Concave Pro-rata Games

Abstract

In this paper, we introduce a family of games called concave pro-rata games. In such a game, players place their assets into a pool, and the pool pays out some concave function of all assets placed into it. Each player then receives a pro-rata share of the payout; i.e., each player receives an amount proportional to how much they placed in the pool. Such games appear in a number of practical scenarios, including as a simplified version of batched decentralized exchanges, such as those proposed by Penumbra. We show that this game has a number of interesting properties, including a symmetric pure equilibrium that is the unique equilibrium of this game, and we prove that its price of anarchy is $\varOmega (n)$ in the number of players. We also show some numerical results in the iterated setting which suggest that players quickly converge to an equilibrium in iterated play.

Appendices

A Numerics

The results of Sect. 1 provide insight into the equilibrium behavior of concave pro-rata games. Here we explore the transient behavior of such games through simulation.

Game Setup. Suppose that the game is played iteratively, and, at each iteration t, player i chooses some action $x_i^t$ as the best response to the actions chosen by the other players in the previous round (denoted as $x_{-i}^{t-1}$), possibly subject to additional constraints. We consider the following scenarios:

• 1. At iteration t, player i takes action equal to the best response to $x_{-i}^{t-1}$.

• 2. At iteration t, player i takes action equal to the best response to $x_{-i}^{t-1}$ subject to a budget constraint ($x_i^t \in [0, M_i]$).

Payoff Functions. For these simulations, we use functions f of the form $f(t) = g(t)- c t$ where $c > 0$ and $g(t) = \frac{\gamma R_2 t}{R_1+\gamma t}$ with $0 < \gamma \le 1$, $R_1, R_2 > 0$. The function g(t) is the forward exchange function for a Uniswap V2 swap pool with reserves $R \in {\text{ R }}^2_+$ and fee parameter $\gamma$ when asset 1 is being tendered and asset 2 is being received. This setting simulates n arbitrageurs competing to maximize their profit, where c denotes the external market price of asset 2. For simulations using a somewhat more simple payoff function, see Appendix B. Note that f is strictly concave and therefore satisfies condition (3), and clearly $f(0) = 0$.

Shared Equilibrium. The (unique) symmetric pure equilibrium strategy is the solution to problem (4). This is easy to compute using the first order optimality conditions for problem (4) given in (6). Plugging in this particular form of f, we obtain the following quadratic equation:

$$\begin{aligned} (cn\gamma ^2)q^2 + q(\gamma ^2 R_2 + 2cnR_1\gamma -\gamma ^2nR_2)+(cnR_1^2-\gamma nR_1R_2) = 0. \end{aligned}$$

(7)

The equilibrium is then given by $x_i = q/n$, for each player $i=1, \dots , n$ where q denotes the positive root of (7).

Best Responses. The best response of player i, given the budget constraint $0 \le x_i \le M_i$ and other players’ strategies $y_i = \textbf{1}^Tx - x_i$, is given by

$$\begin{aligned} \begin{aligned} & \text {maximize} &\,& U(x_i, y_i)\\ & \text {subject to} &\,& x_i \in [0, M_i], \end{aligned} \end{aligned}$$

(8)

with variable $x_i \in {\text{ R }}$. This is a single-variable convex optimization problem that is easily solved in practice by any number of off-the-shelf packages [7, 8]. When $x_i$ is unconstrained, the optimal value of (8) is given by

$$ x_i = \frac{1}{\gamma }\left( \sqrt{\frac{\gamma R_1 R_2 + \gamma ^2 R_2 y}{c}} -R_1\right) - y $$

For more details, the code is available at (anonymized for review).

Simulation Results. In our simulations, we fix $\gamma =0.99$, $R_1 = 200$, $R_2 = 250$, and $c=1$. We average each reported value over 100 trials. In Fig. 1, the intial strategy of each player is drawn uniformly at random from the interval (0, w/n), where w is a value such that $f(w) = 0$.

Fig. 1.

Number of iterations to reach equilibrium versus the number of players in Scenario 1.

Figure 1 illustrates that the number of iterations needed to reach the unique equilibrium, in the absence of budget constraints, scales superlinearly in the number of players. We define the number of iterations to reach equilibrium as the number iterations until the strategy of every player is equal to the unique equilibrium up to the first decimal place; i.e., the first round t such that

$$ \max _i |x_i^t - x^\star | < 0.1, $$

where $x^\star$ denotes the equilibrium strategy.

Fig. 2.

Percent increase in whale strategy and whale profit versus the number of fish when compared to the unconstrained equilibrium strategy and profit.

In Fig. 2, we consider the setting where there is one player who has unlimited budget (whom we will call a whale) and all remaining players have some budget $M_i < q/n$ (these players are referred to as fish). The budgets of the fish are drawn uniformly from the interval $M_i \sim [0, q/n]$ and the initial strategy of each fish is drawn uniformly at random from the interval $[0, M_i]$. The equilibrium strategy chosen by the fish is to use their entire budget, while the whale chooses a strategy in excess of the unconstrained equilibrium strategy and is, as a result, able to extract greater profit. Figure 2 illustrates that the whale chooses an increasingly large strategy and receives an increasing profit as the number of fish increases.

