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Detailed Proofs of Lemmas, Theorems, and Corollaries

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Detailed Proofs of Lemmas, Theorems, and Corollaries Detailed Proofs of Lemmas, Theorems, and Corollaries Dahua Lin John Fisher CSAIL, MIT CSAIL, MIT A List of Lemmas, Theorems, and Theorem 4. The hierarchically bridging Markov Corollaries chain with bk < 1 for k = 0;:::;K − 1, and fk < 1 for k = 1;:::;K is ergodic. If we write the equilibrium For being self-contained, we list here all the lemmas, distribution in form of (αµ0; β1µ1; : : : ; βK µK ), then theorems, and corollaries in the main paper. (S1) µ0 equals the target distribution µ; (S2) for each k ≥ 1, and y 2 Y , µ (y) is proportional to the total Lemma 1. The joint transition matrix P given by k k + probability of its descendant target states (the target Eq.(1) in the main paper has a stationary distribution states derived by filling all its placeholders); (S3) α, in form of (αµ ; βµ ), if and only if X Y the probability of being at the target level, is given by α−1 = 1 + PK (b ··· b )=(f ··· f ). µX QB = µY ; and µY QF = µX : (1) k=1 0 k−1 1 k Corollary 2. If bk=fk+1 ≤ κ < 1 for each k = Under this condition, we have αb = βf. Further, if 1;:::;K, then α > 1 − κ. both PX and PY are both reversible, then P+ is also reversible, if and only if B Proofs µX (x)QB(x; y) = µY (y)QF (y; x); (2) Here, we provide the proofs of the lemmas and theo- for all x 2 X and y 2 Y . rems presented in the paper. Lemma 2. If the condition given above holds, then µX and µY are respectively stationary distributions of B.1 Proof of Lemma 1 QBF and QFB. Moreover, if P+ is reversible, then both QBF and QFB are reversible. Recall that P+ is given by Theorem 1. Given a reversible Markov chain with (1 − b)PX bQB transition matrix P, and " 2 (0; 1=2), then P+ = : fQF (1 − f)PY log(1=(2"))(τ − 1) ≤ tmix(") ≤ log(1=("µmin))τ: (3) Suppose P+ has a stationary distribution in form of (αµX ; βµY ), then Here, τ is called the relaxation time, given by 1/γ∗(P). Theorem 2. Let λ2 be the second largest eigenvalue α(1 − b)µX PX + βfµY QF = αµX ; of a reversible transition matrix P, then αbµX QB + β(1 − f)µY PY = βµY : (6) 2 Φ∗(P)=2 ≤ 1 − λ2 ≤ 2Φ∗(P): (4) Since µX and µY are respectively stationary distribu- tions of PX and PY , i.e. µX PX = µX and µY PY = Theorem 3. The reversible transition matrix P+ as µ , we have given by Eq.(1) in the main paper has Y βfµ QF = αbµ ; and αbµ QB = βfµ : (7) η(b; f) φ Y X X Y · ≤ Φ∗(P+) ≤ maxfb; fg: (5) 2 φ + 1 Note that QB and QF were defined with the condition that each of their rows sums to 1, i.e. QF 1jXj = 1jY j Here, η(b; f) = 2αb = 2βf is the total probability of and QB1jY j = 1jXj. Multiplying 1 to the right of both cross-space transition, φ = minfΦ∗(QBF ); Φ∗(QFB)g. hand sides of either equation results in αb = βf. It Corollary 1. The joint chain P+ is ergodic when the immediately follows that collapsed chains (QBF and QFB) are both ergodic. µ QB = µ ; and µ QF = µ : (8) Lemma 3. Let P be a reversible transition matrix X Y Y X over X, such that P(x; x) ≥ ξ > 0 for each x 2 X For the other direction, we assume QB and QF satisfy then its smallest eigenvalue λn has λn ≥ 2ξ − 1. the conditions above, and αb = βf. Plugging these Detailed Proofs of Lemmas, Theorems, and Corollaries conditions into the left hand sides of the equations B.3 Proof of Theorem 3 in Eq.