SlowGuess/NeurIPS_2023_OCR huggingface dataset:
which satisfies  $\overline{M}(0) = 1$ . It is direct to see that  $M(t) = \exp \left(xS_{t} - \frac{1}{2} x^{2}t\right)$  is a non-negative supermartingale since sub-Gaussian distributions with mean 0 and variance proxy  $\sigma^2 = 1$  satisfy
$$
\forall \lambda \in \mathbb {R}, \quad \mathbb {E} _ {X} [ \exp (s X) ] \leq \exp (\lambda^ {2} / 2).
$$
By Tonelli's theorem, then  $\overline{M}(t)$  is also a non-negative supermartingale of unit initial value.
Let  $i \in [D]$  and consider  $t \in [N_i, N_{i+1})$ . For all  $x$ ,
$$
f _ {N _ {i}, \gamma} (x) \geq \sqrt {\frac {N _ {i}}{t}} f _ {t, \gamma} (x) \geq \frac {1}{\sqrt {1 + \eta}} f _ {t, \gamma} (x)
$$
Direct computations show that
$$
\int f _ {t, \gamma} (x) \exp \left(x S _ {t} - \frac {1}{2} x ^ {2} t\right) d x = \frac {1}{\sqrt {1 + \gamma^ {- 1}}} \exp \left(\frac {S _ {t} ^ {2}}{2 (1 + \gamma) t}\right).
$$
Minoring  $\overline{M}(t)$  by one of the positive terms of its sum, we obtain
$$
\overline {{M}} (t) \geq \frac {1}{D} \frac {1}{\sqrt {(1 + \gamma^ {- 1}) (1 + \eta)}} \exp \left(\frac {S _ {t} ^ {2}}{2 (1 + \gamma) t}\right),
$$
Using Ville's maximal inequality for non-negative supermartingale, we have that with probability greater than  $1 - \delta$ ,  $\log \overline{M}(t) \leq \log (1 / \delta)$ . Therefore, with probability greater than  $1 - \delta$ , for all  $i \in [D]$  and  $t \in [N_i, N_{i+1})$ ,
$$
\frac {S _ {t} ^ {2}}{t} \leq (1 + \gamma) \left(2 \log (1 / \delta) + 2 \log D + \log (1 + \gamma^ {- 1}) + \log (1 + \eta)\right).
$$
Since this upper bound is independent of  $t$ , we can optimize it and choose  $\gamma$  as in Lemma 43.
Lemma 43 (Lemma A.3 in [9]). For  $a, b \geq 1$ , the minimal value of  $f(\eta) = (1 + \eta)(a + \log(b + \frac{1}{\eta}))$  is attained at  $\eta^{\star}$  such that  $f(\eta^{\star}) \leq 1 - b + \overline{W}_{-1}(a + b)$ . If  $b = 1$ , then there is equality.
Therefore, with probability greater than  $1 - \delta$ , for all  $i \in [D]$  and  $t \in [N_i, N_{i+1})$ ,
$$
\begin{array}{l} \frac {S _ {t} ^ {2}}{t} \leq \overline {{W}} _ {- 1} (1 + 2 \log (1 / \delta) + 2 \log D + \log (1 + \eta)) \\ \leq \bar {W} _ {- 1} (1 + 2 \log (1 / \delta) + 2 \log (\log (1 + \eta) + \log n) - 2 \log \log (1 + \eta) + \log (1 + \eta)) \\ = \bar {W} _ {- 1} (2 \log (1 / \delta) + 2 \log (2 + \log n) + 3 - 2 \log 2) \\ \end{array}
$$
The second inequality is obtained since  $D \leq 1 + \frac{\log n}{\log(1 + \eta)}$ . The last equality is obtained for the choice  $\eta^{\star} = e^{2} - 1$ , which minimizes  $\eta \mapsto \log (1 + \eta) - 2\log (\log (1 + \eta))$ . Since  $[n] \subseteq \bigcup_{i \in [D]} [N_i, N_{i+1})$  and  $N_{t,k}(\mu_{t,k} - \mu_k) = \sum_{s \in [N_{t,k}]} X_{s,k}$  (unit-variance), this yields
$$
\mathbb {P} \left(\exists m \leq n, \left| \frac {1}{m} \sum_ {s = 1} ^ {m} X _ {s} \right| \geq \sqrt {\frac {1}{m} \overline {{W}} _ {- 1} (2 \log (1 / \delta) + 2 \log (2 + \log (n)) + 3 - 2 \log 2)}\right) \leq \delta .
$$
Since  $3 - 2\log 2 \leq 2$  and  $\overline{W}_{-1}$  is increasing, taking  $\delta n^{-s}$  yields
$$
\mathbb {P} _ {\nu} \left(\exists t \leq n, \sqrt {N _ {t , k}} | \mu_ {t, k} - \mu_ {k} | \geq \sqrt {2 \tilde {f} _ {1} (n , \delta)}\right) \leq \delta n ^ {- s}.
$$
Doing a union bound over arms yields the result.
