Multiple-Aggregate Games

Abstract

Consider an environment in which individuals are organised into groups, they contribute to the collective action of their group, and are influenced by the collective actions of other groups; there are externalities between groups that are transmitted through the aggregation of groups’ actions. The theory of ‘aggregative games’ has been successfully applied to study games in which players’ payoffs depend only on their own strategy and a single aggregation of all players’ strategies, but the setting just described features multiple aggregations of actions—one for each group—in which the nature of the intra-group strategic interaction may be very different to the inter-group strategic interaction. The aim of this contribution is to establish a framework within which to consider such ‘multiple aggregate games’; present a method to analyse the existence and properties of Nash equilibria; and to discuss some applications of the theory to demonstrate how useful the technique is for analysing strategic interactions involving individuals in groups.

Appendix: Proofs

Proof of Proposition 1

The proof is by simple definition chasing. If x ^∗ is a Nash equilibrium in $\mathcal{G}$, then by definition $x_{i}^{\,j{\ast}} =\tilde{ x}_{i}^{\,j}(\mathbf{X}^{-j{\ast}})$ for all i ∈ I ^j, j ∈ J. But then for each j ∈ J, $X^{j{\ast}} =\sum _{i\in I^{j}}\tilde{x}_{i}^{\,j}(\mathbf{X}^{-j{\ast}}) =\tilde{ X}^{j}(\mathbf{X}^{-j{\ast}})$. Conversely, if $X^{j{\ast}} =\tilde{ X}^{j}(\mathbf{X}^{-j{\ast}})$ for all j ∈ J, then by definition x ^j∗ where $x_{i}^{\,j{\ast}} =\tilde{ x}_{i}^{\,j}(\mathbf{X}^{-j{\ast}})$ for each i ∈ I ^j is a Nash equilibrium in $\mathcal{G}^{j}(\mathbf{X}^{-j{\ast}})$ for all j ∈ J; it then follows that x ^∗ is a Nash equilibrium. □ 

Proof of Proposition 2

The proof is again by definition chasing. First, if x ^j∗ is a Nash equilibrium then x _i ^j∗ = b _i ^j(X _−i ^j∗; X ^−j) for all i ∈ I ^j. This implies x _i ^j∗ = b _i ^j(X ^j∗− x _i ^j∗; X ^−j), and so by definition x _i ^j∗ = r _i ^j(X ^j∗; X ^−j) for all i ∈ I ^j; and therefore $\sum _{i\in I^{j}}r_{i}^{\,j}(X^{j{\ast}};\mathbf{X}^{-j}) = X^{j{\ast}}$. To prove necessity, suppose $\sum _{i\in I^{j}}r_{i}^{\,j}(X^{j{\ast}};\mathbf{X}^{-j}) = X^{j{\ast}}$ and consider the strategy x _i ^j∗ = r _i ^j(X ^j∗; X ^−j). By definition of the replacement function, x _i ^j∗ = b _i ^j(X ^j∗− x _i ^j∗; X ^−j), and since $X^{j{\ast}} =\sum _{i\in I^{j}}r_{i}^{\,j}(X^{j{\ast}};\mathbf{X}^{-j})$ it follows that $X^{j{\ast}}- x_{i}^{\,j{\ast}} =\sum _{h\neq i\in I^{j}}r_{h}^{j}(X^{j{\ast}};\mathbf{X}^{-j}) = X_{-i}^{j{\ast}}$. As such, x _i ^j∗ = b _i ^j(X _−i ^j∗; X ^−j) for all i ∈ I ^j, so x ^j∗ is a Nash equilibrium in $\mathcal{G}^{j}(\mathbf{X}^{-j})$. □ 

Proof of Proposition 3

A player’s share function is the value of σ _i ^j that makes l _i ^j, defined in (5), equal to 0—however, if this is below zero the share function is defined as zero; and if it is above 1 the share function is undefined. First, note that under Assumption 1

$$\displaystyle\begin{array}{rcl} \frac{\partial l_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} & = X^{j}\left [ \frac{\partial ^{2}u_{ i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} + \frac{\partial ^{2}u_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X_{-i}^{j}}\right ] < 0\ \text{ and}& {}\\ \frac{\partial l_{i}^{\,j}} {\partial X^{j}}& =\sigma _{ i}^{\,j} \frac{\partial ^{2}u_{ i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} + [1 -\sigma _{i}^{\,j}] \frac{\partial ^{2}u_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X_{-i}^{j}} < 0.& {}\\ \end{array}$$

The first inequality implies there is at most one σ _i ^j ∈ [0, 1] where l _i ^j = 0 so the share function is indeed a function. Continuity of this function, where it is defined, follows from l _i ^j varying continuously in all its arguments by virtue of the assumed differentiability of utility functions.

If X _i ^j(X ^−j) > 0, by definition, l _i ^j(1, X _i ^j(X ^−j); X ^−j) = 0, so s _i ^j(X _i ^j(X ^−j), X ^−j) = 1 and the monotonicity properties just stated imply that for all X ^j < X _i ^j(X ^−j), l _i ^j > 0 for all σ _i ^j ≤ 1, and therefore the share function is undefined. In addition, again by definition, if $\bar{X}_{i}^{\,j}(\mathbf{X}^{-j}) < \infty$ then $l_{i}^{\,j}(0,\bar{X}_{i}^{\,j}(\mathbf{X}^{-j});\mathbf{X}^{-j}) = 0$ so $s_{i}^{\,j}(\bar{X}_{i}^{\,j}(\mathbf{X}^{-j});\mathbf{X}^{-j}) = 0$ and the monotonicity properties of l _i ^j imply that for all $X^{j} >\bar{ X}_{i}^{\,j}(\mathbf{X}^{-j})$, l _i ^j = 0 only when σ _i ^j < 0, and therefore by definition s _i ^j(X ^j; X ^−j) = 0.

