SlowGuess/NeurIPS_2023_OCR huggingface dataset:
- Case 1.  $s_3 \leq \frac{1}{2} - \varepsilon$ . Then  $u_1(s) = e(s_1) - \frac{1}{4} s_1$ , which is maximized at  $s_1 = \frac{1}{2} + \varepsilon$  and results in a utility of  $\frac{7}{8} - \frac{\varepsilon}{4}$ .
- Case 2.  $s_3 \geq \frac{1}{2} + \varepsilon$ . Then  $u_1(s) = 1 - \frac{1}{4}s_1$ , which is maximized at  $s_1 = 0$  and results in a utility of 1.
- Case 3.  $\frac{1}{2} - \varepsilon \leq s_3 \leq \frac{1}{2} + \varepsilon$ . We consider the intervals in which the best response  $s_1$  to such an  $s_3$  can lie:
-  $s_1 \leq \frac{1}{2} - \varepsilon$ . In this range,  $u_1(\pmb{s}) = e(s_3) - \frac{1}{4} s_1$ , which is maximized at  $s_1 = 0$  and results in a utility of  $e(s_3)$ .
-  $s_1 \geq \frac{1}{2} + \varepsilon$ . In this range,  $u_1(s) = 1 - \frac{1}{4} s_1$ , which is maximized at  $s_1 = \frac{1}{2} + \varepsilon$  and results in a utility of  $\frac{7}{8} - \frac{\varepsilon}{4}$ .
-  $\frac{1}{2} - \varepsilon \leq s_1 \leq \frac{1}{2} + \varepsilon$ . In this range, using the definition of  $e(s_1)$  (eq. 8) we obtain:
$$
u _ {1} (s) = \left(\frac {1 - e (s _ {3})}{2 \varepsilon} - \frac {1}{4}\right) \cdot s _ {1} + (1 - e (s _ {3})) \cdot \left(\frac {1}{2} - \frac {1}{4 \varepsilon}\right) + e (s _ {3}).
$$
Thus  $u_{1}(s)$  is a linear function in  $s_1$  with slope  $\frac{1 - e(s_3)}{2\varepsilon} - \frac{1}{4}$ . If the slope is positive, then the best response in the current interval is  $s_1 = \frac{1}{2} + \varepsilon$ , and gives a utility of  $\frac{7}{8} - \frac{\varepsilon}{4}$ . If the slope is negative, then  $s_1 = \frac{1}{2} - \varepsilon$  is the best response in the current interval and gives a utility of  $e(s_3) - \frac{1}{4} (\frac{1}{2} - \varepsilon)$ . However  $s_1 = 0$  gives a utility of  $e(s_3)$  implying that  $s_1 = \frac{1}{2} - \varepsilon$  cannot be a best response. Finally if the slope is zero, then it must mean that  $e(s_3) = 1 - \frac{\varepsilon}{2}$ , and the utility is  $\frac{\varepsilon}{2} (\frac{1}{2} - \frac{1}{4\varepsilon}) + 1 - \frac{\varepsilon}{2} = \frac{7}{8} - \frac{\varepsilon}{4}$ . However responding with  $s_1 = 0$  gives a utility of  $e(s_3) = 1 - \frac{\varepsilon}{2}$ , which exceeds  $\frac{7}{8} - \frac{\varepsilon}{4}$ , since  $\varepsilon < \frac{1}{16}$ . Thus, the best response does not lie in  $(\frac{1}{2} - \varepsilon, \frac{1}{2} + \varepsilon)$  and  $s_1 = 0$  is the overall best response.