Lemma 44 provides concentration results on the empirical gaps, which are tighter than the ones obtained in Lemma 41.
Lemma 44. Let  $\delta \in (0,1]$  and  $s\geq 0$  .Let  $\overline{W}_{-1}$  defined in Lemma 49. For all  $n > K$  , let
$$
\tilde {f} _ {2} (n, \delta) = \bar {W} _ {- 1} (\log (1 / \delta) + s \log n + 2 \log (2 + \log n) + 2), \tag {28}
$$
$$
\tilde {\mathcal {E}} _ {n, \delta} ^ {2} := \left\{\forall (i, k) \in [ K ] ^ {2} \text {s . t .} i \neq k, \forall t \leq n, \frac {\left| \left(\mu_ {t , i} - \mu_ {t , k}\right) - \left(\mu_ {i} - \mu_ {k}\right) \right|}{\sqrt {1 / N _ {t , i} + 1 / N _ {t , k}}} <   \sqrt {2 \tilde {f} _ {2} (n , \delta)} \right\}. \tag {29}
$$
Then, for all  $n > K$ ,  $\mathbb{P}_{\nu}((\tilde{\mathcal{E}}_{n,\delta}^2)^{\mathbb{C}}) \leq \frac{K(K - 1)}{2}\frac{\delta}{n^s}$ .
Proof. Without loss of generality, we show the result for arms  $(i,k) = (1,2)$ . Let  $(X_{s})_{s\in [n]}$  and  $(Y_{s})_{s\in [n]}$  be i.i.d. observations from two sub-Gaussian distribution with mean 0 and variance proxy  $\sigma^2 = 1$ . Let  $S_{t,k} = \sum_{s\in [t]}\mathbb{1}(I_s = k)X_s$  for all  $k\in \{1,2\}$ . To derived concentration result, we use peeling as in Lemma 42.
$$
\forall k \in \{1, 2 \}, \quad \overline {{M}} _ {k} (t) = \sum_ {i \in [ D ]} w _ {i} \int f _ {N _ {i}, \gamma} (x) \exp \left(x S _ {t, k} - \frac {1}{2} x ^ {2} N _ {t, k}\right) d x,
$$
which satisfies  $\overline{M}_k(0) = 1$  for all  $k\in \{1,2\}$ . As in the proof of Lemma 42, we obtain that  $\overline{M}_1(t)$  and  $\overline{M}_2(t)$  are non-negative supermartingale such that
$$
\forall k \in \{1, 2 \}, \quad \overline {{M}} _ {k} (t) \geq \frac {1}{D} \frac {1}{\sqrt {(1 + \gamma^ {- 1}) (1 + \eta)}} \exp \left(\frac {S _ {t , k} ^ {2}}{2 (1 + \gamma) N _ {t , k}}\right).
$$
where we used  $(i_1,i_2)\in [D]^2$  and consider  $N_{t,k}\in [N_{i_k},N_{i_k + 1})$  for all  $k\in \{1,2\}$ . Let us define  $\overline{M} (t) = \overline{M}_1(t_1)\overline{M}_2(t_2)$ . Then, we have that  $\overline{M} (t)$  is a non-negative supermartingale such that
$$
\overline {{M}} (t) \geq \frac {1}{D ^ {2}} \frac {1}{(1 + \gamma^ {- 1}) (1 + \eta)} \exp \left(\frac {1}{2 (1 + \gamma)} \sum_ {k \in \{1, 2 \}} \frac {S _ {t , k} ^ {2}}{N _ {t , k}}\right)
$$
Using Ville's maximal inequality for non-negative supermartingale, we have that with probability greater than  $1 - \delta$ ,  $\log \overline{M}(t) \leq \log (1 / \delta)$ . Therefore, with probability greater than  $1 - \delta$ , for all  $(i_1, i_2) \in [D]^2$  and  $N_{t,k} \in [N_{i_k}, N_{i_k + 1})$  for all  $k \in \{1, 2\}$ ,
$$
\sum_ {k \in \{1, 2 \}} \frac {S _ {t , k} ^ {2}}{N _ {t , k}} \leq 2 (1 + \gamma) \left(\log (1 / \delta) + 2 \log D + \log (1 + \gamma^ {- 1}) + \log (1 + \eta)\right).
