SlowGuess/NeurIPS_2023_OCR huggingface dataset:
In like manner, we have that the last term satisfies that
$$
E \left[ L _ {m} ^ {t} \left(a _ {m} ^ {t}\right) - \mu_ {a _ {m} ^ {t}} \mid A _ {\epsilon , \delta} \right] \leq \frac {2}{P \left(A _ {\epsilon , \delta}\right) T ^ {4}} \tag {66}
$$
by the same logic as the above and substituting  $i^*$  with  $a_m^t$ , and thus we omit the details here.
We then proceed to bound the second term. Based on the decision rule in Algorithm 2, we have either  $E[U_m^t (i^*) - U_m^t (a_m^t)|A_{\epsilon ,\delta}] < 0$  or  $n_{m,i}(t) < N_{m,i}(t) - K$ . This is equivalent to
$$
\begin{array}{l} E [ U _ {m} ^ {t} (i ^ {*}) - U _ {m} ^ {t} (a _ {m} ^ {t}) | A _ {\epsilon , \delta} ] \\ = E \left[ U _ {m} ^ {t} \left(i ^ {*}\right) - U _ {m} ^ {t} \left(a _ {m} ^ {t}\right) 1 _ {n _ {m, i} (t) \geq N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right] + \\ E \left[ U _ {m} ^ {t} \left(i ^ {*}\right) - U _ {m} ^ {t} \left(a _ {m} ^ {t}\right) 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right] \\ \leq E \left[ U _ {m} ^ {t} \left(i ^ {*}\right) - U _ {m} ^ {t} \left(a _ {m} ^ {t}\right) 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right] \\ \leq E \left[ U _ {m} ^ {t} \left(i ^ {*}\right) 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right]. \tag {67} \\ \end{array}
$$
By definition,  $U_{m}^{t}(i^{*}) = \tilde{\mu}_{i^{*}} + Rad(i^{*},m,t)$  implies that
$$
U _ {m} ^ {t} (i ^ {*}) \leq 1 + R a d (i ^ {*}, m, t)
$$
$$
\begin{array}{l} (6 7) \leq E [ (1 + R a d (i ^ {*}, m, t)) 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} | A _ {\epsilon , \delta} ] \\ \leq \left(1 + \max  R a d \left(i ^ {*}, m, t\right)\right) E \left[ 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right] \\ \leq \left(1 + \max  \left\{\sqrt {C _ {1} \ln T}, C _ {2} \ln T \right\}\right) E \left[ 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right] \\ \end{array}
$$
$$
\begin{array}{l} \frac {1}{M} \sum_ {t = L _ {1} + 1} ^ {T} \sum_ {m = 1} ^ {M} E \left[ U _ {m} ^ {t} \left(i ^ {*}\right) - U _ {m} ^ {t} \left(a _ {m} ^ {t}\right) \mid A _ {\epsilon , \delta} \right] \\ \leq \frac {1}{M} \sum_ {t = L _ {1} + 1} ^ {T} \sum_ {m = 1} ^ {M} (1 + \max  \left\{\sqrt {C _ {1} \ln T}, C _ {2} \ln T \right\}) E \left[ 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right]. \tag {68} \\ \end{array}
$$
Following (53) that only depends on whether clients stay on the same page that relies on the transmission, we obtain
$$
\sum_ {t} E \left[ 1 _ {n _ {m, i} (t) <   N _ {m, i} (t) - K} \mid A _ {\epsilon , \delta} \right] \leq K (K + 2 M) - K + \frac {2}{P \left(A _ {\epsilon , \delta}\right) T ^ {3}}
$$
which immediately leads to
$$
(6 8) \leq (1 + \max  \{\sqrt {C _ {1} \ln T}, C _ {2} \ln T \}) \cdot (K (K + 2 M) - K + \frac {2}{P (A _ {\epsilon , \delta}) T ^ {3}})
$$
Afterwards, we consider the third term and have
$$
\begin{array}{l} E \left[ U _ {m} ^ {t} \left(a _ {m} ^ {t}\right) - L _ {m} ^ {t} \left(a _ {m} ^ {t}\right) \mid A _ {\epsilon , \delta} \right] \\ = E \left[ 2 R a d \left(a _ {m} ^ {t}, m, t\right) \mid A _ {\epsilon , \delta} \right] \tag {69} \\ \end{array}
