SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Compared to the generic analysis of Top Two algorithms [JDB+22], for AdaP-TT, we need to control the sample complexity cost of doubling and forgetting. Due to this reason, we have to pay a multiplicative four-factor: one two-factor due to doubling, and another two-factor due to forgetting. It is possible to show that this cost exists when adapting any "reasonable" BAI algorithm, meaning for any BAI algorithm in which the empirical proportions are converging towards an allocation  $\omega$  such that  $\min_{a}\omega_{a} > 0$ . Those BAI algorithms are "reasonable" because the asymptotic lower bound stipulates that all arms have to be sampled linearly in order to be near optimal.
Let  $\omega \in \Sigma_K$  be any allocation over arms such that  $\min_{a}\omega_{a} > 0$ . Let  $\gamma > 0$ . We denote by  $T_{\mu,\gamma}(\omega)$  the convergence time towards  $\omega$ , which is a random variable quantifying the number of samples required for the global empirical allocations  $N_{n} / (n - 1)$  to be  $\gamma$ -close to  $\omega$  for any subsequent time, namely
$$
T _ {\boldsymbol {\mu}, \gamma} (\omega) \triangleq \inf  \left\{T \geq 1 \mid \forall n \geq T, \left\| \frac {N _ {n}}{n - 1} - \omega \right\| _ {\infty} \leq \gamma \right\}. \tag {29}
$$
Lemma 19 shows that the phase switches of the arms happen in a round-robin fashion, which means that an arm switches phase for a second time after all other arms first switch their own phases.
Lemma 19. Let  $\omega \in \Sigma_K$  such that  $\min_a\omega_a > 0$ . Assume that there exists  $\gamma_{\mu} > 0$  such that for  $\mathbb{E}_{\nu}[T_{\mu ,\gamma}(\omega)] < + \infty$  for all  $\gamma \in (0,\gamma_{\mu})$ , where  $T_{\mu ,\gamma}(\omega)$  is defined in Equation (29). Let  $\eta >0$ . There exists  $\tilde{\gamma}_{\mu}\in (0,\gamma_{\mu})$  such that, for all  $\gamma \in (0,\tilde{\gamma}_{\mu})$ , there exists  $N_{3}\geq T_{\mu ,\gamma}(\omega)$  with  $\mathbb{E}_{\nu}[N_3] < + \infty$  which satisfies
$$
\forall n \geq N _ {3}, \quad \frac {\max  _ {a \in [ K ]} T _ {k _ {n , a}} (a) - 1}{\min  _ {a \in [ K ]} T _ {k _ {n , a}} (a) - 1} \leq 2 + \eta .
$$
Proof. Let  $\eta > 0$ . Let  $\tilde{\gamma}_{\mu} \in (0, \gamma_{\mu})$  such that  $2\max_{a \in [K]} (\omega_a + \gamma) / (\omega_a - \gamma) \leq 2 + \eta$ , which is possible since  $\min_{a} \omega_a > 0$ . Let  $\gamma \in (0, \tilde{\gamma}_{\mu})$ . By assumption, we have  $\mathbb{E}_{\nu}[T_{\mu, \gamma}(\omega)] < +\infty$ . Then, for all  $n \geq T_{\mu, \gamma}(\omega)$ ,
$$
\left\| \frac {N _ {n}}{n - 1} - \omega \right\| _ {\infty} \leq \gamma .
$$
Let  $M \geq T_{\mu, \gamma}(\omega)$ . Let us denote by  $k_M = (k_{M,a})_{a \in [K]}$  the current phases for all arms  $a \in [K]$  at time  $M$ . Then, for all  $n \geq M$  and all  $a \in [K]$ , we have  $N_{n,a} \geq (n - 1)(\omega_a - \gamma)$ . Therefore, taking
$n \geq \max_{a \in [K]} 2^{k_{M,a}} (\omega_a - \gamma)^{-1} + 1$ , we obtain that  $N_{n,a} \geq 2^{k_{M,a}}$  for all  $a \in [K]$ , hence we have  $\max_{a \in [K]} T_{k_{M,a} + 1}(a) \leq n$ . Since  $\min_{a \in [K]} T_{k_{M,a} + 1}(a) \geq M$ , we have
$$
\max  _ {a \in [ K ]} \left| \frac {N _ {T _ {k _ {M , a} + 1} (a) , a}}{n - 1} - \omega_ {a} \right| \leq \gamma .
