Fixed Confidence Best Arm Identification in the Bayesian Setting

Authors: kyoungseok  jang
junpei  komiyama
kazutoshi  yamazaki
Venue: NeurIPS_cc_2024_Conference
Published: 2024-05-13 01:41:33

Abstract: We consider the fixed-confidence best arm identification (FC-BAI) problem in the Bayesian setting. This problem aims to find the arm of the largest mean with a fixed confidence level when the bandit model has been sampled from the known prior. 
Most studies on the FC-BAI problem have been conducted in the frequentist setting, where the bandit model is predetermined before the game starts. 
We show that the traditional FC-BAI algorithms studied in the frequentist setting, such as track-and-stop and top-two algorithms, result in arbitrarily suboptimal performances in the Bayesian setting. 
We also obtain a lower bound of the expected number of samples in the Bayesian setting and introduce a variant of successive elimination that has a matching performance with the lower bound up to a logarithmic factor. Simulations verify the theoretical results.

Main paper: 
 introduction: 
 In many sequential decision-making problems, the learner repeatedly chooses an arm (option) to play with and observes a reward drawn from the unknown distribution of the corresponding arm. One of the most widely-studied instances of such problems is the multi-armed bandit problem [Thompson, 1933 , Robbins, 1952 , Lai, 1987] , where the goal is to maximize the sum of rewards during the rounds. Since the learner does not know the distribution of rewards, they need to explore the different arms, and yet, exploit the arms of the most rewarding arms so far. Different from the classical bandit formulation, there are situations where one is more interested in collecting information rather than maximizing intermediate rewards. The best arm identification (BAI) is a sequential decision-making problem in which the learner is only interested in identifying the arm with the highest mean reward. While the origin of this problem dates back to at least the 1950s [Bechhofer, 1954 , Paulson, 1964 , Gupta, 1977] , recent work in the field of machine learning reformulated the problem [Audibert et al., 2010] . In the BAI, the learner needs to pull arms efficiently for better identification. To achieve efficiency and accuracy, the learner should determine which arm to choose based on the history, when to stop the sampling, and which arm to recommend as the learner's final decision.There are two types of BAI problems depending on the optimization objective. In the fixed-budget (FB) setting [Audibert et al., 2010] , the learner attempts to minimize the probability of error (misidentification of the best arm) given a limited number of arm pulls T . In the fixed confidence (FC) setting [Jamieson and Nowak, 2014] , the learner attempts to minimize the number of arm pulls, subject to a predefined probability of error δ ∈ (0, 1). In this paper, we shall focus on the FC setting, which is useful when we desire a rigorous statistical guarantee. Most of the previous BAI studies focus on the frequentist setting, where the bandit model is chosen adversarially from some hypothesis class beforehand. In this setting, several algorithms, such as Track and Stop [Kaufmann et al., 2016a] and Top-two algorithms [Russo, 2016 , Qin et al., 2017b , Jourdan et al., 2022] , are widely known. These algorithms have an optimal sample complexity, meaning that they are one of the most sample-efficient algorithms among the class of δ-correct algorithms.The sample complexity of these algorithms is problem-dependent. To see this, consider the following example. Example 1. (A/B/C testing) Consider A/B/C testing of web designs. We have three arms (web designs) from which we would like to find the largest retention rate via allocating users to web designs i = 1, 2, 3. If we attempt to find the best arm with confidence δ, we may need a large number of samples (users) when the suboptimality gap (the gap between the retention rate of the best arm and the second best arm) is small because in such a case the identification of the best arm is difficult -the minimum number of samples required is inversely proportional to the square of the suboptimality gap [Kaufmann et al., 2014] . For example, when comparing the testing of retention rates of (0.9, 0.5, 0.1) with (0.9, 0.89, 0.1), the second case requires around 0.9-0.5 0.9-0.89 2 = 1600 times more samples compared to the first case.In practice, the retention rate of 0.89 in the second case may be acceptably good compared to the optimal retention rate of 0.9, and we may stop exploration at the moment the learner identifies a reasonably good arm, which is the first or the second arm in this example. This idea is formalized in several ways. The literature of Ranking and Selection (R&S) usually considers the indifference-zone formulation [Hong et al., 2021] . In the context of best arm identification, a similar notion of ϵ-best answer identification has also been considered [Maron and Moore, 1993 , Even-Dar et al., 2006 , Gabillon et al., 2012 , Kaufmann and Kalyanakrishnan, 2013 , Jourdan et al., 2023] . In these settings, the learner accepts a sub-optimal arm whose means are at most ϵ worse than the mean of the optimal arm. Other related settings include the good arm identification problem [Kano et al., 2019 , Tabata et al., 2020 , Zhao et al., 2023] , where the goal is to identify an arm that exceeds the predefined threshold, and the thresholding bandit problem [Locatelli et al., 2016 , Xu et al., 2019] , where the goal is to identify whether the mean of each arm is above or below the threshold. All these problem settings require an extra parameter, like ϵ or an acceptance threshold, that directly determines the acceptance level. Even though the algorithm's performance depends on this parameter, it is often challenging to determine a reasonable value for it in advance.In this paper, we study an alternative approach based on the Bayesian setting. In particular, we consider the prior distribution on the model parameters. We relax the requirement on the correctness of the best arm identification by using the prior belief. Rather than requiring the frequentist δcorrectness for any model, we require the learner to have marginalized correctness over the prior distribution, which we call Bayesian δ-correctness.We study the fixed confidence BAI (FC-BAI) problem in the Bayesian setting. Our contributions are as follows.• First, we find that in the Bayesian setting, the performance of the traditional frequentist setting-based algorithms, such as Track and Stop and Top-two algorithms, can be arbitrarily worse (Section 3). This is because frequentist approaches spend too many resources when the suboptimality gap is narrow. • Second, we prove that the lower bound of the number of expected samples should attain at least the order of Ω( L(H) 2 δ) as δ → 0 (Section 4). Here L(H) is our novel quantity that represents the sample complexity with respect to the prior distribution H. This order is different from the existing lower bound in the frequentist setting1 , implying that the Bayesian setting is essentially different from the frequentist setting.• Third, we design an algorithm whose expected sample size is upper-bounded by O( L(H) 2 δ log L(H) δ ) (Section 5). Our algorithm is based on the elimination algorithm [Maron and Moore, 1993 , Even-Dar et al., 2006 , Frazier, 2014] , but we add an early stopping criterion to prevent over-commitment of the algorithm for a bandit model with a narrow suboptimality gap. Our algorithm has a matching upper bound up to the logarithmic factor.We also conduct simulation to demonstrate that the sample complexity of frequentist algorithms does indeed diverge in a bandit model with a small suboptimality gap, even in very simple cases (Section 6).To our knowledge, BAI problems studied for the Bayesian setting have been limited to the fixed budget setting [Komiyama et al., 2023 , Atsidakou et al., 2023] . Komiyama et al. [2023] showed that, in the fixed-budget setting, a simple non-Bayesian algorithm has an optimal simple regret up to a constant factor, implying that the advantage the learner could get from the prior is small when the budget is large. This is very different from our fixed-confidence setting, where utilizing the prior distribution is necessary.Several FC-BAI algorithms used Bayesian ideas on the structure of the algorithm, although most of those studies used frequentist settings for measuring the guarantee. The 'Top-Two' type of algorithms are the leading representatives in this direction. The first instance of top-two algorithms, which is called Top-Two Thompson sampling (TTTS), is introduced in the context of Bayesian best arm identification. TTTS requires a prior distribution, and Russo [2016] showed that the sample complexity of posterior convergence of TTTS which is the same as the sample complexity of the frequentist fixed-confidence best arm identification. Subsequent research analyzed the performance of TTTS from the frequentists' viewpoint [Shang et al., 2020] . Later on, the idea of top-two sampling is then extended into many other algorithms, such as Top-Two Transportation Cost [Shang et al., 2020] , Top-Two Expected Improvement (TTEI, Qin et al. 2017b ), Top-Two Upper Confidence bound (TTUCB, Jourdan and Degenne 2022a). Even though some of the top two algorithms adapt a prior, they implicitly solve the optimization that is justified in view of frequentist.Another line of Bayesian sequential decision-making is Bayesian optimization [Srinivas et al., 2010 , Mockus, 2012 , Shahriari et al., 2016 , Jamieson and Talwalkar, 2016 , Frazier, 2018] , where the goal is to find the best arm in Bayesian setting. Note that Bayesian optimization tends to deal with structured identification, especially for Gaussian processes, and most of the algorithms for Bayesian optimization do not have specific stopping criteria. 
problem setup: 
 We study the fixed confidence best arm identification problem (FC-BAI) in a Bayesian setting. In this setup, we have k arms in the set [k] := {1, 2, . . . , k} with unknown distribution P = (P 1 , • • • , P k ) which is drawn from a known prior distribution at time 0, namely H = (H 1 , • • • , H k ). The unknown bandit model P i is a one-parameter distribution, and P is specified by µ := (µ 1 , • • • , µ k ). To simplify the problem, we will focus on the Gaussian case, where each P i is a Gaussian distribution with known variance σ 2 i . Each mean of P i , denoted µ i , is drawn from a known prior Gaussian distribution H i , which can be written as N (m i , ξ 2 i ). At every time step t = 1, 2, • • • , the forecaster chooses an arm A t ∈ [k] and observes a reward X t , which is drawn independently from P At . Since we focus on the Gaussian case, X t ∼ N (µ At , σ 2 At ) conditionally given A t and µ At . After each sampling, the forecaster must decide whether to continue the sampling process or stop sampling and make a recommendationJ ∈ [k]. Let F t = σ(A 1 , X 1 , A 2 , X 2 , • • • , A t , X t) be the σ-field generated by observations up to time t. The algorithm of the forecaster π := ((A t ) t , τ, J) is defined by the following triplet [Kaufmann et al., 2016a] :• A sampling rule (A t ) t , which determines the arm to draw at round t based on the previous history (each A t must be F t-1 measurable).• A stopping rule τ , which means when to stop the sampling (i.e., stopping time with respect to F t ).