SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \left(\frac {N _ {n , i}}{N _ {n , i ^ {\star}}}\right) ^ {2} \geq 1 - \sum_ {j \notin \{i, i ^ {\star} \}} \left(\frac {N _ {n , j}}{N _ {n , i ^ {\star}}}\right) ^ {2} - \tilde {\gamma} \\ \geq 1 - \sum_ {j \notin \{i, i ^ {*} \}} \left(\frac {w _ {j} ^ {\star}}{w _ {i ^ {*}} ^ {\star}} + \tilde {\gamma}\right) ^ {2} - \tilde {\gamma} \\ = \left(\frac {w _ {i} ^ {\star}}{w _ {i \star} ^ {\star}}\right) ^ {2} - \left((K - 2) \tilde {\gamma} + 2 \sum_ {j \notin \{i, i ^ {\star} \}} \frac {w _ {j} ^ {\star}}{w _ {i \star} ^ {\star}} + 1\right) \tilde {\gamma} \\ \end{array}
$$
where the equality uses (9). Therefore, we have shown that for all  $n \geq \max\{N_4, N_6\}$ ,
$$
\frac {N _ {n , i}}{N _ {n , i ^ {\star}}} \geq \frac {w _ {i} ^ {\star}}{w _ {i ^ {\star}} ^ {\star}} \sqrt {1 - \left(\frac {w _ {i ^ {\star}} ^ {\star}}{w _ {i} ^ {\star}}\right) ^ {2} \left((K - 2) \tilde {\gamma} + 2 \sum_ {j \notin \{i , i ^ {\star} \}} \frac {w _ {j} ^ {\star}}{w _ {i ^ {\star}} ^ {\star}} + 1\right) \tilde {\gamma}} \geq \frac {w _ {i} ^ {\star}}{w _ {i ^ {\star}} ^ {\star}} - \gamma ,
$$
where the last inequality holds for  $\tilde{\gamma}$  small enough as a function of  $\gamma$  and  $w^{\star}$ . Therefore, we obtained
$$
\max  _ {i \neq i ^ {\star}} \left| \frac {N _ {n , i}}{N _ {n , i ^ {\star}}} - \frac {w _ {i} ^ {\star}}{w _ {i ^ {\star}} ^ {\star}} \right| \leq \gamma ,
$$
which concludes the proof, i.e.  $\mathbb{E}_{\nu}[T_{\gamma}] < +\infty$
Combining Lemmas 13 and 24 concludes the first part of Theorem 1, i.e.
D.2 Fixed proportions
Using Lemma 9, the  $\beta$ -optimal allocation for  $\varepsilon_0$ -BAI are defined as
$$
w_{\varepsilon_{0},\beta}(\mu):= \operatorname *{arg  max}_{w\in \triangle_{K},w_{i^{\star}} = \beta}\min_{i\neq i^{\star}}\frac{(\mu_{i^{\star}} - \mu_{i} + \varepsilon_{0})^{2}}{1 / \beta + 1 / w_{i}}  .
$$
Since  $|i^{\star}(\mu)| = 1$ , let  $\mu_{\varepsilon_0} = \mu_{\varepsilon_0}(i^{\star})$  where  $\mu_{\varepsilon}(i^{\star})$  as in Lemma 9. Since we have  $T_{\varepsilon_0,\beta}(\mu) = T_{\beta}^{\star}(\mu_{\varepsilon_0})$  and  $w_{\varepsilon_0,\beta}(\mu) = w_{\beta}^{\star}(\mu_{\varepsilon_0})$ , Lemma 6 and 8 yield that  $w_{\varepsilon_0,\beta}(\mu) = \{w_{\beta}^{\star}\}$  is a singleton with unique element denotes by  $w_{\beta}^{\star}$  which satisfies that  $\min_{i\in [K]}w_{\beta ,i}^{\star} > 0$  and
$$
\forall i \in [ K ] \backslash \{i ^ {\star} \}, \quad \frac {\left(\mu_ {i ^ {\star}} - \mu_ {i} + \varepsilon_ {0}\right) ^ {2}}{1 / \beta + 1 / w _ {\beta , i} ^ {\star}} = 2 T _ {\varepsilon_ {0}, \beta} (\mu) ^ {- 1}. \tag {13}
$$
When considering fixed proportions  $\beta$ , the proof strategy is the same as in Appendix D.1. The sole difference lies in the fact that the empirical allocation of the unique best arm is converging towards a fixed  $\beta$ .