$$
Likewise, taking  $n \geq \max_{a \in [K]} 2^{k_{M,a} + 1} (\omega_a - \gamma)^{-1} + 1$ , we obtain that  $N_{n,a} \geq 2^{k_{M,a} + 1}$  for all  $a \in [K]$ , hence we have  $\max_{a \in [K]} T_{k_{M,a} + 2}(a) \leq n$ . Let  $a_1 = \arg \min_{a \in [K]} T_{k_{M,a} + 2}(a)$ . By definition and using Lemma 3, we have
$$
2 ^ {k _ {M, a _ {1}} + 1} = N _ {T _ {k _ {M, a _ {1}} + 2} (a _ {1}), a _ {1}} \leq (T _ {k _ {M, a _ {1}} + 2} (a _ {1}) - 1) (\omega_ {a _ {1}} + \gamma),
$$
$$
\forall a \neq a _ {1}, 2 ^ {k _ {M, a}} \leq N _ {T _ {k _ {M, a _ {1}} + 2} (a _ {1}), a} \leq (T _ {k _ {M, a _ {1}} + 2} (a _ {1}) - 1) (\omega_ {a} + \gamma).
$$
Let  $a_2 = \arg \max_{a \in [K]} T_{k_{M,a} + 2}(a)$ . By definition and using Lemma 3, we have
$$
2 ^ {k _ {M, a _ {2}} + 1} = N _ {T _ {k _ {M, a _ {2}} + 2} (a _ {2}), a _ {2}} \geq (T _ {k _ {M, a _ {2}} + 2} (a _ {2}) - 1) (\omega_ {a _ {2}} - \gamma),
$$
Therefore, combining the above yields
$$
\left(T _ {k _ {M, a _ {2}} + 2} (a _ {2}) - 1\right) \leq \left(T _ {k _ {M, a _ {1}} + 2} (a _ {1}) - 1\right) 2 \frac {\omega_ {a _ {2}} + \gamma}{\omega_ {a _ {2}} - \gamma} \leq \left(T _ {k _ {M, a _ {2}} + 2} (a _ {2}) - 1\right) (2 + \eta),
$$
where the last inequality uses that  $\gamma \in (0, \tilde{\gamma}_{\mu})$  and  $\tilde{\gamma}_{\mu} \in (0, \gamma_{\mu})$  is such that  $2\max_{a \in [K]} (\omega_{a} + \gamma) / (\omega_{a} - \gamma) \leq 2 + \eta$ . We take  $n \geq N_3 = \max_{a \in [K]} T_{k_{M,a} + 2}(a)$ , hence we have  $k_{n,a} \geq k_{M,a} + 2$  for all  $a \in [K]$ . Since  $\mathbb{E}_{\nu}[T_{\mu,\gamma}(\omega)] < +\infty$  (i.e. arms are sampled linearly), it is direct to see that  $\mathbb{E}_{\nu}[\max_{a \in [K]} T_{k_{M,a} + 2}(a)] < +\infty$ . This concludes the proof.
Note that  $T_{\mu, \gamma}$  defined in 26 is such that  $T_{\mu, \gamma} = T_{\mu, \gamma}(\omega_{\mathrm{KL}, \beta}^{\star})$  where  $T_{\mu, \gamma}(\omega)$  as in Equation (29). Lemma 18 showed that  $\mathbb{E}_{\nu}[T_{\mu, \gamma}] < +\infty$  for all  $\gamma > 0$ . Therefore, the condition of Lemma 19 are fulfilled by AdaP-TT.
E.5 Asymptotic expected sample complexity
The final step of the generic analysis of Top Two algorithms  $\left[\mathrm{JDB}^{+}22\right]$  is to invert the private GLR stopping rule by leveraging the convergence of the empirical proportions towards the  $\beta$ -optimal allocation. Compared to the non-private GLR stopping rule, the private threshold in the private GLR stopping rule involves an additive term in  $\mathcal{O}(\log (1 / \delta)^2)$ . This difference is the largest price that we pay to obtain a private BAI algorithm. We defer the reader to Appendix E.7 for a more detailed discussion on it.
Asymptotically  $\beta$ -optimal  $\epsilon$ -DP-FC-BAI algorithm. The inversion of the GLR stopping rule by leveraging the convergence of the empirical proportions towards the  $(\beta)$ -optimal allocation is a generic method used in the BAI literature. Provided this convergence is shown, it only depends on the threshold that ensures  $\delta$ -correctness. More precisely, it only depends on its asymptotic dependence in  $\log(1/\delta)$ . In addition to the multiplicative four-factor, the price of privacy for asymptotically  $\beta$ -optimal BAI algorithms when combined with the non-private GLR stopping rule is a problem dependent multiplicative factor  $1 + \sqrt{1 + \Delta_{\max}^2/(2\epsilon^2)}$  (Lemma 20).