• A decision rule J, which determines the arm the forecaster recommends based on his sampling history (i.e., J is F τ -measurable).In FC-BAI, the forecaster aims to recommend arm J that correctly identifies (one of) the best arm(s) i * (µ) := arg max i∈[k] µ i with probability at least 1 -δ. Since the case of multiple best arms is of measure zero under H, we can focus on µ such that i * (µ) is unique. For the FC-BAI problem in the Bayesian setting, we use the expected probability of misidentification:EQUATIONwhere H µ := {µ is the correct bandit model}. Now we formally define the algorithm of interest as follows:Definition 1. (Bayesian δ-correctness) For a prior distribution H, an algorithm π = ((A t ), τ, J) is said to be Bayesian (H, δ)-correct if it satisfies PoE(π; H) ≤ δ. Let A b (δ, H) be the set of Bayesian (H, δ)-correct algorithms for the prior distribution H.The objective of the FC-BAI problem in the Bayesian setting is to find an algorithm π =((A t ) t , τ, J) ∈ A b (δ, H) that minimizes E µ∼H [τ ]. Terminology Define N i (t) = t-1 s=1 1[A s = i]as the number of times arm i is pulled before timestep t. Let h i be the probability density function of H i . Since we consider Gaussian prior,h i (µ i ) := (1/ √ 2πξ i ) exp(-(µ i -m i ) 2 /(2ξ 2 i )). Let i * , j * : R k → [k]be the best and the second best arm under the input such that for each µ ∈ {x ∈ R k : x i ̸ = x j ∀i ̸ = j}, i * (µ) = arg max i∈[K] µ i and j * (µ) = arg max i∈[K]\{i * (µ)} µ i . Let KL i (a∥b) := (a-b) 2 2σ 2 i represent∈ R k , i.e., E µ = E[•|H µ ] and P µ = P(•|H µ ). We will abuse the notation PoE so that for λ ∈ R k , PoE(π; λ) means PoE(π; λ) := P λ J ̸ = i * (λ)|H λ .Naturally, PoE(π; H) = E µ∼H PoE(π; µ) . Lastly, we introduce the constant L(H) that characterizes the sample complexity in the Bayesian setting. Definition 2. For each i, j ∈ [k], define L ij (H) and L(H) as follows:L(H) := i,j∈[k],i̸ =j L ij (H) where L ij (H) := ∞ -∞ h i (x)h j (x) s:s∈[k]\{i,j} H s (x) dx.This constant has the following interesting property which we call a volume lemma: Lemma 1 (Volume Lemma, informal). For ∆ ∈ (0, 1), letL(H, ∆) := 1 ∆ P µ∼H µ i * (µ) -µ j * (µ) ≤ ∆ . Then, lim ∆→0 + L(H, ∆) = L(H). In particular, for ∆ < L(H) i∈[k] 2(k-1) ξ i , L(H, ∆) ∈ ( 1 2 L(H), 2L(H)).The volume lemma states that the volume of prior where the suboptimality gap is smaller than ∆ is proportional to L(H)∆ when ∆ is small. We will see in Section 3 that such small-gap cases, which require a large amount of exploration to identify the best arm, dominate the Bayesian expectation of the stopping time. Therefore, L(H) defines the Bayesian sample complexity. The formal version of this lemma, which involves some regularity conditions, is shown in Appendix B. Remark 1. Here, we elaborate on how the Bayesian sample complexity is defined. As will be shown in Section 3, for an algorithm to have a finite expected stopping time, it must determine whether the current instance is difficult or not. In particular, if an algorithm tries to identify even the top-O(δ) 'hardest instances' 2 in the prior , the algorithm cannot achieve the finite expected stopping time. By Lemma 1, the suboptimality gap of the top-O(δ) hardest instance is given by L(H)∆ ≈ δ, and the corresponding (frequentist) sample complexity is proportional to ∆ -2 = (L(H)/δ) 2 [Kaufmann et al., 2014] . Such instances constitute an O(δ) fraction of the prior, and thus the Bayesian sample complexity is:O (L(H)) 2 δ 2 × δ = O (L(H)) 2 δ .3 Limitation of traditional frequentist approaches in the Bayesian settingExisting BAI studies mainly focused on the Frequentist δ-correct algorithms which are defined as follows:Definition 3 (Frequentist δ-correctness). An algorithm π = ((A t ), τ, J) is said to be frequentist δ-correct if, for any bandit instance µ ∈ R k such that i * (µ) is unique, it satisfies PoE(π; µ) ≤ δ.Let A f (δ) be the set of all frequentist-δ-correct algorithms.For the frequentist δ-correct algorithms, Garivier and Kaufmann [2016] proved a lower bound for the expected stopping time as follows: for all bandit instance µ∈ R k and for all ((A t ), τ, J) ∈ A f (δ),EQUATIONwhere T * (µ) is a sample complexity function dependent on the bandit instance µ.3 Moreover, many of the known frequentist δ-correct algorithms achieve asymptotic optimality [Garivier and Kaufmann, 2016 , Russo, 2016 , Tabata et al., 2023 , Qin et al., 2017a] , meaning that they are orderwisely tight up to the lower bound on Eq. ( 2) as δ → 0. However, little is known, or at least discussed, about their performance in the Bayesian setting.One can check that a frequentist δ-correct algorithm is also Bayesian δ-correct as well (A f (δ) ⊂ A b (δ, H) for all H). Naturally, our interest is whether or not the most efficient classes of frequentist δ-correct algorithms, such as Tracking algorithms and Top-two algorithms, are efficient in Bayesian settings. Somewhat surprisingly, the following theorem states that any δ-correct algorithm is suboptimal in Bayesian settings. Theorem 2. For all δ > 0, H and ((A t ), τ, J)∈ A f (δ), E µ∼H [τ ] = +∞.Proof of Theorem 2 is found in Appendix C. To illustrate the proof, we will use a two-armed Gaussian instance as an example. 
special case -two armed gaussian case: 
 Here we present one intuitive corollary of the lower bound theorem [Kaufmann et al., 2016a , Garivier and Kaufmann, 2016 , Kaufmann et al., 2016b ] that uses a standard information-theoretic technique.Corollary 3 (Kaufmann et al. 2014) . Let δ ∈ (0, 1). For any frequentist δ-correct algorithm ((A t ), τ, J) and for any fixed mean vector µ= (µ 1 , µ 2 ) ∈ R 2 , E µ [τ ] ≥ d(δ,1-δ) (µ1-µ2) 2 ≥ log 1 2.4δ (µ1-µ2) 2 .In the frequentist setting, Corollary 3 implies the lower bound of E µ [τ ] = Ω(log(δ -1 )/(µ 1 -µ 2 ) 2 ), which is Ω(log(δ -1 )) when we view parameters (µ 1 , µ 2 ) as constants. However, in the Bayesian setting, the algorithm is given the prior distribution H on µ, and thus the stopping time is marginalized over H. In particular, limiting our interest to the case of |µ 1 -µ 2 | < ∆ for small enough ∆ > 0, we can obtain the following lower bound:E µ∼H [τ ] ≥ E µ∼H [τ • 1[|µ 1 -µ 2 | ≤ ∆]] (Since τ is positive r.v.) ≥ E µ∼H [E[τ |µ] • 1[|µ 1 -µ 2 | ≤ ∆]](Law of total expectation)≥ E µ∼H log δ -1 (µ 1 -µ 2 ) 2 • 1[|µ 1 -µ 2 | ≤ ∆] (Corollary 3) ≥ log δ -1 ∆ 2 P µ∼H [|µ 1 -µ 2 | ≤ ∆] ≥ log δ -1 ∆ 2 L(H) 2 ∆ (Lemma 1) = Ω L(H) log δ -1 ∆ .This inequality implies that if we naively use a known frequentist δ-correct algorithm in the Bayesian setting, the expected stopping time will diverge because we can choose an arbitrarily small ∆. The case of a small gap is difficult to identify, and the expected stopping time can be very large for such a case if we aim to identify the best arm for any model. 
lower bound: 
 This section will elaborate on the lower bound of the stopping time in the Bayesian setting. Theorem 4 below states that any Bayesian (H, δ)-correct algorithm requires the expected stopping time of at least Ω( L(H) 2 δ).Theorem 4. Defineσ min = min i∈[k] σ 2 i and N V = L(H) 2 σ 2 min ln 2 16e 4 δ. Let δ < δ L (H) be sufficiently small. 4 Then, for any BAI algorithms π = ((A t ), τ, J), if E µ∼H [τ ] ≤ N V , then PoE(π; H) ≥ δ.In this main body, we will use the two-armed Gaussian bandit model with homogeneous variance condition (i.e. σ 1 = σ 2 = σ) for easier demonstration of the proof sketch. Theorem 4, which is more general in the sense that it can deal with k > 2 arms with heterogeneous variances, is proven in Appendix D.Sketch of the proof, for k = 2: It suffices to show that the following is an empty set:A b (δ, H, N V ) := {π ∈ A b (δ, H) : E µ∼H [τ ] ≤ N V }. Assume that A b (δ, H, N V ) ̸ = ∅ and choose an arbitrary π ∈ A b (δ, H, N V ).We start from the following transportation lemma: Lemma 5 (Kaufmann et al. 2016a , Lemma 1). Let δ ∈ (0, 1). For any algorithm ((A t ), τ, J), anyF τ -measurable event E, any bandit models µ, λ ∈ {(x, y) ∈ R 2 : x ̸ = y} such that i * (µ) ̸ = i * (λ), E µ   2 i=1 KL i (µ i , λ i )N i (τ )   ≥ d(P µ (E), P λ (E)).Note that the above Lemma holds for any algorithm, and thus works for any stopping time τ . Now define ν(µ) as a swapped version of µ ∈ R 2 , which means (ν(µ)) 1 = µ 2 , ν(µ) 2 = µ 1 , and let E = {J ̸ = i * (µ)}, the event that the recommendation of the algorithm is wrong. Substituting λ with ν(µ) from the above equation of Lemma 5 leads toEQUATIONNote that the first inequality comes from the fact that E, the failure event of the bandit model µ, is exactly a success event of ν(µ) in this two-armed case, and the last inequality is from our modified lemma (Lemma 11) from Eq. ( 3) of Kaufmann et al. [2016a] . One can rewrite the above inequality asEQUATIONWe can rewrite the conditions ofA b (δ, H, N V ) as PoE(π; H) = µ∈R 2 PoE(π; µ) dH(µ) ≤ δ and µ∈R 2 E µ [τ] dH(µ) ≤ N V . (Opt0)Using Eq. ( 4) and with some symmetry tricks, we get V 0 ≤ PoE(π; H) whereEQUATIONAlgorithm 1 Successive Elimination with Early-StoppingInput: Confidence level δ, prior H ∆ 0 := δ 4L(H)Initialize the candidate of best armsA(1) = [K]. t = 1 while True do Draw each arm in A(t) once. t ← t + |A(t)|. for i ∈ A(t) do Compute UCB(i, t) and LCB(i, t) from (5). if UCB(i, t) ≤ max j LCB(j, t) then A(t) ← A(t) \ {i}. end if end for if |A(t)| = 1 then Return arm J in A(t). end if Compute ∆safe (t) := max i∈A(t) UCB(i, t) -max i∈A(t) LCB(i, t). if ∆safe (t) ≤ ∆ 0 thenReturn arm J, uniformly sampled from A(t). end if end while Note that on the above two inequalities, only E µ [τ ] is the value that depends on the algorithm π. Now our main idea is that we can relax these two inequalities to the following optimization problem by substituting E µ [τ ] to an arbitrary ñ : R 2 → [0, ∞) as follows:V := inf ñ:R 2 →[0,∞) µ∈R 2 exp - (µ 1 -µ 2 ) 2 2σ 2 ñ(µ) dH(µ) s.t. µ∈R 2 ñ(µ) dH(µ) ≤ N V (Opt2) and V ≤ 2.4V 0 ≤ 2.4δ. Let N := {µ ∈ R 2 : |µ 1 -µ 2 | < ∆ := 8δ L(H) }.From the discussions in Section 3.1, one might notice that N is an important region for bounding E µ∼H [τ ] . We can relax the above (Opt2) to the following version, which focuses more on N :V ′ := inf ñ:R 2 →[0,∞) µ∈N exp - (µ 1 -µ 2 ) 2 2σ 2 ñ(µ) dH(µ) s.t. µ∈N ñ(µ) dH(µ) ≤ N V . (Opt3) One can prove V ′ ≤ V ≤ 2.4δ. Now, if we notice that function x → exp -(µ1-µ2) 2 2σ 2x is a convex function, we can use Jensen's inequality to verify that the optimal solution for (Opt3) is when ñ =N V E µ∼H [1N ]1 N , and when we use N V in Theorem 4, one can get:V ′ ≥ µ∈N exp - ∆ 2 2σ 2 • N V E µ∼H [1 N ] dH(µ) (by optimality of ñ = N V E µ∼H [1N ] ) ≥ µ∈N exp - ∆ 2 • N V σ 2 ∆L(H) dH(µ) ≥ exp - ∆ • N V σ 2 L(H) • ∆L(H) (both by Lemma 1) > 2.4δ, (by definition of N V and ∆)which is a contradiction. This means no algorithm satisfies (Opt1), and the proof is completed. 
main algorithm: 
 This section introduces our main algorithm (Algorithm 1). In short, our algorithm is a modification of the elimination algorithm with the incorporation of the indifference zone technique. Define ∆ 0 := δ 4L(H) which satisfies the following condition, thanks to Lemma 1:P µ∼H (µ i * (µ) -µ j * (µ) ≤ ∆ 0 ) ≤ δ 2 .In each iteration of the while loop of Algorithm 1, the learner selects and observes each arm in the active set. After drawing each arm once, the algorithm calculates the confidence bounds for each arm in the active set using the formula as follows: let Conf(i, t) and μi (t) be the confidence width and the empirical mean of arm i at time t asConf(i, t) := 2σ 2 i log(6(N i (t)) 2 /(( δ 2 2K )π 2 )) N i (t) , μi (t) := t-1 s=1 X s 1[A s = i].Then the upper and lower confidence bounds of arm i at timestep t, denoted as UCB and LCB respectively, can be defined in the following manner:EQUATIONThis confidence bounds ensure that, with high probability, for all t ∈ [T ] and i ∈ [K], µ i ∈ (UCB(i, t), LCB(i, t)) (See Lemma 15 in Appendix for details). After calculating UCB and LCB, the algorithm eliminates arms with UCB smaller than the largest LCB and maintains only arms that could be optimal in the active set A. Up to this point, it follows the traditional elimination approach.The main difference in our algorithm lies in the stopping criterion. At the end of each iteration, the algorithm checks the stopping criterion. Unlike typical elimination algorithms that continue until only one arm remains, we have introduced an additional indifference condition. This condition arises when the suboptimality gap is so small that identifying them would require an excessive number of samples. In such cases, our algorithm stops additional attempts to identify differences between arms in the active set and randomly recommends one from the active set instead. Remark 2. In the context of PAC-(ϵ, δ) identification, Even-Dar et al. [2006, Remark 9 ] introduced a similar approach. The largest difference is that they use the parameter ϵ as a parameter that defines the indifference-zone level, whereas our parameter ∆ 0 is spontaneously derived from the prior H and the confidence level δ without specifying the indifference-zone.Theorem 6 describes the theoretical guarantee of the Algorithm 1.Theorem 6. For δ < 4L(H) • minL(H) i∈[k] k-1 ξ i , min i,j∈[k] ξ i L ij (H) 2, Algorithm 1 which consists of ((A t ), τ, J) has the expected stopping time upper bound as follows:EQUATIONwhere C = 320 π 2 3 + 1 is a universal constant and σ max = max i∈[k] σ i . Here, O(log δ -1 ) is a function of δ and H that is proportional to log δ -1 when we view prior parameters H as constants. Plus, the strategy defined by Algorithm 1 is in A b (δ, H).See Appendix E for the formal proof of Theorem 6. Remark 3. When we compare the lower bound (Theorem 4) with the upper bound of Algorithm 1 (Theorem 6), we can see the algorithm is near-optimal. If we view σ max /σ min as a constant, the bounds are tight up to a log L(H) δ factor.Remark 4. The condition δ < 4L(H) • min L(H)/ i∈[k] k-1 ξi , min i,j∈[k] (ξ i L ij (H)) 2 isonly for cleaner illustration of the regret bound in Theorem 6. The non-asymptotic result, when δ is a moderately large constant, can be found in Appendix E.1.We summarize the general strategy for the proof as follows. By the law of total expectation,E µ∼H [τ ] = E µ∼H E µ [τ ]. Therefore, we first derive a frequentist upper bound of E µ [τ ], and then marginalize it to obtain the expected Bayesian stopping time.First, with the confidence bound defined as Eq. ( 5) we have the following guarantee that the true means for all arms are in the confidence bound interval with high probability.Lemma 7. For any fixed µ ∈ {v ∈ R k : v i ̸ = v j for all i, j ∈ [k]}, let X (µ) := {∀i ∈ [k] and t ∈ N, µ i ∈ (LCB(i, t), UCB(i, t))}. Then, P µ X (µ) ≥ 1 -δ 2 . Now we can rewrite E µ∼H [τ ] as follows: E µ∼H [τ ] =E µ∼H [E µ [τ ]] (Law of Total Expectation) =E µ∼H [E µ [τ 1[X (µ)]]] + E µ∼H [E µ [τ 1[X (µ) c ]]] = i E µ∼H E µ [N i (τ )1[X (µ)]] + E µ∼H [E µ [τ 1[X (µ) c ]]] . (7) Let ∆ i = ∆ i (µ) := (max s∈[k] µ s ) -µ i and R 0 (∆) ≈ ⌈Cσ 2 max • log ∆ -1 ∆ 2 ⌉.For the first term, under X (µ), we can bound N i (τ ) by R 0 (max(∆ 0 , ∆ i )) (Lemma 17 in Appendix E), and integrate it over the prior distribution to obtain the leading factor. For the second term, thanks to the indifference stopping condition ( ∆safe (t) ≤ ∆ 0 ), one can prove that τ is always smaller than R(∆ 0 ) (Lemma 14 in Appendix E), which leads to non-leading term.To check that the expected probability of error is below δ, we have an additional lemma: Lemma 8 (Probability of dropping i * (µ)). For any µ 0 , underH µ0 , X (µ 0 ) ⊂ t {i * (µ 0 ) ∈ A(t)} .This lemma means under the event X (µ), the best arm is never dropped. We can also prove that under the event X (µ), if ∆ i (µ) > ∆ 0 , the sub-optimal arm will eventually be dropped before the algorithm terminates (Lemma 14 in Appendix E). These two facts mean there are only two cases in which the prediction of Algorithm 1 could be wrong.