SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\partial b _ {i j} \Phi (b _ {i}, b _ {- i}) = \ln (a _ {i j}) - \ln (p _ {j}) = \ln (\frac {a _ {i j}}{p _ {j}}),
$$
$$
\partial b _ {i j} \tilde {u} _ {i} (b _ {i}, b _ {i}) = \ln (a _ {i j}) - \ln (p _ {j}) = \ln (\frac {a _ {i j}}{p _ {j}}).
$$
In order to prove Theorem 2, we first define a different associated game denoted  $G'$  that differs from  $\tilde{G}$  only in having a different associated utility function  $u_i' = \sum_j a_{ij} \ln(p_j)$ .
In fact,  $\tilde{G}$  and  $G^{\prime}$  a part of a family of associated games of the market game  $G$ , which have the property that they all share the same best responses to bid profiles (and therefore, also the same Nash equilibria) and for all these games, the function  $\Phi$  is a best-response potential (see [68] for the definition of best-response potential games). Among this family of games, we are particularly interested in the games  $\tilde{G}$  and  $G$  since the former admits  $\Phi$  as an exact potential, and the latter has a particularly simple derivative for its utility  $u_{i}^{\prime}$ , which has a clear economic interpretation:  $\partial_{b_{ij}}u_i'(b) = a_{ij} / p_j$ . This is simply the bang-per-buck of player  $i$  from good  $j$  (see the model section in the main text).
Next, we present several technical lemmas that will assist us in proving Theorem 2 and which will also be useful in our proofs later on.
Lemma 3. For any player  $i$  and fixed  $b_{-i}$ , both  $\tilde{u}_i(b_i, b_{-i})$  and  $u_i'(b_i, b_{-i})$  are strictly concave in  $b_i$ .
Proof. We will show the proof for  $\tilde{u}_i$ . We compute the Hessian and show that it is negative definite. The diagonal elements are
$$
\partial_ {b _ {i j}} ^ {2} \tilde {u} _ {i} (b _ {i}, b _ {i}) = - \frac {1}{p _ {j}},
$$
and all of the off-diagonal elements are
$$
\partial_ {b _ {i k}} \partial_ {b _ {i j}} \tilde {u} _ {i} (b _ {i}, b _ {i}) = 0.
$$
Therefore, the Hessian is a diagonal matrix with all of its elements being negative, and thus,  $\tilde{u}_i$  is strictly concave. The same argument works for  $u_i^\prime$  as well.
Lemma 4. Fix a player  $i$  and any bid profile  $b_{-i} \in S_{-i}$  of the other players, then the following two facts hold.
1.  $b_{i}^{\prime *} = \arg \max_{b_{i}\in S_{i}}u_{i}^{\prime}(b_{i},b_{-i})$  if and only if it holds that  $\forall b_i^\prime \in S_i\sum_j\frac{a_{ij}}{p_j^{'*}} b_{ij}^{\prime *}\geq \sum_j\frac{a_{ij}}{p_j^{'*}} b_{ij}^\prime$ , where  $p_j^{\prime *} = \theta_{ij} + b_{ij}^{\prime *}$ .