Lemma 20. Let  $(\delta, \beta) \in (0,1)^2$ . Assume that there exists  $\gamma_{\mu} > 0$  such that for  $\mathbb{E}_{\nu}[T_{\mu,\gamma}] < +\infty$  for all  $\gamma \in (0,\gamma_{\mu})$ , where  $T_{\mu,\gamma}$  is defined in Equation (26). Combining the private GLR stopping rule (Equation (16)) with private threshold (Equation (4)) yields a  $\delta$ -correct algorithm which satisfies that, for all  $\pmb{\nu}$  with mean  $\pmb{\mu}$  such that  $|a^{\star}(\pmb{\mu})| = 1$ ,
$$
\limsup_{\delta \to 0}\frac{\mathbb{E}_{\boldsymbol{\nu}}\left[\tau_{\delta}\right]}{\log(1 / \delta)}\leq 4T_{\mathrm{KL},\beta}^{\star}(\boldsymbol {\mu})\left(1 + \sqrt{1 + \frac{\Delta_{\max}^{2}}{2\epsilon^{2}}}\right).
$$
Proof. Theorem 4 yields the  $\delta$ -correctness.
Let  $a^\star$  be the unique best arm, i.e.  $a^\star(\pmb{\mu}) = \{a^\star\}$ . Let  $\zeta > 0$ . Using Equation (25) and the continuity of
$$
(\boldsymbol {\mu}, w) \mapsto \min _ {a \neq a ^ {\star} (\boldsymbol {\mu})} \frac {\left(\mu_ {a ^ {\star} (\boldsymbol {\mu})} - \mu_ {a}\right) ^ {2}}{2 \left(1 / w _ {a ^ {\star} (\boldsymbol {\mu})} + 1 / w _ {a}\right)}
$$
yields that there exists  $\gamma_{\zeta} > 0$  such that  $\left\| \frac{N_n}{n - 1} - \omega_\beta^\star \right\|_\infty \leq \gamma_\zeta$  and  $\max_{a \in [K]} |\tilde{\mu}_{k_{n,a} + 1,a} - \mu_a| \leq \gamma_\zeta$  implies that
$$
\forall a \neq a ^ {\star}, \frac {(\tilde {\mu} _ {k _ {n , a} ^ {\star} + 1 , a ^ {\star}} - \tilde {\mu} _ {k _ {n , a} + 1 , a}) ^ {2}}{(n - 1) / N _ {n , a ^ {\star}} + (n - 1) / N _ {n , a}} \geq \frac {2 (1 - \zeta)}{T _ {\mathrm {K L} , \beta} ^ {*} (\boldsymbol {\mu})}
$$
$$
\frac {n - 1}{N _ {n , a ^ {\star}}} + \frac {n - 1}{N _ {n , a}} \leq \frac {\Delta_ {a} ^ {2}}{2} (1 + \zeta) T _ {\mathrm {K L}, \beta} ^ {\star} (\boldsymbol {\mu}).
$$
We choose such a  $\gamma_{\zeta}$ . Let  $\gamma_{\mu} > 0$  be such that for  $\mathbb{E}_{\nu}[T_{\mu,\gamma}] < +\infty$  for all  $\gamma \in (0, \gamma_{\mu})$ , where  $T_{\mu,\gamma}$  is defined in Equation (26). Let  $\eta > 0$ . Let  $\tilde{\gamma}_{\mu} \in (0, \gamma_{\mu})$  as in Lemma 19 for this  $\eta$ . In the following, let us consider  $\gamma \in (0, \min\{\tilde{\gamma}_{\mu}, \gamma_{\zeta}, \beta/4, \Delta_{\min}/4\})$ .
Let  $N_{3} \geq T_{\mu, \gamma}$  with  $\mathbb{E}_{\nu}[N_3] < +\infty$  as Lemma 19 for those  $(\gamma, \eta)$ . Then, we have  $\mathbb{E}_{\nu}[T_{\mu, \gamma}] < +\infty$  and
Since arms are sampled linearly, it is direct to construct  $N_{4} \geq N_{3}$  with  $\mathbb{E}_{\nu}[N_4] < +\infty$  such that, for all  $n \geq N_4$ , we have  $\max_{a \in [K]} \max_{k \in \{k_{n,a}, k_{n,a} + 1\}} |\tilde{\mu}_{k,a} - \mu_a| \leq \gamma$ . Therefore, we have  $\hat{a}_n = a^\star$ .