• Under X (µ) c , both facts cannot guarantee the correct identification. From Lemma 7, P µ [X (µ) c ] ≤ δ 2 for all µ, and thusP µ∼H [X (µ) c ] ≤ δ 2 . • When ∆ i (µ) ≤ ∆ 0 .From Lemma 1 and the definition of ∆ 0 , the probability of drawing such µ from the prior is at most δ/2.Therefore, by union bound, Algorithm 1 has the expected probability of misidentification guarantee smaller than δ 2 + δ/2 < δ. 
simulation: 
 We conduct two experiments to demonstrate that the expected stopping times of frequentist δ-correct algorithms diverge in a Bayesian setting and that the elimination process in Algorithm 1 is necessary for more efficient sampling. In Tables 1 and 2 , each column 'Avg', 'Max', and 'Error' represents the average stopping time, maximum stopping time, and the ratio of the misidentification, respectively.5 More details of these experiments are in Appendix F.We evaluate the empirical performance of our Elimination algorithm (Algorithm 1) by comparing it with other frequentist algorithms such as Top-two Thompson Sampling (TTTS) [Russo, 2016] and Top-two UCB (TTUCB) [Jourdan and Degenne, 2022b] .We design an experiment setup that has k = 2 arms with standard Gaussian prior distribution, which meansm i = 0, ξ i = 1 for all i ∈ [k].We set δ = 0.1 and ran N = 1000 Bayesian FC-BAI simulations to estimate the expected stopping time and success rate.In Table 1 , one can see that the two top-two algorithms exhibit very large maximum stopping time. This supports our theoretical result in Section 3 that the expected stopping time of Frequentist δcorrect algorithms will diverge in the Bayesian setting. We did not check the track and stop algorithm [Garivier and Kaufmann, 2016] because it needs to solve an optimization for each round, but the fact that the expected stopping time of the track and stop is at least half of the TTTS and TTUCB for a small δ implies that the performance of track and stop is similar to that of top-two algorithms.Algorithm 1 shows a significantly smaller average stopping time as well as an average computation time than that of these algorithms. Effect of the elimination process We implemented the modification of Algorithm 1 (denoted as NoElim) that never eliminates an arm from A(t) 6 In this setup, we have k = 10 arms with standard Gaussian prior distribution, which meansm i = 0, ξ i = 1 for all i ∈ [k].We set δ = 0.01 and ran N = 1000 Bayesian FC-BAI simulations.As one can check from Table 2 , elimination of arms helps the efficient use of samples and reduces stopping time and computation time.7 Discussion and future worksWe have considered the Gaussian Bayesian best arm identification with fixed confidence. We show that the traditional Frequentist FC-BAI algorithms do not stop in finite time in expectation, which implies the suboptimality of such algorithms in the Bayesian FC-BAI problem. We have established a lower bound of the Bayesian expected stopping time, which is of order Ω( L(H) 2 δ). Moreover, we have introduced the elimination and early stopping algorithm, which achieves a matching stopping time up to a polylogarithmic factor of L(H) and δ. We conduct simulations to support our results.In the future, we will attempt to tighten the logarithmic and maxi σi mini σi 2 gap between the lower and upper bound, extend the indifference zone strategy for other traditional BAI algorithms in the Bayesian setting, extend our analysis from Gaussian bandit instances to general exponential families, and design a robust algorithm against misspecified priors. N i (t) t-1 s=1 1[A s = i] L(H)See Definition 2 i * (µ), j * (µ) best arm and second best arm ν(µ)alternative model where the top-two means of µ are swappedKL i (•, •) KL divergence between two distributions d(p, q)KL divergence between two Bernoulli distributions with parameters p and q N i (t) number of draws on arm i before time step tB 320 max i∈[k] σ 2 i B 0 π 2 3 + 1 B Θ i {µ ∈ R k : i * (µ) = i} Θ ij {µ ∈ R k : i * (µ) = i, j * (µ) = j} µ \i ( µ \i,j )the vector projection which omits i-th coordinate (i, j-th, respectively) H \i ( H \i,j ) the distribution which omits i-th coordinate (i, j-th, respectively) i (µ) E µ [N i (τ )] D 0 (H)    W (- 1 32 max i∈[k] ξ 3/2 i ) If max i∈[k] ξ i > 3 e 2 2 101 Otherwise (W is the Lambert W function.)D 1 (H) min i̸ =j mi σ 2 i - mj σ 2 j -1 , 1 2σ 2 i + 1 2σ 2 j -2 δ L (H) L(H) 32e 4 • min D 0 (H), D 1 (H), min i∈[k] 1 4m 2 i , L(H) 4(k-1) i∈[k] 1 ξ iWe will use the following formal version of the volume lemma for the proof. The first result is used for the upper bound, and the second result is used for the lower bound.Lemma 9 (Volume Lemma, formal). LetΘ ij := {µ ∈ R k : i * (µ) = i, j * (µ) = j}.1. For any ∆ ∈ (0, 1), define L ij (H, ∆) := 1 δ Θij 1[|µ i -µ j | ≤ ∆] dH(µ).EQUATION))Then,EQUATIONIn particular, when ∆ < L(H)i∈[k] k-1 ξ i , L(H, ∆) ≤ 2L(H).2. (Volume lemma for the lower bound) For small enough positive real number ∆ < min1 4 max i∈[k] m 2 i , D 0 (H) 7 let L ′ ij (H, ∆) := 1 ∆ Θij 1[|µ i -µ j | ≤ ∆]1[|µ i |, |µ j | ≤ 1 √ ∆ ] dH(µ).Then,EQUATIONIn particular, when ∆ < min1 4 max i∈[k] m 2 i , D 0 (H), L(H) i∈[k] 4(k-1) ξ i , L ′ (H, ∆) ∈ 1 2 L(H), 2L(H) .Proof. First, let us prove the upper bound of Eq. ( 8), i.e.L ij (H, δ) ≤ L ij (H) + 1 ξi ∆. Recall for i, j ∈ [k], Θ ij = {µ : i * (µ) = i, j * (µ) = j}. We have Θij 1 [|µ i -µ j | ≤ ∆] dH(µ) = ∞ -∞ µj +∆ µj h i (µ i ) dµ i k̸ =i,j µj -∞ h k (µ k ) dµ k h j (µ j ) dµ j ≤ ∞ -∞ ∆ max 0≤y≤∆ h i (µ j + y) k̸ =i,j µj -∞ h k (µ k ) dµ k h j (µ j ) dµ j ≤ ∆ ∞ -∞ h i (µ j ) + e -1/2 ξ i ∆ k̸ =i,j µj -∞ h k (µ k ) dµ k h j (µ j ) dµ j(by the Lipschitz property of the Gaussian density, e -1/2 /ξ i is the steepest slope of N (m i , ξ 2 i ))≤ ∆        ∞ -∞ h i (µ j ) k̸ =i,j µj -∞ h k (µ k ) dµ k h j (µ j ) dµ j =Lij (H) + e -1/2 ξ i ∆       .Therefore, we verified the upper bound side of Eq. ( 8).For the lower bound, by following the same steps, one can prove Θij 1[|µ i -µ j | ≤ ∆] dH(µ) ≥ (L ij (H) -1 ξi ∆)∆.For the proof of Eq. ( 9), by Chernoff's method we can bound the tail probability as follows:Θij 1 |µ i -m i | > 1 √ ∆ dH(µ) < R k 1 |µ i -m i | > 1 √ ∆ dH(µ) < 2 exp - 1 2∆ξ 2 i . When |m i | < 1 2 √∆ , then we can change the above inequality as:Θij 1 |µ i | > 1 √ ∆ dH(µ) ≤ R k 1 |µ i | > 1 √ ∆ dH(µ) ≤ R k 1 |µ i -m i | > 1 2 √ ∆ dH(µ) < 2 exp - 1 8∆ξ 2 i . Therefore, Θij 1[|µ i -µ j | ≤ ∆]1[|µ i |, |µ j | ≤ 1 √ ∆ ] dH(µ) ≥ Θij 1 [|µ i -µ j | ≤ ∆] dH(µ) - Θij 1 |µ i -m i | > 1 2 √ ∆ dH(µ) - Θij 1 |µ j -m j | > 1 2 √ ∆ dH(µ) ≥ (L i (H) - 1 ξ i ∆)∆ -2 exp - 1 8∆ξ 2 i -2 exp - 1 8∆ξ 2 j ≥ (L i (H) - 2 ξ i ∆)∆ (by ∆ < D 0 (H)) for ∆ < min min i∈[k] 1 4m 2 i , D 0 (H) .C Proof of Theorem 2For ∆ > 0, let Θ i (∆) := {µ ∈ R k : i * (µ) = i, µ i -µ j * (µ) ≤ ∆}.From Lemma 9, we haveP µ∼H i * (µ) = i, µ i -µ j * (µ) ≤ ∆ = j̸ =i P µ∼H i * (µ) = i, j * (µ) = j, µ i -µ j ≤ ∆ ≥ j̸ =i 1 2 L ij (H)∆. Now, for each µ ∈ R k , let ν : R k → R k be a function such that for s ∈ [k], ν(µ) s :=    µ i * (µ) when s = j * (µ) µ j * (µ) when s = i * (µ) µ s Otherwise.For any µ ∈ R k , let E(µ) = {J ̸ = i * (µ)} and ν = ν(µ). By Lemma 1 in Kaufmann et al. [2016a] , for any frequentist δ-correct algorithm π = ((A t ) t , τ, J), we haveEQUATION)For the left side, from the construction of ν,KL s (µ s ||ν s ) := (µ i * (µ) -µ j * (µ) ) 2 2σ 2 s s = i * (µ), j * (µ) 0Otherwise.Since π ∈ A f (δ) and from the definition of E(µ) and ν, P µ (E(µ)) ≤ δ and P ν (E(µ)) ≥ 1 -δ.Overall, we can rewrite Eq. ( 10) to the following simpler form:E µ [N i * (µ) (τ )] (µ i * (µ) -µ j * (µ) ) 2 2σ 2 i * (µ) + E µ [N j * (µ) (τ )] (µ i * (µ) -µ j * (µ) ) 2 2σ 2 j * (µ) ≥ d(δ, 1 -δ) =⇒ E µ [N i * (µ) (τ ) + N j * (µ) (τ )] ≥ 2d(δ, 1 -δ) min s∈[k] σ 2 s (µ i * (µ) -µ j * (µ) ) 2 .Since τ = k s=1 N s (τ ), we can lower bound the expected stopping time when µ is given as follows:EQUATION)Now, when we compute marginal E µ [τ ] over µ, we haveE µ∼H [τ ] = E µ∼H E µ [τ ] (Law of total expectation) ≥ E µ∼H E µ [τ ]1 Θi(∆) ≥ E µ∼H 2d(δ, 1 -δ) min s∈[k] σ 2 s (µ i * (µ) -µ j * (µ) ) 2 1 Θi(∆) (Eq. (11)) ≥ E µ∼H 2d(δ, 1 -δ) min s∈[k] σ 2 s ∆ 2 1 Θi(∆) ≥ j̸ =i L ij (H) d(δ, 1 -δ) min s∈[k] σ 2 s ∆ 2 = j̸ =i L ij (H) d(δ, 1 -δ) min s∈[k] σ 2 s ∆ .Now since ∆ is an arbitrary small positive number, we can conclude that E µ∼H [τ ] diverges.In this subsection, we will prove the following theorem: Theorem 10 (Restatement of Theorem 4). Let δ > 0 be sufficiently small such that δ < δ L (H). For any best arm identification algorithm, ifR k   i∈[k] n i (µ)   dH(µ) ≤ N V , then R k P µ [J ̸ = i * (µ)] dH(µ) ≥ δ.Proof. By Lemma 1 in Kaufmann et al. [2016a] , for any stopping time τ , we haveEQUATION)To modify the RHS of Eq. ( 12), we will use the following lemma:Lemma 11. For any p, q ′ , q ∈ (0, 1) such that q ′ ≤ q, we haveln 1 4(p + q) ≤ d(p, 1 -q ′ ).By using Lemma 11 and the fact that P ν (E(µ)) ≥ 1 -P ν [J ̸ = i * (ν)] by definition, we can transform (12) intoEQUATION)Note that P µ (E(µ)) is exactly the error probability, and we are interested in the marginal error probabilityE µ∼H [P µ (E)]. Let n i (µ) = E µ [N i (τ )],and ∆ is an arbitrary small enough positive variable which we will define later on Eq. ( 19). Then, by rearrangement, we can induce the following inequalitiesR k P µ [J ̸ = i * (µ)] dH(µ) = i∈[k] j̸ =i Θij P µ [J ̸ = i] dH(µ) = 1 2   i∈[k] j̸ =i Θij P µ [J ̸ = i] dH(µ) + j∈[k] i̸ =j Θji P ν(µ) [J ̸ = j]h i (µ j )h j (µ i )   s̸ =i,j h s (µ s )   dµ   (Symmetry of the Lebesgue measure) = 1 2   i∈[k] j̸ =i Θij P µ [J ̸ = i]h i (µ i )h j (µ j ) + P ν(µ) [J ̸ = j]h i (µ j )h j (µ i )   s̸ =i,j h s (µ s )   dµ   ≥ i∈[k] j̸ =i Θij P µ [J ̸ = i * (µ)] + P ν [J ̸ = i * (ν)] 2 min (h i (µ i )h j (µ j ), h i (µ j )h j (µ i )) dµ (AC + BD ≥ A min(C, D) + B min(C, D) = (A + B) min(C, D) for A, B, C, D > 0) ≥ i∈[k] j̸ =i Θij exp (-n i (µ)KL i (µ i , µ j ) -n j (µ)KL j (µ j , µ i )) 8 min 1, h i (µ j )h j (µ i ) h i (µ i )h j (µ j ) h i (µ i )h j (µ j ) dµ(Eq. ( 13))≥ e -4 i∈[k] j̸ =i Θij exp (-n i (µ)KL i (µ i , µ j ) -n j (µ)KL j (µ j , µ i )) 8 1 |µ i |, |µ j | ≤ 1 ∆ dH(µ).