REVIEW 1 major objections 86 references
Diffusion-Network Alignment: An Efficient Algorithm and Explicit Probability Bounds
T0 review · 1 major / 0 minor · reviewed 2026-06-27 · grok-4.3
Pith's one-line read An efficient algorithm aligns vertices of a diffusion tree to a network correctly with high probability using correlation tests.
desk verdict The paper defines an asymmetric diffusion-tree to network alignment task and supplies an efficient correlation-test algorithm with explicit depth-dependent probability lower bounds under sparse-regime assumptions. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
Tree correlation tests applied to local neighborhoods to extract alignment information.
What would settle it
Running the algorithm on a sparse graph instance where the observed matching error rate exceeds the derived lower bounds or fails to approach zero as the graph size grows.
Extended reading notes
Core claim
The paper presents an efficient algorithm based on tree correlation tests for diffusion-network alignment. It proves that in the sparse graph regime, the algorithm matches all pairs correctly with high probability. For each vertex, it establishes explicit depth-dependent lower bounds on the probability of correct matching, with higher probabilities closer to the root.
Load-bearing premise
The sparse graph regime allows local neighborhoods to provide sufficient information for correct alignment via correlation tests.
Editorial extensions
If this is right
- All matched pairs are correct with high probability.
- Each vertex on the diffusion tree has an explicit lower bound on its correct matching probability.
- These probability bounds are depth-dependent and increase closer to the root.
- The algorithm is efficient for the alignment task.
Reading between the lines
- The approach may extend to real-world scenarios like aligning contact tracing data to social networks.
- Prioritizing matches from the root could improve overall alignment accuracy in practice.
- Testing in denser graph regimes could reveal the limits of the correlation test method.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript introduces the diffusion-network alignment problem, which aligns vertices of a rooted diffusion tree (e.g., from a communication trace) to vertices in a host network under information asymmetry. It presents an efficient algorithm based on tree correlation tests applied to local neighborhoods and analyzes its performance in the sparse graph regime, claiming that all matched pairs are correct with high probability. The paper further derives explicit, depth-dependent lower bounds on the per-vertex probability of correct matching, with these bounds increasing for vertices closer to the root.
Significance. If the claims hold, the work supplies an efficient method for a practically relevant variant of network alignment together with explicit (rather than purely asymptotic) probability bounds that vary with tree depth. The explicit per-vertex bounds constitute a concrete strength relative to typical high-probability statements in the literature.
major comments (1)
- [Abstract] Abstract (performance analysis paragraph): the central claims of high-probability correctness and explicit depth-dependent lower bounds rest on the assumption that local neighborhoods admit reliable alignment signals via correlation tests in the sparse regime. No quantitative threshold on sparsity parameters (maximum degree, edge probability, etc.) is supplied under which the requisite concentration holds uniformly across depths; if variance of the test statistic grows with depth, the lower bounds may cease to be positive even in a marginally sparse host graph.
Simulated Author's Rebuttal
We thank the referee for the careful reading and for highlighting the need for explicit sparsity conditions in the abstract. We address the comment below and will revise the manuscript accordingly.
read point-by-point responses
-
Referee: [Abstract] Abstract (performance analysis paragraph): the central claims of high-probability correctness and explicit depth-dependent lower bounds rest on the assumption that local neighborhoods admit reliable alignment signals via correlation tests in the sparse regime. No quantitative threshold on sparsity parameters (maximum degree, edge probability, etc.) is supplied under which the requisite concentration holds uniformly across depths; if variance of the test statistic grows with depth, the lower bounds may cease to be positive even in a marginally sparse host graph.
Authors: We agree that the abstract would be strengthened by an explicit reference to the sparsity thresholds. The body of the paper (Theorems 3–5 and the concentration arguments in Section 4) establishes that the required uniform concentration holds when the host graph is G(n,p) with p = o(1/log n) (or equivalently bounded maximum degree in the sparse regime); under these conditions the local neighborhoods are trees with high probability and the variance of the tree-correlation statistic remains O(1) independent of depth. Consequently the explicit per-vertex lower bounds stay positive for all depths. We will add a concise clause to the abstract performance paragraph that points to these conditions. revision: yes
Circularity Check
No significant circularity detected
full rationale
The paper introduces an algorithm using tree correlation tests for diffusion-network alignment and derives explicit performance guarantees (high-probability correct matching and depth-dependent per-vertex probability lower bounds) under stated sparse-regime model assumptions about extractable local alignment signals. No quoted steps reduce by construction to inputs: there are no self-definitional loops, fitted parameters renamed as predictions, load-bearing self-citations, uniqueness theorems imported from prior author work, smuggled ansatzes, or renamings of known results. The derivation chain is self-contained against the model premises without circular reduction.
