jmlr jmlr2011 jmlr2011-94 knowledge-graph by maker-knowledge-mining

94 jmlr-2011-Theoretical Analysis of Bayesian Matrix Factorization

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Author: Shinichi Nakajima, Masashi Sugiyama

Abstract: Recently, variational Bayesian (VB) techniques have been applied to probabilistic matrix factorization and shown to perform very well in experiments. In this paper, we theoretically elucidate properties of the VB matrix factorization (VBMF) method. Through ﬁnite-sample analysis of the VBMF estimator, we show that two types of shrinkage factors exist in the VBMF estimator: the positive-part James-Stein (PJS) shrinkage and the trace-norm shrinkage, both acting on each singular component separately for producing low-rank solutions. The trace-norm shrinkage is simply induced by non-ﬂat prior information, similarly to the maximum a posteriori (MAP) approach. Thus, no trace-norm shrinkage remains when priors are non-informative. On the other hand, we show a counter-intuitive fact that the PJS shrinkage factor is kept activated even with ﬂat priors. This is shown to be induced by the non-identiﬁability of the matrix factorization model, that is, the mapping between the target matrix and factorized matrices is not one-to-one. We call this model-induced regularization. We further extend our analysis to empirical Bayes scenarios where hyperparameters are also learned based on the VB free energy. Throughout the paper, we assume no missing entry in the observed matrix, and therefore collaborative ﬁltering is out of scope. Keywords: matrix factorization, variational Bayes, empirical Bayes, positive-part James-Stein shrinkage, non-identiﬁable model, model-induced regularization

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Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 2 Full-Bayesian Matrix Factorization (FBMF) and Its Empirical Variant (EFBMF) We use the Gaussian priors on the parameters A and B: φ(U) = φA (A)φB (B), where H φA (A) ∝ exp − ∑ h=1 H φB (B) ∝ exp − ∑ h=1 ah 2 2c2h a = exp − bh 2 2c2h b = exp − −1 tr(ACA A⊤ ) 2 , (2) −1 tr(BCB B⊤ ) . [sent-102, score-1.003]

2 2 (3) Here, ah and bh are the h-th column vectors of A and B, respectively, that is, A = (a1 , . [sent-103, score-0.986]

3 Without loss a b of generality, we assume that the product cah cbh is non-increasing with respect to h. [sent-111, score-0.753]

4 The hyperparameters cah and cbh may be determined so that the Bayes free energy F(V ) is minimized. [sent-123, score-0.801]

5 A,B In the MAP framework, one may determine the hyperparameters cah and cbh so that the Bayes posterior p(A, B|V ) is maximized (equivalently, the negative log posterior is minimized). [sent-132, score-0.816]

6 In our model (1), (2), and (3), the negative log of the Bayes posterior is expressed as − log p(A, B|V ) = ah 2 bh 2 LM log σ2 1 H + ∑ M log c2h + L log c2h + 2 + 2 a b 2 2 h=1 cah cbh 2 H 1 + 2 V − ∑ bh a⊤ h 2σ h=1 + Const. [sent-168, score-2.225]

7 (17) Fro Differentiating Equation (17) with respect to A and B and setting the derivatives to zero, we have the following conditions: ah = bh = bh ah 2 2 −1 σ2 + 2 cah V− −1 σ2 + 2 cbh ∑ h′ =h V− ⊤ b h′ a ⊤ h′ bh , ∑ bh a ⊤ h ′ h′ =h ′ (18) ah . [sent-169, score-4.128]

8 h (20) h h=1 The MAP estimator U MAP is given by U MAP = H ⊤ ∑ γMAP ωb ωa , h h h h=1 where γMAP = max 0, γh − h σ2 cah cbh . [sent-182, score-0.78]

9 (21) The theorem implies that the MAP solution cuts off the singular values less than σ2 /(cah cbh ); otherwise it reduces the singular values by σ2 /(cah cbh ) (see Figure 2). [sent-183, score-0.797]

10 More precisely, Equations (21) and (22) show that the product cah cbh → ∞ is sufﬁcient for MAP to be reduced to ML, which is weaker than both cah , cbh → ∞. [sent-198, score-1.506]

