SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \frac {d}{d \mathbb {E} \alpha^ {i} (x ^ {i}) ^ {2}} \mathbb {E} \mathcal {R} _ {p ^ {\prime}} ^ {i} = \frac {d}{d \mathbb {E} \alpha^ {i} (x ^ {i}) ^ {2}} \mathbb {E} \mathcal {R} _ {p} ^ {i} + \frac {1}{N - 1} \sum_ {j \neq i} C \frac {d}{d \mathbb {E} \alpha^ {i} (x ^ {i}) ^ {2}} \mathbb {E} \| m ^ {j} - \theta^ {s} \| ^ {2} \\ \leq \frac {1}{N ^ {2}} \left((1 - C (N - 1) ^ {2})\right) + C \frac {1}{N ^ {2}} \\ = \frac {1}{N ^ {2}} \left(\left(1 - C \left((N - 1) ^ {2} - 1\right)\right)\right) \\ \end{array}
$$
by inserting 18 and 19.
Similarly, assuming  $b^{j}(x^{j}) = 0$  for all players  $j \neq i$  we get
$$
\begin{array}{l} \frac {d}{d \mathbb {E} \| b ^ {i} (x ^ {i}) \| ^ {2}} \mathbb {E} \mathcal {R} _ {p ^ {\prime}} ^ {i} = \frac {d}{d \mathbb {E} \| b ^ {i} (x ^ {i}) \| ^ {2}} \mathbb {E} \mathcal {R} _ {p} ^ {i} + \frac {1}{N - 1} \sum_ {j \neq i} C \frac {d}{d \mathbb {E} \| b ^ {i} (x ^ {i}) \| ^ {2}} \mathbb {E} \| m ^ {j} - \theta^ {s} \| ^ {2} \\ \leq \frac {1}{N ^ {2}} \left((1 - C (N - 1) ^ {2})\right) + C \frac {1}{N ^ {2}} \\ = \frac {1}{N ^ {2}} \left(\left(1 - C \left((N - 1) ^ {2} - 1\right)\right)\right) \\ \end{array}
$$
by inserting 18 and 20
Thus  $\mathbb{E}\frac{d}{d\mathbb{E}\alpha^i(x^i)^2}\mathcal{R}_{p'}^i$  is negative whenever  $C((N - 1)^2 - 1) > 1$  or  $C > \frac{1}{(N - 1)^2 - 1}$  and  $\frac{d}{d\mathbb{E}\|b^i(x^i)\|^2}\mathbb{E}\mathcal{R}_{p'}^i$  under the same conditions as long as  $b^j(x^j) = 0$  for all other players  $j$ .
The expected penalty paid by each player is zero by symmetry, as every players' paid penalty gets redistributed equally among all other players, such that the payments cancel out in expectation.
Similarly, the calculations for participation incentives are exactly as in D.1, but  $\mathbb{E}\mathcal{R}_{p'}^i$  is now equal to  $(1 - \lambda_i)\left(\frac{\sigma^2}{Nn} +\frac{\sigma_\star^2}{N}\right)$  rather than  $(1 - \lambda_i)\left(\frac{\sigma^2}{Nn} +\frac{\sigma_\star^2}{N}\right) - C\left(\frac{(N - 1)}{Nn}\sigma^2 +\frac{(N - 1)}{N}\sigma_\star^2\right)$  because the expected penalty paid by players is equal to zero at equilibrium.
Lastly, it is easy to see that  $\mathbb{E}\| b^{i}(x^{i})\|^{2}$  has to be nonzero for at least two players at any other equilibrium. Compared to the honest equilibrium, that increases every players' MSE. But the (unpenalized) expected reward for player  $i$  equals  $\frac{\sum_{j\neq i}\|\theta^j - \mu\|^2}{N - 1} -\lambda_i\|\theta^i - \mu\|^2$ , such that the sum of all players' (unpenalized) rewards equals  $\sum_{j}(1 - \lambda_{j})\|\theta^{j} - \mu\|^2$ , which is strictly monotonically decreasing in all players' MSEs when  $\lambda_{i}\geq 1$  for all players. As all players' penalties add up to zero in expectation, the sum of penalized rewards is equally strictly monotonically decreasing in all players' MSEs.