$$
Therefore, with probability greater than  $1 - \delta$ , for all  $(i_1,i_2)\in [D]^2$  and  $N_{t,k}\in [N_{i_k},N_{i_k + 1})$  for all  $k\in \{1,2\}$
$$
\begin{array}{l} \sum_ {k \in \{1, 2 \}} \frac {S _ {t , k} ^ {2}}{N _ {t , k}} \leq 2 \bar {W} _ {- 1} (\log (1 / \delta) + 2 \log D + \log (1 + \eta) + 1) \\ \leq 2 \bar {W} _ {- 1} (\log (1 / \delta) + 2 \log (2 + \log n) + 2) \\ \end{array}
$$
The second inequality is obtained as in Lemma 42 by using that  $D \leq 1 + \frac{\log n}{\log(1 + \eta)}$ ,  $\overline{W}_{-1}$  increasing, taking  $\eta^{\star} = e^{2} - 1$  and using  $3 - 2\log 2 \leq 2$ .
Since  $[n] \subseteq \bigcup_{i \in [D]} [N_i, N_{i+1})$ ,  $N_{t,k} \leq n$  and  $N_{t,k}(\mu_{t,k} - \mu_k) = S_{t,k}$  (unit-variance), this yields
$$
\mathbb {P} \left(\exists t \leq n, \sum_ {k \in \{1, 2 \}} \frac {N _ {t , k}}{2} \left(\mu_ {t, k} - \mu_ {k}\right) ^ {2} \geq \overline {{W}} _ {- 1} \left(\log (1 / \delta) + 2 \log (2 + \log n) + 2\right)\right) \leq \delta .
$$
Let  $C(n,\delta)\coloneqq \overline{W}_{-1}$ $(\log (1 / \delta) + 2\log (2 + \log n) + 2)$ . In the following, we consider that we are under the event,
$$
\mathcal {E} _ {n, \delta} (1, 2) := \left\{\forall t \leq n, \sum_ {k \in \{1, 2 \}} \frac {N _ {t , k}}{2} (\mu_ {t, k} - \mu_ {k}) ^ {2} <   C (n, \delta) \right\},
$$
which has probability at least  $1 - \delta$ . Since it satisfies the above constraint, the quantity  $\mu_{t,1} - \mu_1 - (\mu_{t,2} - \mu_2)$  is upper bounded by
$$
\max  _ {x \in \mathbb {R} ^ {2}} \left\{\mu_ {t, 1} - \mu_ {t, 2} - x _ {1} + x _ {2} \right\} \quad \text {s u b j e c t t o} \quad \sum_ {k \in \{1, 2 \}} \frac {N _ {t , 1}}{2} (\mu_ {t, k} - x _ {k}) ^ {2} \leq C (n, \delta).
$$
Introducing a Lagrange multiplier  $\alpha$ , the above optimization problem is equivalent to
$$
\min  _ {\alpha \geq 0} \max  _ {x \in \mathbb {R} ^ {2}} \left\{\mu_ {t, 1} - \mu_ {t, 2} - x _ {1} + x _ {2} + \alpha \left(C (n, \delta) - \sum_ {k \in \{1, 2 \}} \frac {N _ {t , k}}{2} (\mu_ {t, k} - x _ {k}) ^ {2}\right) \right\}.
$$
Solving for  $x$  by cancelling the derivative yields that  $x_{1} = \mu_{t,1} - \frac{1}{\alpha N_{t,1}}$  and  $x_{2} = \mu_{t,2} + \frac{1}{\alpha N_{t,2}}$ . Therefore, we obtain as solution
$$
\min _ {\alpha \geq 0} \left\{\alpha C (n, \delta) + \frac {1}{2 \alpha} \sum_ {k \in \{1, 2 \}} \frac {1}{N _ {t , k}} \right\} = \sqrt {2 C (n , \delta) \sum_ {k \in \{1 , 2 \}} \frac {1}{N _ {t , k}}},
$$
where the last equality is obtained by solving the optimization since the derivative is null at  $\alpha^{\star} = \sqrt{\sum_{k\in\{1,2\}}\frac{1}{2C(n,\delta)N_{t,k}}}$ . By symmetry, we obtain the same result to upper bound  $\mu_{t,2} - \mu_2 - (\mu_{t,1} - \mu_1)$ . Therefore, under  $\mathcal{E}_{n,\delta}(1,2)$ , we have shown that
$$
\frac {\left| \mu_ {t , 1} - \mu_ {1} - \left(\mu_ {t , 2} - \mu_ {2}\right) \right|}{1 / N _ {t , 1} + 1 / N _ {t , 2}} \leq \sqrt {2 \overline {{W}} _ {- 1} \left(\log \left(1 / \delta\right) + 2 \log \left(2 + \log n\right) + 2\right)}.