2.  $\tilde{b}_i^* = \arg \max_{b_i\in S_i}\tilde{u}_i(b_i,b_{-i})$  if and only if it holds that  $\forall \tilde{b}_i\in S_i\sum_j\ln (\frac{a_{ij}}{\tilde{p}_j^*})\tilde{b}_{ij}^*\geq$ $\sum_{j}\ln (\frac{a_{ij}}{\tilde{p}_{j}^{*}})\tilde{b}_{ij},$  where  $\tilde{p}_j^* = \theta_{ij} + \tilde{b}_{ij}^*$
Proof. We will show the proof for (1), and the proof for (2) is similar. Let  $b_{i}^{*}$  be a best response to  $b_{-i}$  and let  $b_{i}^{\prime}$  be some other strategy. Consider the restriction of  $u_{i}^{\prime}(b_{i})$  to the line segment  $[b_i', b_i^*]$  as follows; define  $f(\xi) = u_{i}^{\prime}(b_{i}(\xi))$  for  $b_{i}(\xi) = b_{i}^{\prime} + \xi (b_{i}^{*} - b_{i}^{\prime})$  where  $\xi \in [0,1]$ . As  $u_{i}^{\prime}$  is strictly concave and  $b_{i}^{*}$  is the unique maximizer of  $u_{i}^{\prime}$ , it holds that  $f$  is strictly concave and monotone increasing in  $\xi$ . Therefore the derivative of  $f$  must satisfy at the maximum point  $\xi = 1$  that  $\frac{d}{d\xi} f(1) \geq 0$ . This is explicitly given by
$$
\frac {d}{d \xi} f (\xi) = \nabla_ {b _ {i}} u _ {i} ^ {\prime} (b _ {i} (\xi)) (b _ {i} ^ {*} - b _ {i} ^ {\prime}).
$$
Therefore, when substituting  $\xi = 1$  in the derivative we get  $b_{i}(1) = b_{i}^{*}$ , and
$$
\begin{array}{l} 0 \leq \frac {d}{d \xi} f (1) = \nabla_ {b _ {i}} u _ {i} ^ {\prime} (b _ {i} ^ {*}) (b _ {i} ^ {*} - b _ {i} ^ {\prime}) \\ = \sum_ {j} \frac {a _ {i j}}{p _ {j} ^ {*}} b _ {i j} ^ {*} - \sum_ {j} \frac {a _ {i j}}{p _ {j} ^ {*}} b _ {i j} ^ {\prime}, \\ \end{array}
$$
which implies  $\sum_{j}\frac{a_{ij}}{p_j^*} b_{ij}'\leq \frac{a_{ij}}{p_j^*} b_{ij}^*$ , as required.
To complete the second direction of the proof of (1), consider  $b_i^*$  for which the expression stated in right hand side of (1) is true for all  $b_i'$ . Then, fix any  $b_i'$  and again consider the restriction of  $u_i'$  to  $[b_i', b_i^*]$ . By direct calculation, as before but in the inverse direction, it holds that  $f(1) \geq 0$ , and as  $u_i'$  and  $f(\xi)$  are strictly concave, it thus must be that  $\frac{d}{d\xi} f(\xi)$  is monotone decreasing in  $\xi$ . Thus, for all  $\xi$  we have  $\frac{d}{d\xi} f(\xi) \geq \frac{d}{d\xi} f(1) \geq 0$ . This must mean that  $\xi = 1$  is the maximizer of  $f(\xi)$  since for all  $\xi$ ,  $\frac{d}{d\xi} f(\xi) \geq 0$  implies that  $f(\xi)$  is monotone increasing and therefore  $u_i'(b_i^*) \geq u_i'(b_i')$ . Finally, note that this holds for any  $b_i'$  and hence  $b_i^*$  must be a global maximum of  $u_i'$ .
Lemma 5. Let  $(c_{j})_{j\in [m]}\in \mathbb{R}^{m}$ , if there exists  $x^{*}\in \Delta^{m}$  (the  $m$ -dimensional simplex) such that  $\forall x\in \Delta^m$  it holds that  $\sum_{j}c_{j}x_{j}\leq \sum_{j}c_{j}x_{j}^{*}:= \alpha$  then:
1. for all  $j$  we have that  $c_{j} \leq \alpha$ , and
2. if  $x_{j}^{*} > 0$  then  $c_{j} = \alpha$ .
Proof. (1) Assume for the sake of contradiction that there exists  $k$  with  $c_k > \alpha$  then  $x = e_k$  (the "one-hot" vector with 1 at the  $k$ 'th coordinate and 0 in all other coordinates) yields  $\sum_j c_j x_j = c_k > \alpha = \sum_j c_j x_j^*$ , a contradiction.