(Lemma 12)For the last inequality, we used the following lemma. The proof for this lemma is found in Subsection D.1.2. In short, we haveR k P µ [J ̸ = i * (µ)] dH(µ) ≥ e -4 i∈[k] j̸ =i Θij exp (-n i (µ)KL i (µ i , µ j ) -n j (µ)KL j (µ j , µ i )) 8 1 |µ i |, |µ j | ≤ 1 ∆ dH(µ) (14) 8 D1(H) := min i̸ =j m i σ 2 i - m j σ 2 j -1 , 1 2σ 2 i + 1 2σ 2 j -2and the following statement is a stronger statement than Theorem 10.If i∈[k] n i (µ) dH(µ) ≤ N V , then (RHS of Eq. ( 14))≥ δ.RHS of Eq. ( 14) is represented in terms of n := (n 1 , • • • , n k ) : R k → [0, ∞) k (the expected number of arm pulls) and does hold for any algorithm given n. To prove the above statement, since 'set of all expected number of arm pulls' is a subset of {ñ : R k → [0, ∞)}, it suffices to show that the optimal value V of the following objectiveV min := inf ñ:R k →[0,∞) k V (ñ) (15) s.t. R k   k s=1 ñs (µ)   dH(µ) ≤ N V where V (ñ) := e -4 i∈[k] j̸ =i Θij exp (-ñ i (µ)KL i (µ i , µ j ) -ñj (µ)KL j (µ j , µ i )) 8 1 |µ i |, |µ j | ≤ 1 ∆ dH(µ)is greater than δ.EQUATIONs.t. i∈[k] j̸ =i Θij   k s=1 ñs (µ)   dH(µ) ≤ N V where V 1 (ñ) := e -4 i∈[k] j̸ =i Θij exp (-ñ i (µ)KL i (µ i , µ j ) -ñj (µ)KL j (µ j , µ i )) 8 dH(µ).To see this, suppose n is an optimal solution to (15). Then, by the constraint of optimization problem (15),R k s∈[k] ns (µ) dH(µ) ≤ N V and since n is a collection of positive functions, i,j∈[k]:i>j Θij ∪ Θji k s=1 ñs (µ) dH(µ) ≤ N V ,which means n satisfies the constraint of ( 16). By the minimality,V min 1 ≤ V 1 (n), and since exp is a positive function, we have V 1 (n) ≤ V (n) = V min . Moreover, V min 1 ≥ V min 2 holds for V min 2 := inf ñ:R k →[0,∞) k V 2 (ñ) (17) s.t. i∈[k] j̸ =i Θij   k s=1 ñs (µ)   dH(µ) ≤ N V where V 2 (ñ) := e -4 i∈[k] j̸ =i Θij exp -ñi (µ) + ñj (µ) ∆2 2 min(σ 2 s ) s∈[k] 8 dH(µ)by using the fact thatKL i (µ i , µ j ) = ∆2 2σ 2 i ≤ ∆2 2 min(σ 2 s ) s∈[k] .Claim 1. We abuse our notation slightly so that H(E) = E dH(µ) for any Lebesgue measurable set E, and let Θ = ∪ i,j∈[k]:i̸ =j Θij (note that all Θij are mutually disjoint except for the measure zero sets). Then, the following n opt is an optimal solutions to (17)n opt s (µ) := N V 2H( Θ) 1 µ ∈ ∪ j̸ =s Θsj ∪ ∪ i̸ =s Θis .Proof. Choose an arbitrary ñ : R k → [0, ∞) k which satisfies the constraint of optimization problem (17). Let Ñ (µ) := s∈[k] ñs (µ). Now, since the function ρ :x → 1 8R exp(-x • ∆2 2 min(σ 2 s ) s∈[k]) is a convex and decreasing function, by Jensen's inequality we can say thatV 2 (ñ) = i∈[k] j̸ =i Θij ρ ñi (µ) + ñj (µ) dH(µ) ≥ i∈[k] j̸ =i H( Θij )ρ 1 H( Θij ) Θij ñi (µ) + ñj (µ) dH(µ) (Jensen's inequality for each integral on Θij ) = H( Θ) • i∈[k] j̸ =i H( Θij ) H( Θ) ρ 1 H( Θij ) Θij ñi (µ) + ñj (µ) dH(µ) ≥ H( Θ) • ρ   i∈[k] j̸ =i H( Θij ) H( Θ) 1 H( Θij ) Θij ñi (µ) + ñj (µ) dH(µ)   (Jensen's inequality) ≥ H( Θ) • ρ    i∈[k] j̸ =i 1 H( Θ) Θij   s∈[k] ñs   dH(µ)    (ñ i + ñj ≤ s∈[k] ñs and ρ is a decreasing function.) ≥ H( Θ) • ρ N V H( Θ) .(Constraint of ( 17))Since ñ is chosen arbitrarily, we can sayV min 2 ≥ H( Θ)ρ N V H( Θ). One can check the above n opt satisfies the constraint in the optimization problem (17) and also satisfies V 2 (n opt ) = H( Θ)ρ( N V H( Θ) ). Therefore, n opt is an optimal solution of optimization problem (17).Using Lemma 9, we can getEQUATION)Now applying Claim 1 and Eq. ( 18) on Optimization (17) implies the following result:EQUATION)) ≥ exp - N V ∆ 2 min(σ 2 s ) s∈[k] L ′ (H, ∆) 8e 4 × L ′ (H, ∆) ∆.(Eq. ( 18))If V ≤ δ is true, then V 2 ≤ δ, which implies exp - N V ∆ 2 min(σ 2 s ) s∈[k] L ′ (H, ∆) 8e 4 ×L ′ (H, ∆) ≤ δ ⇐⇒ N V ≥ 2 min(σ 2 s ) s∈[k] L ′ (H, ∆) ∆ ln L ′ (H, ∆) 8e 4 δ .To make this lower bound greater than 0, ln L ′ (H, ∆) 8e 4 δ > 1. From Lemma 9, we know that for small enough ∆9 ,EQUATIONwe haveN V ≥ min(σ 2 s ) s∈[k] L(H) 2 16e 4 ln 2 (20)which is the inequality we desired.Proof. It is equivalent to prove p + q ≥ 1 4 exp(-d(p, 1 -q ′ )) for any p, q, q ′ ∈ (0, 1) such that q ′ ≤ q. First, if p ≥ 1/4 or q ≥ 1/4 it trivially holds, and thus we assume p < 1/4 and q < 1/4. We haved(p, 1 -q ′ )d(p, 1 -q) (p, q ≤ 1 4 ) ≤ d(p + q, 1 -(p + q)) (p + q < 1 2 )≤ log 1 2.4(p + q) (Eq.( 3) of Kaufmann et al. [2016a] )and transforming this yieldsp + q ≥ 1 2.4 e -(d(p,1-q ′ )) .This completes the proof.Proof. We haveh i (a)h j (b) h i (b)h j (a) = exp - (a -m i ) 2 2σ 2 i + (b -m i ) 2 2σ 2 i - (b -m j ) 2 2σ 2 j + (a -m j ) 2 2σ 2 j = exp - (a -b)(a + b -2m i ) 2σ 2 i - (b -a)(a + b -2m j ) 2σ 2 j = exp 2(a -b) m i σ 2 i - m j σ 2 j -(a -b)(a + b) 1 2σ 2 i + 1 2σ 2 j ≥ exp -2|a -b| m i σ 2 i - m j σ 2 j -|(a -b)(a + b)| 1 2σ 2 i + 1 2σ 2 j ≥ exp -2δ m i σ 2 i - m j σ 2 j -2 √ δ 1 2σ 2 i + 1 2σ 2 j 1 R ′ ij (H, δ) . Define R ′ (H, δ) = min i̸ =j R ′ ij (H, δ)which satisfies the condition of the Ratio lemma. For δ such thatδ < D 1 (H) := min i̸ =j   m i σ 2 i - m j σ 2 j -1 , 1 2σ 2 i + 1 2σ 2 j -2   , 9 ∆ < min D0(H), min i∈[k] 1 4m 2 i , L(H) 4(k-1) i∈[k] 1 ξ i. Check Section G for the condition.we have R ′ (H, δ) ≤ e 4 .For notational convenience, define ∆ s (µ) := µ i * (µ) -µ s for s ∈ [k], ∆(µ) := min s̸ =i * (µ) ∆ s = ∆ j * (µ) . Let A(t) be the subset of arms that have not been eliminated at time t.Since δ < 4L 2 (H) i∈[k] k-1 ξ i , ∆ 0 = δ 4L(H) ≤ L(H) i∈[k] k-1 ξ iand we haveEQUATION)Next, we consider the upper bound estimator such that ∆safe (t) ≥ ∆ holds with high probability. Namely,UCB(i, t), LCB(i, t) = μi (t) ± Conf(i, t), Conf(i, t) = 2σ 2 i log(6(N i (t)) 2 /(( δ 2 2k )π 2 )) N i (t) , ∆safe (t) = max i UCB(i, t) -max j LCB(j, t).From the definition above, we can calculate how many arm pulls the learner needs to narrow down the confidence width. Lemma 13. Define B := 320max i∈[k] σ 2 i and ∆ thr := min log 4√ k δπ , 1 B . Let R 0 (∆) := B log min(∆,∆ thr ) -1min(∆,∆ thr ) 2 . Then, for any ∆ ∈ (0, ∞) and for a timestep t which satisfies N i (t) ≥ R 0 (∆), Conf(i, t) ≤ ∆/4. Proof of Lemma 13. Only for this part of the proof, let ∆ ′ := min(∆, ∆ thr ) for notational convenience. Then,Conf(i, t) = 2σ 2 i log(6(N i (t)) 2 /(( δ 2 2k )π 2 )) N i (t) ≤ 2σ 2 i log(6R 0 (∆) 2 /(( δ 2 2k )π 2 )) R 0 (∆) (assumption on t) = ∆ ′ 2σ 2 i log(6(B × log(∆ ′ -1 ) ∆ ′ 2 ) 2 /(( δ 2 2k )π 2 )) B × log(∆ ′ -1 ) ≤ ∆ ′ 2σ 2 i log(6( B ∆ ′ 3 ) 2 /(( δ 2 2k )π 2 )) Blog(∆ ′ -1 ) (log ∆ ′ -1 ≤ ∆ ′ -1 ) = ∆ ′ 2σ 2 i 6 log(∆ ′ -1 ) + 2 log B + log 12k δ 2 π 2 Blog(∆ ′ -1 ) ≤ ∆ ′ 2σ 2 i 6 B + 2 B + 2 B (Definition of ∆ thr , ∆ thr ≥ ∆ ′ ) = ∆ ′ 20σ 2 i B ≤ 1 4 ∆ ′ ≤ 1 4 ∆.From this lemma, one could induce the following corollary which states that Algorithm 5 always terminates before a certain timestep: Corollary 14. Let τ (and γ) be the stopping time (and the last iteration of the while loop in Algorithm 1, respectively) where Algorithm 1 meets the stopping condition. Then, γ is always bounded by R 0 (∆ 0 ) and τ is uniformly bounded byT 0 := k • R 0 (∆ 0 ).Proof of Corollary 14. Let us assume that τ > T 0 + 1. Then, by Lemma 13, each i ∈ A T0 satisfies Conf(i, T 0 ) ≤ ∆ 0 /4. Let i ucb (t) = arg max i∈A(t) UCB(i, t) and i lcb (t) = arg max i∈A(t) LCB(i, t). From definition,∆safe (T 0 ) = max i∈A(T0) UCB(i, T 0 ) -max i∈A(T0) LCB(i, T 0 ) = UCB(i ucb (T 0 ), T 0 ) -LCB(i lcb (T 0 ), T 0 ) = 2Conf(i ucb (T 0 ), T 0 ) + 2Conf(i lcb (T 0 ), T 0 ) + LCB(i ucb (T 0 ), T 0 ) -UCB(i lcb (T 0 ), T 0 ) ≤ ∆ 0 + LCB(i ucb (T 0 ), T 0 ) -UCB(i lcb (T 0 ), T 0 ) ≤ ∆ 0(Since both arms survived from the elimination phase) which implies ∆safe ≤ ∆ 0 , contradicting τ ≥ T 0 since the algorithm should be terminated by Line 18 at timestep T 0 . Therefore, τ ≤ T 0 .Lemma 14 implies Algorithm 1 always stops before T 0 samples. Morever, the following lemma states that with high probability, true mean µ is in between UCB and LCB for all time steps. Lemma 15. (Uniform confidence bound) The following holds for all i ∈ [k]:EQUATIONEQUATION)Proof of Lemma 15. The following derives the upper bound part, Eq. ( 23). The lower bound is derived by following the same steps.Since each arm is independent of each other, Eq. ( 23) boils down to proveP µ   ∞ s=1 {µ i ≤ UCB(i, t i (s))}   ≥ 1 - δ 2 2k ,wheret i (s) = min{t ∈ N : N i (t) ≥ s} and t i (s) = ∞ if {t ∈ N : N i (t) ≥ s} = ∅. For each event {µ i ≤ U CB i (t i (s))},since each arm pull is independent of each other, by Hoeffding's inequality we haveP µ   1 s s j=1 (X i j -µ i ) ≤ -ϵ   ≤ exp - sϵ 2 2σ 2 0for any ϵ > 0. If we set ϵ = Conf(i, t i (s)), we can transform the above inequality toP µ UCB(i, t i (s)) ≤ µ i ≤ δ 2 2ks 2 • 6 π 2 .By the union bound,P µ   ∞ s=1 {µ i ≤ UCB(i, t i (s))}   ≥ 1 - ∞ s=1 P µ µ i ≥ UCB(i, t i (s)) ≥ 1 - ∞ s=1 δ 2 2ks 2 • 6 π 2 ≥ 1 - ∞ s=1 δ 2 2ks 2 • 6 π 2 = 1 - δ 2 2k ,and the proof is completed.Let us define a good event based on Lemma 15 asEQUATIONWe now prove that, under H µ and under this good event X (µ)• The best arm i * (µ) is always in the active arm set A(t) for all t (Lemma 16),• Each count of the suboptimal arm pull, N i (T 0 ), is bounded by roughly O( log ∆i∆ 2 i ) (Lemma 17). Lemma 16. Let X ′ (µ) = ∞ t=1 {i * (µ) ∈ A(t)} .Then, under H µ , X (µ) ⊂ X ′ (µ) and thereforeP µ [(X ′ (µ)) c ] ≤ δ 2 .Proof of Lemma 16. Suppose that event X (µ) occurs under H µ . Then for all i ∈ [k] and for all t, μi -Conf(i, t) ≤ µ i and μi + Conf(i, t) ≥ µ i . Now, for any i̸ = i * , LCB(i, t) -UCB(i * , t) = μi -Conf(i, t) -(μ i * + Conf(i * , t)) ≤ µ i + Conf(i, t) -Conf(i, t) -(µ i * -Conf(i * , t) + Conf(i * , t)) (Event X occurs) = µ i -µ i * < 0which means when event X occurs, the optimal arm will never be dropped, and thus X ⊂ X ′ . By Lemma 15,P µ (X ′ (µ)) ≥ P µ (X (µ)) ≥ 1 -δ 2 .Lemma 17. For any i ̸ = i * , under H µ we haveEQUATIONand therefore,E µ [N i (T 0 )1[X (µ)]] ≤ R 0 max(∆ i , ∆ 0 ) .Proof of Lemma 17. Only for this part of the proof, let T i := R 0 max(∆ i , ∆ 0 ) for brevity.When ∆ i < ∆ 0 , max(∆ i , ∆ 0 ) = ∆ 0 , which, combined with Corollary 14, implies that {N i (T 0 ) ≥ T i } always holds.For the case of ∆ i > ∆ 0 , suppose that the learner is under the events X (µ) and {i ∈ A(Ti )}. Note that Conf(a, T i ) ≤ ∆ i 4 , ∀a ∈ A(T i ).Then,max a∈A(Ti) LCB a (T i ) -UCB i (T i ) ≥ LCB i * (T i ) -UCB i (T i ) = μi * -Conf i * (T i ) -(μ i + Conf i (T i )) ≥ µ i * -2Conf i * (T i ) -(µ i + 2Conf i (T i )) (X occurs) ≥ µ i * -µ i -∆ i = 0.Therefore, when X occurs, µ i should be eliminated after timestep T i so N i (T 0 ) ≤ T i .Lemmas 16 and 17 guarantee the Bayesian δ-correctness of our Algorithm 1. Theorem 18. (δ-correctness) The Bayesian PoE of Algorithm 1 is at most δ, i.e.,EQUATION)Proof of Theorem 18. Throughout the proof, we use the following results.• The probability that µ i * (µ) -µ j * (µ) ≤ ∆ 0 is at most δ 2 by Eq. ( 21).• The event X fails to hold with probability at most δ 2 by Lemma 15.• When µ i * (µ) -µ j * (µ) > ∆ 0 and event X occurs, by Lemmas 16 and 17, all suboptimal arms will be eliminated before T j * and only the optimal arm will remain in the set. This means E(µ) ⊂ X (µ)∪ {µ i * (µ) -µ j * (µ) ≤ ∆ 0 }.Therefore,PoE(π; H) = E µ∼H P J ̸ = i * (µ)|H µ = E µ∼H E[1(E(µ))|H µ ] ≤ E µ∼H E[1({µ i * (µ) -µ j * (µ) ≤ ∆ 0 }) + 1(X (µ))|H µ ] ≤ E µ∼H 1({µ i * (µ) -µ j * (µ) ≤ ∆ 0 }) + E[1(X (µ))|H µ ] ≤ δ 2 + δ 2 < δ.Finally, Lemma 19 shows the upper bound of the expected stopping time of our algorithm.Lemma 19. We haveE[τ ] ≤ B 0 L(H) 2 δ log L(H) δ + O(log δ -1 ).Proof. We haveE[τ ] = k s=1 E[N s (T 0 )] = k s=1 E[E µ [N s (T 0 )]] = k s=1 E E µ [N s (T 0 )1[X (µ)]] + k s=1 E E µ [N s (T 0 )1[X c (µ)]] ≤ k s=1 E E µ [N s (T 0 )1[X (µ)]] + T 0 • δ 2(Corollary 14 and Lemma 15)and since ∆ 0 < ∆ thr by assumption,EQUATION)Therefore, it remains to compute the scale of the first term,EQUATION(Recall thatΘ i = {µ ∈ R k : µ i ≥ max j̸ =i µ j }).In the following, we calculate each µ∈Θi T s (µ) dH(µ). By assumption, ∆ 0 < ∆ thr . For this part, we define a new notation that for each vector v ∈ R k and i, j∈ [k], v \i (and v \i,j ) is the projection of v to R k-1 (R k-2 , respectively) by omitting i-th coordinate (i, j-th coordinate, respectively), v \i := (v 1 , v 2 , • • • , v i-1 , v i+1 , • • • , v k ) k-1 coordinates . Similarly, for a k-dimensional distribution H and i, j ∈ [k],H \i (and H \i,j ) is the distribution which omits i-th (i, j-th, respectively) coordinate.Case 1: s ̸ = i In this case, by Lemma 17, µ∈Θi T s (µ) dH(µ) = µ∈Θi T s (µ)(1[µ i -µ s ≥ ∆ 0 ] + 1[µ i -µ s < ∆ 0 ]) dH(µ) = µi∈R µi-∆ thr µs=-∞ µ \i,s ∈(-∞,µi) k T s (µ)h i (µ i )h s (µ s ) dH \i,s dµ s dµ i + µi∈R µi-∆0 µs=µi-∆ thr µ \i,s ∈(-∞,µi) k T s (µ)h i (µ i )h s (µ s ) dH \i,s dµ s dµ i + µi∈R µi µs=µi-∆0 µ \i,s ∈(-∞,µi) k T s (µ)h i (µ i )h s (µ s ) dH \i,s dµ s dµ i ≤B • log ∆ -1 thr ∆ 2 thr + µi∈R µi-∆0 µs=µi-∆ thr µ \i,s ∈(-∞,µi) k B log(µ i -µ s ) -1 (µ i -µ s ) 2 h i (µ i )h s (µ s ) dH \i,s dµ s dµ i + µi∈R µi µs=µi-∆0 µ \i,s ∈(-∞,µi) k B log ∆ -1 0 ∆ 2 0 h i (µ i )h s (µ s ) dH \i,s dµ s dµ i (Lemma 17) = µi∈R µi-∆0 µs=-µi-∆ thr B log(µ i -µ s ) -1 (µ i -µ s ) 2 h i (µ i )h s (µ s )   k∈[k]\{i,s} H k (µ i )   dµ s dµ i + B log ∆ -1 0 ∆ 2 0 P(i * (µ) = i, µ i -µ s ≤ ∆ 0 ) + B • log ∆ -1 thr ∆ 2 thr ≤B log ∆ -1 0 µi∈R h i (µ i ) µi-∆0 µs=-∞ 1 (µ i -µ s ) 2 h s (µ s )   k∈[k]\{i,s} H k (µ i )   dµ s dµ i (A) + B log ∆ -1 0 ∆ 2 0 P(i * (µ) = i, µ i -µ s ≤ ∆ 0 ) (Pis) +B log ∆ -1 thr ∆ 2 thr .We first deal with the term (A). The following splits (A) into the sum of two integrals (A1) and (A2):(A) = µi∈R h i (µ i )   k∈[k]\{i,s} H k (µ i )   µi-∆0 µs=-∞ 1 (µ i -µ s ) 2 h s (µ s ) dµ s dµ i = ⌈ 1 2 √ ∆ 0 ⌉-1 l=1 µi∈R h i (µ i )   k∈[k]\{i,s} H k (µ i )   µi-l∆0 µs=µi-(l+1)∆0 1 (µ i -µ s ) 2 h s (µ s ) dµ s dµ i (A1) + µi∈R h i (µ i )   k∈[k]\{i,s} H k (µ i )   µi-⌈ 1 2 √ ∆ 0 ⌉∆0 µs=-∞ 1 (µ i -µ s ) 2 h s (µ s ) dµ s dµ i (A2).For (A1), we haveEQUATIONNow for (A2), we evaluate the inner integral of (A2):(Inner -A2) := 1 √ 2πσ s µi-⌈ 1 2 √ ∆ 0 ⌉∆0 -∞ 1 (µ i -µ s ) 2 exp - (µ s -m s ) 2 2σ 2 s dµ s = 1 √ 2πσ s 1 (µ i -µ s ) exp - (µ s -m s ) 2 2σ 2 s µi-⌈ 1 2 √ ∆ 0 ⌉∆0 -∞ + 1 √ 2πσ s µi-⌈ 1 2 √ ∆ 0 ⌉∆0 -∞ µ s -m s σ 2 s (µ i -µ s ) exp - (µ s -m s ) 2 2σ 2 s dµ s (partial integration) ≤ 1 √ 2πσ s 1 ⌈ 1 2 √ ∆0 ⌉∆ 0 + 1 √ 2πσ 3 s µi-⌈ 1 2 √ ∆ 0 ⌉∆0 -∞ µ i -m s µ i -µ s -1 exp - (µ s -m s ) 2 2σ 2 s dµ s (µ i > µ s ) ≤ 1 √ 2πσ s 1 ⌈ 1 2 √ ∆0 ⌉∆ 0 + 1 √ 2πσ 3 s µi-⌈ 1 2 √ ∆ 0 ⌉∆0 -∞ µ i -m s ⌈ 1 2 √ ∆0 ⌉∆ 0 exp - (µ s -m s ) 2 2σ 2 s dµ s ≤ 1 √ 2πσ s 1 ⌈ 1 2 √ ∆0 ⌉∆ 0 + max   1 σ 2 s µ i -m s ⌈ 1 2 √ ∆0 ⌉∆ 0 , 0   .By integrating above over variable i, we haveEQUATIONTherefore, by Eq. ( 29) and Eq. ( 30),(A) = (A1) + (A2) = 1 ∆ 2 0 • π 2 3 P is + O(∆ -1/2 0 ),and therefore (N s (t) increases only when |A t | > 1, so there should be a competitor)EQUATION≤ j̸ =s µ∈Θsj max i̸ =s R 0 (∆ i (µ), ∆ 0 ) dH(µ) (Lemma 17) ≤ j̸ =s µ∈Θsj R 0 (∆ j (µ), ∆ 0 ) dH(µ)and by the same calculation as Case 1, we obtainµ∈Θsj R 0 (∆ j (µ), ∆ 0 ) dµ ≤ B log ∆ -1 0 ∆ 2 0 P sj π 2 3 + 1 + O(∆ -1/2 0 ),and thereforeEQUATIONFor notational convenience, let B 0 = B • π 2 3 + 1 . From Eq. ( 28), Eq. ( 31), Eq. ( 32), we getEQUATION≤ k i=1 µ∈Θi T s (µ) dH(µ) + E µ∼H E µ [N s (T 0 )1[X (µ) c ]](Eq. ( 28))≤ i̸ =s µ∈Θi T s (µ) dH(µ) + µ∈Θs T s (µ) dH(µ) + E µ∼H E µ [N s (T 0 )1[X (µ) c ]] ≤ B 0 log ∆ -1 0 ∆ 2 0   i̸ =s P is + j̸ =s P sj   + O(∆ -1/2 0 ) + E µ∼H E µ [N s (T 0 )1[X (µ) c ]](Eq. ( 31) and ( 32))EQUATIONNow, the total stopping time is bounded as follows:E[τ ] = E[ k s=1 N s (T 0 )] = B 0 log ∆ -1 0 ∆ 2 0       k s=1 P(i * (µ) ̸ = s, µ i * (µ) -µ s ≤ ∆ 0 ) Psum1 + k s=1 P(i * (µ) = s, µ s -µ j * (µ) ≤ ∆ 0 ) Psum2       + O(∆ -1/2 0 ) + E[τ 1[X c ]].