$$
The same argument as above can be applied for all  $(i,k) \in [K]^2$  such that  $i \neq k$ . A direct union bound yields
$$
\mathbb {P} _ {\nu} ((\tilde {\mathcal {E}} _ {n, \delta} ^ {2}) ^ {\complement}) \leq \sum_ {(i, k) \in [ K ] ^ {2}, i \neq k} \mathbb {P} _ {\nu} \left(\mathcal {E} _ {n, \delta n ^ {- s}} (i, k) ^ {\complement}\right) \leq \frac {K (K - 1)}{2} \frac {\delta}{n ^ {s}},
$$
where we used the inclusion proven above for  $(n, \delta / n^s)$ . This concludes the proof.
Global events Lemma 45 combines the above sequences of concentration events.
Lemma 45. Let  $s > 1$ . Let  $(\mathcal{E}_{n,\delta}^{1})_{n > K}$  and  $(\mathcal{E}_{n,\delta}^{2})_{n > K}$  as in (24) and (25). Let  $(\tilde{\mathcal{E}}_{n,\delta}^{1})_{n > K}$  and  $(\tilde{\mathcal{E}}_{n,\delta}^{2})_{n > K}$  as in (27) and (29). Let us define  $\mathcal{E}_{n,\delta} := \mathcal{E}_{n,\delta}^{1} \cap \mathcal{E}_{n,\delta}^{2}$  and  $\tilde{\mathcal{E}}_{n,\delta} := \tilde{\mathcal{E}}_{n,\delta}^{1} \cap \tilde{\mathcal{E}}_{n,\delta}^{2}$  for all  $n > K$ . Then,
$$
\sum_ {n > K} \mathbb {P} _ {\nu} (\mathcal {E} _ {n, \delta} ^ {\mathbb {C}}) \leq \frac {K (K + 1)}{2} \zeta (s) \delta \quad a n d \quad \sum_ {n > K} \mathbb {P} _ {\nu} (\tilde {\mathcal {E}} _ {n, \delta} ^ {\mathbb {C}}) \leq \frac {K (K + 1)}{2} \zeta (s) \delta .
$$
Proof. It is direct to see that
$$
\sum_ {n > K} \mathbb {P} _ {\nu} \left(\mathcal {E} _ {n, \delta} ^ {\complement}\right) \leq \sum_ {n > K} \frac {K \delta}{n ^ {s}} + \sum_ {n > K} \frac {K (K - 1) \delta}{2 n ^ {s}} = \frac {K (K + 1)}{2} \zeta (s) \delta .
$$
The same proof can be applied to upper bound  $\sum_{n > K}\mathbb{P}_{\nu}(\tilde{\mathcal{E}}_{n,\delta}^{\mathbb{C}})$
H Technicalities
Appendix H gathers existing and new technical results which are used for our proofs.
Key technical result Lemma 4 is the key technical ingredient on which our proofs rely on. It builds on a sequence of "bad" events such that, under each "bad" event, either the leader or the challenger was not often selected as leader or challenger, and shows that the number of times those "bad" events occur is small.
Lemma (Lemma 4). Let  $\delta \in (0,1]$  and  $n > K$ . Let  $(A_{t,\delta}(n,\delta))_{n\geq t > K}$  be a sequence of events and  $(D_{i}(n,\delta))_{i\in [K]}$  be positive thresholds satisfying that, for all  $t\in [n]\setminus [K]$ , under the event  $A_{t,\delta}(n,\delta)$
$$
\exists i _ {t} \in [ K ], \quad T _ {t} (i _ {t}) \leq D _ {i _ {t}} (n, \delta) \quad a n d \quad T _ {t + 1} (i _ {t}) = T _ {t} (i _ {t}) + 1.
