ff Network Analysis via Lasso Penalized D-Trace Loss · 2019. 3. 31. · ff Network Analysis via...
Transcript of ff Network Analysis via Lasso Penalized D-Trace Loss · 2019. 3. 31. · ff Network Analysis via...
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Differential Network Analysis via Lasso PenalizedD-Trace Loss
Ruibin Xi
School of Mathematical Sciences and Center for Statistical SciencePeking University
Joint work with Huili Yuan, Chong Chen and Minghua Deng
April 22, 2018
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Gaussian Graphical Model
In Gaussian graphical model, the precision matrix Θ = Σ−1.
Nonzero elements of Θ correspond to edges in Gaussian graphicalmodel.if x ∼ Np(0,Σ), Θij = 0 iff xi ⊥ xj |{xk, k = i, j} (Wittaker, 1990).
We can impose sparsity on Θ to study the Gaussian graphical model.
(PKU) Penalized D-Trace April 22, 2018 2 / 32
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Gaussian Graphical Model
Meishausen and Buhlmann, P. (2006): neighborhood selectionscheme based on lasso penalized regression
Yuan and Lin (2006) and Friedman et al. (2007) proposed toestimate Θ by minimizing
− log detΘ + tr(ΘΣ) + λ|Θ|1,off
Zhang and Zou (2014) proposed minimizing
LD(Θ, Σ) + λ|Θ|1,off= 1
2 < Θ2, Σ > −tr(Θ) + λ|Θ|1,off
where < A,B >= tr(ATB).
(PKU) Penalized D-Trace April 22, 2018 3 / 32
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Difference of Precision Matrices
Suppose that X1 · · ·XnX ∼ N (0,ΣX), Y1, · · · , YnY ∼ N (0,ΣY ), try toestimate
∆ = ΘX −ΘY = Σ−1X − Σ−1
Y
ΘX may be the gene regulatory network in normal condition, ΘY maybe the gene regulatory network in a perturbed condition.
more interested in the “change” of the network
the change can be measured by ∆ = ΘX −ΘY
usually we can assume the changes are sparse.
(PKU) Penalized D-Trace April 22, 2018 4 / 32
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Difference network
Zhou et al. Nature (1995) find a network attacking mutation at RETin cancer.
Bandyopadhyay et al. Science (2010) profiled genetic interactiondifferences with and without DNA damaging agent.
Guenole et al. Cell (2013) studied genetic interaction changes indifferent conditions..
(PKU) Penalized D-Trace April 22, 2018 5 / 32
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Difference network
Guenole et al. Cell (2013)
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Difference of Precision Matrices
Existing works
Danaher et al. (2014) considered minimizing
−K∑k=1
[nk log det(Θk)−tr(ΣkΘk)
]+λ1
K∑k=1
|Θk|1,off+λ2
∑k<k′
|Θk−Θk′ |1
If K = 2, this can be used for estimating the difference of theprecision matrices. No Theoretical development.
Zhao et al. (2014) proposed estimating ∆ = ΘX −ΘY by solving
argmin∆|∆| subject to |ΣX∆ΣY − ΣX + ΣY |∞ ≤ λn.
Advantage: Do not need to specify the sparsity of ΣX and ΣY .Computational complexity O(p4), very expensive!
(PKU) Penalized D-Trace April 22, 2018 7 / 32
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Difference of Precision Matrices
In Zhao et al. (2014), to solve
argmin∆|∆| subject to |ΣX∆ΣY − ΣX + ΣY |∞ ≤ λn,
they have to transform the problem as
argmin∆|∆| subject to |(ΣX ⊗ ΣY )vec(∆)− vec(ΣX − ΣY )|∞ ≤ λn.
The, the problem can be solved based on a method developed for linearregression (implemented in the R package flare).
(PKU) Penalized D-Trace April 22, 2018 8 / 32
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A new loss function
Estimation based on a new loss function. Idea is to find new loss functionLD(∆|ΣX ,ΣY )
LD is convex in ∆
LD achieves minimum at ΘX −ΘY .
(PKU) Penalized D-Trace April 22, 2018 9 / 32
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A new loss function
We chose
LD(∆|ΣX ,ΣY ) =1
2
(⟨ΣX∆,ΣY ∆⟩+ ⟨ΣY ∆,ΣX∆⟩
)− 2⟨∆,ΣY − ΣX⟩
Note that∂LD∂∆ = ΣX∆ΣY +ΣY ∆ΣX − 2(ΣY − ΣX)
if ∆ = ΘX −ΘY = Σ−1X − Σ−1
Y , we have ∂LD∂∆ (∆) = 0
∂2LD∂∆2 = ΣX ⊗ ΣY +ΣY ⊗ ΣX ⪰ 0
We also call LD(∆|ΣX ,ΣY ) the D-trace loss.
(PKU) Penalized D-Trace April 22, 2018 10 / 32
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Penalized D-trace loss
We use the lasso penalized D-trace loss for the estimation of the differenceof precision matrices
Given X1 · · ·XnX ∼ N (0,ΣX), Y1, · · · , YnY ∼ N (0,ΣY ).