(Eq. ( 34))The final task we have left is bounding (Psum1) and (Psum2). Let us define k * (µ) as the third best arm in µ. For (Psum1),(Psum1) = k s=1 P(i * (µ) ̸ = s, µ i * (µ) -µ s ≤ ∆ 0 ) = k s=1 P(j * (µ) = s, µ i * (µ) -µ s ≤ ∆ 0 ) + k s=1 P(j * (µ) ̸ = s, µ i * (µ) -µ s ≤ ∆ 0 ) ≤ P(µ i * (µ) -µ j * (µ) ≤ ∆ 0 ) + k s=1 P(µ i * (µ) -µ k * (µ) ≤ ∆ 0 ) (When j * (µ) ̸ = s, µ k * (µ) ≥ s) E.2 Proof of Lemmas E.2.1 Proof of Lemma 20Lemma 20. For S + 1 ≤ 1 √ δ and for δ ≤ ξ i L is (H) 2 , we haveP(i * (µ) = i, µ i -µ s ∈ [Sδ, (S + 1)δ]) ≤ 2P(i * (µ) = i, µ i -µ s ≤ δ).Proof. We haveΘi1 [|µ i -µ s | ∈ [Sδ, (S + 1)δ]] dH(µ) = Θ \i µs+(S+1)δ µi=µs+Sδ h i (µ i ) dµ i dH \i (µ \i ) ≤ Θ \i δ h i (µ s ) + e -1/2 ξ i (S + 1)δ dH \i (µ \i )(by Lipschitz property of Gaussian, e -1/2 /ξ i is the steepest slope of N (m i , ξ 2 i ))= δ(Θ \i h i (µ s ) dH(µ \i ) Lis(H) + e -1/2 ξ i (S + 1)δ) ≤ (L is (H) + 1 ξ i (S + 1)δ)δ. When S + 1 < 1 √ δ and √ δ < L is (H)ξ i , we have Θi 1 [|µ i -µ j | ∈ [Sδ, (S + 1)δ]] dH(µ) ≤ 2L is (H)δas intended.We will show that the probability that three or more arms are δ-close is O(δ 2 ). Namely: Lemma 21. We haveP(µ i * (µ) -µ k * (µ) ≤ ∆ 0 ) = O(δ 2 ). Proof. For any i ̸ = j ̸ = k ∈ [k], we have 1 [|µ i -µ j |, |µ i -µ k | ≤ δ] dH(µ) = µ \jk µi+δ µj =µi-δ µi+δ µ k =µi-δ h j (µ j )h k (µ k ) dµ k dµ j dH \jk (µ \jk ) ≤ µ \jk µi+δ µj =µi-δ µi+δ µ k =µi-δ h j (µ i ) + e -1/2 δ ξ j h k (µ i ) + e -1/2 δ ξ k dµ k dµ j dH \jk (µ \jk ) (Lipschitz property of Gaussian) ≤ (2δ) 2 µ \jk h j (µ i )h k (µ i ) + O(δ) dH \jk (µ \jk ) =:Q ijk (H) = O(δ 2 ).The code used in the experiments for this paper can be found at the following link: https:// github.com/jajajang/FC_BAI_Bayes.TTTS We use our theoretical results stated in Section 5 for our stopping criterion. For TTTS, we use Chernoff's stopping rule, as Garivier and Kaufmann [2016] , Jourdan et al. [2022] did. Here is the description of how it works: for each arm i, j ∈[k], let μij (t) := N i (t) N i (t) + N j (t) μi (t) + N j (t) N i (t) + N j (t) μj (t),and defineZ ij (t) := N i (t) • KL i (μ i , μij ) + N j (t) • KL j (μ j , μij ).Now the stopping time is defined as:τ T T T S := inf{t ∈ N : max a∈[k] min b∈[k]\a Z ab (t) ≥ β(t, δ)}for some threshold function β(t, δ), which is defined by the following proposition of Garivier and Kaufmann [2016] :Theorem 22 (Garivier and Kaufmann 2016, Proposition 12) . Let µ be an exponential family bandit model. Let δ ∈ (0, 1) and α > 1. There exists a constant C = C(α, k) such that whatever the sampling strategy, using Chernoff's stopping rule with the thresholdβ(t, δ) = log Ct α δ ensures that for all µ, P µ τ < ∞, J ̸ = i * (µ) ≤ δ.In this theorem, we give an advantage to the stopping time of TTTS by setting α = 1, C = 1.Theoretically, C(α, k) > 1 and C → ∞ as α → 1 + , but we set the threshold smaller than the theoretical guarantee so that TTTS stops earlier. Let ît := arg max i∈[k] μi (t), the empirical best arm at step t. The TTUCB algorithm stops whenmin i̸ = ît μî (t) -μi (t) 1 N ît (t) + 1 Ni(t) ≥ c(t, δ).When the algorithm stops sampling, the TTUCB algorithm recommends the empirical best arm as its final suggestion.Since the computation of C G (x) involves optimization, it is computationally heavy when the number of samples is excessively large (as our Table 1 ). Instead, we approximated C G (x) ≈ x + log x as mentioned in Jourdan and Degenne [2022b] .The NoElim algorithm is shown in Algorithm 2.For all tables in this section, Comp represents the average computation time (second).• We stop additional sampling of TTUCB when its number of samples is over 10 8 because of the time constraint. This means we gave some 'advantage' on TTUCB about expected stopping time (since it makes TTUCB stop earlier than it should.)Result Our algorithm had the average stopping time of 1.1 × 10 5 , while TTUCB had the average stopping time of 7.1 × 10 5 , again proving the superiority of Algorithm 1 over TTUCB in Bayesian settings. L ij (H) := ∞ -∞ h i (x)h j (x) s:s∈[k]\{i,j} H s (x) dx.In Scipy package, there are functions for computing the cumulative function of Gaussian H s (scipy.norm.cdf) and h i (scipy.norm.pdf). Scipy package also supports the numerical integration (scipy.integrate.quad) which we use to numerically compute L ij in our experiments.Codes The codes are in the following GitHub repository: https://github.com/jajajang/ FC_BAI_Bayes.Hardware We used Python 3.7 as our programming language and Macbook Pro M2 16 inch as our hardware.The second result of the Lemma 9 is used for the lower bound. For this result to hold, we need the following two conditions for D 1 :• Proof of Lemma 1, second result: For the proof of Eq. ( 9), we used |m i | ≤ 1 2 √ ∆ . • Proof of Lemma 1, second result: For the proof of Eq. ( 9), we used1 ξi ∆ 2 > 2 exp -1 8∆ξ 2 i + 2 exp -1 8∆ξ 2 j . To satisfy this condition, ∆ < D 0 (H) where D 0 (H) :=    W (- 1 32 max i∈[k] ξ 3/2 i ) If max i∈[k] ξ i > 3 e 2 2 10 1 Otherwise .Here W is the Lambert W function with the principal branch.For the lower bound proof, we consider sufficiently small δ subject to the following constraints on ∆ = 32e 4 L(H) δ:• Two conditions above for Lemma 9.• For the Lemma 12:∆ < D 1 (H) := min i̸ =j mi σ 2 i - mj σ 2 j -1 , 1 2σ 2 i + 1 2σ 2 j -2 . • To make L ′ (H, ∆) ∆ ∈ ( 1 2 L(H), 2L(H)), ∆ ≤ L(H) 4 i∈[k] k-1 ξ i .In summary, Theorem 1 holds for anyδ ≤ δ L := L(H) 32e 4 • min D 0 (H), D 1 (H), min i∈[k] 1 4m 2 i , L(H) 4(k -1) i∈[k] 1 ξi .For the upper bound proof (Theorem 6), we consider δ such that ∆ 0 = δ 4L(H) satisfies the following conditions: • Providing as much information as possible in supplemental material (appended to the paper) is recommended, but including URLs to data and code is permitted. 6. Experimental Setting/Details Question: Does the paper specify all the training and test details (e.g., data splits, hyperparameters, how they were chosen, type of optimizer, etc.) necessary to understand the results? Answer: [Yes] Justification: We specify all the experimental details in Section 6 and Appendix F. Guidelines:• ∆ 0 < L(H) i∈[k] k-1 ξ i to make L(H, ∆ 0 ) • ∆ 0 ≤ 2L(H)∆ 0 by the first result of Lemma 9. • ∆ 0 ≤ min i,j∈[k],i̸ =j (L ij ξ i ) 2 for• The answer NA means that the paper does not include experiments.• The experimental setting should be presented in the core of the paper to a level of detail that is necessary to appreciate the results and make sense of them. • The full details can be provided either with the code, in appendix, or as supplemental material. 
experiment statistical significance: 
 Question: Does the paper report error bars suitably and correctly defined or other appropriate information about the statistical significance of the experiments? Answer: [Yes] Justification: In Section 6 we provide average stopping time, maximum stopping time, and the ratio of misidentification to report our result statistically. Guidelines:• The answer NA means that the paper does not include experiments.• The authors should answer "Yes" if the results are accompanied by error bars, confidence intervals, or statistical significance tests, at least for the experiments that support the main claims of the paper. • The factors of variability that the error bars are capturing should be clearly stated (for example, train/test split, initialization, random drawing of some parameter, or overall run with given experimental conditions).• Examples of negative societal impacts include potential malicious or unintended uses (e.g., disinformation, generating fake profiles, surveillance), fairness considerations (e.g., deployment of technologies that could make decisions that unfairly impact specific groups), privacy considerations, and security considerations. • The conference expects that many papers will be foundational research and not tied to particular applications, let alone deployments. However, if there is a direct path to any negative applications, the authors should point it out. For example, it is legitimate to point out that an improvement in the quality of generative models could be used to generate deepfakes for disinformation. On the other hand, it is not needed to point out that a generic algorithm for optimizing neural networks could enable people to train models that generate Deepfakes faster. • The authors should consider possible harms that could arise when the technology is being used as intended and functioning correctly, harms that could arise when the technology is being used as intended but gives incorrect results, and harms following from (intentional or unintentional) misuse of the technology. • If there are negative societal impacts, the authors could also discuss possible mitigation strategies (e.g., gated release of models, providing defenses in addition to attacks, mechanisms for monitoring misuse, mechanisms to monitor how a system learns from feedback over time, improving the efficiency and accessibility of ML). 
safeguards: 
 Question: Does the paper describe safeguards that have been put in place for responsible release of data or models that have a high risk for misuse (e.g., pretrained language models, image generators, or scraped datasets)? Answer: [NA] Justification: This work poses no such risks. Guidelines:• The answer NA means that the paper poses no such risks.• Released models that have a high risk for misuse or dual-use should be released with necessary safeguards to allow for controlled use of the model, for example by requiring that users adhere to usage guidelines or restrictions to access the model or implementing safety filters. • Datasets that have been scraped from the Internet could pose safety risks. The authors should describe how they avoided releasing unsafe images. • We recognize that providing effective safeguards is challenging, and many papers do not require this, but we encourage authors to take this into account and make a best faith effort. 12. Licenses for existing assets Question: Are the creators or original owners of assets (e.g., code, data, models), used in the paper, properly credited and are the license and terms of use explicitly mentioned and properly respected? Answer: [NA] Justification: This paper does not use existing assets. Guidelines:• The answer NA means that the paper does not use existing assets.• The authors should cite the original paper that produced the code package or dataset.• The authors should state which version of the asset is used and, if possible, include a URL. • The name of the license (e.g., CC-BY 4.0) should be included for each asset.• For scraped data from a particular source (e.g., website), the copyright and terms of service of that source should be provided. • If assets are released, the license, copyright information, and terms of use in the package should be provided. For popular datasets, paperswithcode.com/datasets has curated licenses for some datasets. Their licensing guide can help determine the license of a dataset.38th Conference on Neural Information Processing Systems (NeurIPS 2024).In fact, marginalizing the frequentist sample complexity over the prior distribution leads to an unbounded value.Here the hardness is based on the size of the suboptimality gap. When the suboptimality gap of an instance is small, it is a harder instance for the BAI algorithm to identify the best arm, as mentioned in Example 1.For details about T * (µ), a reader may refer toGarivier and Kaufmann [2016].In particular, δL(H) is defined in Appendix G.We include the computation time in the Appendix F.3See Appendix F.2 for the pseudocode.See Appendix G for the definition of D0(H). 
 