Next, we prove 5.2
Theorem D.2. Consider the modified game with reward
$$
\mathcal {R} ^ {i} = \frac {\sum_ {j \neq i} \| \theta^ {j} - \mu \| ^ {2}}{N - 1} - \lambda_ {i} \| \theta^ {i} - \mu \| ^ {2},
$$
where player  $i$  receives an estimate  $\bar{m} +\sqrt{C}\epsilon^{i}\| m^{i} - \bar{m}\|$  for independent noise  $\epsilon^i$  with mean  $\mathbb{E}\epsilon^i = 0$  and "variance"  $\mathbb{E}\| \epsilon^i\| ^2 = 1$ , instead of the empirical mean  $\bar{m}$ , from the server. Then honesty  $(\alpha^i = 0,b^i = 0,\beta^i = \frac{C}{C + 1})$  is a Nash equilibrium, as long as  $C > \frac{1}{\lambda_i(N - 1)^2 - 1}$  and  $\lambda_{i} > \frac{1}{(N - 1)^{2}}$ . Furthermore, honesty maximizes the sum of all players' rewards among equilibria whenever  $\lambda_{i}\geq 1$  for all players.
Furthermore, for fixed constant  $\lambda_{i} = \lambda$ ,  $\mathbb{E}\left(\| \theta^{i} - \mu \|^{2}\right) \in O\left(\frac{\sigma^{2}}{Nn} + \frac{\sigma_{\star}^{2}}{N}\right)$  whenever  $C = \frac{k}{\lambda(N - 1)^{2} - 1}$  for any constant  $k > 1$ , such that players are incentivized to participate in the penalized game rather than relying on their own estimate, whenever  $N > 2$ , the other  $N - 1$  players participate at the honest equilibrium, and  $\lambda \geq 1$ .
Proof. In order to calculate the reward  $\mathcal{R}^i (\theta^1,\ldots ,\theta^N,\mu) = \frac{\sum_{j\neq i}\|\theta^j - \mu\|^2}{N - 1} -\lambda_i\|\theta^i - \mu \|^2$  for player  $i$ , we have a closer look at the mean squared error incurred by players in the modified game. For
convenience, we set  $\bar{\sigma}^2 \coloneqq \frac{\sigma^2}{n} + \sigma_\star^2$
$$
\begin{array}{l} \mathbb {E} \| \theta^ {i} - \mu \| ^ {2} = \mathbb {E} \| (1 - \beta^ {i}) \left(\bar {m} ^ {i} + \sqrt {C} \| m ^ {i} - \theta^ {s} \| \epsilon^ {i}\right) + \beta^ {i} \bar {x} ^ {i} - \mu \| ^ {2} \\ = \mathbb {E} \| (1 - \beta^ {i}) \bar {m} ^ {i} + \beta^ {i} \bar {x} ^ {i} - \mu \| ^ {2} + \mathbb {E} \| (1 - \beta^ {i}) \sqrt {C} \| m ^ {i} - \theta^ {s} \| \epsilon^ {i} \| ^ {2} \\ + 2 \mathbb {E} <   \left((1 - \beta^ {i}) \sqrt {C} \| m ^ {i} - \theta^ {s} \| \epsilon^ {i}, \left((1 - \beta^ {i}) \bar {m} ^ {i} + \beta^ {i} \bar {x} ^ {i} - \mu\right)\right) > \\ = \left(1 - \beta^ {i}\right) ^ {2} \left(\frac {\bar {\sigma} ^ {2}}{N} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2}\right) \\ + \left(\beta^ {i}\right) ^ {2} \bar {\sigma} ^ {2} + 2 \left(1 - \beta^ {i}\right) \beta^ {i} \frac {\bar {\sigma} ^ {2}}{N} \\ + (1 - \beta^ {i}) ^ {2} C \left(\left(\frac {N - 1}{N}\right) ^ {2} \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \mathbb {E} \| d _ {b} ^ {i} \| ^ {2} + \frac {(N - 1)}{N} \bar {\sigma} ^ {2}\right) \\ + 2 \mathbb {E} <   \left((1 - \beta^ {i}) \sqrt {C} \| m ^ {i} - \theta^ {s} \| \epsilon^ {i}, \left((1 - \beta^ {i}) \bar {m} ^ {i} + \beta^ {i} \bar {x} ^ {i} - \mu\right)\right) > \\ \end{array}