Price of Anarchy. In Sect. 1 we established the order of growth of the price of anarchy. Here we illustrate the price of anarchy numerically for the specific family of payoff functions introduced previously in this section. We again fix $\gamma =0.99, R_1 = 200, R_2 = 250$ and $c=1$. The left plot of Fig. 3 illustrates the optimal payoff function and the equlibrium payoff function as a function of the number of players n while the right plot of Fig. 3 illustrates the price of anarchy as function of n.

Fig. 3.

(Left) Individual payoff of a player versus the number of players. (Right) Ratio of the optimal payoff divided by the equilibrium payoff versus the number of players.

B Additional Numerics

Here we expand on the simulations introduced in Appendix A using a class of utility function that allows us to express many quantities of interest in closed form.

Game Setup. We consider the following three scenarios:

• 1. At iteration t, player i takes action equal to the best response to $x_{-i}^{t-1}$.

• 2. At iteration t, player i takes action equal to the best response to $x_{-i}^{t-1}$ subject to a bounded update constraint ($\vert x_i^t - x_i^{t-1}\vert \le \delta$).

• 3. At iteration t, player i takes action equal to the best response to $x_{-i}^{t-1}$ subject to a budget constraint ($x_i^t \in [0, M_i]$).

Payoff Functions. For these simulations, we use functions f of the form $f(t) = t^\beta -\gamma t$ where $0 < \beta < 1$ and $\gamma > 0$. Note that f is concave as it is the sum of two concave functions and $f(0) = 0$. These functions also satisfy the strict concavity property (3) since

$$ f(\alpha t)=\alpha ^\beta t^\beta -\alpha \gamma t > \alpha t^\beta -\alpha \gamma t = \alpha f(t), $$

for $0 < \alpha < 1$.

Fig. 4.

(Left) Number of iterations to reach equilibrium versus the number of players in Scenario 1. (Right) Number of iterations to reach equilibrium versus $\delta$ in Scenario 2 (with $n=10$ players).

Shared Equilibrium. The (unique) symmetric pure equilibrium strategy is the solution to problem (4). This is easy to compute using the first order optimality conditions for problem (4) given in (6). Plugging in this particular form of f, we have:

$$ (n-1)(q^\beta - \gamma q) + q(\beta q^{\beta - 1} - \gamma ) = 0, $$

which has a solution

$$ q = \left( \frac{\beta + n - 1}{n\gamma }\right) ^{1/(1-\beta )}. $$

The equilibrium is then given by $x_i = q/n$, for each player $i=1, \dots , n$.

Simulation Results. In our simulations, we fix $\beta =0.5$ and $\gamma =0.05$. We average each reported value over 100 trials. In Fig. 4, the intial strategy of each player is drawn uniformly at random from the interval (0, w/n), where w is a value such that $f(w) = 0$.

The left plot of Fig. 4 illustrates that the number of iterations needed to reach the unique equilibrium, in the absence of budget constraints, scales superlinearly in the number of players. The right plot demonstrates that in the scenario of bounded strategy updates, for small values of $\delta$, the number of iterations required to reach equilibrium increases significantly when compared to the unbounded strategy update scenario.

Fig. 5.

Percent increase in whale strategy and whale profit versus the number of fish when compared to the unconstrained equilibrium strategy and profit.

In Fig. 5, we consider the setting where there is one player who has unlimited budget (whom we will call a whale) and all remaining players have some budget $M_i < q/n$ (these players are referred to as fish). The budgets of the fish are drawn uniformly from the interval $M_i \sim [0, q/n]$ and the initial strategy of each fish is drawn uniformly at random from the interval $[0, M_i]$. The equilibrium strategy chosen by the fish is to use their entire budget, while the whale chooses a strategy in excess of the unconstrained equilibrium strategy and is, as a result, able to extract greater profit. Figure 5 illustrates that the whale chooses an increasingly large strategy and receives an increasing profit as the number of fish increases.

Price of Anarchy. The equilibrium payoff can easibly be found to be

$$ \frac{1}{n}f(q) = \bigg {(}\frac{n+\beta -1}{\gamma n}\bigg {)}^{\beta /(1-\beta )} \bigg {(}\frac{1-\beta }{n^2} \bigg {)}. $$

Similarly, it can be show that the optimal payoff conditioned on every agent receving the same payoff is given by

$$ \frac{1}{n} \sup f = \bigg {(} \frac{\beta }{\gamma } \bigg {)}^{\beta /(1-\beta )} \bigg {(}\frac{1-\beta }{n} \bigg {)}. $$

We obtain the price of anarchy by taking the ratio of the equilibrium payoff and the optimal payoff:

$$ \frac{\sup f}{f(q)} = n \bigg {(} \frac{\beta n}{n + \beta - 1} \bigg {)}^{\beta / (1-\beta )}. $$

We again fix $\beta =0.5$ and $\gamma =0.05$. The left plot of Fig. 6 illustrates the optimal payoff function and the equlibrium payoff function as a function of the number of players n while the right plot of Fig. 6 illustrates the price of anarchy as function of n.