Now, to prove (2), assume for the sake of contradiction that there exists  $k$  with  $x_{k}^{*} > 0$  and  $c_{k} < \alpha \Rightarrow c_{k}x_{k}^{*} < \alpha x_{k}^{*}$ . From (1) we have that  $c_{j} \leq \alpha \Rightarrow c_{j}x_{j}^{*} \leq \alpha x_{j}^{*}$ , summing the strict inequality with the weak ones over all  $j$  yields  $\sum_{j}c_{j}x_{j}^{*} < \sum_{j}\alpha x_{j}^{*} = \alpha$ , a contradiction.
Lemma 6. Fix a player  $i$  and any bid profile  $b_{-i} \in S_{-i}$  of the other players, then the following properties of best-response bids hold in the modified games  $\tilde{G}$  and  $G'$ .
1. The support set of  $b_{i}^{*}$ , defined as  $s_{i}^{*} = \{j|b_{ij}^{*} > 0\}$ , is equal to the set  $\{j|a_{ij} > c^{*}\theta_{ij}\}$ , and for every  $j \in s_{i}^{*}$  we have that  $\frac{a_{ij}}{p_j^*} = c^*$ .
2. Best-response bids with respect to the utilities  $u_{i}^{\prime}$  and  $\tilde{u}_{i}$  are equal and unique. That is, in the definition from Lemma 4 we have  $b_{i}^{\prime *} = \tilde{b}_{i}^{*}$  (denoted simply as  $b_{i}^{*}$ ).
3. Best-response bids are given by  $b_{ij}^{*} = \left(\frac{a_{ij}}{c^{*}} -\theta_{ij}\right)^{+}$  for a unique constant  $c^{\ast}\in (0,m / B_i)$ .
Proof. By Lemma 3,  $\tilde{u}_i$  and  $u_i'$  are strictly concave in  $b_i$  for any fixed  $b_{-i}$ , and so each admits a unique maximizer. To see that they are equal, we use Lemma 4 and introduce constants  $c, d$  to obtain
$$
\forall b _ {i} \quad \sum_ {j} \frac {a _ {i j}}{p _ {j} ^ {\prime *}} \frac {b _ {i j}}{B _ {i}} \leq \sum_ {j} \frac {a _ {i j}}{p _ {j} ^ {\prime *}} \frac {b _ {i j} ^ {\prime *}}{B _ {i}} = c, \text {a n d}
$$
$$
\forall b _ {i} \quad \sum_ {j} \ln (\frac {a _ {i j}}{\tilde {p} _ {j} ^ {*}}) \frac {b _ {i j}}{B _ {i}} \leq \sum_ {j} \ln (\frac {a _ {i j}}{\tilde {p} _ {j} ^ {*}}) \frac {\tilde {b} _ {i j} ^ {*}}{B _ {i}} = d,
$$
where  $p_j^{*} = \theta_{ij} + b_{ij}^{*}$  and  $\tilde{p}_j^* = \theta_{ij} + \tilde{b}_{ij}^*$ .
For the ease of exposition, we assume that  $\theta_{ij} > 0$  for all  $j$ . All the results stated below remain valid also when  $\theta_{ij} = 0$  for some  $j$ .
Proof of (1): Applying Lemma 5 to each of those inequalities (once with  $x^{*} = \frac{1}{B_{i}} b_{i}^{\prime *}$  and twice with  $x^{*} = \frac{1}{B_{i}}\tilde{b}_{i}^{*}$ ) and denoting the support sets of  $b_{ij}^{\prime *},\tilde{b}_{ij}^{*}$  as  $s^{\prime *},\tilde{s}^{*}$ , respectively, we obtain the following.  