The above discussion shows that the best response  $f_{1}(s_{2}, s_{3}) \subseteq \{0, \frac{1}{2} + \varepsilon\}$ . By symmetry, the same holds for  $f_{2}$  and  $f_{3}$ . Suppose there exists a Nash equilibrium  $s^{*} = (s_{1}^{*}, s_{2}^{*}, s_{3}^{*})$ . By Proposition 1,  $s^{*} \in f(s^{*})$ . Since the above discussion implies  $s_{3}^{*} \in \{0, \frac{1}{2} + \varepsilon\}$ , we consider two cases:
- Suppose  $s_3^* = 0$ . Then
$$
\begin{array}{l} s _ {3} ^ {*} = 0 \Longrightarrow s _ {1} ^ {*} = \frac {1}{2} + \varepsilon \quad (\text {C a s e 1 f o r a g e n t 1}) \\ \Longrightarrow s _ {2} ^ {*} = 0 \quad (\text {C a s e 2 f o r a g e n t 2}) \\ \Longrightarrow s _ {3} ^ {*} = \frac {1}{2} + \varepsilon , \quad (\text {C a s e 1 f o r a g e n t 3}) \\ \end{array}
$$
Suppose  $s_3^* = \frac{1}{2} +\varepsilon$  . Then
$$
\begin{array}{l} s _ {3} ^ {*} = \frac {1}{2} + \varepsilon \Longrightarrow s _ {1} ^ {*} = 0 \quad (\text {C a s e 2 f o r a g e n t 1}) \\ \Rightarrow s _ {2} ^ {*} = \frac {1}{2} + \varepsilon \quad (\text {C a s e 1 f o r a g e n t 2}) \\ \Longrightarrow s _ {3} ^ {*} = 0, \quad (\text {C a s e 2 f o r a g e n t 3}) \\ \end{array}
$$
which is also a contradiction.
This shows that there is no  $s^*$  such that  $s^* \in f(s^*)$ , implying that the above instance does not admit a Nash equilibrium.
We now prove the fast convergence of best response dynamics.
Theorem 3.2. Let  $G(\mathbf{s})$  be the Jacobian of  $\mathbf{u} : \mathcal{S} \to \mathbb{R}^n$ , i.e.,  $G(\mathbf{s})_{ij} = \frac{\partial^2 u_i(\mathbf{s})}{\partial s_j \partial s_i}$ . Assuming agent utility functions  $u_i$  satisfy
Proof. Observe that  $\pmb{\mu}^t$  is chosen s.t.  $\| g(\pmb{s}^t, \pmb{\mu}^t)\|_2$  is minimized among all  $\mu$  s.t. the updated sample vector  $s^{t+1}$  remains in  $\mathcal{S}$ . Thus:
$$
\left\| g \left(\boldsymbol {s} ^ {t + 1}, \boldsymbol {\mu} ^ {t + 1}\right) \right\| _ {2} \leq \left\| g \left(\boldsymbol {s} ^ {t + 1}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} \tag {10}
$$
Using Taylor's expansion, we have:
$$
g \left(\boldsymbol {s} ^ {t + 1}, \boldsymbol {\mu} ^ {t}\right) = g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) + H \left(\boldsymbol {s} ^ {\prime}, \boldsymbol {\mu} ^ {t}\right) \cdot \left(\boldsymbol {s} ^ {t + 1} - \boldsymbol {s} ^ {t}\right),
$$
where  $H_{ij}(\pmb{s}',\pmb{\mu}^t) = \frac{\partial g(\pmb{s}',\pmb{\mu}^t)}{\partial s_j}$ , and  $\pmb{s}' = \pmb{s}^t + \alpha (\pmb{s}^{t+1} - \pmb{s}^t)$  for some  $\alpha \in [0,1]$ .
By definition,  $g(\pmb{s}^t,\mu^t)_i = \frac{\partial u_i(\pmb{s}^t)}{\partial s_i} +\mu_i^t$ . Thus  $H_{ij}(\pmb {s}',\pmb {\mu}^t) = \frac{\partial^2u_i(\pmb{s}^t)}{\partial s_j\partial s_i} = G_{ij}(\pmb {s}')$ , hence  $H(s^{\prime},\pmb{\mu}^{t}) = G(\pmb{s}^{\prime})$ . The BR dynamics update rule (4) implies  $\pmb{s}^{t + 1} - \pmb {s}^t = \delta^t\cdot g(\pmb {s}^t,\pmb {\mu}^t)$ . We therefore have  $g(\pmb{s}^{t + 1},\pmb {\mu}^t) = (I_{n\times n} + \delta^t\cdot G(\pmb {s}'))\cdot g(\pmb {s}^t,\pmb {\mu}^t)$ . Taking the  $L^2$  norm, we get:
$$
\left\| g \left(\boldsymbol {s} ^ {t + 1}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} ^ {2} = \left\| g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} ^ {2} + \delta_ {t} ^ {2} \cdot \left\| G \left(\boldsymbol {s} ^ {\prime}\right) g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} ^ {2} + 2 \delta_ {t} g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) ^ {T} G \left(\boldsymbol {s} ^ {\prime}\right) g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right), \tag {11}