$$
using the calculations from the proof of C.1 for the first term and D.1 for the second. The last term is equal to zero because:
$$
\begin{array}{l} \mathbb {E} \left(<   (1 - \beta^ {i}) \sqrt {C} \| m ^ {i} - \theta^ {s} \| \epsilon^ {i}, \left((1 - \beta^ {i}) \bar {m} ^ {i} + \beta^ {i} \bar {x} ^ {i} - \mu\right) >\right) \\ = \mathbb {E} \left(<   \epsilon^ {i}, (1 - \beta^ {i}) \sqrt {C} \| m ^ {i} - \theta^ {s} \| \left((1 - \beta^ {i}) \bar {m} ^ {i} + \beta^ {i} \bar {x} ^ {i} - \mu\right) >\right) \\ = \mathbb {E} (<   \epsilon^ {i}, \mathbb {E} \left((1 - \beta^ {i}) \sqrt {C} \| m ^ {i} - \theta^ {s} \| \left((1 - \beta^ {i}) \bar {m} ^ {i} + \beta^ {i} \bar {x} ^ {i} - \mu\right) >\right) \\ = 0 \\ \end{array}
$$
as  $\epsilon^i$  is independent of all the other terms and has mean zero. Simplifying, we obtain
$$
\begin{array}{l} \mathbb {E} \| \theta^ {i} - \mu \| ^ {2} \\ = \left(1 - \beta^ {i}\right) ^ {2} \left(\frac {\bar {\sigma} ^ {2}}{N} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2}\right) + \left(\beta^ {i}\right) ^ {2} \bar {\sigma} ^ {2} + 2 \left(1 - \beta^ {i}\right) \beta^ {i} \frac {\bar {\sigma} ^ {2}}{N} \\ + (1 - \beta^ {i}) ^ {2} C \left(\left(\frac {N - 1}{N}\right) ^ {2} \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right) + \mathbb {E} \| d _ {b} ^ {i} \| ^ {2} + \frac {1}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {(N - 1)}{N} \bar {\sigma} ^ {2}\right) \\ = (\beta^ {i}) ^ {2} \left(\bar {\sigma} ^ {2} + \frac {C (N - 1) - 1}{N} \bar {\sigma} ^ {2} + \frac {1 + C}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2} \right. \\ \left. + C \left(\frac {N - 1}{N}\right) ^ {2} \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right) + C \mathbb {E} \| d _ {b} ^ {i} \| ^ {2}\right) \\ - 2 \beta^ {i} \left(\frac {1 + C}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2} \right. \\ \left. + C \left(\frac {N - 1}{N}\right) ^ {2} \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right) + C \mathbb {E} \| d _ {b} ^ {i} \| ^ {2} + C \frac {(N - 1)}{N} \bar {\sigma} ^ {2}\right) \\ + \frac {1 + C}{N ^ {2}} \sum_ {j \neq i} \mathbb {E} \left(\alpha^ {j} \left(x ^ {j}\right) ^ {2}\right) + \frac {1}{N ^ {2}} \mathbb {E} \| \sum_ {j \neq i} b ^ {j} \left(x ^ {j}\right) \| ^ {2} + \frac {1 - C + C N}{N} \bar {\sigma} ^ {2} \\ + C \left(\frac {N - 1}{N}\right) ^ {2} \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2} + C \mathbb {E} \| d _ {b} ^ {i} \| ^ {2}\right) \\ \end{array}
$$
As in C.1, this yields an optimal value for  $\beta^i$  of
$$
\left(\beta^ {i}\right) ^ {*} = \frac {\left(H _ {i} (\alpha , b) + C \frac {(N - 1)}{N} \bar {\sigma} ^ {2}\right)}{\left(\bar {\sigma} ^ {2} + \frac {C (N - 1) - 1}{N} \bar {\sigma} ^ {2} + H _ {i} (\alpha , b)\right)}
$$