(1)  $\forall j\in s^{\prime *}$  we have  $\frac{a_{ij}}{p_j^{\prime*}} = c$  and  $\forall j\notin s^{\prime *}$  we have  $\frac{a_{ij}}{p_j^{\prime*}}\leq c$ . Therefore,  $\forall j\in s^{\prime *}$  bids are positive and  $c = \frac{a_{ij}}{p_j^{\prime*}} = \frac{a_{ij}}{\theta_{ij} + b_{ij}^{\prime*}} < \frac{a_{ij}}{\theta_{ij}}$  while  $\forall j\notin s^{\prime *}$  the bids are zero, and  $\frac{a_{ij}}{\theta_{ij}}\leq \frac{a_{ij}}{\theta_{ij} + 0} = \frac{a_{ij}}{p_j^{\prime*}}\leq c$ , hence  $s^{\prime *} = \{j|c < \frac{a_{ij}}{\theta_{ij}}\}$ . To prove the second inequality, we use the same argument but with  $d = \ln (\frac{a_{ij}}{\tilde{p}_j^*})$  and so we have that  $\tilde{s}^{*} = \{j|e^{d} < \frac{a_{ij}}{\theta_{ij}}\}$ .
Proof of (2): We will show that  $c = e^d$  and thus obtain that the vectors  $b_{i}^{*}, \tilde{b}_{i}^{*}$  are identical. Assume by way of contradiction that  $c < e^d$ , then  $j \in \tilde{s}^{*} \Rightarrow c < e^{d} < \frac{a_{ij}}{\theta_{ij}} \Rightarrow j \in s'^{*}$ , i.e.,  $\tilde{s}^{*} \subseteq s'^{*}$ . For all  $j \in \tilde{s}^{*}$  it holds that  $\frac{a_{ij}}{p_j'^{*}} = c < e^d = \frac{a_{ij}}{\tilde{p}_j^{*}} \Rightarrow \tilde{p}_j^{*} < p_j'^{*} \Rightarrow \tilde{b}_{ij}^{*} < b_{ij}^{*}$ . Now we sum those inequalities over  $\tilde{s}^{*}$  and extend to the support  $s'^{*}$ . By using the subset relation we proved, we obtain a contradiction:
$$
B _ {i} = \sum_ {j \in \tilde {s} ^ {*}} \tilde {b} _ {i j} ^ {*} <   \sum_ {j \in \tilde {s} ^ {*}} b _ {i j} ^ {\prime *} \leq \sum_ {j \in s ^ {\prime *}} b _ {i j} ^ {\prime *} = B _ {i}.
$$
The case where  $e^d < c$  follows similar arguments with inverse roles of  $s'^*$ ,  $\tilde{s}^*$ . Thus,  $c = e^d$  and  $s'^* = \tilde{s}^*$ , which implies  $\frac{a_{ij}}{p_j'^*} = \frac{a_{ij}}{\tilde{p}_j^*}$ , meaning that the prices are equal as well for all goods purchased. Therefore  $b_i'^* = \tilde{b}_i^*$ .
Proof of (3): Finally, observe that for  $j \in s'^*$ $c = \frac{a_{ij}}{p_j'^*} = \frac{a_{ij}}{\theta_{ij} + b_{ij}'^*} \Rightarrow b_{ij}^* = \frac{a_{ij}}{c^*} - \theta_{ij}$ , while otherwise  $b_{ij}^* = 0$ , and  $\frac{a_{ij}}{c^*} - \theta_{ij} \leq 0$ . For the bounds on  $c$ , notice that by definition it is equal to  $\frac{u_i^*}{B_i}$  and that  $u_i^* \in (0, m)$ , as  $i$  can receive as little as almost zero (by the definition of the allocation mechanism, if  $i$  places a bid on a good it will receive a fraction of this good, no matter how tiny) and receive at most (almost) all the goods.
The intuition of the above Lemma 6 is that it shows a property of the structure of best-response bids. If we consider all the goods sorted by the parameter  $\frac{a_{ij}}{\theta_{ij}}$ , then the best-response bids are characterized by some value  $c^*$  which partitions the goods into two parts: goods that can offer the player a bang-per-buck of value  $c^*$  and those that cannot. The former set of goods is exactly the support  $s^*$ . When a player increases its bid on some good  $j$ , the bang-per-buck offered by that good decreases, so clearly, any good with  $c^* \leq \frac{a_{ij}}{\theta_{ij}}$  cannot be considered in any optimal bundle. Consider the situation where the player has started spending money on goods with  $\frac{a_{ij}}{\theta_{ij}} > c^*$ , and that for some goods  $j$  and  $k$  we have that  $\frac{a_{ij}}{p_j} = \frac{a_{ik}}{\theta_{ik}}$ , then if the player increases the bid on  $j$  without increasing the bid on  $k$ , this means that the bids are not optimal since the player could have received higher bang-per-buck by bidding on  $k$ . The optimal option is a 'water-filling' one: to split the remaining budget and use it to place bids on both  $j$  and  $k$ , yielding equal bang-per-buck for both (as Lemma 6 shows).
