SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Therefore, we can only gain some information about  $i^*$  at rounds in which  $f_t = 2\mathbb{1}_{a_t} - 1$  for some non-empty  $a_t \subset [n]$  and  $y_t = -1$ . In such rounds, under distribution  $\mathcal{D}_{i^*}'$ , the distribution of  $\Delta_t$  is described as follows. Let  $k_t = |a_t|$  denote the cardinality of  $a_t$ . Recall that agent  $(0, \{0\} \cup s, -1)$  breaks ties randomly when choosing  $\Delta_t$  if there are multiple elements in  $a_t \cap s$ . Here are two cases:  $i^* \in a_t$  and  $i^* \notin a_t$ .
1. The case of  $i^{*} \in a_{t}$ : With probability  $p$ , we are sampling from  $\mathcal{D}_{i^{*}}^{\prime \prime}$  and then  $\Delta_t = i^*$ . With probability  $1 - p$ , we are sampling from  $\mathcal{D}_{i^{*}}$ . Conditional on this, with probability  $\frac{1}{k_t}$ , we sample an agent  $(0, \{0\} \cup s_{\sigma, i^{*}}, -1)$  with the permutation  $\sigma$  satisfying that  $\sigma^{-1}(i^{*}) < \sigma^{-1}(j)$  for all  $j \in a_t \setminus \{i^{*}\}$  and thus,  $\Delta_t = 0$ . With probability  $1 - \frac{1}{k_t}$ , there exists  $j \in a_t \setminus \{i^{*}\}$  s.t.  $\sigma^{-1}(j) < \sigma^{-1}(i^{*})$  and  $\Delta_t \neq 0$ . Since all  $j \in a_t \setminus \{i^{*}\}$  are symmetric, we have  $\operatorname{Pr}(\Delta_t = j) = (1 - p)(1 - \frac{1}{k_t}) \cdot \frac{1}{k_t - 1} = \frac{1 - p}{k_t}$ . Hence, the distribution of  $\Delta_t$  is
$$
\Delta_ {t} = \left\{ \begin{array}{l l} j & \text {w . p .} \frac {1 - p}{k _ {t}} \text {f o r} j \in a _ {t}, j \neq i ^ {*} \\ i ^ {*} & \text {w . p .} p \\ 0 & \text {w . p .} \frac {1 - p}{k _ {t}}. \end{array} \right.
$$
2. The case of  $i^* \notin a_t$ : With probability  $p$ , we are sampling from  $\mathcal{D}_{i^*}''$ , we have  $\Delta_t = x_t = 0$ . With probability  $1 - p$ , we are sampling from  $\mathcal{D}_{i^*}$ . Conditional on this, with probability of  $\frac{1}{k_t + 1}$ ,  $\sigma^{-1}(i^*) < \sigma^{-1}(j)$  for all  $j \in a_t$  and thus,  $\Delta_t = x_t = 0$ . With probability  $1 - \frac{1}{k_t + 1}$  there exists  $j \in a_t$  s.t.  $\sigma^{-1}(j) < \sigma^{-1}(i^*)$  and  $\Delta_t \in a_t$ . Since all  $j \in a_t$  are symmetric, we have  $\operatorname{Pr}(\Delta_t = j) = (1 - p)(1 - \frac{1}{k_t + 1}) \cdot \frac{1}{k_t} = \frac{1 - p}{k_t + 1}$ . Hence the distribution of  $\Delta_t$  is
$$
\Delta_ {t} = \left\{ \begin{array}{l l} j & \text {w . p .} \frac {1 - p}{k _ {t} + 1} \text {f o r} j \in a _ {t} \\ 0 & \text {w . p .} p + \frac {1 - p}{k _ {t} + 1}. \end{array} \right.
