SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
I (\Pi ; C _ {1}, \dots , C _ {T}) = \sum_ {j = 1} ^ {T} I (\Pi ; C _ {j} | C _ {<   j}).
$$
by the chain rule of mutual information. We claim that for any  $j$
$$
I (\Pi ; C _ {j} | C _ {<   j}) = I (\Pi ; C _ {j} | C _ {- j}).
$$
First, by data processing and the one-way nature of the protocol
$$
I (\Pi ; C _ {j} | C _ {<   j}) = I (M _ {\leq j}; C _ {j} | C _ {<   j}).
$$
for any  $j$ . Now we just need to show that
$$
I (M _ {\le j}; C _ {j} | C _ {<   j}) = I (\Pi ; C _ {j} | C _ {- j}).
$$
By chain rule of mutual information, we can write the right-hand side as
$$
\begin{array}{l} I (\Pi ; C _ {j} | C _ {- j}) = I (M _ {\leq j}; C _ {j} | C _ {- j}) + I (M _ {> j}; C _ {j} | M _ {\leq j}, C _ {- j}) \\ = I \left(M _ {\leq j}; C _ {j} \mid C _ {- j}\right) + I \left(M _ {> j}; C _ {j} \mid M _ {\leq j}, C _ {> j}\right) \\ \end{array}
$$
Observe that  $M_{>j}$  and  $C_j$  are independent, conditional on  $M_{\leq j}$  and  $C_{>j}$ . Hence,
$$
I \left(M _ {> j}; C _ {j} \mid M _ {\leq j}, C _ {> j}\right) = 0
$$
and this proves the claim.
Let  $\Pi_b$  be the distribution of the protocol transcript when the input is fixed to be  $b \in \{0,1\}^n$  and  $\oplus$  denote the binary XOR. Now we can bound
$$
\begin{array}{l} I (\Pi ; C _ {1}, \dots , C _ {T}) = \sum_ {j = 1} ^ {T} I (\Pi ; C _ {j} | C _ {<   j}) \\ = \sum_ {j = 1} ^ {T} I (\Pi ; C _ {j} | C _ {- j}) \\ \geq \frac {1}{8} \frac {1}{2 ^ {T}} \sum_ {b \in \{0, 1 \} ^ {T}} \sum_ {j = 1} ^ {T} d _ {\mathrm {T V}} ^ {2} \left(\Pi_ {b \oplus e _ {j}}, \Pi_ {b}\right) \\ \geq \frac {1}{8} \frac {1}{2 ^ {T}} \sum_ {b \in \{0, 1 \} ^ {T}} \sum_ {j: b _ {j} = 0} d _ {\mathrm {T V}} ^ {2} \left(\Pi_ {b \oplus e _ {j}}, \Pi_ {b}\right). \tag {E.2} \\ \end{array}
$$
Conditioned on an input  $b \in \{0, 1\}^T$ , let  $k = |\{i : b_i = 0\}|$  and assume for the sake of a contradiction that
$$
\sum_ {i: b _ {i} = 0} d _ {\mathrm {T V}} \left(\Pi_ {b \oplus e _ {i}}, \Pi_ {b}\right) = k p, \tag {E.3}
$$
where  $p < \frac{1 - c}{2}$ . Let  $p_i = d_{\mathrm{TV}}(\Pi_{b \oplus e_i}, \Pi_b)$  for every player  $i \in [T]$ . Lemma E.4 implies that the protocol can be equivalently implemented such that if the other players receive  $b_{-i}$ , player  $i$  only looks at their input with probability  $p_i$ . If the player  $i$  does not look at their bit, then their message  $M_i$  is independent of their input bit. Let  $Y_i$  denote the indicator random variable for the event that player  $i$  looks at their input in this equivalent protocol.
It follows from our assumption (E.3) that if the input is  $b$ , then  $\mathbb{E}\left[\sum_{i:b_i = 0}Y_i\right] = \sum_i p_i = kp$ . By the definition of  $Y_{i}$ , if for any set  $S$ ,  $Y_{i} = 0$  for all  $i\in S$ , then all players in  $S$  do not look at their input bits. Let  $E_S$  denotes the event that  $Y_{i} = 0$  for all  $i\in S$ , for some  $S\subseteq \{i:b_i = 0\}$ . Then since the players in  $S$  do not look at their input bits,
$$
d _ {\mathrm {T V}} \left(\Pi_ {b \oplus e _ {S}} | E _ {S}, \Pi_ {b} | E _ {S}\right) = 0.
$$
In particular, using this and the law of total probability, we get that
$$
\begin{array}{l} d _ {\mathrm {T V}} \left(\Pi_ {b \oplus e _ {S}}, \Pi_ {b}\right) = \Pr (E _ {S}) \cdot d _ {\mathrm {T V}} \left(\Pi_ {b \oplus e _ {S}} \mid E _ {S}, \Pi_ {b} \mid E _ {S}\right) + \Pr (\overline {{E _ {S}}}) \cdot d _ {\mathrm {T V}} \left(\Pi_ {b \oplus e _ {S}} \mid \overline {{E}} _ {S}, \Pi_ {b} \mid \overline {{E}} _ {S}\right) \\ \leq \Pr (\overline {{E _ {S}}}). \tag {E.4} \\ \end{array}
$$
By our assumption,  $\mathbb{E}\left[\sum_{i:b_i = 0}Y_i\right] = kp$  for  $p < \frac{1 - c}{2}$ . Applying Lemma E.5, we obtain that there exists a set  $S \subseteq \{i : b_i = 0\}$  with  $|S| = ck$  such that  $\operatorname{Pr}(E_S) \geq e^{-k / \gamma_c - 1}$ . For any  $k < (T - 2)\gamma_c < T - 2$ , we have  $\operatorname{Pr}(E_S) > e\delta$ , and so  $\operatorname{Pr}(\overline{E_S}) < 1 - e\delta$ . By Eqn. (E.4),  $d_{\mathrm{TV}}(\Pi_{b \oplus e_S}, \Pi_b) < 1 - e\delta$ . Observe that  $b \oplus e_S$  differs from  $b$  by having  $|S| = ck$  more 1's; and they have the same value at all other coordinates. Recall that in a typical single-coordinate YES instance, there are  $T - M$  number of 1's, which is  $T / 2 - M$  more than a typical NO instance. Now suppose this
gap  $T / 2 - M < ck$ ; then solving BIASDETECT $_1$  is at most as hard as distinguishing  $b$  and  $b \oplus e_S$ . Hence, if we choose  $c'$  such that  $M = c'T > T / 2 - ck$ , then the protocol  $\Pi$  fails with probability greater than  $\delta$ . This is a contradiction.
Thus, for any  $b$  such that  $ck = c\cdot |\{i:b_i = 0\} | > T / 2 - M$
$$
\sum_ {i: b _ {i} = 0} d _ {\mathrm {T V}} (\Pi_ {b \oplus e _ {i}}, \Pi_ {b}) \geq \Omega \left(\frac {(1 - c) T}{2}\right).
$$
From (E.2) and Jensen's inequality,
$$
I (\Pi ; C _ {1}, \dots , C _ {T}) \geq \Omega \left((1 - c) ^ {2} T\right).
$$