Convergence time Let  $\gamma > 0$ . Let us define the convergence time  $T_{\varepsilon_0,\beta,\gamma}$ , which is a random variable quantifies the number of samples required for the empirical allocations  $N_n$  to be  $\gamma$ -close to  $w_\beta^\star$ :
$$
T _ {\varepsilon_ {0}, \beta , \gamma} := \inf  \left\{T \geq 1 \mid \forall n \geq T, \left\| \frac {N _ {n}}{n - 1} - w _ {\beta} ^ {\star} \right\| _ {\infty} \leq \gamma \right\}. \tag {14}
$$
Lemma 25 gives a sufficient condition for asymptotic  $\beta$ -optimality. The case  $\varepsilon_0 = \varepsilon_1$  is a direct consequence of existing methods, e.g. Theorem 2 in [21] or Theorem 3 in [32].
Lemma 25. Let  $\varepsilon_0 > 0$ ,  $\varepsilon_1 \geq 0$  and  $\delta \in (0,1)$ . Assume that there exists  $\gamma_1(\mu) > 0$  such that for all  $\gamma \in (0,\gamma_1(\mu)], \mathbb{E}_{\nu}[T_{\varepsilon_0,\beta,\gamma}] < +\infty$ . Using the threshold (4) in the stopping rule (3) with slack  $\varepsilon_1$  yields an algorithm such that, for all  $\nu \in \mathcal{D}^K$  such that  $|i^{\star}(\mu)| = 1$ ,
Proof. The proof is the same of the ones of Lemma 13. The sole difference is that we have  $w_{\beta, i^{\star}}^{\star} = \beta$  by definition.
Sufficient exploration Lemma 26 shows that all arms are sufficiently explored and that the leader is the unique best arm for  $n$  large enough.
Lemma 26. There exist  $N_{1}$  with  $\mathbb{E}_{\nu}[N_1] < +\infty$  such that for all  $n \geq N_{1}$  and all  $i \in [K]$ ,  $N_{n,i} \geq \sqrt{n / K}$  and  $B_{n}^{EB} = i^{\star}$ .
Proof. Since Lemmas 16 and 17 hold, we can use the intermediate results of Lemma 18. It is direct to see that the proof can be conducted similarly by noting that, for all  $i \neq j$ ,
$$
\min  \left\{\beta_ {t} (i, j), 1 - \beta_ {t} (i, j) \right\} = \min  \left\{\beta , 1 - \beta \right\} > 0.
$$
Convergence towards optimal allocation Combining Lemma 26 with the proof of Lemma F.8 in [20] yields Lemma 27.
Lemma 27. Let  $\gamma >0$ . There exists  $N_{2}$  with  $\mathbb{E}_{\mu}[N_2] < +\infty$  such that for all  $n\geq N_2$
$$
\left| \frac {N _ {n , i ^ {\star}}}{n - 1} - \beta \right| \leq \gamma .
$$
Combining Lemmas 26 and 27 with the proof of Lemma F.9 in [20] yields Lemma 28.
Lemma 28. Let  $\gamma >0$ . There exists  $N_{5}$  with  $\mathbb{E}_{\nu}[N_5] < +\infty$  such that for all  $n\geq N_5$  and all  $i\neq i^{\star}$ ,
$$
\frac {N _ {n , i}}{n - 1} \geq w _ {\beta , i} ^ {\star} + \gamma \quad \Longrightarrow \quad C _ {n} ^ {T C \varepsilon_ {0}} \neq i.
$$
Combining Lemmas 27 and 28 with the proof of Lemma F.10 in [20] yields Lemma 29.
Lemma 29. Let  $\gamma >0$ . There exists  $N_{6}$  with  $\mathbb{E}_{\nu}[N_6] < +\infty$  such that for all  $n\geq N_6$
$$
\left\| \frac {N _ {n}}{n - 1} - w _ {\beta} ^ {\star} \right\| _ {\infty} \leq \gamma .
$$
Lemma 30 is a direct corollary of Lemma 29.
Lemma 30. Let  $\varepsilon_0 > 0$ ,  $\beta \in (0,1)$ ,  $\gamma > 0$  and  $T_{\varepsilon_0,\beta,\gamma}$  as in (14). Under the  $EB-TC_{\varepsilon_0}$  sampling rule with fixed proportions  $\beta$ , we have  $\mathbb{E}_{\nu}[T_{\varepsilon_0,\beta,\gamma}] < +\infty$ .
Combining Lemmas 25 and 30 concludes the second part of Theorem 1, i.e.
E Non-asymptotic analysis
Let  $\varepsilon_0 > 0$ ,  $\beta \in (0,1)$  and  $\delta \in (0,1)$ . In this section, we provide a non-asymptotic analysis of  $\mathrm{EB - TC}_{\varepsilon_0}$  with fixed proportions  $\beta$  and slack  $\varepsilon_0 > 0$  when combined with the stopping rule (3) with parameters  $(\varepsilon_0,\delta)$ . In practice, we will mostly be interested in the case  $\beta = 1 / 2$ . First, we prove Lemma 5 in Appendix E.1. Then, we detail the proof of Theorem 2 in Appendix E.2.