$$
By the strong concavity assumption, for a constant  $\lambda > 0$ ,  $G + \lambda \cdot I_{n\times n}$  is negative semi-definite, i.e.,  $v^{T}(G + \lambda \cdot I_{n\times n})v\leq 0$  for any  $v\in \mathbb{R}^n$ . With  $v = g(\pmb {s}^t,\pmb {\mu}^t)$ , we have:
$$
g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) ^ {T} G \left(\boldsymbol {s} ^ {\prime}\right) g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) \leq - \lambda \cdot \| g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) \| _ {2} ^ {2}. \tag {12}
$$
Next we use the fact that the  $L^2$  norm  $\| A\| _2$  of an  $n\times n$  matrix  $A$  is bounded by its Frobenius norm  $\| A\| _F$ :
$$
\| A \| _ {2} := \sup  _ {x \neq 0} \frac {\| A x \| _ {2}}{\| x \| _ {2}} \leq \| A \| _ {F} := \sqrt {\sum_ {i} \sum_ {j} | A _ {i j} | ^ {2}}
$$
By the bounded derivatives assumption, we have  $|G(s')_{ij}| \leq L$ , which implies that  $\| G(s')\|_F = \sqrt{\sum_i \sum_j L^2} = nL$ . This gives:
$$
\left\| G \left(\boldsymbol {s} ^ {\prime}\right) g \left(\boldsymbol {s} ^ {t}, \mu^ {t}\right) \right\| _ {2} \leq n L \left\| g \left(\boldsymbol {s} ^ {t}, \mu^ {t}\right) \right\| _ {2}. \tag {13}
$$
Using (12) and (13) in (11), we get:
$$
\left\| g \left(\boldsymbol {s} ^ {t + 1}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} ^ {2} = \left(1 + \delta_ {t} ^ {2} \cdot n ^ {2} L ^ {2} - 2 \delta^ {t} \lambda\right) \cdot \left\| g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} ^ {2},
$$
Since  $\delta^t = \frac{\lambda}{n^2L^2}$ , the above equation together with (10) gives:
Using  $(1 - x)^{r} \leq e^{-x r}$  repeatedly we obtain that:
$$
\left\| g \left(\boldsymbol {s} ^ {t}, \boldsymbol {\mu} ^ {t}\right) \right\| _ {2} \leq e ^ {- \frac {\lambda^ {2}}{2 n ^ {2} L ^ {2}} \cdot t} \cdot \left\| g \left(\boldsymbol {s} ^ {0}, \boldsymbol {\mu} ^ {0}\right) \right\| _ {2}.
$$
Thus if we want the error  $\| g(\pmb{s}^t,\pmb{\mu}^t)\|_2\leq \varepsilon, T = \frac{2n^2L^2}{\lambda^2}\log \left(\frac{\|\pmb{g}(\pmb{s}^0,\pmb{\mu}^0)\|_2}{\varepsilon}\right)$  iterations suffice, as claimed.
B Appendix to Section 4
Proof. Using the quadratic formula, we see that  $\beta^{*}$  given by:
$$
\beta^ {*} = \frac {A n (n - 2) + C - \sqrt {\left(A n (n - 2) + C\right) ^ {2} - 4 A C (n - 1) ^ {2}}}{2 C} \tag {14}
$$
We first argue  $\beta^{*}$  is real, by showing  $(An(n - 2) + C)^{2} - 4AC(n - 1)^{2} \geq 0$ . This is equivalent to showing  $q(y) := (y + n(n - 2))^2 - 4(n - 1)^2y \geq 0$ , where  $y = C / A$ . Expanding  $q$ , we have  $q(y) = y^2 - 2(n^2 - 2n + 2)y + n^2(n - 2)^2$ . The roots of  $q$  are:
$$
y _ {1}, y _ {2} = \frac {2 (n ^ {2} - 2 n + 2) \pm \sqrt {4 (n ^ {2} - 2 n + 2) ^ {2} - 4 n ^ {2} (n - 2) ^ {2}}}{2} = (n ^ {2} - 2 n + 2) \pm 2 (n - 1),
$$
i.e.,  $y_{1} = (n - 2)^{2}$  and  $y_{2} = n^{2}$ . Since  $q(y)$  has a positive leading coefficient, we have that  $q(y) \geq 0$  for all  $y \geq y_{2} = n^{2}$ . Thus it remains to show that  $y = C / A \geq n^{2}$ . To see this, we use the AM-HM inequality:
$$
\frac {C}{n} = \frac {c _ {1} + \cdots + c _ {n}}{n} \geq \frac {n}{\frac {1}{c _ {1}} + \cdots + \frac {1}{c _ {n}}} = \frac {n}{A}, \tag {15}
$$
implying  $C / A \geq n^2$  as desired. This shows that the root  $\beta^{*}$  of equation (6) is real, hence well-defined. We now show  $0 \leq \beta^{*} \leq 1 - 1 / n$ . From (14), we see:
$$
\begin{array}{l} \beta^ {*} = \frac {A n (n - 2) + C - \sqrt {(A n (n - 2) + C) ^ {2} - 4 A C (n - 1) ^ {2}}}{2 C} \\ \geq \frac {A n (n - 2) + C - \sqrt {(A n (n - 2) + C) ^ {2}}}{2 C} = 0 \\ \end{array}
$$
Further, from (14) we also have:
$$
\begin{array}{l} \beta^ {*} = \frac {A n (n - 2) + C - \sqrt {(A n (n - 2) + C) ^ {2} - 4 A C (n - 1) ^ {2}}}{2 C} \\ \leq \frac {A n (n - 2) + C}{2 C} = \frac {C n (n - 2) / n ^ {2} + C}{2 C} = 1 - \frac {1}{n}, \\ \end{array}
$$
where we used  $A / C \leq 1 / n^2$  (15) in the last inequality. This concludes the proof of Lemma 1.
Proof. When  $0 \leq \beta \leq 1$ , the program (5) is a convex program for general convex cost functions. Since  $u_{i}(\cdot)$  is concave, a proof similar to the proof of Theorem 3.1 shows the existence of a Nash equilibrium.
We now show the welfare-maximizing property. For simplicity, we only consider feasible strategies where each agent participates in the mechanism, i.e.,  $s_i > 0$ . Let  $\rho_i$  and  $\lambda_i$  as the dual variables to the first and second constraints respectively for each  $i$ , and let  $S = \| s \|_1$ . Writing the KKT conditions and eliminating all  $\rho_i$ , we get that a NE  $(b^*, s^*)$  together with dual variables  $\lambda^*$  satisfies:
$$
\forall i: \quad \frac {\partial u _ {i} \left(b _ {i} ^ {*} , S ^ {*}\right)}{\partial S} = (1 - \beta) \cdot c _ {i} \cdot \left(\frac {\partial u _ {i} \left(b _ {i} ^ {*} , S ^ {*}\right)}{\partial b _ {i}} + \lambda_ {i} ^ {*}\right) \quad (\text {f r o m s t a t i o n a r i t y c o n d i t i o n s}) \tag {16}
$$
$$
\forall i: \lambda_ {i} ^ {*} \geq 0 \quad (\text {d u a l f e a s i b i l i t y}) \tag {17}
$$
$$
\forall i: \lambda_ {i} ^ {*} \cdot b _ {i} = 0 \quad (\text {c o m p l i m e n t a r y s l a c k n e s s}) \tag {18}
$$
Now we turn to the  $p$ -mean welfare maximizing solution which is an optimal solution to the following program.
$$
\text {s . t .} \quad \forall i: b _ {i} + (1 - \beta) c _ {i} \left(s _ {i}\right) + \frac {\beta}{n - 1} \sum_ {j \neq i} c _ {j} \left(s _ {j}\right) = B _ {i} \tag {19}
$$
The following lemma establishes that (19) is a convex program. For ease of readability we defer its proof to B.1.
Lemma 2. For  $\beta \in [0,1]$  and  $p\leq 1$  , the program (19) is convex.
We can now write the KKT conditions of program (19). By letting  $\mu_{i}$  and  $\gamma_{i}$  denote the dual variables corresponding to the first and second constraints respectively for each  $i$  and  $S = \| s\| _1$ , the KKT conditions (considering only solutions with  $s_i > 0$ ) are:
$$
\forall i: \left(\sum_ {j} u _ {j} ^ {p}\right) ^ {1 / p - 1} \sum_ {k} u _ {k} ^ {p - 1} \frac {\partial u _ {k}}{\partial S} = c _ {i} \cdot \left[ \mu_ {i} (1 - \beta) + \frac {\beta}{n - 1} \sum_ {k \neq i} \mu_ {k} \right] \quad (\text {s t a t i o n a r i t y}) \tag {20}
$$
$$
\forall i: \left(\sum_ {j} u _ {j} ^ {p}\right) ^ {1 / p - 1} u _ {i} ^ {p - 1} \frac {\partial u _ {i}}{\partial b _ {i}} = \mu_ {i} - \gamma_ {i} \quad \text {(s t a t i o n a r i t y)} \tag {21}
$$
$$
\forall i: \quad \gamma_ {i} \geq 0 \tag {22}
$$
$$
\forall i: \quad \gamma_ {i} \cdot b _ {i} = 0 \tag {23}
$$
Since KKT conditions are sufficient for optimality, to prove Theorem 4.1 it suffices to show that for an NE  $(\pmb{b}^{*},\pmb{s}^{*})$ , there exist dual variables  $\pmb{\mu}^{*}$  and  $\gamma^{*}$  which satisfy (20)-(23) for  $\beta = \beta^{*}$ .