11 µah , µbh , Σah , and Σbh satisfy ⊤ 1 µah = 2 Σah V − ∑ µbh′ µ⊤ ′ ah σ h′ =h µ bh = µ bh , 1 Σb V − ∑ µbh′ µ⊤ ′ ah σ2 h h′ =h (23) µ ah , Σ ah = 1 σ2 µ bh 2 + tr(Σbh ) + c−2 ah Σ bh = 1 σ2 µ ah 2 + tr(Σah ) + c−2 bh (24) −1 −1 IM , (25) IL . [sent-209, score-5.499]

12 When √ γh > Mσ2 , γVB (= µah µbh ) is bounded as h max 0, 1 − Mσ2 γ2 h γh − σ2 M/L cah cbh ≤ γVB < 1 − h Mσ2 γ2 h γh . [sent-218, score-0.753]

13 Theorem 2 states that, in the limit of cah cbh → ∞, the lower bound agrees with the upper bound and we have  2  max 0, 1 − Mσ γ if γh > 0, h VB 2 γh (29) lim γh = cah cbh →∞  0 otherwise. [sent-221, score-1.516]

14 Let us compare the behavior of the VB solution (29) with that of the MAP solution (21) when cah cbh → ∞. [sent-227, score-0.779]

15 In contrast, VB offers PJS-type regularization even when cah cbh → ∞. [sent-229, score-0.753]

16 (30) NAKAJIMA AND S UGIYAMA γVB is monotone decreasing with respect to cah cbh , and is lower-bounded as h γVB > h lim cah cbh →∞ γVB = h √ Mσ2 . [sent-239, score-1.516]

17 cah cbh Since σ4 = γMAP , h c2h c2h a b γVB > h VB has a stronger shrinkage effect than MAP in terms of the vanishing condition of singular values. [sent-241, score-0.806]

18 We can derive another upper bound of γVB , which depends on hyperparameters cah and cbh (its h proof is also included in Appendix C): Theorem 4 When γh > √ Mσ2 , γVB is upper-bounded as h γVB ≤ h 1− Lσ2 γ2 h 1− Mσ2 γ2 h · γh − σ2 . [sent-242, score-0.776]

19 cah cbh (31) √ When L = M and γh > Mσ2 , the lower bound in Equation (28) and the upper bound in Equation (31) agree with each other. [sent-243, score-0.753]

20 Thus, we have an analytic-form expression of γVB as follows: h  2  max 0, 1 − Mσ γ2 γVB = h h  0 γh − σ2 cah cbh if γh > 0, (32) otherwise. [sent-244, score-0.753]

21 + 4σ2 M − γVB + h cah cbh σ2 γVB + h cah cbh γVB + h 2 σ2 cah cbh (35) (36) 3. [sent-246, score-2.259]

22 3 EMAPMF In the EMAPMF framework, the hyperparameters cah and cbh are determined so that the Bayes posterior p(A, B|V ) is maximized (equivalently, the negative log posterior is minimized). [sent-247, score-0.816]

23 c2h = b L c2h = a (37) (38) Alternating Equations (18), (19), (37), and (38), one may learn the parameters A, B and the hyperparameters cah , cbh at the same time. [sent-250, score-0.768]

24 (2007), EMAPMF does not work properly since its objective (17) is unbounded from below at ah , bh = 0 and cah , cbh → 0. [sent-252, score-1.747]

25 Differentiating Equation (39) with respect to c2h and a c2h and setting the derivatives to zero, we obtain the following optimality conditions: b c2h = a c2h = b 2 + tr(Σ ah ) , M 2 + tr(Σ ) bh . [sent-256, score-0.986]

26 L µ ah µ bh (40) (41) Here, we observe the invariance of Equation (39) with respect to the transform −1/2 1/2 (µah , µbh , Σah , Σbh , c2h , c2h ) → (sh µah , sh a b µbh , sh Σah , s−1 Σbh , sh c2h , s−1 c2h ) a h h b (42) for any {sh ∈ R; sh > 0, h = 1, . [sent-257, score-1.05]

27 cbh (43) Then, Equations (40) and (41) yield c2h = a c2h = b ( µ ah 2 + tr(Σ µ bh 2 + tr(Σ bh )) ( µ ah LM 2 + tr(Σ )) ( µ ah bh LM 2 + tr(Σ bh )) ah )) ( , (44) . [sent-262, score-4.315]