$$
Then, we have  $\sum_{t = K + 1}^{n}\mathbb{1}\left(A_{t,\delta}(n,\delta)\right)\leq \sum_{i\in [K]}D_{i}(n,\delta).$
Proof. Using the inclusion of events given by the assumption on  $(A_{t,\delta}(n,\delta))_{n\geq t > K}$ , we obtain
$$
\begin{array}{l} \sum_ {t = K + 1} ^ {n} \mathbb {1} \left(A _ {t, \delta} (n, \delta)\right) \leq \sum_ {t = K + 1} ^ {n} \mathbb {1} \left(\exists k _ {t} \in [ K ], T _ {t} \left(k _ {t}\right) \leq D _ {k _ {t}} (n, \delta), T _ {t + 1} \left(k _ {t}\right) = T _ {t} \left(k _ {t}\right) + 1\right) \\ \leq \sum_ {i \in [ K ]} \sum_ {t = K + 1} ^ {n} \mathbb {1} \left(T _ {t} (i) \leq D _ {i} (n, \delta), T _ {t + 1} (i) = T _ {t} (i) + 1\right) \leq \sum_ {i \in [ K ]} D _ {i} (n, \delta). \\ \end{array}
$$
The second inequality is obtained by union bound. The third inequality is direct since the number of times one can increase by one a quantity that is positive and bounded by  $D_{i}(n,\delta)$  is at most  $D_{i}(n,\delta)$ .
Tracking Lemma 46 provide general results for the  $K(K - 1)$  tracking procedures both for IDS and fixed proportions, which includes the ones of Lemma 2. The main theoretical argument behind those results is based on applying Theorem 6 in [12].
Lemma 46. For all  $n > K$ ,  $i \in [K]$ ,  $j \neq i$ , we have
$$
- 1 / 2 \leq N _ {n, j} ^ {i} - (1 - \bar {\beta} _ {n} (i, j)) T _ {n} (i, j) \leq 1 \quad i. e. \quad - 1 \leq \left(T _ {n} (i, j) - N _ {n, j} ^ {i}\right) - \bar {\beta} _ {n} (i, j) T _ {n} (i, j) \leq 1 / 2
$$
$$
\frac {N _ {n , i} ^ {i} - (K - 1) / 2}{\max _ {j \neq i} \bar {\beta} _ {n} (i , j)} \leq \sum_ {j \neq i} T _ {n} (i, j) \leq \frac {N _ {n , i} ^ {i} + K - 1}{\min _ {j \neq i} \bar {\beta} _ {n} (i , j)}.
$$
Let  $\beta \in (0,1)$ . For fixed proportions  $\beta$ , we have
$$
N _ {n, i} \geq \min  \left\{\beta , 1 - \beta \right\} \left(T _ {n} (i) - \frac {3 (K - 1)}{2}\right).
$$
Proof. We have  $K(K - 1)$  independent two-arms C-Tracking between the challenger and the leader. Theorem 6 in [12] yields the result. The second inequality is a simple re-ordering.
To obtain the second part, direct manipulations yield that
$$
\sum_ {j \neq i} T _ {n} (i, j) \leq \sum_ {j \neq i} \frac {T _ {n} (i , j) - N _ {n , j} ^ {i} + 1}{\bar {\beta} _ {n} (i , j)} \leq \frac {\sum_ {j \neq i} (T _ {n} (i , j) - N _ {n , j} ^ {i}) + K - 1}{\min _ {j \neq i} \bar {\beta} _ {n} (i , j)},
$$
which allows to conclude since  $N_{n,i}^{i} = \sum_{j\neq i}(T_{n}(i,j) - N_{n,j}^{i})$ . The lower bound is obtained similarly.
The third is a direct consequence of the first part since
$$
\begin{array}{l} N _ {n, i} = \sum_ {j \neq i} \left(T _ {n} (i, j) - N _ {n, j} ^ {i}\right) + \sum_ {j \neq i} N _ {n, i} ^ {j} \leq \beta \sum_ {j \neq i} T _ {n} (i, j) + (1 - \beta) \sum_ {j \neq i} T _ {n} (j, i) + 3 (K - 1) / 2 \\ \geq \beta \sum_ {j \neq i} T _ {n} (i, j) + (1 - \beta) \sum_ {j \neq i} T _ {n} (j, i) - 3 (K - 1) / 2. \\ \end{array}
$$
Having shown
$$
\max  _ {i \in [ K ]} \left| N _ {n, i} - \left(\beta \sum_ {j \neq i} T _ {n} (i, j) + (1 - \beta) \sum_ {j \neq i} T _ {n} (j, i)\right) \right| \leq 3 (K - 1) / 2,
$$
it is direct to conclude.