for  $H_{i}(\alpha ,b) = \frac{1 + C}{N^{2}}\sum_{j\neq i}\mathbb{E}(\alpha^{j}(x^{j})^{2}) + \frac{1}{N^{2}}\mathbb{E}\| \sum_{j\neq i}b^{j}(x^{j})\|^{2} + C(\frac{N - 1}{N})^{2}\mathbb{E}(\alpha^{i}(x^{i})^{2}) + C\mathbb{E}\| d_{b}^{i}\|^{2}$ . Now, calculating the derivative of  $\mathbb{E}\mathcal{R}^i (\theta^1,\dots ,\theta^N,\mu)$  with respect to  $\mathbb{E}(\alpha^i (x^i)^2)$  yields:
$$
\begin{array}{l} \frac {d}{d \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right)} \mathbb {E} \mathcal {R} ^ {i} \left(\theta^ {1}, \dots , \theta^ {N}, \mu\right) = \frac {d}{d \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right)} \frac {\sum_ {j \neq i} \| \theta^ {j} - \mu \| ^ {2}}{N - 1} - \frac {d}{d \mathbb {E} \left(\alpha^ {i} \left(x ^ {i}\right) ^ {2}\right)} \lambda_ {i} \| \theta^ {i} - \mu \| ^ {2} \\ = \sum_ {j \neq i} \frac {(1 - \beta^ {j}) ^ {2}}{(N - 1) N ^ {2}} ((1 + C)) - (1 - \beta^ {i}) ^ {2} (\lambda_ {i} C \frac {(N - 1) ^ {2}}{N ^ {2}}) \\ \end{array}
$$
Similarly, assuming  $b^{j}(x^{j}) = 0$  for all other players  $j \neq i$ , we get
$$
\begin{array}{l} \frac {d}{d \mathbb {E} \| b ^ {i} (x ^ {i}) \| ^ {2}} \mathbb {E} \mathcal {R} ^ {i} (\theta^ {1}, \dots , \theta^ {N}, \mu) = \frac {d}{d \mathbb {E} \| b ^ {i} (x ^ {i}) \| ^ {2}} \frac {\sum_ {j \neq i} \| \theta^ {j} - \mu \| ^ {2}}{N - 1} - \frac {d}{d \mathbb {E} \| b ^ {i} (x ^ {i}) \| ^ {2}} \lambda_ {i} \| \theta^ {i} - \mu \| ^ {2} \\ = \sum_ {j \neq i} \frac {(1 - \beta^ {j}) ^ {2}}{(N - 1) N ^ {2}} ((1 + C)) - (1 - \beta^ {i}) ^ {2} (\lambda_ {i} C \frac {(N - 1) ^ {2}}{N ^ {2}}) \\ \end{array}
$$
Both are negative, whenever  $\beta^j = \beta^i \neq 1$  and  $\lambda_i C(N - 1)^2 > 1 + C$  or  $C(\lambda_i (N - 1)^2 - 1) > 1$ , which is true whenever  $C > \frac{1}{\lambda_i (N - 1)^2 - 1}$  and  $\lambda_i > \frac{1}{(N - 1)^2}$ .
But for  $\alpha^j = b^j = 0$  for all players  $j$ , the formula for  $(\beta^i)^*$  simplifies to
$$
(\beta^ {i}) ^ {*} = \frac {C \frac {(N - 1)}{N} \bar {\sigma} ^ {2}}{\bar {\sigma} ^ {2} + \frac {C (N - 1) - 1}{N} \bar {\sigma} ^ {2}} = \frac {C \frac {(N - 1)}{N}}{1 + \frac {C (N - 1) - 1}{N}} = \frac {C \frac {(N - 1)}{N}}{\frac {N - 1}{N} + \frac {C (N - 1)}{N}} = \frac {C}{1 + C},
$$
for all players. In particular, we have  $(\beta^i)^* < 1$  for  $C > 0$ , such that  $\beta^j = \beta^i \neq 1$  and the derivatives with respect to both  $\alpha$  and  $b$  are negative for all players, turning  $\alpha^j = b^j = 0$ ,  $(\beta^i) = \frac{C}{1 + C}$  into a Nash equilibrium.
We can now upper bound  $\mathbb{E}\| \theta^i -\mu \| ^2$  at the honest equilibrium by considering it at the suboptimal  $\beta^i = 0$ :
$$
\mathbb {E} \| \theta^ {i} - \mu \| ^ {2} \leq \frac {\bar {\sigma} ^ {2}}{N} + C \frac {N - 1}{N} \bar {\sigma} ^ {2},
$$
which is  $O\left(\frac{\bar{\sigma}^2}{N}\right)$  as long as  $C$  is in  $O\left(\frac{1}{N}\right)$ , which is the case for constant  $\lambda_i = \lambda$ ,  $C = \frac{k}{\lambda(N - 1)^2 - 1}$  and any  $k > 1$ .
