"estimating tree-structured covariance matrices via mixed-integer programming"
DESCRIPTION
セミナーで紹介したネタ論文．スライドはあわてて作ったので見映えは良くない．TRANSCRIPT
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T ;d
• T
1 1
1 1 1 1
• B
• Bij
i j
1 1
1 1 1 1
T ;d
• B
• Bij
i j
1 1
1 1 1 1
T ;d
1+1
1 1
1 1 1 1
T ;d
• B
• Bij
i j
1 1
1 1 1 1
T ;d
• B
• Bij
i j
1
11 1
1 1 1 1
T ;d
• B
• Bij
i j
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
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
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
• 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
• 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
• T B.
– B
– B T
T
T
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B
B
B
B
B
B
B
B
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
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
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
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
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
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
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
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
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
• dk
d0=0
d1=1 d2=2
d3=2 d4=3 d5=1 d6=1
• vk
– p (2p -1)
v0
v1 v2
v3 v4 v5 v6
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
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
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
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
2.9 0.9 0 0
0.9 3.9 0 0
0 0 3.1 2
0 0 2 3.5
B T
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T B
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
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
T B
2.9 0.9 0 0
0.9 3.9 0 0
0 0 3.1 2
0 0 2 3.5
T B
T B
d0 = 0
T B
2.9 0.9
0.9 3.9
0.9 = d1
T B
2.9 0.9
0.9 3.9
d0 = 0d1 = 0.9
T B
2.9 0.9
0.9 3.9
0.9
1 1
1 1
T B
2 0
0 3
0.9
1 1
1 1
T B
d0 = 0d1 = 0.9
T B
d0 = 0d1 = 0.9
2 0
0 3
T B
2 2 = d3
T B
3
3 = d4
T B
d0 = 0d1 = 0.9
d3 = 2 d4 = 3
2
3
T B
d0 = 0d1 = 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
V D
VDV
D : a p p diagonal matrix
All the entries are non-negative.
v0
V : a p (2p -1) Boolean matrix
v0
V has the ‘partition property’.
v0
V contains the vector of all ones as a column.
V has the ‘partition property’.
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’.
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’.
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’.
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’.
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
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.
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• S
subject to
minimise B
(difference between B and S )
BT
S Y
(B is tree-structured)
subject to
minimise D,V
(difference between B and S )
(B is tree-structured)
BT
S Y
V
B D
B S
V
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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
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subject to
minimise d
(B is tree-structured)
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B
∵
∵
∵
F
F
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∈ ℝ𝑝
N
S
B
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