))(1−TV(Q1 , Q2 ))2n ρ(θ, ˆ θ Q∈Q 8 0 if x = y For homology, use the trivial metric. ρ(x, y) = 1 if x = y ˆ = H(M )) ≥ 1 (1 − TV(Q1 , Q2 ))2n Rn = inf sup Q (Hn ˆ H Q∈Q 8
- 44. The lower bound requires two manifolds that are geometrically close but topologically distinct.
- 45. The lower bound requires two manifolds that are geometrically close but topologically distinct. B = balld (0, 1 − τ ) A = B balld (0, 2τ )
- 46. The lower bound requires two manifolds that are geometrically close but topologically distinct. B = balld (0, 1 − τ ) A = B balld (0, 2τ ) M1 = ∂(B ) τ M2 = ∂(A ) τ
- 47. The lower bound requires two manifolds that are geometrically close but topologically distinct. B = balld (0, 1 − τ ) A = B balld (0, 2τ ) M1 = ∂(B ) τ M2 = ∂(A ) τ The overlap
- 48. It suffices to bound the total variation distance.
- 49. It suffices to bound the total variation distance. Total Variation Distance: TV(Q1 , Q2 ) = sup |Q1 (A) − Q2 (A)| A ≤ a max{vol(M1 M2 ), vol(M2 M1 )} ≤ Cd aτ d
- 50. It suffices to bound the total variation distance. Total Variation Distance: TV(Q1 , Q2 ) = sup |Q1 (A) − Q2 (A)| A ≤ a max{vol(M1 M2 ), vol(M2 M1 )} ≤ Cd aτ d Minimax Risk: 1 2n 1 d 2n 1 −2Cd aτ d n Rn ≥ (1 − TV(Q1 , Q2 ) ≥ (1 − Cd aτ ) ≥ e 8 8 8
- 51. It suffices to bound the total variation distance. Total Variation Distance: TV(Q1 , Q2 ) = sup |Q1 (A) − Q2 (A)| A ≤ a max{vol(M1 M2 ), vol(M2 M1 )} ≤ Cd aτ d Minimax Risk: 1 2n 1 d 2n 1 −2Cd aτ d n Rn ≥ (1 − TV(Q1 , Q2 ) ≥ (1 − Cd aτ ) ≥ e 8 8 8 Sampling Rate: 1 d 1 n( ) ≥ log τ ε
- 52. The upper bound uses a union of balls to estimate the homology of M.
- 53. The upper bound uses a union of balls to estimate the homology of M.
- 54. The upper bound uses a union of balls to estimate the homology of M.
- 55. The upper bound uses a union of balls to estimate the homology of M. 1 Take a union of balls.
- 56. The upper bound uses a union of balls to estimate the homology of M. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 57. The upper bound uses a union of balls to estimate the homology of M. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 58. The upper bound uses a union of balls to estimate the homology of M. 0 Denoise the data. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 59. The upper bound uses a union of balls to estimate the homology of M. 0 Denoise the data. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 60. The upper bound uses a union of balls to estimate the homology of M. 0 Denoise the data. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 61. The upper bound uses a union of balls to estimate the homology of M. 0 Denoise the data. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 62. The upper bound uses a union of balls to estimate the homology of M. 0 Denoise the data. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the
- 63. The upper bound uses a union of balls to estimate the homology of M. 0 Denoise the data. 1 Take a union of balls. 2 resulting Cech complex. Compute the homology of the To prove: The density is bounded from below near M and from above far from M.
- 64. Many fundamental problems are still open.
- 65. Many fundamental problems are still open. 1 Is the reach the right parameter?
- 66. Many fundamental problems are still open. 1 Is the reach the right parameter? 2 What about manifolds with boundary?
- 67. Many fundamental problems are still open. 1 Is the reach the right parameter? 2 What about manifolds with boundary? 3 Homotopy equivalence?
- 68. Many fundamental problems are still open. 1 Is the reach the right parameter? 2 What about manifolds with boundary? 3 Homotopy equivalence? 4 How to choose parameters?
- 69. Many fundamental problems are still open. 1 Is the reach the right parameter? 2 What about manifolds with boundary? 3 Homotopy equivalence? 4 How to choose parameters? 5 Are there efficient algorithms?
- 70. Thank you.