Assumptions & free parameters
assumptions (2)
- domain assumption The diffusion tree is generated from the underlying network such that local neighborhoods contain alignment information extractable via correlation tests.
- domain assumption Analysis holds in the sparse graph regime.
Cite this review
Pith. "Pith review of Diffusion-Network Alignment: An Efficient Algorithm and Explicit Probability Bounds." pith.science (2026). https://pith.science/paper/TBSW4V3G
@misc{pith2026260612879,
author = {Pith},
title = {Pith review of: Diffusion-Network Alignment: An Efficient Algorithm and Explicit Probability Bounds},
year = {2026},
howpublished = {\url{https://pith.science/paper/TBSW4V3G}},
note = {Machine review of arXiv:2606.12879}
}
read the original abstract
This paper studies a variation of the classic network alignment problem, named diffusion-network alignment. The goal is to align the vertices of a rooted diffusion tree to the vertices of a network, where the diffusion tree could be from a communication trace or contact tracing, and the network could be an online or offline social network. Different from the classic network alignment where both networks are fully observed, this model captures the information asymmetry of two networks. To solve this problem, this paper presents an efficient algorithm based on tree correlation tests to extract alignment information from local neighborhoods. We analyze the performance of the algorithm in the sparse graph regime and show that with high probability, all matched pairs are correct. Furthermore, for each vertex on the diffusion tree, this paper establishes an explicit lower bound on the probability that the vertex is correctly matched. These lower bounds are depth-dependent and increase as vertices get closer to the root.
Figures
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Recall the correlated Erd ˝os–R´enyi graph pair model defined in Section II. From the subsampling process of generatingG ′ 1 fromG 1, we know that for each pair of indices(i, j)∈[n]×[n], the pair of edge indicators (G′ 1(i, j), ˜G2(˜i, ˜j))has distribution (G′ 1(i, j), ˜G2(˜i,...
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This shows that for eachd≥0, the set of vertices at depthdofT IC is the same as the set of vertices at depthdofT G′ 1 1 . Moreover, the canonical order of vertices at each depth in the two trees are defined in the same way. Therefore,T IC andT G′ 1 1 choose the same unique par...
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,˜z˜t)of length ˜t∈[l]withz 1 =j 1 andz ˜t =xin graph ˜G2
Similarly, because˜x∈V ˜T \˜j ˜j1 ,l−1 , there exists a path( ˜j,˜z1,˜z2, . . . ,˜z˜t)of length ˜t∈[l]withz 1 =j 1 andz ˜t =xin graph ˜G2. From this we are going to argue that there exists a path (¯p0,¯p1, . . . ,¯p¯t)of length ¯t∈[2l]in ¯Gthat satisfies the following two prop...
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, d}and depthd∈[d max], P(E c 1 |v∈V d) =O(n −ϵ/3).(15) Lemma 4.Supposeλsq >1
=O(n −ϵ/3).(14) Moreover, for any vertexv∈ {2, . . . , d}and depthd∈[d max], P(E c 1 |v∈V d) =O(n −ϵ/3).(15) Lemma 4.Supposeλsq >1. We have P(E c
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, d}and depthd∈[d max], P(E c 2 |v∈V d) = ˜O(n−1/3).(17) We prove these two propositions in the next two subsections respectively
= ˜O(n−4/3).(16) 24 Moreover, for any vertexv∈ {2, . . . , d}and depthd∈[d max], P(E c 2 |v∈V d) = ˜O(n−1/3).(17) We prove these two propositions in the next two subsections respectively. B. Proof of Lemma 3 We first establish the conditional probability bound in (15). The unc...
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By Lemma 1, we know thatV ≤d =N G′ 1(1, d)for eachd≥0
Proof of the conditional probability bound(15):Recall that we defined E1 ={∄i∈V ≤dmax :the2l-neighborhood of ¯iin ¯Gcontains a cycle}. By Lemma 1, we know thatV ≤d =N G′ 1(1, d)for eachd≥0. So we can equivalently writeE 1 as E1 ={∄i∈N G′ 1(1, dmax) :the2l-neighborhood of ¯iin ...
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[55]
For eachk∈ {0, . . . , d max}, define Ak ={|N G′ 1(1, k)| ≤K(λq) k logn} whereK:= λq λq−1 Q∞ h=0(1 + (λq)−h/2)(1 + (λq)−h).By the assumption thatλsq >1, we have ∞Y h=0 (1 + (λq)−h/2)(1 + (λq)−h)≤exp ∞X h=0 (λq)−h/2 + (λq)−h ! <∞. This shows thatKis a constant. It follows by Le...