Fig. 6.

(Left) Individual payoff of a player versus the number of players. (Right) Ratio of the optimal payoff divided by the equilibrium payoff versus the number of players.

C Relaxing Strict Concavity

We do not, in fact, need strict concavity in the proofs above. Instead, we only need that f has ‘some curvature’ at 0. Specifically, it suffices that for all t and $t'$ such that $0 < t < t'$, we have

$$ f(t) > \frac{f(t')}{t'} t. $$

Written in English, this is the condition that the chord from 0 to t always lies strictly below the function. This condition is sometimes difficult to confirm for general functions f, so we will show that this is equivalent to the (potentially simpler-to-handle) property that all supergradients at 0 lie strictly above the function at all points. We will show that, for any concave function $f: {\text{ R }}_+ \rightarrow {\text{ R }}$ with $f(0) = 0$, the following two statements are equivalent: (a) there is some $s' > 0$ and $\alpha \in {\text{ R }}$ such that for every s with $0 \le s \le s'$ we have

$$ f(s) = \alpha s, $$

and (b) there exists some $0 < t < t'$ such that

$$\begin{aligned} \frac{f(t)}{t} = \frac{f(t')}{t'}. \end{aligned}$$

(9)

The statement above follows from the negation of both (a) and (b). This equivalence has a simple interpretation: if the point (0, 0) is collinear with any other two points on the graph of f, $\{(s, f(s)) \mid s > 0\}$, then the function f is a piecewise function with a linear segment starting at 0. The converse of this is that if the function f has no linear segment around 0 (i.e., every linear overestimator around 0 lies strictly above f) then any chord must lie strictly below the function.

Proof. The forward implication is very easy: pick $t' = s'$ and let t be any $0 < t < s'$, then we have

$$ \frac{f(t')}{t'} = \alpha = \frac{f(t)}{t}. $$

Now we’ll consider the reverse implication. Given $0 < t < t'$ satisfying (9), we will show that, for any $0 \le s \le t$ we have

$$ f(s) = \frac{f(t)}{t}s, $$

which satisfies the original claim with $\alpha = f(t)/t$. First, it is easy to show that

$$\begin{aligned} f(s) \ge \frac{f(t)}{t}s, \end{aligned}$$

(10)

since

$$ f(s) = f\left( \frac{s}{t}t + \left( 1-\frac{s}{t}\right) 0\right) \ge \frac{s}{t}f(t), $$

where the inequality follows from the concavity of f and the fact that $f(0) = 0$. We will now show that any function f satisfying (10) strictly, i.e.,

$$\begin{aligned} f(s) > \frac{f(t)}{t}s, \end{aligned}$$

(11)

for some $0 < s < t$ cannot be concave. The result follows from the contrapositive. To see this, let $0 < \gamma \le 1$ such that $t = \gamma s + (1-\gamma )t'$, then

$$ \gamma f(s) + (1-\gamma )f(t') > \gamma \frac{f(t)}{t}s + (1-\gamma )\frac{f(t)}{t}t' = f(t) = f(\gamma s + (1-\gamma )t'), $$

so f cannot be concave. The inequality follows directly from conditions (9) and (11), and both the first and second equalities follow from the definition of $\gamma$.

D Rosen Condition

Pro-rata games, even concave ones, do not satisfy the Rosen condition [10] for the uniqueness of equilibria in concave games. The Rosen condition for uniqueness is that if, there exists some $z \ge 0$ with $z \ne 0$ such that

$$ \varPhi (x) = \begin{bmatrix} z_1 \partial _1 U_1(x)\\ \vdots \\ z_n \partial _n U_n(x) \end{bmatrix} $$

is a strictly monotone operator; i.e., for any $x \ne y$ we have

$$ (y - x)^T(\varPhi (y) - \varPhi (x)) > 0, $$

then there is a unique equilibrium that is also pure. (Here, $\partial _i$ denotes the ith partial derivative.) This is a common condition used to prove the uniqueness of pure equilibria in games. We will show that this condition does not hold in general for concave pro-rata games, even under most ‘niceness’ assumptions such as strict concavity or even strong concavity and differentiability.

Setting $2x = y = \textbf{1}$ then the condition can be written as (using the definition of U)

$$ (\textbf{1}^Tz/2)((1/n)(f'(n) - f'(n/2)) + (1-1/n)(f(n) - 2f(n/2))) > 0, $$

but this can be rewritten (since $\textbf{1}^Tz > 0$)

$$ (1/n)(f'(n) - f'(n/2)) + (1-1/n)(f(n) - 2f(n/2)) > 0, $$

which is clearly not true for all concave functions f, since picking $f(t) = \min \{t, 3n\}$ suffices. (A mollifying argument would show that this also gives a reasonable counterexample even in the case that f is strictly concave and differentiable.) A more direct counterexample that is differentiable and strictly concave is $f(t) = (4n)^2 - (4n - t)^2$, which is slightly more difficult to verify.