28 (45) One may learn the parameters A, B and the hyperparameters cah , cbh by applying Equations (44) and (45) after every iteration of Equations (23)–(26) (this gives a local minimum of Equation (39) at convergence). [sent-263, score-0.768]

29 Thus, even when the hyperparameters cah and cbh are learned from data by EVB, the same upper bound as the ﬁxed-hyperparameter case in VB holds. [sent-279, score-0.768]

30 Furthermore, from Equation (32), we can conﬁrm that the upper bound (49) is equivalent to the VB solution when cah cbh = γh /M. [sent-288, score-0.766]

31 If γh ≥ 2 Mσ and ϕ(γh ) ≤ 0, then the EVB estimator of cah cbh is given by cEVB cEVB = ah bh γh ρ+ . [sent-293, score-1.766]

32 The EVB posterior is obtained by Corollary 1 with ah bh (c2h , c2h ) = cEVB cEVB , cEVB cEVB . [sent-295, score-1.005]

33 a ah ah b bh bh Furthermore, when γh ≥ √ 7Mσ, it holds that ϕ(γh ) < 0. [sent-296, score-1.972]

34 Note that γEVB is theoretically bounded as ah bh 2 = 2σ2 = γEVB ≤ γEVB ≤ γEVB = √ 7σ2 ≈ 2. [sent-570, score-0.986]

35 3, EMAPMF always results in the trivial solution, A, B = 0 and cah , cbh → 0. [sent-690, score-0.753]

36 The model (1) and the priors (2) and (3) are invariant under the following parameter transformation 1/2 −1/2 (ah , bh , cah , cbh ) → (sh ah , sh 1/2 −1/2 bh , sh cah , sh cbh ) for any {sh ∈ R; sh > 0, h = 1, . [sent-702, score-2.99]

37 Let us double Equation (17) and neglect some constant terms which are irrelevant to its minimization with respect to {ah , bh }H : h=1 L MAP ({ah , bh }H ) h=1 H = ∑ h=1 ah 2 bh 2 + 2 c2h cbh a H 1 + 2 V − ∑ bh a ⊤ h σ h=1 2 . [sent-789, score-2.608]

38 h h (62) h=1 There exists at least one minimizer that can be written as ah = ah ωah , (63) bh = bh ωbh , (64) where {ah , bh } are scalars such that γh = ah bh ≥ 0. [sent-805, score-3.394]

39 , H} such that cah cbh = cah′ cbh′ if and only if h and h′ belong to the same group (i. [sent-814, score-0.753]

40 , H, MAP Lh (ah , bh ) = a2 b2 h + 2h c2h cbh a + 1 (γ − ah bh )2 . [sent-829, score-1.774]

41 2 h σ This can be written as MAP Lh (ah , bh ) = b2 h c2h a ah cah − bh cbh 2 + 1 σ2 ah bh − γh − σ2 cah cbh 2 + σ2 2γh − 2 2 cah cbh cah cbh . [sent-830, score-5.401]

42 (65) The third term is constant with respect to ah and bh . [sent-831, score-0.986]

43 The ﬁrst nonnegative term vanishes by setting the ratio ah /bh to ah cah = bh cbh (or bh = 0). [sent-832, score-2.725]

44 (66) Minimizing the second term in Equation (65), which is quadratic with respect to the product ah bh (≥ 0), we can easily obtain Equation (21), which completes the proof. [sent-833, score-0.994]

45 a b H = ∑ h=1 µ ah − log |Σah | + 2 + tr(Σ c2h a H 1 + 2 V − ∑ µ bh µ ⊤ ah σ h=1 1 H + 2 ∑ σ h=1 µ ah ah ) 2 − log |Σbh | + µ bh 2 + tr(Σ bh ) c2h b 2 Fro tr(Σbh ) + tr(Σah ) µbh 2 + tr(Σah )tr(Σbh ) . [sent-838, score-3.547]

46 Given ﬁxed {(Σah , Σbh )}, the objective function (67) is of the same form as Equation (60) if we replace {(c2h , c2h )} in Equation (60) with {(c′2 , c′2 )} deﬁned by a ah bh b c′2 = ah 1 tr(Σbh ) + 2 cah σ2 c′2 bh 1 tr(Σah ) + 2 σ2 cbh = −1 , (70) . [sent-853, score-2.733]