In the following, we consider a sub-Gaussian bandit with distribution  $\nu \in \mathcal{D}^K$  having mean parameter  $\mu \in \mathbb{R}^K$ . In particular, our analysis holds when several arms have the largest mean  $\mu_{\star}$ .
E.1 Proof of Lemma 5
Let  $s \geq 0$ . For all  $n > K$  and  $\delta \in (0,1]$ , let  $\mathcal{E}_{n,\delta} = \mathcal{E}_{n,\delta}^{1} \cap \mathcal{E}_{n,\delta}^{2}$  with  $(\mathcal{E}_{n,\delta}^{1})_{n > K}$  and  $(\mathcal{E}_{n,\delta}^{2})_{n > K}$  as in (24) and (25), i.e.
$$
\begin{array}{l} \mathcal {E} _ {n, \delta} ^ {1} = \left\{\forall k \in [ K ], \forall t \leq n, | \mu_ {t, k} - \mu_ {k} | <   \sqrt {\frac {2 f _ {1} (n , \delta)}{N _ {t , k}}} \right\}, \\ \mathcal {E} _ {n, \delta} ^ {2} = \left\{\forall (i, k) \in [ K ] ^ {2} \text {s . t .} i \neq k, \forall t \leq n, \frac {\left| (\mu_ {t , i} - \mu_ {t , k}) - (\mu_ {i} - \mu_ {k}) \right|}{\sqrt {1 / N _ {t , i} + 1 / N _ {t , k}}} <   \sqrt {2 f _ {2} (n , \delta)} \right\}, \\ \end{array}
$$
where  $f_{2}(x,\delta) = \log (1 / \delta) + (1 + s)\log (x)$  and  $f_{2}(x,\delta) = \log (1 / \delta) + (2 + s)\log (x)$ .
Lemma 31 shows that when there are arms with strictly higher true mean than the one of the leader, then at least one of those arms is undersampled.
Lemma 31. Under  $\mathcal{E}_{n,\delta}^{1}$ , for all  $t\in [n]\setminus [K]$  let  $B_{t}^{EB} = k$ . Then,
$$
\forall i \neq k, \quad \mathbb {1} (\mu_ {i} > \mu_ {k}) \min  \{N _ {t, k}, N _ {t, i} \} \leq \frac {8 f _ {1} (n , \delta)}{(\mu_ {i} - \mu_ {k}) ^ {2}}.
$$
Proof. Under  $\mathcal{E}_{n,\delta}^{1}$ , for all  $t\in [n]\setminus [K]$ , let  $B_{t}^{\mathrm{EB}} = k$ . Then, for all  $i\neq k$ , we have
$$
\begin{array}{l} \mu_ {i} - \sqrt {\frac {2 f _ {1} (n , \delta)}{\min  \left\{N _ {t , k} , N _ {t , i} \right\}}} \leq \mu_ {i} - \sqrt {\frac {2 f _ {1} (n , \delta)}{N _ {t , i}}} \leq \mu_ {t, i} \leq \mu_ {t, k} \leq \mu_ {k} + \sqrt {\frac {2 f _ {1} (n , \delta)}{N _ {t , k}}} \\ \leq \mu_ {k} + \sqrt {\frac {2 f _ {1} (n , \delta)}{\operatorname* {m i n} \left\{N _ {t , k} , N _ {t , i} \right\}}}. \\ \end{array}
$$
Re-ordering the above equations for  $i$  such that  $\mu_{i} > \mu_{k}$  yields the result.
Lemma 32. Let  $\varepsilon \geq 0$ ,  $\Delta_{\mu}(\varepsilon) = \min_{k\notin \mathcal{I}_{\varepsilon}(\mu)}\Delta_{k}$  and  $C_{\mu ,\varepsilon_0}(\varepsilon) = \max \{2\Delta_\mu (\varepsilon)^{-1} - \varepsilon_0^{-1},\varepsilon_0^{-1}\}^2$  Let  $A_{\varepsilon_0,\varepsilon ,i} = 2 / \Delta_\mu (\varepsilon)^2$  for all  $i\in i^{\star}(\mu)$ ,  $A_{\varepsilon_0,\varepsilon ,i} = C_{\mu ,\varepsilon_0}(\varepsilon)$  for all  $i\in \mathcal{I}_{\varepsilon}(\mu)\setminus i^{\star}(\mu)$ , otherwise  $A_{\varepsilon_0,\varepsilon ,i} = \max \{C_{\mu ,\varepsilon_0}(\varepsilon),2 / \Delta_i^2\}$ . For all  $n > K$ , under event  $\mathcal{E}_{n,\delta}$ , for all  $t\in [n]\setminus [K]$  such that  $B_t^{EB}\notin \mathcal{I}_\varepsilon (\mu)$ , there exists  $i_t\in [K]$  such that
$$
T _ {t} (i _ {t}) \leq \frac {4 f _ {2} (n , \delta)}{\min  \left\{\beta , 1 - \beta \right\}} A _ {\varepsilon_ {0}, \varepsilon , i _ {t}} + 3 (K - 1) / 2 \quad a n d \quad T _ {t + 1} (i _ {t}) = T _ {t} (i _ {t}) + 1.