 Appendix:  


Decision: This paper considers a Bayesian formulation of the classic fixed confidence best-arm identification problem. In particular, they consider the special case of known independent Gaussian priors for each arm, and Gaussian rewards with known variance. Given the overlapping support of the priors, there is a non-zero chance of an arbitrarily small minimum gap which blows up the sample complexity of naively applied delta-PAC algorithms for the frequentist setting. The main contributions are a novel lower bound for this setting and an algorithm that matches it up to log factors.

The goals of the paper are clear and they are well-executed. Perhaps the biggest weakness of this paper is the assumption that the prior distributions are Gaussian. As the critical identified quantity in the sample complexity denoted as L(H) is derived directly from this assumption, it is unlikely to be meaningful in the general prior case. Indeed, if the prior is a uniformly random permutation of a fixed set of separated means, L(H) has no impact on the sample complexity of this problem (see Chen, Li, Qiao 2016 and Simchowitz and Jamieson 2016). As some reviewers suggested, the authors would improve the paper by adding a longer discussion motivating L(H) and under what prior conditions it is a relevant quantity.

Reviews: review: paper_summary: paper_summary: The paper studies the problem of FC-BAI; the goal is to find the arm with largest mean with a given probability of correctness. It analyzes FC-BAI for a Gaussian bandit model where the arms reward mean is sampled from a known prior and the reward variances are known. It proves other (frequentist) FC-BAI algorithms could fail at this setting. Then it proves a lower bound for the sample complexity of this problem. Finally, an elimination algorithm is introduced which is near-optimal. The main difference of this algorithm with previous algorithms is in its less conservative stopping criteria, where it stops if the remaining arms have small enough optimality gap.
strength_weakness: strength_weakness: *Contribution*: The paper considers a small subset of bandit models, Gaussian reward with know prior and variance. This limits applicability of the algorithm and theoretical results. As also mentioned in the discussion section, it would be great if they could generalize the algorithm to deal with misspecified or under-specified prior (see [1]), or unknown variance.