Methodology Lemma 47 is a standard result to upper bound the expected sample complexity of an algorithm, e.g. see Lemma 1 in [11]. This is a key method extensively used in the literature.
Lemma 47. Let  $(\mathcal{E}_n)_{n > K}$  be a sequence of events and  $T(\delta) > K$  be such that for  $n\geq T(\delta)$ ,  $\mathcal{E}_n\subseteq \{\tau_\delta \leq n\}$ . Then,  $\mathbb{E}_{\nu}[\tau_{\delta}]\leq T(\delta) + \sum_{n > K}\mathbb{P}_{\nu}(\mathcal{E}_n^{\complement})$ .
Proof. Since the random variable  $\tau_{\delta}$  is positive and  $\{\tau_{\delta} > n\} \subseteq \mathcal{E}_n^{\mathbb{C}}$  for all  $n\geq T(\delta)$ , we have
$$
\mathbb {E} _ {\nu} \left[ \tau_ {\delta} \right] = \sum_ {n \geq 0} \mathbb {P} _ {\nu} \left(\tau_ {\delta} > n\right) \leq T (\delta) + \sum_ {n \geq T (\delta)} \mathbb {P} _ {\nu} \left(\mathcal {E} _ {n} ^ {\complement}\right),
$$
which concludes the proof by adding positive terms.
Lemma 48 is a key method to upper bound the probability of error of an algorithm.
Lemma 48. Let  $\varepsilon \geq 0$  and  $(\mathcal{E}_{n,\delta})_{n > K,\delta \in (0,1)}$  be a sequence of events such that  $\mathbb{P}_{\nu}(\mathcal{E}_{n,\delta}^{\complement}) \leq C\delta$  with  $C > 0$ . Suppose that  $T_{\varepsilon}(\delta) > K$  is such that for  $n > T_{\varepsilon}(\delta)$ ,  $\mathcal{E}_{n,\delta} \subseteq \{\hat{i}_n \in \mathcal{I}_{\varepsilon}(\mu)\}$ . Then,
$$
\mathbb {P} _ {\nu} \left(i _ {n} \notin \mathcal {I} _ {\varepsilon} (\mu)\right) \leq C \inf  \left\{\delta \mid n > T _ {\varepsilon} (\delta) \right\}.
$$
Proof. Using the assumption, we have  $\mathbb{P}_{\nu}(\hat{i}_n \notin \mathcal{I}_{\varepsilon}(\mu)) \leq \mathbb{P}_{\nu}(\mathcal{E}_{n,\delta}^{\complement}) \leq C\delta$  for all  $n > T_{\varepsilon}(\delta)$ . Taking the infimum yields the result.
Inversion results Lemma 49 gathers properties on the function  $\overline{W}_{-1}$ , which is used in the literature to obtain concentration results.
Lemma 49 ([22]). Let  $\overline{W}_{-1}(x) = -W_{-1}(-e^{-x})$  for all  $x\geq 1$ , where  $W_{-1}$  is the negative branch of the Lambert  $W$  function. The function  $\overline{W}_{-1}$  is increasing on  $(1, + \infty)$  and strictly concave on  $(1, + \infty)$ . In particular,  $\overline{W}_{-1}'(x) = \left(1 - \frac{1}{\overline{W}_{-1}(x)}\right)^{-1}$  for all  $x > 1$ . Then, for all  $y\geq 1$  and  $x\geq 1$ ,
$$
\overline {{W}} _ {- 1} (y) \leq x \quad \Longleftrightarrow \quad y \leq x - \log (x).
$$
Moreover, for all  $x > 1$
$$
x + \log (x) \leq \overline {{W}} _ {- 1} (x) \leq x + \log (x) + \min  \left\{\frac {1}{2}, \frac {1}{\sqrt {x}} \right\}.
$$
Lemma 50 provides an ordering on the thresholds for tighter concentration results. It leverages properties of  $\overline{W}_{-1}$ .