Let  $\kappa \in (0,1)$ . Let  $n \geq N_4 / \kappa$  and  $(k_{n,a})_{a \in [K]}$  be the current phases at time  $n$ . Combining the above, we have  $\hat{a}_n = a^\star$  and
$$
\max  _ {a \in [ K ]} | \tilde {\mu} _ {k _ {n, a} + 1, a} - \mu_ {a} | \leq \gamma , \quad \left\| \frac {N _ {n}}{n - 1} - \omega_ {\beta} ^ {\star} \right\| _ {\infty} \leq \gamma \quad \text {a n d} \quad \frac {\max  _ {a \in [ K ]} T _ {k _ {n , a}} (a) - 1}{\min  _ {a \in [ K ]} T _ {k _ {n , a}} (a) - 1} \leq 2 + \eta .
$$
Let  $a_1 = \arg \min_{a \in [K]} T_{k_{n,a}}(a)$  and  $a_2 = \arg \max_{a \in [K]} T_{k_{n,a}}(a)$ . Therefore, we obtain, for all  $a \neq a^{\star}$ ,
$$
\begin{array}{l} \frac {(\tilde {\mu} _ {k _ {n , \hat {a} _ {n}} + 1 , \hat {a} _ {n}} - \tilde {\mu} _ {k _ {n , a} + 1 , a}) ^ {2}}{1 / \tilde {N} _ {k _ {n , \hat {a} _ {n}} + 1 , \hat {a} _ {n}} + 1 / \tilde {N} _ {k _ {n , a} + 1 , a}} = \frac {(\tilde {\mu} _ {k _ {n , a ^ {\star}} + 1 , a ^ {\star}} - \tilde {\mu} _ {k _ {n , a} + 1 , a}) ^ {2}}{1 / N _ {T _ {k _ {n , a ^ {\star}}} (a ^ {\star}) , a ^ {\star}} + 1 / N _ {T _ {k _ {n , a}} (a) , a}} \\ \geq \frac {(\tilde {\mu} _ {k _ {n , a ^ {\star}} + 1 , a ^ {\star}} - \tilde {\mu} _ {k _ {n , a} + 1 , a}) ^ {2}}{1 / N _ {T _ {k _ {n , a _ {1}}} (a _ {1}) , a ^ {\star}} + 1 / N _ {T _ {k _ {n , a _ {1}}} (a _ {1}) , a}} \\ \geq (\min _ {a \in [ K ]} T _ {k _ {n, a}} (a) - 1) \frac {2 (1 - \zeta)}{T _ {\mathrm {K L} , \beta} ^ {\star} (\pmb {\mu})}. \\ \end{array}
$$
Similarly, we can show that, for all  $a \neq a^{\star}$ ,
$$
\begin{array}{l} \frac {1}{\tilde {N} _ {k _ {n , a ^ {\star}} + 1 , a ^ {\star}}} + \frac {1}{\tilde {N} _ {k _ {n , a} + 1 , a}} = \frac {1}{N _ {T _ {k _ {n , a ^ {\star}}} (a ^ {\star}) , a ^ {\star}}} + \frac {1}{N _ {T _ {k _ {n , a}} (a) , a}} \\ \leq \frac {1}{N _ {T _ {k _ {n} , a _ {1}} (a _ {1}) , a ^ {\star}}} + \frac {1}{N _ {T _ {k _ {n} , a _ {1}} (a _ {1}) , a}} \\ \leq \frac {1}{\min  _ {a \in [ K ]} T _ {k _ {n , a}} (a) - 1} \frac {\Delta_ {a} ^ {2}}{2} (1 + \zeta) T _ {\mathrm {K L}, \beta} ^ {\star} (\boldsymbol {\mu}) \\ \leq \frac {1}{\min _ {a \in [ K ]} T _ {k _ {n , a}} (a) - 1} \frac {\Delta_ {\max} ^ {2}}{2} (1 + \zeta) T _ {\mathrm {K L}, \beta} ^ {\star} (\boldsymbol {\mu}). \\ \end{array}
$$
Let  $(c_k)_{k\in \mathbb{N}}$  as in Equation (14). Using Lemma 3, we obtain, for all  $a\neq a^{\star}$
$$
\begin{array}{l} 2 c _ {\left(k _ {n, a ^ {\star}} + 1\right) \left(k _ {n, a} + 1\right)} \left(\tilde {N} _ {k _ {n, a ^ {\star}} + 1, a ^ {\star}}, \tilde {N} _ {k _ {n, a} + 1, a}, \delta / 2\right) \leq 8 \log \left(4 + \left(\max  _ {b \in [ K ]} k _ {n, b} - 1\right) \log 2\right) \\ + 4 \mathcal {C} _ {G} \left(\log (1 / \delta) / 2 + s \log \left(\max  _ {b \in [ K ]} k _ {n, b} - 1\right) + \log \left(2 (K - 1) \zeta (s) ^ {2}\right) / 2\right) \\ \end{array}