[1] Azizi, J., Kveton, B., Ghavamzadeh, M. and Katariya, S. 2023. Meta-Learning for Simple Regret Minimization. Proceedings of the AAAI Conference on Artificial Intelligence. 37, 6 (Jun. 2023), 6709-6717. DOI:https://doi.org/10.1609/aaai.v37i6.25823.
questions: questions: 1. In Thm 4, the constants in the lower bound seem very small ($\frac{1}{16 e^4} \approx \frac{1}{2^8}$), so it could entail a trivial lower bound in most cases, specially since it is the lower bound to the sample complexity (natural number). What is the actual value of this lower bound in some of your example simulations?
limitations: limitations: Yes.
paper_summary: paper_summary: In this paper, the focus is on the fixed-confidence best-arm identification problem within a Bayesian setting. The objective is to determine the arm with the largest mean with a certain confidence, given a known prior distribution. Existing work in FC-BAI is mostly in the frequentist setting. This paper shows that the popular algorithms used in the frequentist setting are not optimal in this setting. Additionally, a lower bound for this setting is derived, using a sampling complexity constant and the confidence level. Lastly, a successive elimination method that matches the order of the derived lower bound along with experiments comparing the performance of popular frequentist algorithms and this algorithm are presented.
strength_weakness: strength_weakness: 1. 47: Example 1 lacks proof or reference for the number of samples.