47 2611 NAKAJIMA AND S UGIYAMA where VB Lh (µah , µbh , σ2h , σ2h ) = −M log σ2h + a a b − µ2 + Lσ2h µ2h + Mσ2h a a b b − L log σ2h + h 2 b c2h cbh a 2 1 γ µ µ + 2 µ2h + Mσ2h a a 2 h ah bh σ σ µ2h + Lσ2h . [sent-866, score-1.377]

48 (83) σ2 + cah cbh (M − L) cah cbh + σ2 + cah cbh (M − L) cah cbh 2 Next, we investigate the positive stationary points, assuming that µah = 0, µbh = 0. [sent-877, score-3.081]

49 γh − cah cbh γ2 h The following lemma also holds (its proof is given in Appendix G. [sent-891, score-0.783]

50 We neglect such solutions, because they almost surely do not exist; Equations (70), (71), (35), and (36) together imply that any pair {(h, h′ ); h = h′ } such that max(γVB , γVB ) > 0 and h h′ c′ h c′ h = c′ h′ c′ ′ can exist only when cah cbh = cah′ cbh′ and γh = γh′ (i. [sent-913, score-0.753]

51 a a b b H = ∑ log h=1 c2 c2M µah 2 + tr(Σah ) µbh 2 + tr(Σbh ) ah b + + log h + |Σah | c2h |Σbh | c2h a b H 1 + 2 V − ∑ µ bh µ ⊤ ah σ h=1 H + 1 ∑ σ2 h=1 µ ah 2 2 Fro tr(Σbh ) + tr(Σah ) µbh 2 + tr(Σah )tr(Σbh ) . [sent-919, score-2.144]

52 We divide the domain (105) into two regions (see Figure 12): ˚ R = (µah , µbh , σ2h , σ2h , c2h , c2h ) ∈ R2 × R2 × R2 ; cah cbh ≤ κ , a a ++ ++ b b ˘ R = (µah , µbh , σ2h , σ2h , c2h , c2h ) ∈ R2 × R2 × R2 ; cah cbh > κ . [sent-951, score-1.506]

53 a a b b ˚ (µah ,µbh ,σ2 ,σ2 ,c2 ,c2 )∈R ah bh ah bh (109) ˘ and the inﬁmum over R , EVB ˘ Lh = ˘ (µah ,µbh ,σ2 ,σ2 ,c2 ,c2 )∈R ah bh ah bh ˚ Rigorously speaking, no minimizer over R exists. [sent-953, score-3.963]

54 To make discussion simple, we approximate ˚ R by its subregion with an arbitrary accuracy; for any ε (0 < ε < κ), we deﬁne an ε-margin subregion ˚ of R : ˚ ˚ Rε = (µah , µbh , σ2h , σ2h , c2h , c2h ) ∈ R ; cah cbh ≥ ε . [sent-954, score-0.753]

55 We evaluate the difference between the objectives: EVB ˚ EVB ∆h (˘ ah , µbh , σ2h , σ2h , c2h , c2h ) = Lh (˘ ah , µbh , σ2h , σ2h , c2h , c2h ) − L h . [sent-989, score-1.138]

56 ): Lemma 25 ∆h (˘ ah , µbh , σ2h , σ2h , c2h , c2h ) is upper-bounded as µ ˘ ˘ a ˘ b ˘a ˘b ∆h (˘ ah , µbh , σ2h , σ2h , c2h , c2h ) < Mψ(α, β), µ ˘ ˘ a ˘ b ˘a ˘b (127) where ψ(α, β) = log β + α log β − (1 − α) + (1 − α) + α 2 1− √ (α+ α+1) β − β, (128) L , (129) M γ2 (130) β = h2. [sent-993, score-1.158]

57 In this case, Equation (122) leads to cah cbh = ˘ ˘ γh ρ+ . [sent-1006, score-0.753]

58 cah cbh ˘ ˘ σ Corollary 1 provides the exact values for the positive stationary points (˘ ah , µbh , σ2h , σ2h ), given µ ˘ ˘a ˘b 2 , c2 ) = (c c , c c ). [sent-1008, score-1.391]

59 2 This is minimized when dh = γh cah cbh ,5 and the minimum coincides to the right-hand side of Equation (61), which completes the proof. [sent-1043, score-0.774]