With the above lemmas, we are now ready to prove Theorem 2.
Proof. (Theorem 2): We start by making the following claim.
Claim: The set of Market equilibria is equal to the set of Nash equilibria in the game  $G'$ .
Proof. By definition,  $b^{*}$  is a Nash equilibrium of  $G'$  if and only if for every  $i$  it holds that  $b_{i}^{*} = \arg \max_{b_{i} \in S_{i}} u_{i}'(b_{i}, b_{-i}^{*})$ , where by Lemma 3, for any fixed  $b_{-i}$ , the bid profile  $b_{i}^{*}$  is unique. By Lemma 4, we have that for  $x_{ij}^{*} = b_{ij}^{*} p_{j}^{*}$  and any other  $x_{ij}' = b_{ij}' p_{j}^{*}$ ,  $u_{i}'(x_{i}^{*}) \geq u_{i}'(x_{i}')$ , if and only if  $(X^{*}, p^{*})$  is a market equilibrium (market clearing and budget feasibility hold trivially). That is, the set of Nash equilibria of the game  $G'$  corresponds to the set of market equilibria (i.e., every bid profile  $b^{*}$  which is a market equilibrium must be a Nash equilibrium of  $G'$ , and vice versa).
Then, by Lemma 6, best responses by every player  $i$  to any bid profile  $b_{-i}$  of the other players with respect to  $\tilde{u}_i$  and with respect to  $u_i'$  are the same. Therefore, every Nash equilibrium in one game must be a Nash equilibrium in the other. Thus, we have that Nash equilibria of the game  $\tilde{G}$  are market equilibria, and vice versa – every market equilibrium must be Nash equilibrium of  $\tilde{G}$ . Finally, at a Nash equilibrium, no player can unilaterally improve their utility, so no improvement is possible to the potential, and in the converse, if the potential is not maximized, then there exists some player with an action that improves the potential, and so by definition their utility function as well, thus contradicting the definition of a Nash equilibrium. Therefore, we have that every bid profile that maximizes the potential is a Nash equilibrium of  $\tilde{G}$  and a market equilibrium (and vice versa).
Appendix B Best Response Dynamics
We start with the following characterizations of best-response bids in the games  $\tilde{G}$  and  $G^{\prime}$ .
Lemma 7. Fix  $\theta_{i}$  and let  $b_{i}^{*}$  be  $i$ 's best response to  $\theta_{i}$  with support  $s^{*}$ . Define  $c_{s} = \frac{\sum_{j \in s} a_{ij}}{B_{i} + \sum_{j \in s} \theta_{ij}}$  for every subset  $s \subseteq [m]$ . Let  $c^{*}$  be as described in Lemma 6. Then, it holds that  $c^{*} = c_{s^{*}} \geq c_{s}$  for all  $s \subseteq [m]$ . Furthermore, if  $s^{*} \not\subset s$  then  $c_{s^{*}} > c_{s}$ .