In terms of participation incentives, we again look at the difference in rewards obtained by player  $i$  in both cases:
$$
\begin{array}{l} \mathbb {E} \mathcal {R} ^ {i} \left(\theta^ {1}, \dots , \theta^ {N}, \mu\right) - \mathbb {E} \mathcal {R} ^ {i} \left(\theta^ {1} (N - 1), \dots , \theta^ {i} (1), \dots , \theta^ {N} (N - 1), \mu\right) (23) \\ = \mathbb {E} \left(\frac {\sum_ {j \neq i} \| \theta^ {j} (N) - \mu \| ^ {2}}{N - 1} - \lambda_ {i} \| \theta^ {i} (N) - \mu \| ^ {2}\right) (24) \\ - \mathbb {E} \left(\frac {\sum_ {j \neq i} \| \theta^ {j} (N - 1) - \mu \| ^ {2}}{N - 1} - \lambda_ {i} \| \theta^ {i} (1) - \mu \| ^ {2}\right) \\ = (1 - \lambda_ {i}) \mathbb {E} \| \theta^ {i} (N) - \mu \| ^ {2} - \mathbb {E} \left(\frac {\sum_ {j \neq i} \| \theta^ {j} (N - 1) - \mu \| ^ {2}}{N - 1} - \lambda_ {i} \| \theta^ {i} (1) - \mu \| ^ {2}\right) \\ \end{array}
$$
It is obvious that  $\mathbb{E}\| \theta^i (N) - \mu \| ^2 <  \mathbb{E}\| \theta^j (N - 1) - \mu \| ^2\leq \mathbb{E}\| \theta^i (1) - \mu \| ^2$  at the honest equilibrium because of symmetry and as players could otherwise improve their reward by setting  $\beta^i = 1$ , which
was shown to be suboptimal in the proof of 5.2. This has two implications: First, 23 is positive for  $\lambda_{i} = 1$ . Second, the derivative of 23 with respect to  $\lambda_{i}$  is always positive. Combined, this implies that 23 is positive for all  $\lambda_{i} \geq 1$ .
Lastly, it is easy to see that  $\mathbb{E}\| b^{i}(x^{i})\|^{2}$  has to be nonzero for at least two players at any other equilibrium. Compared to the honest equilibrium, that increases every players' MSE. But the expected reward for player  $i$  equals  $\frac{\sum_{j\neq i}\|\theta^j - \mu\|^2}{N - 1} -\lambda_i\|\theta^i - \mu \|^2$ , such that the sum of all players' rewards equals  $\sum_{j}(1 - \lambda_{j})\| \theta^{j} - \mu \|^{2}$ , which is strictly monotonically decreasing in all players' MSEs when  $\lambda_{i}\geq 1$  for all players.
E Proofs on stochastic gradient descent
Outlook In this section we prove Theorem 6.1. To this end, we first present the formal definitions of the assumptions on  $f$  that we make. Next, we prove two results which bound the difference between the performance of a model resulting from a corrupted optimization scheme (in which players send corrupted estimates) and the performance of a model resulting from honest participation (i.e. from vanilla SGD). In particular, Theorem E.1 provides such a result for a simple penalization scheme, which then easily extends to the penalties presented in Section 6 (Theorem E.2). Next, we combine these results with classic bounds on the distance of the plain SGD trajectory to the minimum value of  $f$  (Lemma E.1), to provide an upper bound on the difference between the performance of the corrupted trajectory and the minimum value of  $f$  in Theorem E.1. Finally, we show how this last result can be extended to general Lipschitz utilities in Theorem E.2, thereby proving the result from the main text.
Definitions First we formally state our assumptions on the function  $f$ .
Definition 2. A function  $f: W \subset \mathbb{R}^n \to \mathbb{R}^d$  is called  $L$ -Lipschitz with respect to given norms  $\| \cdot \|_n$  and  $\| \cdot \|_d$  if for all  $x, y \in W$
$$
\| f (x) - f (y) \| _ {d} \leq L \| x - y \| _ {n}.
$$
Definition 3. A continuously differentiable function  $f: W \subset \mathbb{R}^n \to \mathbb{R}$  is called  $B$ -smooth if its gradient  $\nabla f: W \to \mathbb{R}^n$  is  $B$ -Lipschitz with respect to the euclidean norm.
