"Estimating Tree-Structured Covariance Matrices via Mixed-Integer Programming"
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"Estimating Tree-Structured Covariance Matrices via Mixed-Integer Programming"
- 5. • •
- 6. • • • •
- 7. • •
- 8. • • • • •
- 20. T ;d • T 1 1 1 1 1 1
- 21. • B • Bij i j 1 1 1 1 1 1 T ;d
- 22. • B • Bij i j 1 1 1 1 1 1 T ;d
- 23. 1+1 1 1 1 1 1 1 T ;d • B • Bij i j
- 24. 1 1 1 1 1 1 T ;d • B • Bij i j
- 25. 1 11 1 1 1 1 1 T ;d • B • Bij i j
- 26. 1+1 1 0 0 1 1+1 0 0 0 0 1+1 1 0 0 1 1+1 1 1 1 1 1 1 T ;d • B • Bij i j
- 27. T • T 1 1 1 1 1 1 B T T 1+1 1 0 0 1 1+1 0 0 0 0 1+1 1 0 0 1 1+1
- 28. T • T 1 1 1 2 1 1 1+1 1 0 0 1 1+2 0 0 0 0 1+1 1 0 0 1 1+1 B T
- 29. • T 1 2 1 1 1 1 1+1 1 0 0 1 1+1 0 0 0 0 2+1 2 0 0 2 2+1 T B T
- 30. • T 0.9+2 0.9 0 0 0.9 0.9+3 0 0 0 0 2+1.1 2 0 0 2 2+1.5 T 0.9 2 2 3 1.1 1.5 B T
- 31. • T B. – B – B T
- 33. T T
- 36. B
- 38. B
- 39. B
- 40. B
- 41. B
- 42. B
- 43. B
- 44. B
- 45. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0
- 46. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1
- 47. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 2
- 48. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2
- 49. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 3
- 50. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1
- 51. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
- 52. B 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 2 0 0 2 2 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
- 53. 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 2 0 0 2 2 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
- 54. • dk d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
- 55. • vk – p (2p -1) v0 v1 v2 v3 v4 v5 v6
- 57. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 VDVT d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
- 58. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 VkVk B d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
- 59. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 B = VDVT d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
- 60. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 B = VDVT d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
- 61. 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 B T ? T B
- 62. T B 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 min (Bij) = 0 = d0 B
- 63. T B 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B 0
- 64. T B 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5
- 65. T B
- 66. T B d0 = 0
- 67. T B 2.9 0.9 0.9 3.9 0.9 = d1
- 68. T B 2.9 0.9 0.9 3.9 d0 = 0 d1 = 0.9
- 69. T B 2.9 0.9 0.9 3.9 0.9 1 1 1 1
- 70. T B 2 0 0 3 0.9 1 1 1 1
- 71. T B d0 = 0 d1 = 0.9
- 72. T B d0 = 0 d1 = 0.9 2 0 0 3
- 73. T B 2 2 = d3
- 74. T B 3 3 = d4
- 75. T B d0 = 0 d1 = 0.9 d3 = 2 d4 = 3 2 3
- 76. T B d0 = 0 d1 = 0.9 d3 = 2 d4 = 3 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 d2 = 2 d5 = 1.1 d6 = 1.5 TB
- 77. V D VDV
- 78. D : a p p diagonal matrix All the entries are non-negative.
- 79. v0 V : a p (2p -1) Boolean matrix
- 80. v0 V has the ‘partition property’.
- 81. v0 V contains the vector of all ones as a column. V has the ‘partition property’.
- 82. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
- 83. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
- 84. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
- 85. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
- 86. For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’. v0 v1 v2 v3 v4 v5 v6
- 87. D A p p diagonal matrix with all entries non-negative V A p (2p -1) Boolean matrix with the ‘partition property‘ VDV BT B is a p p matrix spanned by vkvk in bijective correspondence with its rooted binary tree T with p leaves.
- 89. • S subject to minimise B (difference between B and S ) BT S Y (B is tree-structured)
- 90. subject to minimise D,V (difference between B and S ) (B is tree-structured) BT S Y
- 91. V B D
- 92. B S V
- 93. • • • • 0 0 0 0 0 0 0 0 d1 0 0 0 0 0 0 0 d2 0 0 0 0 0 0 0 d3 0 0 0 0 0 0 0 d4 0 0 0 0 0 0 0 d5 0 0 0 0 0 0 0 d6 d1 d2 d3 d4 d5 d6 B b
- 94. • •
- 95. • • • • subject to minimise d (B is tree-structured)
- 96. • •
- 100. B
- 101. ∵ ∵ ∵
- 110. F
- 111. F
- 124. ?
- 125. ?
- 126. ?
- 134. ∈ ℝ 𝑝 N
- 135. S
- 136. B
- 147. • – – •
- 156. • • •