Proof. Let  $b_{i}^{*}$  be a best response to  $\theta_{i}$  with support  $s^{*}$ . By Lemma 6 we have that  $b_{ij}^{*} = (\frac{a_{ij}}{c^{*}} - \theta_{ij})^{+}$ . By summing over  $s^{*}$  we obtain that  $B_{i} = \sum_{j \in s^{*}} \frac{a_{ij}}{c^{*}} - \theta_{ij}$ . Rearranging yields  $c^{*} = \frac{\sum_{j \in s^{*}a_{ij}}}{B_{i} + \sum_{j \in s^{*}}\theta_{ij}}$  which is  $c_{s^{*}}$  by definition. Now we prove that  $c^{*} = \max_{s \subseteq [m]} c_{s}$ . A key observation to the proof is that, by Lemma 6, if  $j \in s^{*}$  then  $c^{*}\theta_{ij} < a_{ij}$  and otherwise  $c^{*}\theta_{ij} \geq a_{ij}$ .
For a set  $s'$  distinct from  $s^*$  we have two cases:
Case (1):  $s^* \notin s'$ . Consider a bid profile  $b_i'$  that for every good  $j$  in  $s' \cap s^*$  (if the intersection is not empty) places a bid higher by  $\epsilon > 0$  than  $b_{ij}^*$  and distributes the rest of  $i$ 's budget uniformly between all other goods in  $s$ :
$$
b _ {i j} ^ {\prime} = \left\{ \begin{array}{l l} b _ {i j} ^ {*} + \epsilon & \text {i f} j \in s ^ {\prime} \cap s ^ {*}, \\ \hline B _ {i} - \frac {\sum_ {j \in s ^ {\prime} \cap s ^ {*}} (b _ {i j} ^ {*} + \epsilon)}{| s ^ {\prime} \setminus s ^ {*} |} & \text {o t h e r w i s e}. \end{array} \right.
$$
For  $\epsilon$  small enough, we have  $\sum_{j\in s^{\prime}}b_{ij}^{\prime} = B_{i}$  and the support of  $b_{i}^{\prime}$  is indeed  $s^\prime$ . For every  $j\in s^{*}\cap s^{\prime}$  we have  $b_{ij}^{\prime} > b_{ij}^{*}$  and by adding  $\theta_{ij}$  to both sides we obtain  $p_j^{\prime} > p_j^*$ ; multiplying both sides by  $c^*$  yields (i)  $c^{*}p_{j}^{\prime} > c^{*}p_{j}^{*} = a_{ij}$ , where the equality is by Lemma 6, while for every  $j\in s^{\prime}\backslash s^{*}$  it holds that  $c^{*}\theta_{ij}\geq a_{ij}$  by which adding  $c^{*}b_{ij}^{\prime}$  to the left hand side only increases it and implies (ii)  $c^{*}p_{j}^{\prime} > a_{ij}$ . Summing over inequalities (i) and (ii) for all  $j$  appropriately, we obtain  $c^{*}\sum_{j\in s^{\prime}}p_{j}^{\prime} > \sum_{j\in s^{\prime}}a_{ij}$ , observe that  $\sum_{j\in s^{\prime}}p_{j}^{\prime} = \sum_{j\in s^{\prime}}(b_{ij}^{\prime} + \theta_{ij}) = B_{i} + \sum_{j\in s^{\prime}}\theta_{ij}$ , and thus by division, we obtain the result:  $c^{*} > \frac{\sum_{j\in s^{\prime}}a_{ij}}{B_{i} + \sum_{j\in s^{\prime}}\theta_{ij}} = c_{s^{\prime}}$ .
Case (2):  $s^* \subset s'$ . In this case, the idea used above can not be applied since adding  $\epsilon$  to every bid  $b_{ij}^*$  would create bids  $b_{ij}'$  that exceed the budget  $B_i$ . As stated above, the equality  $c^* = \frac{\sum_{j \in s^*} a_{ij}}{B_i + \sum_{j \in s^*} \theta_{ij}}$  holds where the sums are taken over all members of  $s^*$ , by rearranging we get  $c^* B_i + c^* \sum_{j \in s^*} \theta_{ij} = \sum_{j \in s^*} a_{ij}$ . For all  $j \in s' \setminus s^*$  it holds that  $c^* \theta_{ij} \geq a_{ij}$  and by summing those inequalities for all  $j$
and adding the equality above we obtain:  $c^{*}B_{i} + c^{*}\sum_{j\in s^{\prime}}\theta_{ij}\geq \sum_{j\in s^{\prime}}a_{ij}$  Rearranging yields the result:  $c^{*}\geq \frac{\sum_{j\in s^{\prime}}a_{ij}}{B_{i} + \sum_{j\in s^{\prime}}\theta_{ij}} = c_{s^{\prime}}$ .