Lemma 50. Let  $s \geq 0$ . Let  $\overline{W}_{-1}$  defined in Lemma 49. For all  $\delta \in (0,1]$  and  $n \geq 1$ , let us denote by
$$
\tilde {f} _ {1} (n, \delta) = \frac {1}{2} \overline {{W}} _ {- 1} (2 \log (1 / \delta) + 2 s \log n + 2 \log (2 + \log n) + 2),
$$
$$
\tilde {f} _ {2} (n, \delta) = \bar {W} _ {- 1} (\log (1 / \delta) + s \log n + 2 \log (2 + \log n) + 2).
$$
Then, we have  $\tilde{f}_1(n,\delta)\leq \tilde{f}_2(n,\delta)$
Proof. Let  $a > 0$  and  $b \geq 1$ . Let us define  $f(x) = \frac{1}{x}\overline{W}_{-1}(ax + b)$ . Using Lemma 49, we have
$$
f ^ {\prime} (x) = - \frac {1}{x ^ {2}} \left(\overline {{W}} _ {- 1} (a x + b) - \frac {a x}{1 - \overline {{W}} _ {- 1} (a x + b) ^ {- 1}}\right).
$$
Therefore, we obtain  $f'(x) \leq 0$  if and only if  $\overline{W}_{-1}(ax + b) \geq ax + 1$ . Since  $\overline{W}_{-1}(ax + b) \geq ax + b$  and  $b \geq 1$ , the function  $f(x)$  is decreasing for all  $x \geq 1$ . Therefore,  $\frac{1}{2}\overline{W}_{-1}(2x + b) \leq \overline{W}_{-1}(a + b)$ . Applying this result with  $a = \log(1/\delta) + s\log n$  and  $b = 2\log(2 + \log n) + 2 \geq 1$  for all  $n \geq 1$ , we obtain  $\tilde{f}_1(n,\delta) \leq \tilde{f}_2(n,\delta)$ .
Lemma 51 is an inversion result to upper bound a time which is implicitly defined. It is a direct consequence of Lemma 49.
Lemma 51. Let  $\overline{W}_{-1}$  defined in Lemma 49. Let  $A > 0$ ,  $B > 0$  such that  $B / A + \log A > 1$  and
$$
C (A, B) = \sup  \left\{x \mid x <   A \log x + B \right\},
$$
$$
C (A, B, D) = \sup  \left\{x \mid x <   A \bar {W} _ {- 1} \left(2 \log (2 + \log x) + D\right) + B \right\}.
$$
Then,  $C(A,B) < h_1(A,B)$  with  $h_1(z,y) = z\overline{W}_{-1}(y / z + \log z)$  and  $C(A,B) < Ah_2(A,B,D) + B$  where
$$
h _ {2} (x, y, z) = \inf  \left\{u \mid u - \log u - 2 \log (2 + \log (x u + y)) \geq z \right\}.
$$
Proof. Since  $B / A + \log A > 1$ , we have  $C(A, B) \geq A$ , hence
$$
C (A, B) = \sup  \left\{x \mid x <   A \log (x) + B \right\} = \sup  \left\{x \geq A \mid x <   A \log (x) + B \right\}.
$$
Using Lemma 49 yields that
$$
x \geq A \log x + B \iff \frac {x}{A} - \log \left(\frac {x}{A}\right) \geq \frac {B}{A} + \log A \iff x \geq A \overline {{W}} _ {- 1} \left(\frac {B}{A} + \log A\right).
$$
By changing the variable, we obtain
$$
C (A, B, D) = B + A \sup  \left\{y \mid y <   \overline {{W}} _ {- 1} \left(2 \log \left(2 + \log (A y + B)\right) + D\right) \right\}
$$
Likewise, Lemma 49 yields that
$$
y \geq \overline {{W}} _ {- 1} \left(2 \log \left(2 + \log (A y + B)\right) + D\right) \iff y - \log y - 2 \log \left(2 + \log (A y + B)\right) \geq D.
$$
Lemma 52 is an inversion result to upper bound a probability which is implicitly defined based on times that are implicitly defined.
Lemma 52. Let  $\overline{W}_{-1}$  defined in Lemma 49. Let  $A > 0$ ,  $B > 0$ ,  $C > 0$ ,  $E > 0$ ,  $\alpha > 0$ ,  $\beta > 0$  and
$$
D _ {A, B, C} (\delta) = \sup  \left\{x \mid x \leq A \left(\log (1 / \delta) + C \log x\right) + B \right\},
$$
$$
D _ {A, B, C, E, \alpha , \beta} (\delta) = \sup  \left\{x \mid x \leq \frac {A}{\alpha} \bar {W} _ {- 1} \left(\alpha \left(\log (1 / \delta) + C \log (\beta + \log x) + E)\right) + B \right. \right\}.