$$
Proof. Let  $\Delta_{i} = \mu_{\star} - \mu_{i}$  and  $\Delta_{\mathrm{max}} = \max_{i\in [K]}\Delta_{i}$ . When  $\varepsilon \geq \Delta_{\mathrm{max}}$ , we have  $\mathcal{I}_{\varepsilon}(\mu)^{\complement} = \emptyset$ , hence the above result holds trivially since the event  $B_t^{\mathrm{EB}} \notin \mathcal{I}_{\varepsilon}(\mu)$  cannot happen. Let  $\varepsilon \in [0, \Delta_{\mathrm{max}})$ , i.e.  $\mathcal{I}_{\varepsilon}(\mu)^{\complement} \neq \emptyset$ . We will consider in two distinct cases since
$$
\left\{B _ {t} \notin \mathcal {I} _ {\varepsilon} (\mu) \right\} = \left\{B _ {t} \notin \mathcal {I} _ {\varepsilon} (\mu), C _ {t} \in i ^ {\star} (\mu) \right\} \cup \left\{B _ {t} \notin \mathcal {I} _ {\varepsilon} (\mu), C _ {t} \notin i ^ {\star} (\mu) \right\}.
$$
Case 1. Let  $t \in [n] \setminus [K]$  such that  $(B_t^{\mathrm{EB}}, C_t^{\mathrm{TC}\varepsilon_0}) = (i, j)$  with  $i \notin \mathcal{I}_{\varepsilon}(\mu)$  and  $j \in i^\star(\mu)$ . Using Lemmas 46 and 31, we obtain
$$
\min  \left\{\beta , 1 - \beta \right\} \left(\min  \left\{T _ {t} (i), T _ {t} (j) \right\} - 3 (K - 1) / 2\right) \leq \min  \left\{N _ {t, i}, N _ {t, j} \right\} \leq \frac {8 f _ {1} (n , \delta)}{\Delta_ {i} ^ {2}} \leq \frac {8 f _ {2} (n , \delta)}{\Delta_ {i} ^ {2}},
$$
which can be rewritten as
$$
\min \{T _ {t} (i), T _ {t} (j) \} \leq \frac {8 f _ {2} (n , \delta)}{\min \{\beta , 1 - \beta \} \Delta_ {i} ^ {2}} + 3 (K - 1) / 2.
$$
Let us define  $\Delta_{\mu}(\varepsilon) = \min_{k \notin \mathcal{I}_{\varepsilon}(\mu)} \Delta_k$ , and
$$
\forall i \notin \mathcal {I} _ {\varepsilon} (\mu), \quad D _ {\varepsilon , i} = 2 / \Delta_ {i} ^ {2} \quad \text {a n d} \quad \forall i \in i ^ {\star} (\mu), \quad D _ {\varepsilon , i} = 2 / \Delta_ {\mu} (\varepsilon) ^ {2}.
$$
The above shows that there exists  $k_{t}\in \mathcal{I}_{\varepsilon}(\mu)^{\complement}\cup i^{\star}(\mu)$  such that
$$
T _ {t} (k _ {t}) \leq \frac {4 f _ {2} (n , \delta)}{\min  \left\{\beta , 1 - \beta \right\}} D _ {\varepsilon , k _ {t}} + 3 (K - 1) / 2 \quad \text {a n d} \quad T _ {t + 1} (k _ {t}) = T _ {t} (k _ {t}) + 1.