$$
To measure the information obtained from  $\Delta_t$ , we will use the KL divergence of the distribution of  $\Delta_t$  under the data distribution  $\mathcal{D}_{i^*}'$  from that under a benchmark data distribution. We use the average distribution over  $\{\mathcal{D}_i' | i \in [n]\}$ , which is denoted by  $\overline{\mathcal{D}} = \frac{1}{n} \sum_{i \in n} \mathcal{D}_i'$ . The sampling process is equivalent to drawing  $i^* \sim \mathrm{Unif}([n])$  first and then sampling from  $\mathcal{D}_{i^*}'$ . Under  $\overline{\mathcal{D}}$ , for any  $j \in a_t$ , we have
$$
\begin{array}{l} \Pr \left(\Delta_ {t} = j\right) = \Pr \left(i ^ {*} \in a _ {t} \backslash \{j \}\right) \Pr \left(\Delta_ {t} = j \mid i ^ {*} \in a _ {t} \backslash \{j \}\right) + \Pr \left(i ^ {*} = j\right) \Pr \left(\Delta_ {t} = j \mid i ^ {*} = j\right) \\ + \Pr (i ^ {*} \notin a _ {t}) \Pr (\Delta_ {t} = \mathbf {e} _ {j} | i ^ {*} \notin a _ {t}) \\ = \frac {k _ {t} - 1}{n} \cdot \frac {1 - p}{k _ {t}} + \frac {1}{n} \cdot p + \frac {n - k _ {t}}{n} \cdot \frac {1 - p}{k _ {t} + 1} = \frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}, \\ \end{array}
$$
$$
\begin{array}{l} \Pr \left(\Delta_ {t} = 0\right) = \Pr \left(i ^ {*} \in a _ {t}\right) \Pr \left(\Delta_ {t} = 0 \mid i ^ {*} \in a _ {t}\right) + \Pr \left(i ^ {*} \notin a _ {t}\right) \Pr \left(\Delta_ {t} = 0 \mid i ^ {*} \notin a _ {t}\right) \\ = \frac {k _ {t}}{n} \cdot \frac {1 - p}{k _ {t}} + \frac {n - k _ {t}}{n} \cdot (p + \frac {1 - p}{k _ {t} + 1}) = \frac {(n + 1) (1 - p)}{n (k _ {t} + 1)} + \frac {(n - k _ {t}) p}{n}. \\ \end{array}
$$
Thus, the distribution of  $\Delta_t$  under  $\overline{D}$  is
$$
\Delta_ {t} = \left\{ \begin{array}{l l} j & \text {w . p .} \frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n} \text {f o r} j \in a _ {t} \\ 0 & \text {w . p .} \frac {(n + 1) (1 - p)}{n (k _ {t} + 1)} + \frac {(n - k _ {t}) p}{n}. \end{array} \right.
$$
We denote this distribution by  $\overline{P}(a_t)$ . Next we will compute the KL divergence of  $P_{\notin}(a_t)$  and  $P_{\in}(a_t)$  from  $\overline{P}(a_t)$ . Since  $p = \frac{\varepsilon}{16n^2} \leq \frac{1}{16n^2}$ , we have  $\frac{(nk_t - 1)(1 - p)}{nk_t(k_t + 1)} + \frac{p}{n} \leq \frac{1 - p}{k_t + 1}$  and  $\frac{(n + 1)(1 - p)}{n(k_t + 1)} + \frac{(n - k_t)p}{n} \leq \frac{1}{k_t} + p$ . We will also use  $\log(1 + x) \leq x$  for  $x \geq 0$  in the following calculation. For any  $i^* \in a_t$ , we have
$$
\begin{array}{l} \mathrm {D} _ {\mathrm {K L}} (\bar {P} (a _ {t}) \| P _ {\in} (a _ {t}, i ^ {*})) \\ = (k _ {t} - 1) \left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}\right) \log \left(\left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}\right) \cdot \frac {k _ {t}}{1 - p}\right) \\ + \left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}\right) \log \left(\left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}\right) \cdot \frac {1}{p}\right) \\ + \left(\frac {(n + 1) (1 - p)}{n \left(k _ {t} + 1\right)} + \frac {(n - k _ {t}) p}{n}\right) \log \left(\left(\frac {(n + 1) (1 - p)}{n \left(k _ {t} + 1\right)} + \frac {(n - k _ {t}) p}{n}\right) \cdot \frac {k _ {t}}{1 - p}\right) \\ \leq \left(k _ {t} - 1\right) \left(\frac {\left(n k _ {t} - 1\right) (1 - p)}{n k _ {t} \left(k _ {t} + 1\right)} + \frac {p}{n}\right) \log \left(\frac {1 - p}{k _ {t} + 1} \cdot \frac {k _ {t}}{1 - p}\right) + \frac {1 - p}{k _ {t} + 1} \log \left(1 \cdot \frac {1}{p}\right) \\ + \left(\frac {1}{k _ {t}} + p\right) \cdot \log \left(1 + p k _ {t}\right) \\ \leq 0 + \frac {1}{k _ {t} + 1} \log \left(\frac {1}{p}\right) + \frac {2}{k _ {t}} \cdot p k _ {t} = \frac {1}{k _ {t} + 1} \log \left(\frac {1}{p}\right) + 2 p. \tag {22} \\ \end{array}
$$
For  $P_{\notin}(a_t)$ , we have
$$
\begin{array}{l} \mathrm {D} _ {\mathrm {K L}} (\bar {P} (a _ {t}) \| P _ {\notin} (a _ {t})) \\ = k _ {t} \left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}\right) \log \left(\left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} (k _ {t} + 1)} + \frac {p}{n}\right) \cdot \frac {k _ {t} + 1}{1 - p}\right) \\ + \left(\frac {(n + 1) (1 - p)}{n \left(k _ {t} + 1\right)} + \frac {(n - k _ {t}) p}{n}\right) \log \left(\left(\frac {(n + 1) (1 - p)}{n \left(k _ {t} + 1\right)} + \frac {(n - k _ {t}) p}{n}\right) \cdot \frac {1}{p + \frac {1 - p}{k _ {t} + 1}}\right) \\ \leq k _ {t} \left(\frac {(n k _ {t} - 1) (1 - p)}{n k _ {t} \left(k _ {t} + 1\right)} + \frac {p}{n}\right) \log \left(\frac {1 - p}{k _ {t} + 1} \cdot \frac {k _ {t} + 1}{1 - p}\right) \\ + \left(\frac {1}{k _ {t}} + p\right) \log \left(\left(\frac {(n + 1) (1 - p)}{n \left(k _ {t} + 1\right)} + \frac {(n - k _ {t}) p}{n}\right) \cdot \frac {1}{p + \frac {1 - p}{k _ {t} + 1}}\right) \\ = 0 + \left(\frac {1}{k _ {t}} + p\right) \log \left(1 + \frac {1 - p \left(k _ {t} ^ {2} + k _ {t} + 1\right)}{n \left(1 + k _ {t} p\right)}\right) \\ \leq \left(\frac {1}{k _ {t}} + p\right) \frac {1}{n \left(1 + k _ {t} p\right)} = \frac {1}{n k _ {t}}. \tag {23} \\ \end{array}
$$
Lower bound of the information Now we adopt the similar framework used in the proofs of Theorem 7 and 8. For notation simplicity, for all  $i \in [n]$ , let  $\mathbf{P}_i$  denote the dynamics of  $(x_1, f_1, \Delta_1, y_1, \widehat{y}_1, \ldots, x_T, f_T, \Delta_T, y_T, \widehat{y}_T)$  under  $\mathcal{D}_i'$  and  $\overline{\mathbf{P}}$  denote the dynamics under  $\overline{\mathcal{D}}$ . Let  $B_t$  denote the event of  $\{f_t = 2\mathbb{1}_{a_t} - 1$  for some non-empty  $a_t \subset [n]\}$ . As discussed before, for any  $a_t$ , conditional on  $\neg B_t$  or  $y_t = +1$ ,  $(x_t, \Delta_t, y_t, \widehat{y}_t)$  are identical in all  $\{\mathcal{D}_i' | i \in [n]\}$ , and therefore, also identical in  $\overline{\mathcal{D}}$ . We can only obtain information at rounds when  $B_t \wedge (y_t = -1)$  occurs.