$$
Likewise, we obtain, for all  $a \in [K]$ ,
$$
\begin{array}{l} \frac {1}{\tilde {N} _ {k _ {n , a ^ {\star}} + 1 , a ^ {\star}} \epsilon^ {2}} \log \left(\frac {2 K (k _ {n , a ^ {\star}} + 1) ^ {s} \zeta (s)}{\delta}\right) ^ {2} + \frac {1}{\tilde {N} _ {k _ {n , a} + 1 , a} \epsilon^ {2}} \log \left(\frac {2 K (k _ {n , a} + 1) ^ {s} \zeta (s)}{\delta}\right) ^ {2} \\ \leq \frac {\Delta_ {\operatorname* {m a x}} ^ {2}}{2 \epsilon^ {2}} \frac {(1 + \zeta) T _ {\mathrm {K L} , \beta} ^ {\star} (\boldsymbol {\mu})}{\min  _ {b \in [ K ]} T _ {k _ {n , b}} (b) - 1} \left(\log (1 / \delta) + s \log \left(\max  _ {b \in [ K ]} k _ {n, b} + 1\right) + \log (2 K \zeta (s))\right) ^ {2} \\ \end{array}
$$
Let us denote by  $T_{k_n + 1}^+ = \max_{b\in [K]}T_{k_{n,b} + 1}(b)$ ,  $T_{k_n + 2}^+ = \max_{b\in [K]}T_{k_{n,b} + 2}(b)$ ,  $T_{k_n + 1}^- = \min_{b\in [K]}T_{k_{n,b} + 1}(b)$ ,  $T_{k_n}^- = \min_{b\in [K]}T_{k_{n,b}}(b)$ . Let  $T$  be a time such that  $T\geq T_{k_n + 1}^+ \geq \kappa T$ . Using Lemmas 3 and 19, we have
$$
\left(k _ {n, b} - 1\right) \log 2 = \log N _ {T _ {k _ {n, b}} (b), b} \leq \log T _ {k _ {n, b}} (b) \leq \log T _ {k _ {n}} ^ {+} \leq \log T _ {k _ {n}} ^ {-} + \log (2 + \eta).
$$
Using the private GLR stopping rule (Equation (16)), we have
$$
\begin{array}{l} \min  \left\{\tau_ {\delta}, T \right\} - \kappa T \leq \sum_ {T \geq T _ {k _ {n}} ^ {+} \geq \kappa T} (T _ {k _ {n} + 2} ^ {+} - T _ {k _ {n} + 1} ^ {+}) \mathbb {1} \left(\tau_ {\delta} > T _ {k _ {n} + 1} ^ {+}\right) \\ \leq \sum_ {T \geq T _ {k _ {n}} ^ {+} \geq \kappa T} (T _ {k _ {n} + 2} ^ {+} - T _ {k _ {n} + 1} ^ {+}) \mathbb {1} \left(\exists a \neq a ^ {\star}, \frac {(\tilde {\mu} _ {k _ {n , a ^ {\star}} + 1 , a ^ {\star}} - \tilde {\mu} _ {k _ {n , a} + 1 , a}) ^ {2}}{1 / \tilde {N} _ {k _ {n , a ^ {\star}} + 1 , a ^ {\star}} + 1 / \tilde {N} _ {k _ {n , a} + 1 , a}} \right. \\ <   2 c _ {\epsilon , k _ {n, a ^ {\star}} + 1, k _ {n, a} + 1} \left(\tilde {N} _ {k _ {n, a ^ {\star}} + 1, a ^ {\star}}, \tilde {N} _ {k _ {n, a} + 1, a}, \delta\right) \Bigg) \\ \leq \sum_ {T \geq T _ {k _ {n}} ^ {+} \geq \kappa T} (T _ {k _ {n} + 2} ^ {+} - T _ {k _ {n} + 1} ^ {+}) \mathbb {1} \left((T _ {k _ {n}} ^ {-} - 1) \frac {1 - \zeta}{T _ {\mathrm {K L} , \beta} ^ {\star} (\boldsymbol {\mu})} <   8 \log (4 + \log T _ {k _ {n}} ^ {-} + \log (2 + \eta)) \right. \\ + 4 \mathcal {C} _ {G} (\log (1 / \delta) / 2 + s \log (2 + \log_ {2} T _ {k _ {n}} ^ {-} + \log_ {2} (2 + \eta)) + \log (2 (K - 1) \zeta (s) ^ {2}) / 2) \\ + \frac {\Delta_ {\operatorname* {m a x}} ^ {2}}{2 \epsilon^ {2}} \frac {(1 + \zeta) T _ {\mathrm {K L} , \beta} ^ {\star} (\boldsymbol {\mu})}{T _ {k _ {n}} ^ {-} - 1} \left(\log (1 / \delta) + s \log (2 + \log_ {2} T _ {k _ {n}} ^ {-} + \log_ {2} (2 + \eta)) + \log (2 K \zeta (s))\right) ^ {2}\left. \right), \\ \end{array}