Let  $\alpha := (\sum_{j} u_{j}(b_{j}^{*}, \boldsymbol{s}^{*})^{p})^{1/p-1} \sum_{k} u_{k}(b_{j}^{*}, \boldsymbol{s}^{*})^{p-1} \frac{\partial u_{k}(b_{k}^{*}, \boldsymbol{s}^{*})}{\partial S}$ , i.e., the common value of the equality (20) at the NE  $(\boldsymbol{b}^{*}, \boldsymbol{s}^{*})$ . The equation (20) then becomes  $\alpha \cdot c_{i}^{-1} = \mu_{i}(1 - \beta) + \frac{\beta}{n-1} \sum_{k \neq i} \mu_{k}$ . Summing these over all  $i$  and letting  $T = \sum_{j} \mu_{j}$ , we obtain:
$$
\alpha \cdot (\sum_ {i} c _ {i} ^ {- 1}) = \sum_ {i} [ \mu_ {i} (1 - \beta) + \frac {\beta}{n - 1} \sum_ {k \neq i} \mu_ {k} ] = T.
$$
Putting this back in (20), we obtain the following expression for  $\mu_i^*$ , which can be computed from the NE  $(\pmb{b}^{*},\pmb{s}^{*})$  with  $T = \alpha \cdot (\sum_{i}c_{i}^{-1})$ :
$$
\mu_ {i} ^ {*} = \frac {\frac {T c _ {i} ^ {- 1}}{\sum_ {i} c _ {i} ^ {- 1}} - \frac {\beta T}{n - 1}}{1 - \frac {\beta n}{n - 1}}. \tag {24}
$$
Recall that the NE  $(\pmb{b}^{*},\pmb{s}^{*})$  satisfies (16)-(18) for some dual variables  $\lambda^{*}$ . We define  $\gamma_{i}^{*}$  as follows:
$$
\gamma_ {i} ^ {*} = \mu_ {i} ^ {*} \cdot \left(\frac {\lambda_ {i} ^ {*}}{\lambda_ {i} ^ {*} + \frac {\partial u _ {i} \left(b _ {i} ^ {*} , s\right)}{\partial b _ {i}}}\right) \tag {25}
$$
The next lemma proves Theorem 4.1.
Lemma 3. A NE  $(\pmb{b}^{*},\pmb{s}^{*})$  with  $\mu^{*}$  and  $\gamma^{*}$  defined by (24) and (25) satisfy the KKT conditions (20)-(23) of program (19).
Proof. First observe that at the NE,  $(1 - \beta)c_{i}\cdot \left(\frac{\partial u_{i}(b_{i}^{*},S^{*})}{\partial b_{i}} +\lambda_{i}^{*}\right) = \frac{\partial u_{i}(b_{i}^{*},S^{*})}{\partial S} >0$  by assumption. Since  $\beta \in (0,1)$  and  $c_{i} > 0$ , we have  $\frac{\partial u_i(b_i^*,S^*)}{\partial b_i} +\lambda_i^* >0$ . Together with  $\lambda_{i}^{*}\geq 0$  (17), this shows  $\gamma_{i}^{*}\geq 0$  thus satisfying dual feasibility (22).
Next we show complimentary slackness (23) holds. For any  $i$ ,  $\lambda_i^* \cdot b_i = 0$  due to (18). Then by the definition of  $\gamma_i^*$ , we have  $\gamma_i^* \cdot b_i = 0$  for all  $i$ .