In such rounds, we know that  $x_{t}$  is always 0,  $f_{t}$  is fully determined by history (possibly with external randomness, which does not depend on data distribution),  $y_{t} = -1$  and  $\widehat{y}_t$  is fully determined by  $\Delta_t$ $(\widehat{y}_t = +1$  iff.  $\Delta_t \neq 0$ ).
Therefore, conditional the history  $H_{t-1} = (x_1, f_1, \Delta_1, y_1, \widehat{y}_1, \ldots, x_{t-1}, f_{t-1}, \Delta_{t-1}, y_{t-1}, \widehat{y}_{t-1})$  before time  $t$ , we have
$$
\begin{array}{l} \mathrm {D} _ {\mathrm {K L}} \left(\overline {{\mathbf {P}}} \left(x _ {t}, f _ {t}, \Delta_ {t}, y _ {t}, \widehat {y} _ {t} \mid H _ {t - 1}\right) \| \mathbf {P} _ {i} \left(x _ {t}, f _ {t}, \Delta_ {t}, y _ {t}, \widehat {y} _ {t} \mid H _ {t - 1}\right)\right) \\ = \overline {{\mathbf {P}}} \left(B _ {t} \wedge y _ {t} = - 1\right) \mathrm {D} _ {\mathrm {K L}} \left(\overline {{\mathbf {P}}} \left(\Delta_ {t} \mid H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1\right) \| \mathbf {P} _ {i} \left(\Delta_ {t} \mid H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1\right)\right) \\ = 6 \varepsilon \overline {{\mathbf {P}}} \left(B _ {t}\right) \mathrm {D} _ {\mathrm {K L}} \left(\overline {{\mathbf {P}}} \left(\Delta_ {t} \mid H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1\right) \| \mathbf {P} _ {i} \left(\Delta_ {t} \mid H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1\right)\right), \tag {24} \\ \end{array}
$$
For any algorithm that can successfully identify  $i$  under the data distribution  $\mathcal{D}_i$  with probability  $\frac{3}{4}$  for all  $i\in [n]$ , then  $\mathbf{P}_{\mathcal{D}_i}(i_{\mathrm{out}} = i)\geq \frac{3}{4}$  and  $\mathbf{P}_{\mathcal{D}_j}(i_{\mathrm{out}} = i)\leq \frac{1}{4}$  for all  $j\neq i$ . Recall that  $\mathcal{D}_i$  and  $\mathcal{D}_i^\prime$  are very close when the mixture parameter  $p$  is small. Combining with Eq (21), we have
$$
\begin{array}{l} | \mathbf {P} _ {i} (i _ {\mathrm {o u t}} = i) - \mathbf {P} _ {j} (i _ {\mathrm {o u t}} = i) | \\ \geq \left| \mathbf {P} _ {\mathcal {D} _ {i}} (i _ {\text {o u t}} = i) - \mathbf {P} _ {\mathcal {D} _ {j}} (i _ {\text {o u t}} = i) \right| - \left| \mathbf {P} _ {\mathcal {D} _ {i}} (i _ {\text {o u t}} = i) - \mathbf {P} _ {i} (i _ {\text {o u t}} = i) \right| - \left| \mathbf {P} _ {\mathcal {D} _ {j}} (i _ {\text {o u t}} = i) - \mathbf {P} _ {j} (i _ {\text {o u t}} = i) \right| \\ \geq \frac {1}{2} - \frac {1}{4} = \frac {1}{4}. \\ \end{array}
$$
$$
\operatorname {T V} \left(\mathbf {P} _ {i}, \mathbf {P} _ {j}\right) \geq \left| \mathbf {P} _ {i} \left(i _ {\text {o u t}} = i\right) - \mathbf {P} _ {j} \left(i _ {\text {o u t}} = i\right) \right| \geq \frac {1}{4}. \tag {25}