$$
Let  $T_{\zeta}(\delta)$  defined as the largest deterministic time such that the above condition is satisfied when replacing  $T_{k_n}^-$  by  $(1 - \kappa)T$ . Let  $k_{\delta}$  be the largest random vector of phases such that that  $T_{k_{\delta} + 1}^{+} \leq T_{\zeta}(\delta)$  almost surely, hence  $T_{k_{\delta} + 2}^{+} > T_{\zeta}(\delta)$  almost surely. Then, using the above yields that  $\tau_{\delta} \leq T_{k_{\delta} + 2}^{+}$  almost surely, hence
$$
\limsup_{\delta \to 0}\frac{\mathbb{E}_{\boldsymbol{\nu}}[\tau_{\delta}]}{\log(1 / \delta)}\leq \limsup_{\delta \to 0}\frac{\mathbb{E}_{\boldsymbol{\nu}}[T_{k_{\delta} + 2}^{+}]}{\log(1 / \delta)}\leq (2 + \eta)^{2}\limsup_{\delta \to 0}\frac{\mathbb{E}_{\boldsymbol{\nu}}[T_{k_{\delta} + 1}^{+}]}{\log(1 / \delta)}\leq (2 + \eta)^{2}\limsup_{\delta \to 0}\frac{T_{\zeta}(\delta)}{\log(1 / \delta)},
$$
where the second inequality uses Lemma 19 twice, i.e.  $T_{k_{\delta} + 2}^{+} \leq (2 + \eta)T_{k_{\delta} + 2}^{-} \leq (2 + \eta)^{2}T_{k_{\delta} + 1}^{+}$ , and the last one used the definition of  $k_{\delta}$  and that  $T_{\zeta}(\delta)$  is deterministic.
Since we are only interested in upper bounding  $\lim_{\delta \to 0} \sup_{\delta \to 0} \frac{T_{\zeta}(\delta)}{\log(1 / \delta)}$ , we can safely drop the second orders terms in  $T$  and  $\log(1 / \delta)$ . This allows us to remove the terms in  $\mathcal{O}(\log \log T)$  and in  $\mathcal{O}(\log \log (1 / \delta))$ . Using that  $\mathcal{C}_G(x) = x + \mathcal{O}(\log x)$ , tedious manipulations yield that
$$
\lim  _ {\delta \rightarrow 0} \sup  _ {\log (1 / \delta)} \frac {T _ {\zeta} (\delta)}{\log (1 / \delta)} \leq \frac {T _ {\mathrm {K L} , \beta} ^ {\star} (\boldsymbol {\mu})}{1 - \kappa} D _ {\zeta} (\mu , \epsilon),
$$
$$
D _ {\zeta} (\mu , \epsilon) = \sup  \left\{x \mid x ^ {2} <   \frac {2}{1 - \zeta} x + \frac {1 + \zeta}{1 - \zeta} \frac {\Delta_ {\max } ^ {2}}{2 \epsilon^ {2}} \right\} \leq \frac {1}{1 - \zeta} \left(1 + \sqrt {1 + (1 - \zeta^ {2}) \frac {\Delta_ {\max } ^ {2}}{2 \epsilon^ {2}}}\right).
$$
The last inequality uses that  $x^{2} - 2bx - c < 0$  for all  $x \in [0, b(1 + \sqrt{1 + c / b^2}))$ . Therefore, we have shown that
$$
\limsup_{\delta \to 0}\frac{\mathbb{E}_{\boldsymbol{\nu}}[\tau_{\delta}]}{\log(1 / \delta)}\leq (2 + \eta)^{2}\frac{T_{\mathrm{KL},\beta}^{\star}(\boldsymbol{\mu})}{(1 - \kappa)(1 - \zeta)}\left(1 + \sqrt{1 + (1 - \zeta^{2})\frac{\Delta_{\max}^{2}}{2\epsilon^{2}}}\right).