$$
Case 2. Let  $t \in [n] \setminus [K]$  such that  $(B_t^{\mathrm{EB}}, C_t^{\mathrm{TC}\varepsilon_0}) = (i, j)$  with  $i \notin \mathcal{I}_{\varepsilon}(\mu)$  and  $j \notin i^{\star}(\mu)$ . Let  $i_0 \in i^{\star}(\mu)$ . Using the TC challenger, we obtain
$$
\begin{array}{l} \frac {\varepsilon_ {0} - \Delta_ {i}}{\sqrt {1 / N _ {t , i} + 1 / N _ {t , i _ {0}}}} + \sqrt {2 f _ {2} (n , \delta)} \geq \frac {\mu_ {t , i} - \mu_ {t , i _ {0}} + \varepsilon_ {0}}{\sqrt {1 / N _ {t , i} + 1 / N _ {t , i _ {0}}}} \geq \frac {\mu_ {t , i} - \mu_ {t , j} + \varepsilon_ {0}}{\sqrt {1 / N _ {t , i} + 1 / N _ {t , j}}} \\ \geq \varepsilon_ {0} \sqrt {\min  \left\{N _ {t , i} , N _ {t , j} \right\} / 2}. \\ \end{array}
$$
Using Lemma 31, we obtain
$$
\frac {1}{1 / N _ {t , i} + 1 / N _ {t , i _ {0}}} \leq \min  \left\{N _ {t, i}, N _ {t, i _ {0}} \right\} \leq \frac {8 f _ {1} (n , \delta)}{\Delta_ {i} ^ {2}} \leq \frac {8 f _ {2} (n , \delta)}{\Delta_ {i} ^ {2}}.
$$
By distinguishing between  $\varepsilon_0 > \Delta_i$  and  $\varepsilon_0 \leq \Delta_i$  and using that  $\Delta_i > 0$ , we have
$$
\frac {\varepsilon_ {0} - \Delta_ {i}}{\sqrt {1 / N _ {t , i} + 1 / N _ {t , i _ {0}}}} + \sqrt {2 f _ {2} (n , \delta)} \leq \max \{2 \varepsilon_ {0} / \Delta_ {i} - 1, 1 \} \sqrt {2 f _ {2} (n , \delta)}.
$$
Using Lemma 46 to lower bound  $\min \{N_{t,i}, N_{t,j}\}$  and reordering, we have shown that
$$
\min  \{T _ {t} (i), T _ {t} (j) \} \leq \max  \left\{\left(\frac {2}{\Delta_ {i}} - \frac {1}{\varepsilon_ {0}}\right) ^ {2}, \frac {1}{\varepsilon_ {0} ^ {2}} \right\} \frac {4 f _ {2} (n , \delta)}{\min  \{\beta , 1 - \beta \}} + 3 (K - 1) / 2.
$$
Let us define  $C_{\mu, \varepsilon_0}(\varepsilon) = \max \{2\Delta_\mu(\varepsilon)^{-1} - \varepsilon_0^{-1}, \varepsilon_0^{-1}\}^2$ . The above shows that, there exists  $k_t \notin i^\star(\mu)$  such that
$$
T _ {t} (k _ {t}) \leq \frac {4 f _ {2} (n , \delta)}{\min  \left\{\beta , 1 - \beta \right\}} C _ {\mu , \varepsilon_ {0}} (\varepsilon) + 3 (K - 1) / 2 \quad \text {a n d} \quad T _ {t + 1} (k _ {t}) = T _ {t} (k _ {t}) + 1.
$$
Summary. Let us define  $(A_{\varepsilon_0,\varepsilon ,i})_{i\in [K]}$  as in the statement of Lemma 32. Under  $\mathcal{E}_{n,\delta}$ , we have show that, when  $B_{t}\notin \mathcal{I}_{\varepsilon}(\mu)$ , there exists  $i_t\in [K]$  such that
$$
T _ {t} \left(k _ {t}\right) \leq \frac {4 f _ {2} (n , \delta)}{\operatorname* {m i n} \left\{\beta , 1 - \beta \right\}} A _ {\varepsilon , \varepsilon_ {0}, i _ {t}} + 3 (K - 1) / 2 \quad \text {a n d} \quad T _ {t + 1} \left(k _ {t}\right) = T _ {t} \left(k _ {t}\right) + 1 \}.
$$
For all  $n > K$ , under event  $\mathcal{E}_{n,\delta}$ , combining Lemma 4 and 32 for  $A_{t}(n,\delta) = \{B_{t}^{\mathrm{EB}} \notin \mathcal{I}_{\varepsilon}(\mu)\}$  and  $D_{i}(n,\delta) = \frac{4f_{2}(n,\delta)}{\min\{\beta,1 - \beta\}} A_{\varepsilon_{0},\varepsilon,i} + 3(K - 1)/2$  yields that
$$
\sum_ {t = K + 1} ^ {n} \mathbb {1} \left(B _ {t} ^ {\mathrm {E B}} \notin \mathcal {I} _ {\varepsilon} (\mu)\right) \leq \frac {4 f _ {2} (n , \delta)}{\min  \left\{\beta , 1 - \beta \right\}} H _ {\mu , \varepsilon_ {0}} (\varepsilon) + 3 K (K - 1) / 2.