$$
$$
\begin{array}{l} \mathbb {E} _ {i \sim \mathrm {U n i f} ([ n ])} \left[ \mathrm {T V} ^ {2} (\mathbf {P} _ {i}, \mathbf {P} _ {(i + 1) \bmod n}) \right] \leq 4 \mathbb {E} _ {i \sim \mathrm {U n i f} ([ n ])} \left[ \mathrm {T V} ^ {2} (\mathbf {P} _ {i}, \overline {{\mathbf {P}}}) \right] \\ \leq 2 \mathbb {E} _ {i} \left[ \mathrm {D} _ {\mathrm {K L}} (\overline {{\mathbf {P}}} \| \mathbf {P} _ {i}) \right] \quad \text {(P i n s k e r ' s i n e q)} \\ = 2 \mathbb {E} _ {i} \left[ \sum_ {t = 1} ^ {T} \mathrm {D} _ {\mathrm {K L}} \left(\overline {{\mathbf {P}}} \left(x _ {t}, f _ {t}, \Delta_ {t}, y _ {t}, \widehat {y} _ {t} \mid H _ {t - 1}\right) \| \mathbf {P} _ {i} \left(x _ {t}, f _ {t}, \Delta_ {t}, y _ {t}, \widehat {y} _ {t} \mid H _ {t - 1}\right)\right) \right] \quad (\text {C h a i n r u l e}) \\ \leq 1 2 \varepsilon \mathbb {E} _ {i} \left[ \sum_ {t = 1} ^ {T} \overline {{\mathbf {P}}} \left(B _ {t}\right) \mathrm {D} _ {\mathrm {K L}} \left(\overline {{\mathbf {P}}} \left(\Delta_ {t} \mid H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1\right) \| \mathbf {P} _ {i} \left(\Delta_ {t} \mid H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1\right)\right) \right] \tag {ApplyEq(24)} \\ \leq \frac {1 2 \varepsilon}{n} \sum_ {t = 1} ^ {T} \overline {{\mathbf {P}}} (B _ {t}) \sum_ {i = 1} ^ {n} \mathrm {D} _ {\mathrm {K L}} (\overline {{\mathbf {P}}} (\Delta_ {t} | H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1) \| \mathbf {P} _ {i} (\Delta_ {t} | H _ {t - 1}, B _ {t} \wedge y _ {t} = - 1)) \\ = \frac {1 2 \varepsilon}{n} \mathbb {E} _ {a _ {1: T} \sim \overline {{\mathbf {P}}}} \left[ \sum_ {t = 1} ^ {T} \mathbb {1} (B _ {t}) \left(\sum_ {i: i \in a _ {t}} \mathrm {D} _ {\mathrm {K L}} (\bar {P} (a _ {t}) \| P _ {\in} (a _ {t})) + \sum_ {i: i \notin a _ {t}} \mathrm {D} _ {\mathrm {K L}} (\bar {P} (a _ {t}) \| P _ {\notin} (a _ {t}))\right) \right] \\ \leq \frac {1 2 \varepsilon}{n} \mathbb {E} _ {a _ {1: T} \sim \overline {{\mathbf {P}}}} \left[ \sum_ {t: \mathbb {I} (B _ {t}) = 1} \left(\sum_ {i: i \in a _ {t}} \left(\frac {1}{k _ {t} + 1} \log (\frac {1}{p}) + 2 p\right) + \sum_ {i: i \notin a _ {t}} \frac {1}{n k _ {t}}\right) \right] \quad (\text {A p p l y E q (2 2), (2 3)}) \\ \leq \frac {1 2 \varepsilon}{n} \sum_ {t = 1} ^ {T} \left(\log \left(\frac {1}{p}\right) + 2 n p + 1\right) \\ \leq \frac {1 2 T \varepsilon (\log (1 6 n ^ {2} / \varepsilon) + 2)}{n}. \\ \end{array}
$$
Combining with Eq (25), we have that there exists a universal constant  $c$  such that  $T \geq \frac{cn}{\varepsilon(\log(n / \varepsilon) + 1)}$ .