$$
Putting (65, 66, 68, 69) all together, we deduce that
$$
\begin{array}{l} (6 4) \leq L _ {1} + \frac {1}{M} \sum_ {t = L _ {1} + 1} ^ {T} \sum_ {m = 1} ^ {M} \left(\frac {2}{P \left(A _ {\epsilon , \delta}\right) T ^ {4}} + E [ 2 R a d \left(a _ {m} ^ {t}, m, t\right) \mid A _ {\epsilon , \delta} ] + \frac {2}{P \left(A _ {\epsilon , \delta}\right) T ^ {4}}\right) \\ + \left(1 + \max  \left\{\sqrt {C _ {1} \ln T}, C _ {2} \ln T \right\}\right) \cdot \left(K (K + 2 M) - K + \frac {2}{P \left(A _ {\epsilon , \delta}\right) T ^ {3}}\right) \\ \leq L _ {1} + \frac {4}{P (A _ {\epsilon , \delta}) T ^ {3}} + \frac {1}{M} \sum_ {t > L _ {1}} \sum_ {m} (E [ 2 R a d (a _ {m} ^ {t}, m, t) | A _ {\epsilon , \delta} ]) + \\ (1 + \max  \{\sqrt {C _ {1} \ln T}, C _ {2} \ln T \}) \cdot (K (K + 2 M) - K + \frac {2}{P (A _ {\epsilon , \delta}) T ^ {3}}). \\ \end{array}
$$
Meanwhile, we observe that by definition
$$
\begin{array}{l} \frac {1}{M} \sum_ {t > L _ {1}} \sum_ {m} (E [ 2 R a d (a _ {m} ^ {t}, m, t) | A _ {\epsilon , \delta} ]) \\ = \frac{1}{M}\sum_{i}\sum_{m}\sum_{\substack{a^{t}_{m} = i\\ t > L_{1}}}(E[2Rad(i,m,t)|A_{\epsilon ,\delta}]) \\ = \frac {1}{M} \sum_ {i} \sum_ {m} \sum_ {\substack {a _ {m} ^ {t} = i \\ t > L _ {1}}} E [ 2 \sqrt {\frac {C _ {1} \ln T}{n _ {m , i} (t)}} + \frac {C _ {2} \ln T}{n _ {m , i} (t)} | A _ {\epsilon , \delta} ]. \tag{70} \\ \end{array}
$$
By the sum of the Harmonic series, we have
$$
\sum_{\substack{a_{m,i}^{t} = i\\ t > L_{1}}}\frac{C_{2}\ln T}{n_{m,i}(t)}\leq C_{2}\ln T\ln n_{m,i}(T) + C_{2}\ln T\leq C_{2}(\ln T)^{2} + C_{2}\ln T. \tag{71}
$$
Meanwhile, by the Cauchy-Schwartz inequality, we obtain
$$
\begin{array}{l} \sum_{\substack{a_{t}^{i} = i\\ t > L_{1}}}\sqrt{\frac{C_{1}\ln T}{n_{m,i}(t)}} \\ \leq \sqrt {C _ {1} \ln T} \sqrt {\left(\sum_ {t} 1\right) \left(\sum_ {t} \left(\sqrt {\frac {1}{n _ {m , i} (t)}}\right) ^ {2}\right)} \\ \leq \sqrt {C _ {1} \ln T} \sqrt {T (\ln T + 1)} \\ \end{array}
$$
where the last inequality again uses the result on the Harmonic series as in (71).
Therefore, the cumulative value can be bounded as
$$
\begin{array}{l} (7 0) \leq \frac {1}{M} \sum_ {i} \sum_ {m} \left(C _ {2} (\ln T) ^ {2} + C _ {2} \ln T + \sqrt {C _ {1} \ln T} \sqrt {T (\ln T + 1)}\right) \\ = K \left(C _ {2} (\ln T) ^ {2} + C _ {2} \ln T + \sqrt {C _ {1} \ln T} \sqrt {T (\ln T + 1)}\right) \\ \end{array}
$$
Using the result of (70), we have
$$
\begin{array}{l} (6 4) \leq L _ {1} + \frac {4}{P \left(A _ {\epsilon , \delta}\right) T ^ {3}} + K \left(C _ {2} (\ln T) ^ {2} + C _ {2} \ln T + \sqrt {C _ {1} \ln T} \sqrt {T (\ln T + 1)}\right) + \\ (1 + \max  \{\sqrt {C _ {1} \ln T}, C _ {2} \ln T \}) \cdot (K (K + 2 M) - K + \frac {2}{P (A _ {\epsilon , \delta}) T ^ {3}}) \\ = O \left(\max  \left\{\sqrt {T} \ln T, (\ln T) ^ {2} \right\}\right) \\ \end{array}
$$
G Choices of parameter  $c_{0}$  in Theorem 2
Parameter  $c_{0}$  We note that  $c_{0}$  is a pre-specified parameter which are different in different settings. The choices of  $c_{0}$  are as follows. Meanwhile, we need to study whether the possible choices of  $c_{0}$  explode in terms of the order of  $T$ .