$$
where we used that  $\sum_{i\in [K]}A_{\varepsilon_0,\varepsilon ,i} = H_{\mu ,\varepsilon_0}(\varepsilon)$  where  $H_{\mu ,\varepsilon_0}(\varepsilon)$  is defined in (5). To conclude the proof of Lemma 5, we use that
$$
\sum_ {t = K + 1} ^ {n} \mathbb {1} \left(B _ {t} ^ {\mathrm {E B}} \notin \mathcal {I} _ {\varepsilon} (\mu)\right) = n - 1 - \sum_ {i \in \mathcal {I} _ {\varepsilon} (\mu)} \sum_ {j} T _ {n} (i, j).
$$
Let  $\tilde{f}_1(n,\delta)$  and  $\tilde{f}_2(n,\delta)$  defined as in (26) and (28). Using Lemma 50, we have  $\tilde{f}_1(n,\delta) \leq \tilde{f}_2(n,\delta)$ . Therefore, it is direct to see that Lemma 33 can be proven with the same proof as for Lemma 5 based on the concentration event  $\tilde{\mathcal{E}}_{n,\delta} = \tilde{\mathcal{E}}_{n,\delta}^{1} \cap \tilde{\mathcal{E}}_{n,\delta}^{2}$  with  $(\tilde{\mathcal{E}}_{n,\delta}^{1})_{n > K}$  and  $(\tilde{\mathcal{E}}_{n,\delta}^{2})_{n > K}$  as in (27) and (29).
Lemma 33. Let  $\delta \in (0,1]$ ,  $\varepsilon \geq 0$  and  $H_{\mu,\varepsilon_0}(\varepsilon)$  as in (5). For all  $n > K$ , under the event  $\tilde{\mathcal{E}}_{n,\delta}$ ,
$$
\sum_ {i \in \mathcal {I} _ {\varepsilon} (\mu)} \sum_ {j} T _ {n} (i, j) \geq n - 1 - \left(\frac {4 H _ {\mu , \varepsilon_ {0}} (\varepsilon)}{\min  \{\beta , 1 - \beta \}} \tilde {f} _ {2} (n, \delta) + 3 K (K - 1) / 2\right).
$$
Let  $s > 1$ . For all  $n > K$ , let  $\mathcal{E}_n = \mathcal{E}_{n,1}^1 \cap \mathcal{E}_{n,1}^2$  with  $(\mathcal{E}_{n,\delta}^{1})_{n > K}$  and  $(\mathcal{E}_{n,\delta}^{2})_{n > K}$  as in (24) and (25). Using Lemma 45, we know that  $\sum_{n > K} \mathbb{P}_{\nu}(\mathcal{E}_n^{\complement}) \leq \frac{K(K + 1)}{2} \zeta(s)$ .
Let  $\varepsilon_0 > 0$ ,  $\varepsilon_1 \geq \varepsilon_0$  and  $\delta \in (0,1)$ . Since  $\varepsilon_1 \geq \varepsilon_0$ , and using the definition of the stopping rule (3), it is direct to see that  $\mathbb{E}_{\nu}[\tau_{\varepsilon_1,\delta}] \leq \mathbb{E}_{\nu}[\tau_{\varepsilon_0,\delta}]$ .
Suppose that we have constructed a time  $T(\delta) > K$  be such that for  $n \geq T(\delta)$ ,  $\mathcal{E}_n \subset \{\tau_{\varepsilon_0, \delta} \leq n\}$ . Then, Lemma 47 yields
$$
\mathbb {E} _ {\nu} \left[ \tau_ {\varepsilon_ {0}, \delta} \right] \leq T (\delta) + \frac {K (K + 1)}{2} \zeta (s).
$$
We will construct an infinite number of times  $\{T(\delta, u)\}_{u \in \mathcal{U}}$  such that the above property holds, hence taking the infimum yields
$$
\mathbb {E} _ {\nu} \left[ \tau_ {\varepsilon_ {0}, \delta} \right] \leq \inf  _ {u \in \mathcal {U}} T (\delta , u) + \frac {K (K + 1)}{2} \zeta (s).
$$
This proof strategy is the one used to prove Theorem 5. Theorem 2 is a corollary of Theorem 5. Note that the notation of both theorems differ slightly since we provide a shorter statement in the main content. The key difference lies in the refined analysis used to clip  $\min_{j\neq i}w_{\varepsilon_0,1 / 2}(\mu ,i)_j$  by a fixed value  $x\in [0,(K - 1)^{-1}]$  for all  $i\in \mathcal{I}_{\varepsilon}(\mu)$ .