$$
Letting  $\kappa$ ,  $\eta$  and  $\zeta$  goes to zero yields that
$$
\lim _ {\delta \to 0} \sup _ {\log (1 / \delta)} \frac {\mathbb {E} _ {\pmb {\nu}} [ \tau_ {\delta} ]}{\log (1 / \delta)} \leq 4 T _ {\mathrm {K L}, \beta} ^ {\star} (\pmb {\mu}) \left(1 + \sqrt {1 + \frac {\Delta_ {\max} ^ {2}}{2 \epsilon^ {2}}}\right) .
$$
Concluding the proof of Theorem 5. Combining Lemmas 14, 18, 19 and 20 concludes the proof of Theorem 5. We restrict the result to instances such that  $\min_{a\neq b}|\mu_a - \mu_b| > 0$  in order for Lemma 14 to hold. Note that this is an artifact of the asymptotic proof which could be alleviated with more careful considerations.
Asymptotically optimal  $\epsilon$ -DP-FC-BAI algorithm. While Lemma 20 is derived for BAI algorithms converging towards the  $\beta$ -optimal allocation  $\omega_{\mathrm{KL},\beta}^{\star}(\pmb{\mu})$ , it is direct to see that a similar inversion results can be obtained for BAI algorithms that converge towards the unique optimal allocation  $\omega_{\mathrm{KL}}^{\star}(\pmb{\mu}) = \{w^{\star}\}$  defined as
$$
\omega_ {\mathrm {K L}} ^ {\star} (\boldsymbol {\mu}) \triangleq \underset {\omega \in \Sigma_ {K}} {\arg \max } \underset {a \neq a ^ {\star}} {\min } \frac {\Delta_ {a} ^ {2}}{1 / \omega_ {a ^ {\star}} + 1 / \omega_ {a}}. \tag {30}
$$
$$
\forall a \neq a ^ {\star}, \quad \frac {\Delta_ {a} ^ {2}}{1 / \omega_ {a ^ {\star}} ^ {\star} + 1 / \omega_ {a} ^ {\star}} = 2 T _ {\mathrm {K L}} ^ {\star} (\boldsymbol {\mu}) ^ {- 1}. \tag {31}
$$
In addition to the multiplicative four-factor, the price of privacy for asymptotically optimal BAI algorithms when combined with the non-private GLR stopping rule is a problem dependent multiplicative factor  $1 + \sqrt{1 + \Delta_{\max}^2 / (2\epsilon^2)}$  (Lemma 21). We omit the proof since it is the same as the one of Lemma 20.
Lemma 21. Let  $\delta \in (0,1)$  Assume that there exists  $\gamma_{\mu} > 0$  such that for  $\mathbb{E}_{\nu}[T_{\mu,\gamma}(\omega^{\star})] < +\infty$  for all  $\gamma \in (0,\gamma_{\mu})$ , where  $T_{\mu,\gamma}(w)$  is defined in Equation (29) and  $\omega^{\star}$  is defined in Equation (30). Combining the private GLR stopping rule (Equation (16)) with private threshold (Equation (4)) yields a  $\delta$ -correct algorithm which satisfies that, for all  $\pmb{\nu}$  with mean  $\mu$  such that  $|a^{\star}(\pmb{\mu})| = 1$ ,
$$
\limsup_{\delta \to 0}\frac{\mathbb{E}_{\boldsymbol{\nu}}\left[ \tau_{\delta}\right]}{\log(1 / \delta)}\leq 4T_{\mathrm{KL}}^{\star}(\boldsymbol {\mu})\left(1 + \sqrt{1 + \frac{\Delta_{\max}^{2}}{2\epsilon^{2}}}\right)  .
$$
E.6 Connection to the lower bound
In this section, we compare the sample complexity lower bound of Corollary 1 with the sample complexity upper bound of Theorem 5.
Simplification of the upper bound. The asymptotic expected sample complexity of AdaP-TT (Theorem 5) is upper bounded by
$$
\begin{array}{l} \lim  _ {\delta \rightarrow 0} \sup  _ {\log (1 / \delta)} \frac {\mathbb {E} _ {\boldsymbol {\nu}} [ \tau_ {\delta} ]}{\log (1 / \delta)} \leq 4 T _ {\mathrm {K L}, \beta} ^ {\star} (\boldsymbol {\mu}) \left(1 + \sqrt {1 + \frac {\Delta_ {\operatorname* {m a x}} ^ {2}}{2 \epsilon^ {2}}}\right) \\ \leq_ {(a)} 4 T _ {\mathrm {K L}, \beta} ^ {\star} (\boldsymbol {\mu}) \left(2 + \frac {\Delta_ {\max}}{\sqrt {2} \epsilon}\right) \\ \end{array}
$$
where  $T_{\mathrm{KL},\beta}^{\star}(\pmb {\mu})$  is the  $\beta$ -characteristic time for Gaussian bandits, and (a) is due to the sub-additivity of the square root.