Remark (2). The regret reads
$$
E [ R _ {T} | A _ {\epsilon , \delta} ] \leq L + C _ {1} \sum_ {i \neq i ^ {*}} ([ \frac {4 \log T}{\Delta_ {i} ^ {2}} ]) + (K - 1) (2 (K ^ {2} + M K) + \frac {2 \pi^ {2}}{3} + K ^ {2} + (2 M - 1) K)
$$
with  $L$  denoted as  $L = \max \{L_1, L_2, L_3\} = \max \{a_1, a_2, a_3, \frac{b_1}{c_0}, \frac{b_2}{c_0}, \frac{b_3}{c_0}\}$  and  $C_1$  denoted as  $\max \{\frac{e}{1 - c_0}, f\}$ , where parameters  $a_1, a_2, a_3, b_1, b_2, b_3, e, f$  are specified as
$$
a _ {1} = \frac {\ln \frac {2 T}{\epsilon}}{2 \delta^ {2}}
$$
$$
b _ {1} = 4 K \log_ {2} T
$$
$$
a _ {2} = \frac {\ln \frac {\delta}{1 0}}{\ln p ^ {*}} + 2 5 \frac {1 + \lambda}{1 - \lambda} \frac {\ln \frac {2 T}{\epsilon}}{2 \delta^ {2}}
$$
$$
b _ {2} = 4 K \log_ {2} T
$$
$$
a _ {3} = \frac {\ln \frac {\delta}{1 0}}{\ln p ^ {*}} + 2 5 \frac {1 + \lambda}{1 - \lambda} \frac {\ln \frac {2 T}{\epsilon}}{2 \delta^ {2}}
$$
$$
b _ {3} = \frac {K \ln (\frac {M T}{\epsilon})}{\ln (\frac {1}{1 - c})}
$$
$$
e = 1 6 \frac {4 (M + 2)}{3 M}
$$
$$
f = 1 6 (M + 2) \left(1 + 4 M d _ {m, t} ^ {2}\right).
$$
This function of  $c_{0}$  is non-differentiable which brings additional challenges and requires a case-by-case analysis.
Let  $a = \{a_1, a_2, a_3\}$  and  $b = \{b_1, b_2, b_3\}$ . Then continue with the decision rule as in the previous discussion.
- Case 1: there exists  $c_0$  such that  $a \geq \frac{b}{c_0}$ , i.e.  $c \geq \frac{b}{a}$  and  $\frac{b}{a} \leq 1$ . Then  $R_T$  is monotone increasing in  $c$  due to  $C_1$  and  $c_0 = \frac{b}{a}$  gives us the optimal regret  $R_T^1$ .
Case 2: if  $a \leq \frac{b}{c_0}$ , i.e.  $c_0 \leq \frac{b}{a}$
- if  $\frac{e}{1 - c_0} < f$ , i.e.  $c_0 \leq 1 - \frac{e}{f}$ , then  $c_0 = \min \left\{\frac{b}{a}, 1 - \frac{e}{f}\right\}$  is the minima.
- else we have  $c_0 \geq 1 - \frac{e}{f}$
* if  $1 - \frac{e}{f} > \frac{b}{a}$ , it leads to contradiction and this can not be the case.
\* else  $1 - \frac{e}{f}\leq \frac{b}{a},$  we obtain
$$
R _ {T} \leq \frac {b}{c _ {0}} + \frac {e}{1 - c _ {0}} \sum_ {i \neq i ^ {*}} ([ \frac {4 \log T}{\Delta_ {i} ^ {2}} ]) + (K - 1) (2 (K ^ {2} + M K) + \frac {2 \pi^ {2}}{3} + K ^ {2} + (2 M - 1) K)
$$
which implies that the optimal choice of  $c_0$  is  $\frac{\sqrt{b}}{\sqrt{b} + \sqrt{e\sum_{i\neq i^*}([\frac{4\log T}{\Delta_i^2}]})}$ .
- if  $1 - \frac{e}{f} \leq \frac{\sqrt{b}}{\sqrt{b} + \sqrt{e\sum_{i \neq i^*}(\left[\frac{4\log T}{\Delta_i^2}\right])}} \leq \frac{b}{a}$ , this gives us the final choice of  $c_0$  and the subsequent local optimal regret  $R_T^2$ .
-  $\operatorname{elif}\frac{\sqrt{b}}{\sqrt{b} + \sqrt{e\sum_{i\neq i^{*}}(\left[\frac{4\log T}{\Delta_{i}^{2}}\right])}} < 1 - \frac{e}{f}$ , the optimal choice of  $c_0$  is  $1 - \frac{e}{f}$  and the subsequent local optimal regret is  $R_T^2$
- else the optimal choice of  $c_0$  is  $\frac{b}{a}$  and the subsequent local optimal regret is  $R_T^2$ .