And so  $c^*$  is obtained as the maximum over all  $c_s$ , as required.
Lemma 8. The function  $BR_{i}: S_{-i} \to S_{i}$  which maps  $b_{-i}$  to its best response  $b_{i}^{*}$  is continuous.
Proof. By Lemma 6, best-response bids are given by  $b_{ij}^{*} = \max \left\{\frac{a_{ij}}{c^{*}} - \theta_{ij}, 0\right\}$ , with support  $s_i^*$ . We wish to show that  $b_{i}^{*}$  is a continuous in  $b_{-i}$ . We do so by showing that  $b_{ij}^{*}$  is obtained by a composition of continuous functions. As  $\theta_{i}$  is a sum of elements from  $b_{-i}$ , it suffices to prove continuity in the variable  $\theta_{i}$ . The expression for  $b_{ij}^{*}$  is the maximum between zero and a continuous function of  $\theta_{ij}$ , which is continuous in  $\theta_{i}$ , and so we are left to prove that  $\frac{a_{ij}}{c^{*}} - \theta_{ij}$  is continuous in  $\theta_{i}$ . More specifically, it suffices to show that  $c^{*}$  as defined in Lemma 6 is continuous in  $\theta_{i}$ .
By Lemma 7,  $c^*$  is obtained as the maximum over all  $c_s$  functions, where each is a continuous function itself in  $\theta_i$ , and thus  $c^*$  is continuous in  $\theta_i$ .
To prove Theorem 1 on the convergence of best-response dynamics we use the following known result (for further details see [46]).
Theorem (Jensen 2009 [46]): Let  $G$  be a best-reply potential game with single-valued, continuous best-reply functions and compact strategy sets. Then any admissible sequential best-reply path converges to the set of pure strategy Nash equilibria.
Proof. (Theorem 4):  $\tilde{G}$  is a potential game, which is a stricter notion than being a best-reply potential game (i.e., every potential game is also a best-reply potential game). By Lemma 6, best replies are unique, and so the function  $BR_{i}$  is single valued. Furthermore, Lemma 8 shows that it is also a continuous function. By definition, for every  $i$  the strategy set  $S_{i}$  is compact, and so their product  $S$  is compact as well. Note that while the best-reply function with respect to the standard utility function is formally undefined when the bids of all other players on some good  $j$  are zero, what we need is for the best-reply function to the associated utility to be continuous, and this is indeed the case. Admissibility of the dynamics is also guaranteed by the liveness constraint on adversarial scheduling of the dynamics, and thus by the theorem cited above, best-reply dynamics converges to the set of Nash equilibria of  $\tilde{G}$ . Since every element in this set is market equilibrium (by Theorem 2) and equilibrium prices are unique (see the model section in the main text), we have that any dynamic of the prices are guaranteed to converge to equilibrium prices. Furthermore for a generic market there is a unique market equilibrium (by Theorem 5) and convergence to the set in fact means convergence to the point  $b^{*}$ , the market-equilibrium bid profile.
Proof. (Proposition 1): Fix a player  $i$ , fix any bid profile  $b_{-i}$  of the other players and let  $b_i^*$  be  $i$ 's best response to  $b_{-i}$ , by Lemma 6,  $b_{ij}^* = \left(\frac{a_{ij}}{c^*} - \theta_{ij}\right)^+$  for  $c^*$  being a unique constant. We present a simple algorithm which computes  $c^*$  and has a run-time of  $\mathcal{O}(m \log(m))$ .