For  $\beta = 1 / 2$ , [Rus16] showed that  $T_{\mathrm{KL},1 / 2}^{\star}(\pmb {\mu})\leq 2T_{\mathrm{KL}}^{\star}(\pmb {\mu})$
On the other hand, [GK16] showed that  $H(\pmb{\mu}) \leq T_{\mathrm{KL}}^{\star}(\pmb{\mu}) \leq 2H(\pmb{\mu})$ , where  $H(\pmb{\mu}) \triangleq \sum_{a \in [K]} 2\Delta_a^{-2}$  with  $\Delta_{a^{\star}} = \Delta_{\min}$ .
Plugging these two inequalities in the upper bound of Theorem 5 with  $\beta = 1 / 2$  gives that
$$
\lim  _ {\delta \rightarrow 0} \sup  _ {\log (1 / \delta)} \frac {\mathbb {E} _ {\boldsymbol {\nu}} [ \tau_ {\delta} ]}{\log (1 / \delta)} \leq 8 T _ {\mathrm {K L}, 1 / 2} ^ {\star} (\boldsymbol {\mu}) + 1 6 H (\boldsymbol {\mu}) \frac {\Delta_ {\max }}{\sqrt {2} \epsilon}
$$
Since we consider Bernoulli distributions, we know that  $0 < \Delta_{\mathrm{min}} \leq \Delta_{\mathrm{max}} < 1$ . If we restrict ourselves to instances such that all the gaps have the same order of magnitude (Condition 1): there exists a constant  $C \geq 1$  such that  $\Delta_{\mathrm{max}} \leq C \Delta_{\mathrm{min}}$ .
For such instances, we obtain
$$
\begin{array}{l} \lim  _ {\delta \rightarrow 0} \sup  _ {\log (1 / \delta)} \frac {\mathbb {E} _ {\boldsymbol {\nu}} [ \tau_ {\delta} ]}{\log (1 / \delta)} \leq 8 T _ {\mathrm {K L}, 1 / 2} ^ {\star} (\boldsymbol {\mu}) + 1 6 H (\boldsymbol {\mu}) \frac {C \Delta_ {\min }}{\sqrt {2} \epsilon} \\ \leq 8 T _ {\mathrm {K L}, 1 / 2} ^ {\star} (\boldsymbol {\mu}) + 1 6 \sqrt {2} \frac {C}{\epsilon} \left(\frac {1}{\Delta_ {\min }} + \sum_ {a = 2} ^ {K} \frac {1}{\Delta_ {a}}\right) \\ \end{array}
$$
where the last inequality is due to  $H(\pmb{\mu})\Delta_{\min} \leq \frac{2}{\Delta_{\min}} + \sum_{a=2}^{K} \frac{2}{\Delta_a}$ .
Finally using that  $a + b \leq 2\max(a, b)$ , we get that
$$
\lim  _ {\delta \rightarrow 0} \sup  _ {\log (1 / \delta)} \frac {\mathbb {E} _ {\boldsymbol {\nu}} [ \tau_ {\delta} ]}{\log (1 / \delta)} \leq c \max  \left\{T _ {\mathrm {K L}, 1 / 2} ^ {\star} (\boldsymbol {\mu}), \frac {C}{\epsilon} \left(\frac {1}{\Delta_ {\min }} + \sum_ {a = 2} ^ {K} \frac {1}{\Delta_ {a}}\right)\right\}
$$
for the universal constant  $c = 45.26$ .
The lower bound for Bernoulli instances. For Bernoulli instances, Corollary 1 gives that the lower bound of the expected sample complexity of any  $\delta$ -correct  $\epsilon$ -global DP BAI strategy is
$$
\lim  _ {\delta \rightarrow 0} \sup  _ {\log (1 / \delta)} \frac {\mathbb {E} _ {\boldsymbol {\nu}} [ \tau_ {\delta} ]}{\log (1 / \delta)} \geq \max  \left\{T _ {\mathrm {K L}} ^ {\star} (\boldsymbol {\nu}), \frac {1}{6 \epsilon} \left(\frac {1}{\Delta_ {\min }} + \sum_ {a = 2} ^ {K} \frac {1}{\Delta_ {a}}\right)\right\}.
$$
where we use Proposition 1 to replace  $T_{\mathrm{TV}}^{\star}(\nu)$  and  $T_{\mathrm{KL}}^{\star}(\nu)$  is the characteristic time for Bernoulli bandits.