Definition 4. A differentiable function  $f: W \subset \mathbb{R}^n \to \mathbb{R}$  is called  $m$ -strongly convex if for all  $x, y \in W$
$$
f (x) \geq f (y) + \nabla f (y) ^ {t} (x - y) + \frac {m}{2} \| x - y \| ^ {2},
$$
where  $\| \cdot \|$  denotes the euclidean norm.
We start with proving a weaker version of 6.1 with constant learning rates in which player's paid penalties do not get redistributed:
Proposition E.1. Assume  $f$  is  $B$ -smooth and  $L$ -Lipschitz with respect to the euclidean norm on  $W$  and  $m$ -strongly convex on  $\mathbb{R}^d$ . Also assume that for all  $i, t$  the gradient noise  $e_t^i$  is  $B'$ -Lipschitz with respect to the euclidean norm with probability one and that the constant learning rate  $\gamma$  fulfills  $0 < \gamma < \frac{2m}{B^2 + B'^2}$ . Then for the penalized game with reward
$$
\mathcal {R} _ {p} ^ {i} (\theta_ {T + 1} ^ {1}, \ldots , \theta_ {T + 1} ^ {N}, f) = \left(\frac {1}{N - 1} \sum_ {j \neq i} f (\theta_ {T + 1} ^ {j})\right) - \sum_ {t = 1} ^ {T} C _ {t} \| m _ {t} ^ {i} - \frac {1}{N} \sum_ {j} m _ {t} ^ {j} \| ^ {2},
$$
any player's best response strategy fulfills  $\alpha_{t}^{i}\leq \frac{LNc^{\frac{T - t}{2}}\gamma}{C_{t}(N - 1)^{2}}\leq \frac{LN\gamma}{C_{t}(N - 1)^{2}}$  for  $c = (1 + \gamma^2 (B^2 +B'^2) - 2\gamma m)$ , independent of other players' strategies.
Given  $\epsilon > 0$ , the expected absolute change in function values  $f(\theta_T^j)$  due to noise added by players playing best responses compared to full honesty can be bounded by  $\frac{1}{1 - \sqrt{c}} L\gamma \epsilon$  by choosing  $C_t \geq \frac{LNc^{\frac{T - t}{2}}\gamma}{\epsilon(N - 1)^2} < \frac{LN\gamma}{\epsilon(N - 1)^2}$ . The total penalties paid by player  $i$  can then be bounded by  $\frac{1}{1 - \sqrt{c}} \frac{L\gamma}{(N - 1)} \left( \frac{G^2}{\epsilon} + \epsilon \right) \leq$  for a global bound on the "variance" of the gradient estimates  $\| e_t^i (\theta) \|^2 \leq G^2$ .
Proof. We compare two trajectories  $\theta_t$  and  $\theta_t'$  starting at the same  $\theta_0$  and sharing the same realizations for the noise variables  $e_t^j$  and  $\xi_t^j$  in which player  $i$  employs different agressiveness schedules  $\alpha_t^i$  and  $(\alpha_t^i)'$  with squared difference  $\delta_t^i = (\alpha_t^i - (\alpha_t^i)')^2$ . We define  $\bar{e}_t = \frac{1}{N} \sum_i e_t^i$ . Then:
$$
\begin{array}{l} \mathbb {E} \| \theta_ {t + 1} - \theta_ {t + 1} ^ {\prime} \| ^ {2} = \mathbb {E} \| \Pi_ {W} (\theta_ {t} - \gamma_ {t} (\nabla f (\theta_ {t}) + \bar {e} _ {t} (\theta_ {t}) + \frac {1}{N} \sum_ {j \neq i} \alpha_ {t} ^ {j} \xi_ {t} ^ {j} + \frac {1}{N} \alpha_ {t} ^ {i} \xi_ {t} ^ {i})) \\ - \Pi_ {W} \left(\theta_ {t} ^ {\prime} - \gamma_ {t} (\nabla f \left(\theta_ {t} ^ {\prime}\right) + \bar {e} _ {t} \left(\theta_ {t} ^ {\prime}\right) + \frac {1}{N} \sum_ {j \neq i} \alpha_ {t} ^ {j} \xi_ {t} ^ {j} + \frac {1}{N} \left(\alpha_ {t} ^ {i}\right) ^ {\prime} \xi_ {t} ^ {i})\right) \| ^ {2} \\ \leq \mathbb {E} \| \theta_ {t} - \gamma_ {t} (\nabla f (\theta_ {t}) + \bar {e} _ {t} (\theta_ {t}) + \frac {1}{N} \sum_ {j \neq i} \alpha_ {t} ^ {j} \xi_ {t} ^ {j} + \frac {1}{N} \alpha_ {t} ^ {i} \xi_ {t} ^ {i}) \\ - \theta_ {t} ^ {\prime} - \gamma_ {t} (\nabla f (\theta_ {t} ^ {\prime}) + \bar {e} _ {t} (\theta_ {t} ^ {\prime}) + \frac {1}{N} \sum_ {j \neq i} \alpha_ {t} ^ {j} \xi_ {t} ^ {j} + \frac {1}{N} (\alpha_ {t} ^ {i}) ^ {\prime} \xi_ {t} ^ {i}) \| ^ {2} \\ = \mathbb {E} \| \theta_ {t} - \theta_ {t} ^ {\prime} \| ^ {2} + \gamma_ {t} ^ {2} \mathbb {E} \| \nabla f (\theta_ {t}) - \nabla f \left(\theta_ {t} ^ {\prime}\right) \| ^ {2} + \gamma_ {t} ^ {2} \mathbb {E} \| \bar {e} _ {t} (\theta_ {t}) - \bar {e} _ {t} \left(\theta_ {t} ^ {\prime}\right) \| ^ {2} \\ + 2 \gamma_ {t} \mathbb {E} | <   \theta_ {t} - \theta_ {t} ^ {\prime}, \nabla f (\theta_ {t} ^ {\prime}) - \nabla f (\theta_ {t}) > + \frac {\gamma_ {t} ^ {2}}{N ^ {2}} \mathbb {E} \| (\alpha_ {t} ^ {i} - (\alpha_ {t} ^ {i}) ^ {\prime}) \xi_ {t} ^ {i} \| ^ {2} \\ - 2 \gamma_ {t} \mathbb {E} <   \bar {e} _ {t} (\theta_ {t}) - \bar {e} _ {t} \left(\theta_ {t} ^ {\prime}\right), \theta_ {t} - \theta_ {t} ^ {\prime} - \gamma_ {t} \left(\nabla f (\theta_ {t}) - \nabla f \left(\theta_ {t} ^ {\prime}\right)\right) > \\ \end{array}
$$
Where the first inequality follows from the well-known 1-Lipschitzness of projections onto convex closed sets with respect to the euclidean norm (Balashov & Golubev (2012)) and the  $\xi^i$  terms factor because of their zero mean and independence of the other variables. Similarly, the last term turns out to equal zero because:
$$
\begin{array}{l} \mathbb {E} <   \bar {e} _ {t} \left(\theta_ {t}\right) - \bar {e} _ {t} \left(\theta_ {t} ^ {\prime}\right), \theta_ {t} - \theta_ {t} ^ {\prime} - \gamma_ {t} \left(\nabla f \left(\theta_ {t}\right) - \nabla f \left(\theta_ {t} ^ {\prime}\right)\right) > \\ = \mathbb {E} [ \mathbb {E} [ <   \bar {e} _ {t} (\theta_ {t}) - \bar {e} _ {t} (\theta_ {t} ^ {\prime}), \theta_ {t} - \theta_ {t} ^ {\prime} - \gamma_ {t} (\nabla f (\theta_ {t}) - \nabla f (\theta_ {t} ^ {\prime})) > | \theta_ {t}, \theta_ {t} ^ {\prime} ] ] \\ = \mathbb {E} [ <   \mathbb {E} [ \bar {e} _ {t} (\theta_ {t}) - \bar {e} _ {t} (\theta_ {t} ^ {\prime}) | \theta_ {t}, \theta_ {t} ^ {\prime} ], \theta_ {t} - \theta_ {t} ^ {\prime} - \gamma_ {t} (\nabla f (\theta_ {t}) - \nabla f (\theta_ {t} ^ {\prime})) > ] \\ = \mathbb {E} [ <   0, \theta_ {t} - \theta_ {t} ^ {\prime} - \gamma_ {t} (\nabla f (\theta_ {t}) - \nabla f (\theta_ {t} ^ {\prime})) > ] = 0 \\ \end{array}
$$
as  $\mathbb{E}\bar{e}_t(\theta) = 0$  for any fixed  $\theta$ . Correspondingly,