60 2622 T HEORETICAL A NALYSIS OF BAYESIAN M ATRIX FACTORIZATION Lemma 7 guarantees that cah γh ωah , cbh cbh γh ωbh , cah ah = bh = give a solution for the problem (135) for any (so far unknown) set of {γh }, which completes the proof. [sent-1060, score-2.513]

61 Further, (a ) (bh ) }) = ∞, (a ) (bh ) }) = ∞, lim L VB ({ah , bh , τm h , τl (a ) τm h →0 lim L VB ({ah , bh , τm h , τl (a ) τm h →∞ for any (h, m), because (x − log x) ≥ 1 for any x > 0, limx→+0 (x − log x) = ∞, and limx→∞ (x − (b ) log x) = ∞. [sent-1077, score-0.884]

62 (142) µbh µah = γh − η2 + σ2 (M − L) + h Subtraction of Equation (142) from Equation (141) gives 2σ2 (M − L)µah µbh + 2σ4 Lµb Mµah − 2 h 2 µ cah bh cbh µah = 2σ2 (M − L)γh , which is equivalent to Equation (88). [sent-1106, score-1.17]

63 7 Proof of Lemma 13 Squaring both sides of Equation (86) (which are positive) and substituting Equation (87) into it, we have γ2 + h + σ2 cah cbh cah cbh δh + cah cb δh γh h σ4 σ2 L − 1− 2 c2h c2h γh a b 1− σ2 M 2 γh γ2 h = 0. [sent-1121, score-1.958]

64 (146) Multiplying both sides of Equation (88) by δh (> 0) and solving it with respect to δh , we obtain δh = (M − L)(γh − γh ) + LM (M − L)2 (γh − γh )2 + 4σ c2 c2 4 ah bh 2σ2 Mc−2 ah (147) as a positive solution. [sent-1122, score-1.567]

65 γ2 cah cbh h The left-hand side can be factorized as κ2 − LMσ4 γ−2 γ2 − κ + h h κ2 − LMσ4 γ2 − κ − h > 0, (148) where κ= σ4 (L + M)σ2 + 2 2 . [sent-1140, score-0.762]

66 The partial derivatives of Equation (73) are given by VB µa 1 ∂Lh = 2h + 2 ∂µah cah −γh µbh + (µ2h + Lσ2h )µah b b VB 1 ∂Lh µb = 2h + 2 ∂µbh cbh −γh µah + (µ2h + Mσ2h )µbh a a σ2 σ2 , . [sent-1145, score-0.753]

67 Let us lower-bound Equation (94) by ignoring the positive h term 4σ4 LM/(c2h c2h ): a b LM −(M − L)2 (γh − γh ) + (L + M) (M − L)2 (γh − γh )2 + 4σ c2 c2 4 q1 (γh ) = > ah bh 2LM (M − L)2 (γh − γh )2 −(M − L)2 (γh − γh ) + (L + M) 2LM = 1− L (γh − γh ). [sent-1165, score-0.998]

68 Let us upper-bound Equation (94) by using the relation h x2 + y2 ≤ x2 + y2 + 2xy ≤ x + y for x, y ≥ 0: LM −(M − L)2 (γh − γh ) + (L + M) (M − L)2 (γh − γh )2 + 4σ c2 c2 4 q1 (γh ) = ≤ ah bh 2LM −(M − L)2 (γh − γh ) + (L + M) 2LM √ LM L (γh − γh ) + = 1− M 2LMcah cbh σ2 (L + M) L . [sent-1171, score-1.357]

69 (γh − γh ) + √ = 1− M LMcah cbh 2σ2 (L + M) 2630 √ LM c h bh 2 (M − L)(γh − γh ) + 2σa c T HEORETICAL A NALYSIS OF BAYESIAN M ATRIX FACTORIZATION We also upper-bound Equation (95) by adding a non-negative term (M − L)σ2 Lcah cbh √ σ2 LM 1 + cah cbh γh . [sent-1172, score-1.912]

70 Then we can obtain a lower-bound of γh : γh ≥ γlo , h where γlo is the larger solution of the following equation: h 2 L(γlo ) + (M − L)γh + h + σ2 (L + M) M/L cah cbh γlo h σ4 M(M − L) M/L σ2 L M 2 σ4 −M 1− 2 + γh cah cbh Lc2h c2h γh a b 1− σ2 M 2 γh = 0. [sent-1173, score-1.519]