(6) results in (αµX ; βµY )P+ = (αµX ; βµY ), To prove theorem 3, we first establish a lemma on flow which implies that (αµX ; βµY ) is a stationary distri- bution of P+. The proof of the first part is done. decomposition, and then accomplish the proof based on the lemma. Next, we show the second part of the lemma, which is about the reversibility. Let µ+ = (αµX ; βµY ). Under B.3.1 A Lemma on Flow Decomposition the condition that PX and PY are both reversible, we 0 have for each x; x 2 X, For the joint chain P+, we analyze its bottleneck ratio by decomposing the flows. Consider a partition of the 0 0 µ+(x)P+(x; x ) = α(1 − b)µX (x)PX (x; x ); union space X [ Y into two parts: A [ B (with A ⊂ X 0 0 0 0 c c c c µ+(x )P+(x ; x) = α(1 − b)µX (x )PX (x ; x); (9) and B ⊂ Y ) and A [ B (with A = X=A and B = Y=B). The flow between them comprises three parts: 0 0 0 and thus µ+(x)P+(x; x ) = µ+(x )P+(x ; x) (due c c c c to the reversibility of PX ). Likewise, we can get F(A; A ) + F(B; B ) + (F(A; B ) + F(B; A )): 0 0 0 µ+P+(y; y ) = µ+(y )P(y ; y). Hence, P+ is re- Here, F(A; Ac) is the flow within X, F(B; Bc) is the versible if and only if µ+(x)P+(x; y) = µ+(y)P+(y; x) for each x 2 X and y 2 Y . This can be expanded as flow within Y , and F(A; Bc) + F(B; Ac) is the flow between X and Y . The first two are inherited from αbµX (x)QB(x; y) = βfµY (y)QF (y; x); (10) the original Markov chains. We focus on the third one, which reflects the effect of bridging. For this part which holds if and only if µX (x)QB(x; y) = of flow, we derive the following lemma by decomposing µY (y)QF (y; x) (under the condition αb = βf). The it along multiple paths. proof is completed. Lemma 4. Given arbitrary partition of X [ Y into A [ B and Ac [ Bc as described above, we have B.2 Proof of Lemma 2 c c F(A; B ) + F(B; A ) ≥ αb · Φ∗(QBF )µX (A); (13) If (αµX ; βµY ) is a stationary distribution of P+, then Eq.(8) holds. Thus, c when µX (A) ≤ µX (A ), and µX QBQF = µY QF = µX ; (11) c c F(A; B ) + F(B; A ) ≥ βf · Φ∗(QFB)µY (B); (14) implying that µX is a stationary distribution of c when µY (B) ≤ µY (B ). QBQF . Similarly, we can show that µY is a stationary distribution of Q Q . F B Proof. To analyze the flow F(A; Bc) + F(B; Ac), we Furthermore, if P+ is reversible, according to further decompose it along multiple paths. Based on Lemma 1, we have µX (x)QB(x; y) = µY (y)QF (y; x) Eq.(1) in the main paper, we have for each x 2 X and y 2 Y . Then for any x; x0 2 X, c X X F(A; B ) = αbµX (x)QB(x; y) (15) 0 X 0 µX (x)QBF (x; x ) = µX (x) QB(x; y)QF (y; x ) x2A y2Bc y2Y X X X 0 = αb µX (x)QB(x; y) QF (y; x ) X 0 = µX (x)QB(x; y)QF (y; x ) x2A y2Bc x02X y2Y X 0 In this way, we decompose the flow into a sum of the = µY (y)QF (y; x)QF (y; x ) 0 terms in form of µX (x)QB(x; y)QF (y; x ), which we y2Y call the path weight along x ! y ! x0, denoted by 0 Similarly, we can get !(x; y; x ). We can then rewrite F(A; B) as 0 0 X 0 X X X 0 µX (x )QBF (x ; x) = µY (y)QF (y; x )QF (y; x): F(A; B) = αb !(x; y; x ): (16) y2Y x2A y2B x02X Hence, For conciseness, we use !(A; B; C) to denote the sum of paths traveling from A, via B, and ending up in C, 0 0 0 µX (x)QBF (x; x ) = µX (x )QBF (x ; x): (12) P P P 0 i.e. x2A y2B x02C !(x; y; x ). Then, we have This implies that Q is reversible. Likewise, we can BF F(A; Bc) = αb(!(A; Bc;A) + !(A; Bc;Ac)); (17) show that QFB is reversible under the condition that F(Ac;B) = αb(!(Ac; B; A) + !(Ac; B; Ac)): (18) P+ is reversible. The proof is completed. Dahua Lin, John Fisher As F is symmetric for a reversible chain, we have Note that η=2 = αb = βf and α + β = 1. Thus F(B; Ac) = F(Ac;B), and thus F αF + βF s ≥ s s ≥ ηφ/2: (25) c c c c c F(A; B ) + F(B; A ) ≥ αb(!(A; B ;A ) + !(A ; B; A)) µ+(A [ B) αµX (A) + βµY (B) = αb !(A; Y; Ac): (19) Case 2. µX (A) < 1=2 and µY (B) > 1=2. On the other hand, we note that Given arbitrary κ > 2, there are two possibilities: c X X X 0 !(A; Y; A ) = !(x; y; x ) Case 2.1. 1/κ ≤ µX (X) < 1=2 and µY (B) > 1=2. x2A y2Y x02Ac Then 1 X X X 0 F ≥ αb · φµ (A) > αbφ. (26) = µX (x)QB(x; y)QF (y; x ) s X κ x2A y2Y x02Ac Recall that µ+(A [ B) ≤ 1=2.
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