Let ΣX and ΣY be the sample covariance matrices.
Estimate ∆ by minimizing
LD(∆|ΣX , ΣY ) + λ|∆|1
= 12
(⟨ΣX∆, ΣY ∆⟩+ ⟨ΣY ∆, ΣX∆⟩
)− 2⟨∆, ΣY − ΣX⟩+ λ|∆|1
How to solve this?
(PKU) Penalized D-Trace April 22, 2018 11 / 32
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Penalized D-trace loss
The augmented Lagrange:
L(∆1,∆2,∆3,Λ1,Λ2,Λ3)
=1
2
(⟨ΣY ∆1, ΣX∆1⟩+ ⟨ΣX∆2, ΣY ∆2⟩
)− ⟨∆1, ΣY − ΣX⟩ − ⟨∆2, ΣY − ΣX⟩+ λ∥∆3∥1+
ρ
2∥∆1 −∆2∥2F +
ρ
2∥∆2 −∆3∥2F +
ρ
2∥∆3 −∆1∥2F
+ ⟨Λ1,∆1 −∆2⟩+ ⟨Λ2,∆2 −∆3⟩+ ⟨Λ3,∆3 −∆1⟩.
Iteratively update ∆1,∆2,∆3,Λ1,Λ2,Λ3
(PKU) Penalized D-Trace April 22, 2018 12 / 32
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Penalized D-trace loss
Taking partial derivative about ∆1 and setting it as zero, we get
∂L
∂∆1= ΣY ∆1ΣX + 2ρ∆1 − ρ(∆2 +∆3) + Λ1 − Λ3 − (ΣY − ΣX)
= 0
We get the equation of the form A∆B + ξ∆ = C, with ξ = 2ρ andC = ρ(∆2 +∆3) + Λ3 − Λ1 + ΣY − ΣX .
(PKU) Penalized D-Trace April 22, 2018 13 / 32
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Lemma (1)
Assume that A,B are symmetric and semidefinite matrices, C is a symmetric matrix and ρ is areal number. Let G(A,B,C, ρ) be the solution to the equation
A∆B + ρ∆ = C,
thenG(A,B,C, ρ) = UA[D ◦ (UT
ACUTB )]UB ,
where A = UAΣAUTA , B = UBΣBUT
B and Dij = 1σAj σB
i +ρ.
(PKU) Penalized D-Trace April 22, 2018 14 / 32
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The proof of lemma 1
Proof.First, we make a vectorization for the original equation.
vec(C) = (A⊗B)vec(∆) + (ρI ⊗ I)vec(∆)
=((UAΣAUT
A )⊗ (UBΣBUTB ) + ρI ⊗ I
)vec(∆)
= (UA ⊗ UB)(ΣA ⊗ ΣB + ρI ⊗ I)(UTA ⊗ UT
B )vec(∆)
(1)
Since (UA ⊗ UB)(UTA ⊗ UT
B ) = I, it is easy to get
vec(∆) = (UA ⊗ UB)(ΣA ⊗ ΣB + ρI ⊗ I)−1(UTA ⊗ UT
B )vec(C)
= (UA ⊗ UB)(ΣA ⊗ ΣB + ρI ⊗ I)−1vec(UTACUT
B )
= (UA ⊗ UB)vec(D ◦ (UTACUT
B ))
= vec(UA[D ◦ (UTACUT
B )]UB)
(2)
So∆ = UA[D ◦ (UT
ACUTB )]UB .
(PKU) Penalized D-Trace April 22, 2018 15 / 32
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Penalized D-trace loss
Fixing ∆1,∆2,Λ1,Λ2,Λ3, the part of the augmented Lagrange involving∆3 is ρ∥∆3∥F + λ|∆3|1− < ∆3, ρ∆2 + ρ∆3 + Λ2 − Λ3 >
Lemma (2)
Let S(A, λ) = argmin∆12∥∆∥2F + λ∥∆∥1 − ⟨X,A⟩, then
S(A, λ)ij =
Aij − λ Aij > λ
Aij + λ Aij < −λ
0 otherwise
(PKU) Penalized D-Trace April 22, 2018 16 / 32
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Algorithm
Require: ΣY , ΣX , ρ, λ1: Initialize ∆0
1,∆02,∆
03,Λ
01,Λ
02,Λ
03, k = 0
2: while Stop condition do3: ∆k+1
1 = G(ΣY , ΣX , ρ∆k2 + ρ∆k
3 + (ΣY − ΣX) + Λk3 − Λk
1, 2ρ)4: ∆k+1
2 = G(ΣX , ΣY , ρ∆k+11 + ρ∆k
3 + (ΣY − ΣX) + Λk1 − Λk
2, 2ρ)
5: ∆k+13 = S(
ρ∆k+11 +ρ∆k+1
2 +Λk2−Λk
32ρ , λ
2ρ)
6: Λk+11 = Λk
1 + ρ(∆k+11 −∆k+1
2 )7: Λk+1
2 = Λk2 + ρ(∆k+1
2 −∆k+13 )
8: Λk+13 = Λk
3 + ρ(∆k+13 −∆k+1
1 )9: end while
10: return ∆k3
(PKU) Penalized D-Trace April 22, 2018 17 / 32
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Theoretical Results
Theorem (1)
Assume that Xi, Yj (i = 1, · · · , nX , j = 1, · · · , nY ) are sub-Gaussian.Assume that max{∥Σ∗
X∥∞, ∥Σ∗Y ∥∞} ≤ M and s < p. Under an
irrepresentability condition, if λn is chosen properly, with probability largerthan 1− 2/pη−2 (η > 2), we have
∥∆−∆∗∥∞ ≤ MG
{η log p+ log 4
min (nX , nY )
}1/2
,
∥∆−∆∗∥F ≤ MG
{η log p+ log 4
min (nX , nY )
}1/2
s1/2.