Finally, we show that equations (20) and (21) are satisfied for a specific choice of  $\beta = \beta^{*}$ . Together, (20) and (21) imply that an optimal solution to program (19) satisfies:
$$
\forall i: \sum_ {k} \left(\mu_ {k} - \gamma_ {k}\right) \cdot \frac {\partial u _ {k} / \partial S}{\partial u _ {k} / \partial b _ {k}} = c _ {i} \cdot \left[ \mu_ {i} (1 - \beta) + \frac {\beta}{n - 1} \sum_ {k \neq i} \mu_ {k} \right] \tag {26}
$$
The choice of  $\gamma_{i}^{*}$  from equation 25 implies that  $\mu_{i}^{*} - \gamma_{i}^{*} = \mu_{i}^{*} \cdot \left(\frac{\partial u_{i}(b_{i}^{*},\mathbf{s}) / \partial b_{i}}{\partial u_{i}(b_{i}^{*},\mathbf{s}) / \partial b_{i} + \lambda_{i}^{*}}\right)$ . Moreover at the NE, equation (16) implies that:
$$
\begin{array}{l} \left(\mu_ {i} ^ {*} - \gamma_ {i} ^ {*}\right) \cdot \frac {\partial u _ {i} \left(b _ {i} ^ {*} , \boldsymbol {s}\right) / \partial S}{\partial u _ {i} \left(b _ {i} ^ {*} , \boldsymbol {s}\right) / \partial b _ {i}} = \mu_ {i} ^ {*} \cdot \left(\frac {\partial u _ {i} \left(b _ {i} ^ {*} , \boldsymbol {s}\right) / \partial b _ {i}}{\partial u _ {i} \left(b _ {i} ^ {*} , \boldsymbol {s}\right) / \partial b _ {i} + \lambda_ {i} ^ {*}}\right) \cdot (1 - \beta) c _ {i} \cdot \left(1 + \frac {\lambda_ {i} ^ {*}}{\partial u _ {i} \left(b _ {i} ^ {*} , \boldsymbol {s}\right) / \partial b _ {i}}\right) \\ = \mu_ {i} ^ {*} \cdot (1 - \beta) c _ {i}. \\ \end{array}
$$
Using the above in (26), it only remains to be argued that  $\pmb{\mu}^{*}$ ,  $\pmb{b}^{*}$  and  $\pmb{s}^{*}$  satisfy:
$$
\forall i: (1 - \beta) \cdot \sum_ {k} \mu_ {k} ^ {*} \cdot c _ {k} = c _ {i} \cdot [ \mu_ {i} ^ {*} (1 - \beta) + \frac {\beta}{n - 1} \sum_ {k \neq i} \mu_ {k} ^ {*} ] = \alpha ,
$$
for  $\beta = \beta^{*}$ . By plugging in the value of  $\mu_{i}^{*}$  from (24) and using  $\alpha = T\cdot (\sum_{k}c_{k}^{-1})^{-1}$ , we get:
$$
(1 - \beta) \cdot \sum_ {k} \left\{\frac {T e _ {k} ^ {- 1} \left(\sum_ {i} c _ {i} ^ {- 1}\right) ^ {- 1} - \frac {\beta T}{n - 1}}{1 - \frac {\beta n}{n - 1}} \right\} \cdot c _ {k} = T \cdot \left(\sum_ {k} c _ {k} ^ {- 1}\right) ^ {- 1}.
$$
Let us define  $A := (\sum_{i} c_{i}^{-1})^{-1}$  and  $C := \sum_{i} c_{i}$ . Manipulating the above expression, the above equation then becomes:
$$
C \beta^ {2} - (A n (n - 2) + C) \beta + A (n - 1) ^ {2} = 0,
$$
which is true for  $\beta = \beta^{*}$  since it is exactly the definition of  $\beta^{*}$  (Definition 2).
Thus for  $\beta = \beta^{*}$ , the NE  $(\pmb{b}^{*},\pmb{s}^{*})$  with dual variables  $\mu$  and  $\gamma$  as defined in (24) and (25) respectively satisfy the KKT conditions of program (19).
B.1 Proof of Lemma 2
Proof. For  $\beta \in [0,1]$  the constraints of program 19 are convex since  $c_{i}(\cdot)$  are convex functions. It remains to be shown that the objective  $W_{p}(\boldsymbol{b},\boldsymbol{s})\coloneqq (\sum_{i}u_{i}(b_{i},\| s\|_{1})^{p})^{1 / p}$  to be maximized is concave.
We use the following standard fact about the concavity of composition of functions (see e.g. Boyd and Vandenberghe [2004], Page 86).