Theorem 5. Let  $\varepsilon_0 > 0$ ,  $\varepsilon_1 \geq \varepsilon_0$  and  $\delta \in (0,1)$ . Using the threshold (4) in the stopping rule (3) with error  $\varepsilon_1$ , the  $EB-TC_{\varepsilon_0}$  algorithm with fixed proportions  $\beta = 1/2$  is  $(\varepsilon_1, \delta)$ -PAC and satisfies that, for all  $\nu \in \mathcal{D}^K$ ,  $\mathbb{E}_{\nu}[\tau_{\varepsilon_1,\delta}] \leq \mathbb{E}_{\nu}[\tau_{\varepsilon_0,\delta}]$  and
$$
\mathbb {E} _ {\nu} \left[ \tau_ {\varepsilon_ {0}, \delta} \right] \leq \inf  _ {(\varepsilon , x) \in [ 0, \varepsilon_ {0} ] \times [ 0, (K - 1) ^ {- 1} ]} \max  \left\{T _ {\mu , \varepsilon_ {0}} (\delta , \varepsilon , x) + 1, S _ {\mu , \varepsilon_ {0}} (\varepsilon , x) \right\} + \zeta (s) \frac {K (K + 1)}{2},
$$
$$
\begin{array}{l} T _ {\mu , \varepsilon_ {0}} (\delta , \varepsilon , x) = \sup  \left\{n \mid n - 1 \leq \frac {2 (1 + \gamma) ^ {2} \sum_ {i \in \mathcal {I} _ {\varepsilon} (\mu)} T _ {\varepsilon_ {0} , 1 / 2} (\mu , i)}{(1 - x) ^ {d _ {\mu , \varepsilon_ {0}} (\varepsilon , x)}} \right. \\ \left. \left(\sqrt {c (n - 1 , \delta)} + \sqrt {(2 + s) \log n}\right) ^ {2} \right\}, \\ \end{array}
$$
$$
S _ {\mu , \varepsilon_ {0}} (\varepsilon , x) = h _ {1} \left(\frac {4 (2 + s) (1 + \gamma^ {- 1})}{a _ {\mu , \varepsilon_ {0}} (\varepsilon , x) v _ {\mu , \varepsilon_ {0}} (\varepsilon)} H _ {\mu , \varepsilon_ {0}} (\varepsilon), \frac {(1 + \gamma^ {- 1}) (3 K ^ {2} / 4 + 1)}{a _ {\mu , \varepsilon_ {0}} (\varepsilon , x) v _ {\mu , \varepsilon_ {0}} (\varepsilon)} + 1\right),
$$
$$
v _ {\mu , \varepsilon_ {0}} (\varepsilon) = \frac {\min  _ {i \in \mathcal {I} _ {\varepsilon} (\mu)} T _ {\varepsilon_ {0} , 1 / 2} (\mu , i)}{\sum_ {i \in \mathcal {I} _ {\varepsilon} (\mu)} T _ {\varepsilon_ {0} , 1 / 2} (\mu , i)}
$$
$$
a _ {\mu , \varepsilon_ {0}} (\varepsilon , x) = \min  _ {i \in \mathcal {I} _ {\varepsilon} (\mu)} (1 - x) ^ {d _ {\mu , \varepsilon_ {0}} (x, i)} \max  \left\{\min  _ {j \neq i} w _ {\varepsilon_ {0}, 1 / 2} (\mu , i) _ {j}, x / 2 \right\},
$$
$$
d _ {\mu , \varepsilon_ {0}} (\varepsilon , x) = \max  _ {i \in \mathcal {I} _ {\varepsilon} (\mu)} d _ {\mu , \varepsilon_ {0}} (x, i) \quad w i t h \quad d _ {\mu , \varepsilon_ {0}} (x, i) = | \{j \neq i \mid w _ {\varepsilon_ {0}, 1 / 2} (\mu , i) _ {j} <   x / 2 \} |  ,
$$
where  $\gamma \in (0,1 / 2]$  and  $s > 1$  are analysis parameters.  $T_{\varepsilon_0,1 / 2}(\mu ,i)$  and  $w_{\varepsilon_0,1 / 2}(\mu ,i)$  are defined in (1),  $H_{\mu ,\varepsilon_0}(\varepsilon)$  in (5),  $\zeta$  is the Riemann  $\zeta$  function and  $h_1(z,y) = z\overline{W}_{-1}(\log (z) + y / z)$  is defined in Lemma 51 and satisfies that  $h_1(z,y)\approx x\log (z) + y + \log (\log (z) + y / z)$ .