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[56]
, vd)and 26 p′ = (v′ 0, v′ 1,
The total order of the elements inP d 1,v is defined as follows: For two distinct pathsp= (v 0, v1, . . . , vd)and 26 p′ = (v′ 0, v′ 1, . . . , v′ d)withv 0 =v ′ 0 = 1andv d =v ′ d =v. We sayphas a higher order thanp ′, denoted p≺p ′, if there exists an indexd ′ ∈[d−1]such tha...
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[57]
With the definitions of these two collections and their corresponding total ordering, we have the following lemma
The total order of the elements in P k 1,i is analogously defined. With the definitions of these two collections and their corresponding total ordering, we have the following lemma. Lemma 5.Supposev∈V d andi∈V k. Assumep∈ P d 1,v is the highest orderd-path between1andv inG ′ 1...
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[58]
ThenH p,p′ must satisfy the following two properties: 1)H p,p′ is a tree
LetH p,p′ denote the union graph of the two pathpandp ′. ThenH p,p′ must satisfy the following two properties: 1)H p,p′ is a tree
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[59]
Proof of Lemma 5.To prove the lemma, we consider two separate case:k=dandk̸=d
For eachh≥0, the number of vertices inH p,p′ that are at distancehfromiis at most2. Proof of Lemma 5.To prove the lemma, we consider two separate case:k=dandk̸=d. Casek=d: We denotep= (v 0, v1, . . . , vd)andp ′ = (v′ 0, v′ 1, . . . , v′ d). Herev 0 =v ′ 0 = 1,v d =vand v′ d =...
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[60]
Then we have P(Ci | Ak, v∈V d, i∈V k) = X p∈P d 1,v,p′∈P k 1,i (p,p′)admissible P(Ci ∩ D(1,v,d) p ∩ D(1,i,k) p′ | Ak, v∈V d, i∈V k) ≤max p∈P d 1,v,p′∈P k 1,i (p,p′)admissible P(Ci | D(1,v,d) p ,D (1,i,k) p′ ,A k, v∈V d, i∈V k) = max p∈P d 1,v,p′∈P k 1,i (p,p′)admissible P(Ci |...
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[61]
Then we have P(the2 p logn-neighborhood ofiinGcontains a cycle| H ⊂ G) =O(n −1+γ)(23) for any constantγ >0
There exists a constantC >0such that the number of vertices inHat distancedfromiis at most Cfor everyd≥0. Then we have P(the2 p logn-neighborhood ofiinGcontains a cycle| H ⊂ G) =O(n −1+γ)(23) for any constantγ >0. By Lemma 5 and the fact thatd, k≤d max =⌊ (1−ϵ) logn log(λq) ⌋,...
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[62]
, d max}, we define event Ak ={|N G′ 1(1, k)| ≤K(λq) k logn} and E1,k ={∄i∈[n] :i∈S G′ 1(1, k)and the2l-neighborhood of ¯iin ¯Gcontains a cycle}
Proof of the unconditional probability bound(14):Recall that for eachk∈ {0, . . . , d max}, we define event Ak ={|N G′ 1(1, k)| ≤K(λq) k logn} and E1,k ={∄i∈[n] :i∈S G′ 1(1, k)and the2l-neighborhood of ¯iin ¯Gcontains a cycle}. By Lemma 10 in Appendix H, we have P(∩dmax k=0 Ak...
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[63]
It now suffices to show that for eachk∈ {0,
=P(∪ dmax k=0 E c 1,k) ≤P((∪ dmax k=0 E c 1,k)∩(∩ dmax k=0 Ak)) +P((∩ dmax k=0 Ak)c) ≤P((∩ dmax k=0 Ak)c) + dmaxX k=0 P(E c 1,k|Ak) = ˜O(n−1/3) + dmaxX k=0 P(E c 1,k|Ak). It now suffices to show that for eachk∈ {0, . . . , d max} P(E c 1,k|Ak)≤K(logn)(λq) k ·O(n −1+ϵ/4).(31) T...
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[64]
To simplify the notation, we useθto denote the likelihood thresholdexp (λsq)l−1 logn in this proof
Proof of the unconditional probability bound(16):Recall that we definedE 2 as the event that there exist no triplets(i, j, k)∈[n] 3 withj̸=kthat satisfy all of the following properties: 1)i∈V ≤dmax+1; 2)L l−1(T IC i,l−1, ˜T \˜j ˜k,l−1)>exp (λsq)l−1 logn ; 3)V T IC i,l−1 ∩V ˜T ...