Proposition 2. Let  $h: \mathbb{R}^n \to \mathbb{R}$  and  $g_i: \mathbb{R}^k \to \mathbb{R}$  and let  $f: \mathbb{R}^n \to \mathbb{R}$  be given by  $f(x) = h(g(x)) = h(g_1(x), \ldots, g_n(x))$ . Then  $f$  is concave if  $h$  is concave,  $h$  is non-decreasing in each argument and  $g_i$  are concave.
Note that  $W_{p}(\boldsymbol {b},\boldsymbol {s}) = h(g(\boldsymbol {b},\boldsymbol {s}))$  , where  $h(x_{1},\ldots ,x_{n}) = (\sum_{i}x_{i}^{p})^{1 / p}$  and  $g_{i}(\pmb {b},\pmb {s}) = u_{i}(\pmb {b},\pmb {s})$
We now observe that:
-  $h$  is non-decreasing in each argument. This is because:
$$
\frac {\partial h}{\partial x _ {i}} = h ^ {1 - p} x _ {i} ^ {p - 1} \geq 0.
$$
-  $h$  is concave. Using the above, we can compute the Hessian  $H$  given by:
$$
H _ {i j} = \frac {\partial^ {2} h}{\partial x _ {j} \partial x _ {i}} = \left\{ \begin{array}{l l} (1 - p) h ^ {1 - 2 p} (x _ {i} x _ {j}) ^ {p - 1} & \mathrm {(i f} i \neq j) \\ (1 - p) h ^ {1 - 2 p} x _ {i} ^ {p - 2} \cdot (x _ {i} ^ {p} - h ^ {p}) & \mathrm {(i f} i = j) \end{array} \right.
$$
Thus for any  $v\in \mathbb{R}^n$  , we have:
$$
\begin{array}{l} v ^ {T} H v = \sum_ {i} \sum_ {j} v _ {i} H _ {i j} v _ {j} \\ = (1 - p) h ^ {1 - 2 p} \cdot \left(\sum_ {i} v _ {i} \sum_ {j \neq i} H _ {i j} v _ {j} + \sum_ {i} v _ {i} ^ {2} H _ {i i}\right) \\ = (1 - p) h ^ {1 - 2 p} \cdot \left(\sum_ {i} v _ {i} x _ {i} ^ {p - 1} \cdot \left(\left(\sum_ {j} v _ {j} x _ {j} ^ {p - 1}\right) - v _ {i} x _ {i} ^ {p - 1}\right) + \sum_ {i} v _ {i} ^ {2} \left(x _ {i} ^ {2 p - 2} - h ^ {p} x _ {i} ^ {p - 2}\right)\right) \\ = (1 - p) h ^ {1 - 2 p} \cdot \left(\left(\sum_ {i} v _ {i} x _ {i} ^ {p - 1}\right) ^ {2} - \sum_ {i} \left(v _ {i} x _ {i} ^ {p - 1}\right) ^ {2} + \sum_ {i} v _ {i} ^ {2} x _ {i} ^ {2 p - 2} - \sum_ {i} v _ {i} ^ {2} h ^ {p} x _ {i} ^ {p - 2}\right) \\ = (1 - p) h ^ {1 - 2 p} \cdot \left(\left(\sum_ {i} v _ {i} x _ {i} ^ {p - 1}\right) ^ {2} - \left(\sum_ {i} v _ {i} ^ {2} x _ {i} ^ {p - 2}\right) \left(\sum_ {j} x _ {j} ^ {p}\right)\right) \\ \leq 0, \\ \end{array}
$$
Algorithm 1 FedBR-BG
1: Input: Number of iterations in game  $H$ , number of iterations of gradient descent  $T$ , learning rate  $\alpha$ , step size  $\delta$ , data increasing interval  $\Delta s$   
2: Output: Model weights  $\theta^T$ , individual contributions  $s$   
3: for  $h = 1, 2, \dots, H$  do  
4: Server sends  $\theta^t$  to agents;  
5: for  $t = 0, 1, \dots, T - 1$  do  
6: for  $i \in [n]$  in parallel do  
7:  $i$  computes  $\nabla_{\theta^t} \mathcal{L}_i(\theta^t)$  on its local dataset  $\mathcal{D}_i$ ;  
8:  $i$  sends  $\nabla_{\theta^t} \mathcal{L}_i(\theta^t)$  to server;  
9: end for  
10: Server aggregates the gradients following  $\nabla_{\theta^t} \mathcal{L}(\theta^t) \leftarrow \frac{1}{\sum_{i \in [n]} |\mathcal{D}_i|} \sum_{i \in [n]} |\mathcal{D}_i| \cdot \nabla_{\theta^t} \mathcal{L}_i(\theta^t)$ ;  