Proof. Let  $\varepsilon \in [0,\varepsilon_0]$  be an analysis parameter, and  $I_{\varepsilon} = |\mathcal{I}_{\varepsilon}(\mu)|$ . Let  $v\in \hat{\Delta}_{I_{\varepsilon}}$  defined as
$$
\forall i \in \mathcal {I} _ {\varepsilon} (\mu), \quad v _ {i} = \frac {T _ {\varepsilon_ {0} , 1 / 2} (\mu , i)}{\sum_ {j \in \mathcal {I} _ {\varepsilon} (\mu)} T _ {\varepsilon_ {0} , 1 / 2} (\mu , j)}, \quad \text {h e n c e} \quad \min  _ {i \in \mathcal {I} _ {\varepsilon} (\mu)} v _ {i} = v _ {\mu , \varepsilon_ {0}} (\varepsilon).
$$
The main technical result on which our proof relies on is Lemma 34, which we will prove afterwards.
Lemma 34. There exist  $D_0 > 0$  and  $T_{\mu} > 0$  such that for all  $n \geq T_{\mu}$  such that  $\mathcal{E}_n \cap \{n < \tau_{\varepsilon_1,\delta}\}$  holds true, there exists  $i_0 \in \mathcal{I}_{\varepsilon}(\mu)$  and  $t \leq n$  with  $B_t^{EB} = i_0$ , which satisfies
$$
1 / N _ {t, i _ {0}} ^ {i _ {0}} + 1 / N _ {t, C _ {t} ^ {T C \varepsilon_ {0}}} ^ {i _ {0}} \leq \frac {D _ {0}}{v _ {i _ {0}} (n - 1)} \left(1 / \beta + 1 / w _ {\beta} ^ {\star} \left(i _ {0}\right) _ {C _ {t} ^ {T C \varepsilon_ {0}}}\right). \tag {15}
$$
Before proving Lemma 34, we show how it can be used to conclude. Based on Lemma 34, let  $D_0 > 0$ ,  $T_{\mu} > 0$ ,  $n > T_{\mu}$  such that  $\mathcal{E}_n \cap \{n < \tau_{\varepsilon_1,\delta}\}$  holds true,  $i_0$  and  $t \leq n$  such that as  $B_t^{\mathrm{EB}} = i_0$  and
$$
1 / N _ {t, i _ {0}} ^ {i _ {0}} + 1 / N _ {t, C _ {t} ^ {\mathrm {T C} \varepsilon_ {0}}} ^ {i _ {0}} \leq \frac {D _ {0}}{v _ {i _ {0}} (n - 1)} \left(1 / \beta + 1 / w _ {\beta} ^ {\star} (i _ {0}) _ {C _ {t} ^ {\mathrm {T C} \varepsilon_ {0}}}\right).
$$
Using the stopping rule (3) with error  $\varepsilon_0$ ,  $t \leq n < \tau_{\varepsilon_1, \delta}$ ,  $B_t^{\mathrm{EB}} = \hat{i}_t = i_0$  and the definition of the  $\mathrm{TC}_{\varepsilon_0}$  challenger in (2) with slack  $\varepsilon_0$ , we obtain
$$
\begin{array}{l} \sqrt {2 c (n - 1 , \delta)} \geq \min  _ {i \neq i _ {0}} \frac {\mu_ {t , i _ {0}} - \mu_ {t , i} + \varepsilon_ {0}}{\sqrt {1 / N _ {t , i _ {0}} + 1 / N _ {t , i}}} \\ \geq \frac {\mu_ {i _ {0}} - \mu_ {C _ {t} ^ {\mathrm {T C} \varepsilon_ {0}}} + \varepsilon_ {0}}{\sqrt {1 / N _ {t , i _ {0}} + 1 / N _ {t , C _ {t} ^ {\mathrm {T C} \varepsilon_ {0}}}}} - \sqrt {2 (2 + s) \log n} \\ \geq \sqrt {\frac {1 / \beta + w _ {\beta} ^ {\star} (i _ {0}) _ {C _ {t} ^ {\mathrm {T C} \varepsilon_ {0}}} ^ {- 1}}{1 / N _ {t , i _ {0}} ^ {i _ {0}} + 1 / N _ {t , C _ {t} ^ {\mathrm {T C} \varepsilon_ {0}}} ^ {i _ {0}}}} \sqrt {2 T _ {\varepsilon_ {0} , \beta} (\mu , i _ {0}) ^ {- 1}} - \sqrt {2 (2 + s) \log n} \\ \geq \sqrt {\frac {2 (n - 1)}{D _ {0} \sum_ {i \in \mathcal {I} _ {\varepsilon} (\mu)} T _ {\varepsilon_ {0} , \beta} (\mu , i)}} - \sqrt {2 (2 + s) \log n}. \\ \end{array}
$$
The third inequality is obtained since  $\mathcal{E}_n$  holds. The fourth inequality is obtained since  $N_{t,i}^{i_0}\geq N_{t,i}^{i_0}$  for all  $i\in [K]$  and by using Lemmas 8 and 9. The last inequality is obtained by Lemma 34 and using the definition of  $v_{i_0}$ .