71 γ2 h This can be factorized as γlo − 1 − h σ2 M/L σ2 M γh + cah cbh γ2 h Lγlo + M 1 − h σ2 M M/L σ2 L γh + cah cbh γ2 h Thus, the larger solution of this equation, γlo = 1 − h σ2 M/L σ2 M , γh − cah cbh γ2 h gives the lower-bound in Equation (28). [sent-1174, score-2.281]

72 The coefﬁcient of the second term of Equation (146), σ2 cah cbh cah cbh δh + cah cb δh , h is minimized when δh = cah . [sent-1175, score-2.327]

73 cbh Then we can obtain another upper-bound of γh : ′up γh ≤ γh , ′up where γh is the larger solution of the following equation: ′up (γh )2 + 2σ2 cah cbh ′up γh + σ2 L σ4 − 1− 2 c2h c2h γh a b 2631 1− σ2 M 2 γh = 0. [sent-1176, score-1.137]

74 NAKAJIMA AND S UGIYAMA This can be factorized as ′up γh − 1− σ2 L γ2 h 1− σ2 M σ2 γh + cah cbh γ2 h ′up × γh + 1− σ2 L γ2 h 1− σ2 M σ2 γh + cah cbh γ2 h = 0. [sent-1178, score-1.515]

75 Thus, the larger solution of this equation, ′up γh = 1− σ2 L γ2 h 1− σ2 M σ2 , γh − cah cbh γ2 h gives the upper-bound in Equation (31). [sent-1179, score-0.766]

76 σ2 (154) where λa,k (cah cbh ) = 1 2M(cah cbh )k − + λb,k (cah cbh ) = 1 2L(cah cbh )k σ2 − cah cbh (M − L) cah cbh σ2 − cah cbh (M − L) cah cbh 2 + 4Mσ2 , σ2 + cah cbh (M − L) cah cbh − + σ2 + cah cbh (M − L) cah cbh 2 + 4Lσ2 . [sent-1185, score-7.508]

77 Note that λa,k > 0, λb,k > 0 for any k, and that Equation (154) depends on c2h and c2h only through a b their product cah cbh . [sent-1186, score-0.753]

78 2M 2L T HEORETICAL A NALYSIS OF BAYESIAN M ATRIX FACTORIZATION Since they are increasing with respect to x, λa,1 and λb,1 are decreasing with respect to cah cbh . [sent-1189, score-0.753]

79 Further, λa,1 and λb,1 are upper-bounded as λa,1 (cah cbh ) < λb,1 (cah cbh ) < lim λa,1 (cah cbh ) = lim λ′ (x) = 1, a,1 lim λb,1 (cah cbh ) = lim λ′ (x) = 1. [sent-1190, score-1.524]

80 (156) cah cbh →+0 Similarly, using the same decreasing mapping, we have λ′ (x) · λ′ (x) = a,0 b,0 σ2 (x + (L + M)) − 2LM (x + (L + M))2 − 4LM . [sent-1192, score-0.753]

81 Since this is decreasing with respect to x and lower-bounded by zero, λa,0 λb,0 is increasing with respect to cah cbh and lower-bounded as λa,0 (cah cbh ) · λb,0 (cah cbh ) > lim cah cbh →+0 λa,0 (cah cbh ) · λb,0 (cah cbh ) = lim λ′ (x) · λ′ (x) = 0. [sent-1193, score-3.01]

82 a,0 b,0 x→∞ Therefore, the third term in Equation (154) is increasing with respect to cah cbh , and lower-bounded as LMλa,0 λb,0 LMλa,0 λb,0 > lim = 0. [sent-1194, score-0.763]

83 2 cah cbh →+0 σ σ2 (157) Now we have found that Equation (154) is increasing with respect to cah cbh , because it consists ˚ of the increasing terms. [sent-1195, score-1.506]

84 Equations (114) and (115) minimize cah cbh over Rε when Equation (43) is adopted. [sent-1196, score-0.753]

85 σ4 h Using this bound, we have 1− σ2 L γ2 h 1− σ2 M σ2 γh − cah cbh ´ ´ γ2 h γ2 − (L + M)σ2 + h > LMσ4 − γ2 h γ2 − (L + M)σ2 h > 0. [sent-1232, score-0.753]