2. The authors do not discuss how they derived the constant L(H) and its significance. Since this constant represents sampling complexity and is not a traditional term, it is harder to compare the complexity to existing work.

3. 171: In Chapter 3, the authors claim there is a known lower bound for the expected stopping time without providing any reference.

4. 196: The derivation of the inequalities in Section 3.1 is not clear.

5. 242-243: In Chapter 5, there is no explanation for why this particular confidence width was considered.

6. In Chapter 6, there is no discussion on why Algorithm 1 shows higher error percentages than the other two algorithms.

7. 313: there are two dots instead of one.
questions: questions: - 267-268: In Remark 3, can the authors explain if in the general case the result matches with the lower bound?
- 309-313: The authors state that the expected stopping time of the track-and-stop method is at least half of the TTS and TTUCB methods, which according to Table 1, is on average at least 300. This number is less than that for Algorithm 1, so can the authors explain why they didn’t compare?
limitations: limitations: The research was done only for the Gaussian distribution.
paper_summary: paper_summary: This paper considers a best arm identification problem in Bayesian multi-armed bandit setting. The arms' distributions are generated according to the unknown prior, and the probability of error is averaged over this prior distribution. The paper makes two key contributions: first, a lower bound characterizing the fundamental hardness of BAI in the Bayesian setting (where the L(H) parameter is identified as playing a key role), and second, and achievability part that proposes a successive elimination-based algorithm (with early stopping in case the arm instance has very close best arms). The algorithm achieves within a logarithmic factor of the lower bound.
strength_weakness: strength_weakness: There is no glaring weakness. It appears to be obvious in hindsight that "very close" instances do not occur with substantial probability, when we have a Gaussian prior, making the hard frequentist instances statistically irrelevant in the Bayesian setting. This makes the BAI problem easier in terms of the expected stopping time. If the prior model creates close instances with non-vanishing probability, I suppose we'd be back to the frequentist performance.
questions: questions: The definition of $\mathcal H_\mu$ at the beginning of page 4 may not be precise...is it an event or a $\sigma$-algebra? In the former case, with continuous Gaussian priors, it is not clear if the event has non-zero probability, which makes all the conditional probabilities technically dubious. Please clarify.
limitations: limitations: Limitations are acknowledged in the conclusions reasonably.
paper_summary: paper_summary: This paper considers the best arm identification (BAI) problem in the fixed confidence setting and in the Bayesian setting, where the mean rewards of each arm is drawn from a known prior distribution. The authors formulate the problem and discuss the related literature. The authors provide the lower bound of the expected sample complexity of finding the optimal arm, which is proportional $L(\mathbb{H})/\delta$ where $L(\mathbb{H})$ is defined by the authors to characterize the sample complexity. Then the authors show the current optimal algorithms in the frequentist setting: the top two and tack-and-stop algorithms are suboptimal in the Bayesian setting. The authors then propose a new algorithm based on the successive elimination algorithm in the frequentist setting, showing that this algorithm reaches the optimal sample complexity up to a logarithmic factor. The authors provide the upper bound of the algorithm on the expected sample complexity. Finally the authors run two simulations to show that the top-two algorithm is indeed suboptimal comparing with Algorithm 1, and also show the superior performance of algorithm1 compared to the no-elimination version of it.
strength_weakness: strength_weakness: 1. The assumption of the arms may be stringent, i.e. all arms are from the Gaussian distribution. The current literature considers many more general distributions, such as sub-gaussian, exponential family, or even non-parametric cases.
2. The confidence interval is not optimized to the tightest confidence interval which is suggested in the lil-UCB paper from Jamieson: "lil' UCB : An Optimal Exploration Algorithm for Multi-Armed Bandits"
3. The lower bound proof (and possibly the upper bound proof) uses standard techniques from the literature. Perhaps the authors could point out the difficulty or challenge in the proofs.
4. The upper bound is not optimal compared to the lower bound, with an extra logarithmic factor. Is it possible to reduce this gap using tighter confidence bound or do some modification to the algorithm, such as modification of top-two or tack-and-stop instead of modify the SE algorithm (which is not optimal in the frequentist setting)
5. The simulations are limited to just two sets of experiments. The baselines or compared algorithms are just 2/1 in the respective setting. And it might be better to include different $\delta$ values since the authors consider the fixed confidence setting and change priors as well.
questions: questions: 1. Is it possible to use tighter confidence intervals to get a better upper bound?
2. Is it trivial or direct to consider similar algorithms when the distribution of the arms is more general, like subgaussian, or non-parametric?
3. Is it possible to modify top-two or tack-and-stop to make it better in the Bayesian setting?
4. Just out of curiosity, is it possible to modify LUCB or lil-UCB to this setting so that it achieves similar performance when the authors modify the successive elimination algorithm? Since the former two algorithms have better sample complexity in the frequentist setting.
5. Can the upper bound be modified to a high probability bound instead of an expectation bound on the sample complexity? Would it be challenging or trivial?
limitations: limitations: See weaknesses and questions.

Comments: title: Thank you for your response.
Details on experiment (W5)
comment: I appreciate your technical clarification. I keep my score.
Thank you for providing detailed explanations in response to the points I raised in my review. I appreciate the effort and clarity with which you addressed my concerns.
After reviewing your responses, I understand and acknowledge the explanations provided. While your rebuttal has clarified several aspects of your work, I believe that a score of 6 is an appropriate reflection of the significance of your research. At this point, I do not plan on changing my score.
I wish you the best of luck with your submission.
Thanks to the authors for addressing all my questions and concerns. The explanations are pretty convincing and I have bumped my score. Good luck.
**Experiment Setting**

- Number of arms $k=10$
- Prior means: we sample 10 random numbers before the experiment starts, and set them as prior means. Here's the list of prior means: [-0.053  0.528 -0.332 -0.368 -0.273  0.909 0.418 -1.17  0.873 -0.405]
- Prior variance: we sample 10 random numbers before the experiment starts, and set them as prior means. Here's the list of prior std: [0.604 1.477 1.163 0.988 0.560 0.513 0.997 1.332 0.828 0.833]
- Instance variance: we sample 10 random numbers before the experiment starts, and set them as prior means. Here's the list of instance std: [1.498, 1.262, 1.485, 0.963, 1.375, 0.969, 1.357, 1.238, 1.088, 0.699]
- Number of experiments: 500
- We stop additional sampling of TTUCB when its number of samples is over $10^8$ because of the time constraint. This means we gave some 'advantage' on TTUCB about expected stopping time (since it makes TTUCB stop earlier than it should.)