Algorithm 1 Compute  $c^*$
Require:  $a_{i},B_{i},\theta_{i}$
Sort the values  $a_i, \theta_i$  according to  $\frac{a_{ij}}{\theta_{ij}}$  in a descending order. If there are goods with  $\theta_{ij} = 0$ , sort them separately according to  $a_{ij}$  and place them as a prefix (lower indices) before the other sorted goods. Equal values are sorted in a lexicographical order.
Set:  $a\gets 0,\theta \gets 0,c_s\gets 0,c^*\gets 0$    
for  $j = 1,\ldots ,m$  do  
 $a\gets a + a_{ij},\theta \gets \theta +\theta_{ij}$ $c_{s}\leftarrow \frac{a}{\theta + B_{i}}$ $c^{*}\gets \max \{c^{*},c_{s}\}$   
end for  
return  $c^*$
To see that this process indeed reaches  $c^*$ , assume w.l.o.g. that the goods are sorted by  $\frac{a_{ij}}{\theta_{ij}}$  in a descending order. For ease of exposition, assume  $\theta_{ij} > 0$  for all  $j$ ; the case with  $\theta_{ij} = 0$  for some goods is similar. By Lemma 6 we have  $s^* = \{j | a_{ij} > c^* \theta_{ij}\}$ . And so, if  $k < j$  and  $j \in s^*$  then  $k \in s^*$ , since in this case  $\frac{a_{ik}}{\theta_{ik}} > \frac{a_{ij}}{\theta_{ij}} > c^*$ . Therefore,  $s^*$  must be one of the following sets: [1], [2], [3], ..., [m]. By Lemma 7 we have  $c^* = \max_{s \subseteq [m]} c_s$ . For any set mentioned, the algorithm computes  $c_s = \frac{\sum_{j \in s} a_{ij}}{B_i + \sum_{j \in s} \theta_{ij}}$  and finds the maximal among all such  $c_s$ , and therefore it finds  $c^*$ .
As for the running time of the algorithm, it is dominated by the running time of the sorting operation which is  $\mathcal{O}(m\log (m))$
After proving that the best response to a bid profile can be computed efficiently, we can prove now that proportional response, applied by a single player while all the other players' bids are held fix, converges in the limit to that best response.
Proof. (Proposition 2): Fix a player  $i$  and fix any bid profile  $b_{-i}$  of the other players, let  $b_i^*$  be the best response of  $i$  to  $b_{-i}$  with support  $s^*$  and let  $(b_i^t)_{t=1}^\infty$  be a sequence of consecutive proportional responses made by  $i$ . That is,  $b_i^{t+1} = f_i(b_i^t)$ . We start the proof with several claims proving that any sub-sequence of  $(b_i^t)_{t=1}^\infty$  cannot converge to any fixed point of  $f_i$  other than  $b_i^*$ . After establishing this, we prove that the sequence indeed converges to  $b_i^*$ .
Claim 1: Every fixed point of Proportional Response Dynamic has equal 'bang-per-buck' for all goods with a positive bid. That is, if  $b_{i}^{**}$  is a fixed point of  $f_{i}$  then  $\frac{a_{ij}}{p_j^{**}} = \frac{u_i^{**}}{B_i}$  for every good  $j$  with  $b_{ij}^{**} > 0$ , where  $u_{i}^{**}$  is the utility achieved for  $i$  with the bids  $b_{i}^{**}$ .
Proof: By substituting  $b_{i}^{**}$  into the PRD update rule, we have
$$
\begin{array}{l} b _ {i} ^ {* *} = f _ {i} \left(b _ {i} ^ {* *}\right) \\ \Longleftrightarrow \forall j \quad b _ {i j} ^ {* *} = \frac {a _ {i j} / p _ {j} ^ {* *}}{u _ {i} ^ {* *} / B _ {i}} b _ {i j} ^ {* *} \\ \Longleftrightarrow \text {e i t h e r} b _ {i j} ^ {* *} = 0 \text {o r} a _ {i j} / p _ {j} ^ {* *} = u _ {i} ^ {* *} / B _ {i}. \\ \end{array}
$$
Claim 2: The following properties of  $b_{i}^{*}$  hold.