where GA, GB, δ, CG, MG are constants depending on M , s, κΓ, α, σXand σY .
(PKU) Penalized D-Trace April 22, 2018 18 / 32
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The irrepresentability condition
Γ(ΣX ,ΣY ) =12(ΣX ⊗ ΣY +ΣY ⊗ ΣX).
S = {(i, j) : ∆∗i,j = 0} is the support of ∆∗.
Γ∗ = Γ(Σ∗X ,Σ∗
Y )
the irrepresentability condition
maxe∈Sc
∥Γ∗e,S(Γ
∗S,S)
−1∥1< 1− α
(PKU) Penalized D-Trace April 22, 2018 19 / 32
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The irrepresentability condition
Zhao et al. (2014) assumes a stronger condition, which implies
maxi =j
|Γ∗ij | ≤ min
jΓ∗jj(2s)
−1
Let
A =
(1 1/21/2 1
),
Σ∗X = Ip and Σ∗
Y = diag{A, Ip−2}.Γ∗ = Γ(Σ∗
X ,Σ∗Y ) satisfies the irrepresentability condition, but not the
above condition.
(PKU) Penalized D-Trace April 22, 2018 20 / 32
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Theoretical Results
Define M(∆) = {sgn(∆j,k) : j = 1, . . . , p, k = 1, . . . , p}.
Theorem (2)
Under the same conditions and notations in Theorem (1), if
minj,k:∆∗
j,k =0| ∆∗
j,k | ≥ 2MG
{η log p+ log 4
min (nX , nY )
}1/2
for some η > 2 and, then M(∆) = M(∆∗) with probability 1− 2/pη−2.
(PKU) Penalized D-Trace April 22, 2018 21 / 32
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Simulation
n = 100.
ΘX = Σ−1X was defined as 0.5|i−j|
∆: around p/4 nonzero elements.
data were generated by Gaussian distribution
Compare with Fused Graphic Lasso with λ1 = 0 and Zhao et al.(2014).
(PKU) Penalized D-Trace April 22, 2018 22 / 32
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Simulation
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
p= 100
1−TN
TP
DTL
FGL
L1−M
(PKU) Penalized D-Trace April 22, 2018 23 / 32
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Simulation
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
p= 200
1−TN
TP
DTL
FGL
(PKU) Penalized D-Trace April 22, 2018 24 / 32
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Simulation
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
p= 500
1−TN
TP
DTL
FGL
(PKU) Penalized D-Trace April 22, 2018 25 / 32
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Simulation
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.2
0.4
0.6
0.8
1.0
p= 1000
1−TN
TP
DTL
FGL
(PKU) Penalized D-Trace April 22, 2018 26 / 32
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Real Data Analysis
Colorectal cancer patients from TCGA
two groups: 77 microsatellite instable (MSI) patients, 122microsatellite stable (MSS) patients
Expression data of the genes in DNA mismatch repair pathway
(PKU) Penalized D-Trace April 22, 2018 27 / 32
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Real Data Analysis
●
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AXIN2
MLH1
BIRC5
PIK3CB
PIK3CG
(a)
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●●
APC
AXIN2
MLH1
BIRC5
TGFB3
PIK3CBPIK3CG
(b)
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APC2
AKT3
AXIN2
APC
MLH1
CYCS
PIK3CB
TGFB2
(c)
Figure: (a): D-trace loss estimate under LF -norm; (b): Fused graphical lassounder LF -norm (c): the L1-minimization estimate under LF -norm.
(PKU) Penalized D-Trace April 22, 2018 28 / 32
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Real Data Analysis
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Figure: Boxplots of somatic mutation numbers in patients with/without a AXIN2or PIK3CG mutation; The y-axis is in log10 scale.
(PKU) Penalized D-Trace April 22, 2018 29 / 32
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Conclusion
Proposed a new loss function, the D-trace loss, for the estimation ofthe difference of precision matrices.
Proposed an efficient greedy algorithm for the penalized D-trace loss
Developed asymptotic results
(PKU) Penalized D-Trace April 22, 2018 30 / 32
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Acknowledgement
NSFC
The Recruitment Program of Global Youth Experts of China
National Key Basic Research Program of China
(PKU) Penalized D-Trace April 22, 2018 31 / 32
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Thank you for your attention!
(PKU) Penalized D-Trace April 22, 2018 32 / 32