- Compare  $R_T^1$  and  $R_T^2$  and choose the  $c_0$  associated with the smaller value.
The possible choices of  $c_{0}$  are  $\{\frac{b}{a}, \min \{\frac{b}{a}, 1 - \frac{e}{f}\}$ ,  $\frac{\sqrt{b}}{\sqrt{e} \sum_{i \neq i^{*}} \left( \left[ \frac{4 \log T}{\Delta_{i}^{2}} \right] \right)} \}$ , i.e.
$$
\begin{array}{l} c _ {0} = \{\frac {b _ {1} + b _ {2} + b _ {3}}{a _ {1} + a _ {2} + a _ {3}}, \min  \{\frac {b _ {1} + b _ {2} + b _ {3}}{a _ {1} + a _ {2} + a _ {3}}, 1 - \frac {1 6 \frac {4 (M + 2)}{3 M}}{1 6 (M + 2) (1 + 4 M d _ {m , t} ^ {2})} \}, \frac {\sqrt {b}}{\sqrt {b} + \sqrt {e \sum_ {i \neq i ^ {*}} ([ \frac {4 \log T}{\Delta_ {i} ^ {2}} ])}} \} \\ = \left\{\frac {b _ {1} + b _ {2} + b _ {3}}{a _ {1} + a _ {2} + a _ {3}}, \min  \left\{\frac {b _ {1} + b _ {2} + b _ {3}}{a _ {1} + a _ {2} + a _ {3}}, 1 - \frac {1 6 \frac {4 (M + 2)}{3 M}}{1 6 (M + 2) (1 + 4 M d _ {m , t} ^ {2})} \right\}, \frac {\sqrt {b}}{\sqrt {b} + \sqrt {e \sum_ {i \neq i ^ {*}} \left(\left[ \frac {4 \log T}{\Delta_ {i} ^ {2}} \right]\right)}} \right\} \\ \begin{array}{r l r} & & {8 K \log T + \frac {K \ln (\frac {M T}{\epsilon})}{\ln (\frac {1}{1 - c})}} \\ & & {= \frac {\ln \frac {2 T}{\epsilon}}{2 \delta^ {2}} + 2 \frac {\ln \frac {\delta}{1 0}}{\ln p ^ {*}} + 5 0 \frac {1 + \lambda}{1 - \lambda} \frac {\ln \frac {2 T}{\epsilon}}{2 \delta^ {2}},} \end{array} \\ \end{array}
$$
$$
\begin{array}{l} \min \bigl\{\frac{8K\log T + \frac{K\ln(\frac{MT}{\epsilon})}{\ln(\frac{1}{1 - c})}}{\frac{\ln\frac{2T}{\epsilon}}{2\delta^{2}} + 2\frac{\ln\frac{\delta}{10}}{\ln p^{*}} + 50\frac{1 + \lambda}{1 - \lambda}\frac{\ln\frac{2T}{\epsilon}}{2\delta^{2}}},1 - \frac{16\frac{4(M + 2)}{3M}}{16(M + 2)(1 + 4Md_{m,t}^{2})}\bigr \} , \\ \frac {\sqrt {8 K \log T + \frac {K \ln (\frac {M T}{\epsilon})}{\ln (\frac {1}{1 - c})}}}{\sqrt {8 K \log T + \frac {K \ln (\frac {M T}{\epsilon})}{\ln (\frac {1}{1 - c})}} + \sqrt {e \sum_ {i \neq i ^ {*}} ([ \frac {4 \log T}{\Delta_ {i} ^ {2}} ])}} \\ \end{array}
$$
which implies the choice of  $c_0$  is between  $\frac{\sqrt{8K\log T + \frac{K\ln(\frac{MT}{\epsilon})}{\ln(\frac{1}{1 - c})}}}{\sqrt{8K\log T + \frac{K\ln(\frac{MT}{\epsilon})}{\ln(\frac{1}{1 - c})}} + \sqrt{e\sum_{i\neq i^*}([\frac{4\log T}{\Delta_i^2}])}}$  and  $1 - \frac{16\frac{4(M + 2)}{3M}}{16(M + 2)(1 + 4Md_{m,t}^2)}$ . Meanwhile, we observe that the choice of  $c_0$  satisfies
$$
E \left[ R _ {T} \mid A _ {\epsilon , \delta} \right] \leq R _ {T} \left(1 - \frac {1 6 ^ {\frac {4 (M + 2)}{3 M}}}{1 6 (M + 2) (1 + 4 M d _ {m , t} ^ {2})}\right) = O (\log T).
$$