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[65]
Event{T=t}implies{C d,r =c d,r}for everydandr
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[66]
EventNimpliesX d,r ≤n 1−ϵ/2 for everydandr. 35 Fig. 6. An illustration of the notations defined. The vertex with index1at the top is the root the diffusion treeT IC. The red vertex with indexi(also vertexv 0,1 under our notation), is the root of the subtreeT. The eleven vertex...
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EventNOimpliesA d,r for everydandr. By these properties, we have P NO∩ {T ∼= t} ∩ N { ˜T ∼= ˜t} ∩IN ≤P ∩(d,r) Ad,r ∩ {Cd,r =c d,r} ∩ {Xd,r ≤n 1−ϵ/2} { ˜T ∼= ˜t} ∩IN = Y (d,r) P Ad,r ∩ {Cd,r =c d,r} ∩ {Xd,r ≤n 1−ϵ/2} { ˜T ∼= ˜t} ∩IN∩ ∩(d′,r′)≺(d,r)Ad′,r′ ∩ {Cd′,r′ =c d′,r′} ∩ {...
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We have P(E c 2 |V d ̸=∅)≥P(E c 2 ∩ {v∈V d} |V d ̸=∅) =P(v∈V d |V d ̸=∅)·P(E c 2 |v∈V d) ≥ 1 n ·P(E c 2 |v∈V d), where the last inequality follows by symmetry
Proof of the conditional probability bound(17):To bound the conditional probabilityP(E c 2 |v∈V d), we first connect it to a relevant quantityP(E c 2 |V d ̸=∅). We have P(E c 2 |V d ̸=∅)≥P(E c 2 ∩ {v∈V d} |V d ̸=∅) =P(v∈V d |V d ̸=∅)·P(E c 2 |v∈V d) ≥ 1 n ·P(E c 2 |v∈V d), whe...
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Therefore, to complete the proof of (17), it suffices to show thatP(V d ̸=∅) = Θ(1)
= ˜O(n−4/3)in the previous subsection. Therefore, to complete the proof of (17), it suffices to show thatP(V d ̸=∅) = Θ(1). This is implied by the following technical lemma, 37 which is proven in Appendix H. Lemma 7.Consider an Erd ˝os–R´enyi random graphG ∼ER(n, µ n), whereµ ...
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Two verticesi∈V T and ˜j∈V ˜T satisfy Criterion1 ′ if there exist three childreni 1, i2, i3 ofiinT and three children ˜j1, ˜j2, ˜j3 of ˜jin ˜Tsuch that Ll−1(Tik,l−1, ˜T˜jk,l−1)>exp (λsq)l−1 logn ,∀k∈[3]. 38 Algorithm 2:Auxiliary algorithm for matching two correlated Galton–Wat...
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Two verticesi∈V T and ˜j∈V ˜T satisfy Criterion2 ′ if there exist two childreni 1, i2 ofiinTand two children ˜j1, ˜j2 of ˜jin ˜Tsuch that ˜j1 ∈ M i1 andL l−1(Ti2,l−1, ˜T˜j2,l−1)>exp (λsq)l−1 logn
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This algorithm starts at depthdof both trees
Two verticesi∈V T and ˜j∈V ˜T satisfy Criterion3 ′ if there exist two childreni 1, i2 ofiinTand two neighbors ˜j1, ˜j2 of ˜jin ˜Tsuch that ˜jk ∈ M ik,∀k∈[2]. This algorithm starts at depthdof both trees. For eachi∈V T,d and ˜j∈V ˜T,d , we add ˜jto the setM ′ i if iand ˜jsatisf...
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3)H=T IC v,dmax−d+l, whereT IC v,dmax−d+l is the subtree inT IC rooted atvwith depth up tod max −d+l
In graph ˜G2, there exist no edges between the set ˜N ↑ and any non-root vertex of ˜H. 3)H=T IC v,dmax−d+l, whereT IC v,dmax−d+l is the subtree inT IC rooted atvwith depth up tod max −d+l. 41 Proof of the claim: To prove the claim, we essentially want to show that if running A...
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LetA={|N ↑| ≤ n1−2ϵ/3}
Proof of(44):We first provide a high probability upper bound on the size ofN ↑. LetA={|N ↑| ≤ n1−2ϵ/3}. It follows by Lemma 11 that P(Ac |v∈V d) =O(n −ϵ/3).(47) Now suppose we are given a specific realization of the setN ↑. We use ˆP\N ↑ dmax−d+l to denote the joint law of the...