11: Server updates  $\theta^{t+1}$  following  $\theta^{t+1} \leftarrow \theta^t - \alpha \cdot \nabla_{\theta^t} \mathcal{L}(\theta^t)$ ;  
12: end for  
13: for  $i \in [n]$  in parallel do  
14:  $\frac{\partial u_i}{\partial s_i} \leftarrow \frac{a(\sum_i s_i + \Delta s) - a(\sum_i s_i)}{\Delta s} - (1 - \beta) c_i$   
15: if ( $s_i = 0$  and  $\frac{\partial u_i}{\partial s_i} < 0$ ) or ( $s_i = \tau_i$  and  $\frac{\partial u_i}{\partial s_i} > 0$ ) then  
16:  $s_i^{h+1} \leftarrow s_i^h$ ;  
17: else  
18:  $s_i^{h+1} = s_i^h + \delta \cdot \frac{\partial u_i}{\partial s_i}$ ;  
19: end if  
20: end for  
21: end for
since  $p \leq 1$ ,  $h \geq 0$ , and by the Cauchy-Schwarz inequality  $(\sum_{i} a_i \cdot b_i)^2 \leq (\sum_{i} a_i^2) \cdot (\sum_{i} b_i^2)$  with  $a_i = v_i x_i^{p/2-1}$  and  $b_i = x_i^{p/2-1}$ . Thus  $H$  is negative semi-definite and hence  $h$  is concave.
- For each  $i$ ,  $g_{i}(\pmb{b}, \pmb{s}) = u_{i}(\pmb{b}, \pmb{s})$  is concave.
Using Proposition 2 and the fact that  $W_{p}(\boldsymbol{b}, \boldsymbol{s}) = h(g(\boldsymbol{b}, \boldsymbol{s}))$  we conclude that  $W_{p}(\boldsymbol{b}, \boldsymbol{s})$  is concave.
C Distributed Algorithms
In this section, we present the distributed algorithms of our two mechanisms, FedBR and FedBR-BG.
Algorithm 2 FedBR
Input: Number of iterations in game  $H$ , number of iterations of gradient descent  $T$ , learning rate  $\alpha$ , step size  $\delta$ , data increasing interval  $\Delta s$
Output: Model weights  $\theta^T$ , individual contributions  $s$
for  $h = 1,2,\dots ,H$  do
Server sends  $\theta^t$  to agents;
for  $i\in [n]$  in parallel do
$i$  computes  $\nabla_{\theta^t}\mathcal{L}_i(\theta^t)$  on its local dataset  $\mathcal{D}_i$ ;
$i$  sends  $\nabla_{\theta^t}\mathcal{L}_i(\theta^t)$  to server;
Server aggregates the gradients following
$$
\nabla_ {\theta^ {t}} \mathcal {L} \left(\theta^ {t}\right) \leftarrow \frac {1}{\sum_ {i \in [ n ]} \left| \mathcal {D} _ {i} \right|} \sum_ {i \in [ n ]} \left| \mathcal {D} _ {i} \right| \cdot \nabla_ {\theta^ {t}} \mathcal {L} _ {i} \left(\theta^ {t}\right);
$$
Server updates  $\theta^{t + 1}$  following
$$
\theta^ {t + 1} \leftarrow \theta^ {t} - \alpha \cdot \nabla_ {\theta^ {t}} \mathcal {L} (\theta^ {t});
$$
$$
\frac {\partial u _ {i}}{\partial s _ {i}} \leftarrow \frac {a (\sum_ {i} s _ {i} + \Delta s) - a (\sum_ {i} s _ {i})}{\Delta s} - c _ {i}
$$
if  $(s_i = 0$  and  $\frac{\partial u_i}{\partial s_i} < 0)$  or  $(s_i = \tau_i$  and  $\frac{\partial u_i}{\partial s_i} > 0)$  then
$$
s _ {i} ^ {h + 1} \leftarrow s _ {i} ^ {h};
$$
$$
s _ {i} ^ {h + 1} = s _ {i} ^ {h} + \delta \cdot \frac {\partial u _ {i}}{\partial s _ {i}};
$$