86 Substituting Equations (76), (77), (119), and (120) into (164), we have 1 EVB H = 2  1 σ2 a  γ −2µ h µ  h ah bh  σ2  µa   − c4h a  γh −2µah µbh σ2 1 σ2 b h 0 h 0 µb − c4h bh 2636 µa − c4h ah 0 M 2c4 a h 0  0  µb  − c4h  bh  . [sent-1244, score-2.389]

87 σ2 (166) Similarly from Equation (75), we obtain µ2h b = σ2 h b By using Equations (78), (84), (159), (165), and (166), we obtain 1 1 EVB H = 4 4 2 cah cbh Since γ2 γh γh h − 2 4 c4 cah bh 2σ M δ−2 Lδ2 + 4 cah c4h b + LM γh γh − γ2 h σ4 . [sent-1248, score-1.934]

88 (167) √ M δ−2 Lδ2 2 LM + 4 ≥ 2 2 cah c4h cah cbh b for any δ2 > 0, Equation (167) is upper-bounded by √ γ2 1 γh γh LM LM 1 EVB h ≤ 4 4 − 2 2 2 + 4 γh γh − γ2 H h 4 c4 2 σ cah cbh cah bh σ cah cbh √ √ γh LM LM 1 1 = 4 4 − 2 + 2 2 c2 2 c2 σ σ cah cbh cah bh cah bh √ LM γh − 2 γh . [sent-1249, score-5.791]

89 Therefore, Equation (168) is written as √ √ 1 LM LM 1 ´ EVB ≤C H γh − 2 γh , + 2 2 c2 2 σ σ cah ´bh ´ 2637 (168) NAKAJIMA AND S UGIYAMA with a positive factor γh C= 4 4 cah cbh ´ ´ √ LM 1 − 2 2 c2 σ cah ´bh ´ . [sent-1251, score-1.529]

90 cah cbh ´ ´ 1− Mσ2 γ2 h γh At the last inequality, we neglected the negative last term in the curly braces. [sent-1253, score-0.753]

91 Using Equation (163), we have 1 ´ EVB < −C′ ( f (γh ) − g(γh )), H 2 (169) where γ2C h , 2σ2 cah cbh ´ ´ √ √ ( M − L)2 σ2 f (γh ) = 1 − γ2 h C′ = g(γh ) = 2 1− Lσ2 γ2 h × 1− 1− + (L + M)σ2 1− γ2 h 2 − 4LMσ4 , γ4 h Mσ2 γ2 h (L + M)σ2 γ2 h 2638 + 1− (L + M)σ2 γ2 h 2 − 4LMσ4 . [sent-1254, score-0.753]

92 2 Lσ L cah cbh ˘ ˘ σ σ2 x2 − y2 > x − y for x > y > 0, Equation (122) yields c2h c2h ˘a ˘b √ γ2 − (L + M + LM)σ2 h . [sent-1267, score-0.753]

93 a b Since the lower-bound in Equation (28) is nondecreasing with respect to cah cbh , substituting Equation (173) into Equation (28) yields   σ2 M γh − γh ≥ max 0, 1 − 2  γh It holds that − σ2 M >− γh σ2 M . [sent-1282, score-0.766]

94 Let us consider a partially minimized objective function: ˜ EVB Lh (cah cbh ) = min (µah ,µbh ,σ2 ,σ2 ) a b h EVB Lh (µah , µbh , σ2h , σ2h , cah cbh , cah cbh ). [sent-1289, score-1.885]

95 cah cbh →0 (176) Figure 13 depicts the partially minimized objective function (175) when L = M = H = 1, σ2 = 1, and V = 1. [sent-1291, score-0.761]

96 In this case, the objective function (175) has a stationary point at cah cbh = 1. [sent-1305, score-0.826]

97 In this case, there exists a large positive stationary point (which is a local minimum) at cah cbh ≈ 1. [sent-1310, score-0.83]

98 37, as well as a small positive stationary point (which is a local maximum) at cah cbh ≈ 0. [sent-1311, score-0.83]

99 The convergence Lh (cah cbh ) → L + M (= 2) as cah cbh → 0 is observed (see Equation (176)). [sent-1329, score-1.124]

100 As Lemma 28 states, a large positive stationary point at cah cbh ≈ 2. [sent-1334, score-0.83]

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