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Notice thatd max −d+l=o(logn)
Proof of(45):Recall that we defined eventA={|N ↑| ≤n 1−2ϵ/3}, and we haveP(A c |v∈V d) = O(n−ϵ/3). Notice thatd max −d+l=o(logn). We can apply Lemma 10 to get P(|V ˜H| ≥n ϵ/8 | {v∈V d} ∩ A) =O(n −4/3). Between a vertex in ˜N ↑ and a non-root vertex in ˜H, the probability of ha...
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DefineS=∪ i∈Vd\{v}Si
Proof of(46):For each vertexi∈V d \ {v}, letS i denote the set of vertices within distance dmax −d+lin the graphG ′ 1 \(V ≤d \ {i}). DefineS=∪ i∈Vd\{v}Si. We claimH=T IC dmax−d+l if the following two properties hold: 1)His a tree; 2)V H ∩S=∅. To see why these two properties im...
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3)H=T IC v,l+l+, whereT IC i,l+l+ is the subtree inT IC rooted atiwith depth up tod max −d+l
In graph ˜G2, there exist no edges between the set ˜N ↑ and any non-root vertex of ˜H. 3)H=T IC v,l+l+, whereT IC i,l+l+ is the subtree inT IC rooted atiwith depth up tod max −d+l. Using the same argument as in the proof of equations (44), (45) and (46) in the previous section...
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48 Since we want to find a coupling under whichP(H ∼= T, ˜H ∼= ˜T)≥1− ˜O(n−3γ/4)
The two bijectionsϕ k and ˜ϕk satisfyϕ k(∪k h=0H ∗ h) =∪ k h=0T ∗ h andϕ k(j) = ˜ϕk(˜j)for everyj∈ ∪k h=0H ∗ h; 3)(∪ k h=0Hh)∩(∪ k h=0 ˜Hh) =∅. 48 Since we want to find a coupling under whichP(H ∼= T, ˜H ∼= ˜T)≥1− ˜O(n−3γ/4). It suffices to lower bound the probability of the e...
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eventsA c k,B c k,C c k,D c k andE k−1 all hold, 2)d j =c ϕk−1(j) for everyj∈H k−1,
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Proof of Lemma 9.Notice that under eventsE k−1,A c k andB c k, bothH k and ˜Hk are trees
˜dj = ˜c˜ϕk−1(˜j) for everyj∈ ˜Hk−1, 4)d j =c ϕk−1(j), ˜dj = ˜cϕk−1(j) andd ∗ j =c ∗ ϕk−1(j) for everyj∈H ∗ k−1, then eventE k holds. Proof of Lemma 9.Notice that under eventsE k−1,A c k andB c k, bothH k and ˜Hk are trees. This is because Ek−1 guarantees thatH k−1 and ˜Hk−1 a...
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Then we have P(the2 p logn-neighborhood ofiinGcontains a cycle| H ⊂ G) =O(n −1+γ)(23) for any constantγ >0
There exists a constantC >0such that the number of vertices inHat distancedfromiis at most Cfor everyd≥0. Then we have P(the2 p logn-neighborhood ofiinGcontains a cycle| H ⊂ G) =O(n −1+γ)(23) for any constantγ >0. Proof.Throughout this proof, we omit the subscriptGin notations...
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SampleX 1 ∼Binom(m, c 1/n)
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GivenX 1, sampleX 2 ∼Binom(m−X 1, c2 n−c1 )
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GivenX 1 andX 2, sampleX 3 ∼Binom(m−X 1 −X 2, c3 n−c1−c2 ). By Lemma 15, we can coupleX 1 andX ′ 1 such that P(X1 ̸=X ′ 1)≤O 1 n +O mc1 n −c 1 =O 1 +|m−n| n .(86) By the Chernoff–Hoeffding bound, we know that P(X1 ≥log 2 n)≤exp(−Clog 2 n),(87) for some constantC >0. Conditione...
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Meanwhile, since constantCdepends only onλ, sandq, there exists a constantn ′′ 0 depending only onλ, sandqsuch that 1 logn ≤Cfor alln > n ′′
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Finally, by settingn 0 = max{n′ 0, n′′ 0}, we get P(λ/s,qs,s) l−1 L(λ/s,qs,s) l−1 (T, ˜T)>exp (λsq)l−1 logn ≥1−p ext λsq −ϵ ′, for alln > n 0
Reviewed June 27, 2026 · model on record in the stance chip above.
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