$$
$$
\inf  \left\{\delta \mid x > D _ {A, B, C} (\delta) \right\} \leq x ^ {C} \exp \left(- \frac {x - B}{A}\right),
$$
$$
\inf  \left\{\delta \mid x > D _ {A, B, C, E, \alpha , \beta} (\delta) \right\} \leq e ^ {E} \left(\alpha \frac {x - B}{A}\right) ^ {1 / \alpha} (\beta + \log x) ^ {C} \exp \left(- \frac {x - B}{A}\right).
$$
Proof. Direct manipulations yield that
$$
x > D _ {A, B, C} (\delta) \iff x > A (\log (1 / \delta) + C \log x) + B \iff \delta <   x ^ {C} \exp \left(- \frac {x - B}{A}\right).
$$
Likewise, using Lemma 49, we obtain
$$
\begin{array}{l} x > D _ {A, B, C, E, \alpha , \beta} (\delta) \iff \alpha \frac {x - B}{A} > \bar {W} _ {- 1} \left(\alpha \left(\log (1 / \delta) + C \log (\beta + \log x) + E)\right) \right. \\ \Longleftrightarrow \frac {x - B}{A} - \frac {1}{\alpha} \log \left(\alpha \frac {x - B}{A}\right) > \log (1 / \delta) + C \log (\beta + \log x) + E \\ \Longleftrightarrow \delta <   e ^ {E} \left(\alpha \frac {x - B}{A}\right) ^ {1 / \alpha} (\beta + \log x) ^ {C} \exp \left(- \frac {x - B}{A}\right). \\ \end{array}
$$
I Multiplicative setting
Let  $\nu \in \mathcal{D}^K$  with mean parameters  $\mu \in (\mathbb{R}_+^\star)^K$ . Compared to the additive setting, we only consider strictly positive mean parameters. In many applications, this assumption is natural. In other applications, there is a known lower bound on the value of the true mean parameters, hence we can simply translate the problem by adding this lower bound. We also note that the problem of identifying which arms are above the threshold 0 is the thresholding bandit problem, which has been extensively studied in the literature. Overall, while this assumption might seem restrictive, we think it is rather mild. Since the means are strictly positive, the multiplicative error  $\varepsilon$  has also to be positive and strictly smaller than 1. Given a multiplicative error  $\varepsilon \in [0,1)$ , the set of multiplicative  $\varepsilon$ -good arms are denoted by  $\mathcal{T}_{\varepsilon}^{\mathrm{mul}}(\mu) \coloneqq \{i \in [K] \mid \Delta_i \leq \mu_\star \varepsilon\}$ . We will refer to this problem as multiplicative  $\varepsilon$ -BAI.
First, we discuss the characteristic times involved in the fixed-confidence multiplicative  $\varepsilon$ -BAI (Appendix I.1). Second, for  $\varepsilon_0 \in (0, 1)$ , we present the  $\mathrm{EB-TC}_{\varepsilon_0}^{\mathrm{m}}$  with IDS or fixed  $\beta$  proportions (Appendix I.2). Finally, we prove that  $\mathrm{EB-TC}_{\varepsilon_0}^{\mathrm{m}}$  is asymptotically (resp.  $\beta$ -)optimal when using IDS (resp. fixed  $\beta$ ) when combined with the appropriate GLR stopping rule (Appendix I.3).
I.1 Characteristic times
Let  $\nu \in \mathcal{D}^K$  with mean parameters  $\mu \in (\mathbb{R}_+^\star)^K$ . We assume in the following that there exists a unique best arm, i.e.  $i^\star (\mu) = \{i^\star \}$ . Let  $\varepsilon \in [0,1)$ . The  $(\beta)$ -characteristic times for the fixed confidence multiplicative  $\varepsilon$ -BAI setting with Gaussian bandits  $\mathcal{N}(\mu, 1)$  are defined as  $T_{\varepsilon}^{\mathrm{mul}}(\mu) = \min_{i\in \mathcal{I}_{\varepsilon}^{\mathrm{mul}}(\mu),\beta \in (0,1)}T_{\varepsilon ,\beta}^{\mathrm{mul}}(\mu ,i)$  and  $T_{\varepsilon ,\beta}^{\mathrm{mul}}(\mu) = \min_{i\in \mathcal{I}_{\varepsilon}^{\mathrm{mul}}(\mu)}T_{\varepsilon ,\beta}^{\mathrm{mul}}(\mu ,i)$  where