Where it is positive and defined, implicit differentiation of the first-order condition that defines the share function, (5), gives

$$\displaystyle{ \frac{\partial s_{i}^{\,j}} {\partial X^{j}} = -\frac{\frac{\partial l_{i}^{\,j}} {\partial X^{j}}} {\frac{\partial l_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} } < 0, }$$

where the inequality follows from the deductions above.

If X _i ^j(X ^−j) = 0 the share function is defined for all X ^j > 0 where it takes values in the compact set [0, 1], so (taking subsequences if necessary) the limit as X ^j → 0 will exist, which is denoted $\bar{s}_{i}^{\,j}(\mathbf{X}^{-j})$. □ 

Proof of Proposition 4

If $\bar{X}_{i}^{\,j}(\mathbf{X}^{-j}) < \infty$ the share function of individual i ∈ I ^j is equal to zero for all $X^{j} \geq \bar{ X}_{i}^{\,j}(\mathbf{X}^{-j})$. If not, then since it is assumed that $\lim _{x_{i}^{\,j}\rightarrow \infty }\frac{\partial u_{i}^{\,j}(x_{ i}^{\,j},\cdot;\cdot )} {\partial x_{i}^{\,j}} < 0$ the first-order condition (5) can hold as X ^j → ∞ only if $\lim _{X^{j}\rightarrow \infty }\sigma _{i}^{\,j}X^{j} < \infty$ which requires σ _i ^j → 0, implying the share function vanishes in the large X ^j limit. This implies there is an $\bar{x}^{\,j}(\mathbf{X}^{-j})$ such that $\sum _{i\in I^{j}}s_{i}^{\,j}(X^{j};\mathbf{X}^{-j}) < 1$ for all $X^{j} >\bar{ x}^{\,j}(\mathbf{X}^{-j})$. The function $\sum _{i\in I^{j}}s_{i}^{\,j}(X^{j};\mathbf{X}^{-j})$ is continuous and strictly decreasing in X ^j for all $\max _{i\in I^{j}}\{\underline{X}_{i}^{\,j}(\mathbf{X}^{-j})\} < X^{j} <\bar{ x}^{\,j}(\mathbf{X}^{-j})$, and is therefore equal to 1 for at most one value of X ^j. If X _i ^j(X ^−j) > 0 for any i ∈ I ^j then $\sum _{i\in I^{j}}s_{i}^{\,j}(\max _{i\in I^{j}}\{\underline{X}_{i}^{\,j}(\mathbf{X}^{-j})\},\mathbf{X}^{-j}) \geq 1$ and so there is a unique value of X ^j where $\sum _{i\in I^{j}}s_{i}^{\,j}(X^{j},\mathbf{X}^{-j}) = 1$. If X _i ^j(X ^−j) = 0 for all i ∈ I ^j then the aggregate share function is defined for all X ^j > 0, with $\lim _{X^{j}\rightarrow 0}\sum _{i\in I^{j}}s_{i}^{\,j}(X^{j},\mathbf{X}^{-j}) =\sum _{i\in I^{j}}\bar{s}_{i}^{\,j}(\mathbf{X}^{-j})$. As such, the existence of a (unique) Nash equilibrium requires $\sum _{i\in I^{j}}\bar{s}_{i}^{\,j}(\mathbf{X}^{-j}) > 1$. □ 

Proof of Proposition 5

For $\mathbf{X}^{-j} \in \mathbb{R}_{+}^{N-1}\setminus \bar{\mathcal{X}}^{j}$, the group best response is implicitly defined by (7). Continuity of the group best response follows from continuity of individual share functions in each of its arguments, which follows from the assumed differentiability of utility functions. With apology⁹ implicit differentiation of (7) gives

$$\displaystyle{ \frac{\partial \tilde{X}^{j}(\mathbf{X}^{-j})} {\partial X^{k}} = -\frac{\sum _{i\in I^{j}} \frac{\partial s_{i}^{\,j}} {\partial X^{k}}} {\sum _{i\in I^{j}}\frac{\partial s_{i}^{\,j}} {\partial X^{j}}}. }$$

(10)

Where an individual’s share function is positive, recall that it is defined by the first-order condition l _i ^j(σ _i ^j, X ^j; X ^−j) = 0 as in (5). As deduced previously,

$$\displaystyle{ \frac{\partial s_{i}^{\,j}} {\partial X^{j}} = -\frac{\frac{\partial l_{i}^{\,j}} {\partial X^{j}}} {\frac{\partial l_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} } = \frac{\sigma _{i}^{\,j} \frac{\partial ^{2}u_{ i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} + [1 -\sigma _{i}^{\,j}] \frac{\partial ^{2}u_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X_{-i}^{j}}} {X^{j}\left [ \frac{\partial ^{2}u_{i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} - \frac{\partial ^{2}u_{i}^{\,j}} {\partial x_{i}^{\,j}\partial X_{-i}^{j}}\right ]} < 0 }$$

under Assumption 1. In addition,

$$\displaystyle{ \frac{\partial s_{i}^{\,j}} {\partial X^{k}} = -\frac{\frac{\partial l_{i}^{\,j}} {\partial X^{k}}} {\frac{\partial l_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} } = - \frac{ \frac{\partial ^{2}u_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X^{k}}} {X^{j}\left [ \frac{\partial ^{2}u_{i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} - \frac{\partial ^{2}u_{i}^{\,j}} {\partial x_{i}^{\,j}\partial X_{-i}^{j}}\right ]}. }$$

Since the denominator is negative under Assumption 1, $\mathop{\mathrm{sgn}}\nolimits \{ \frac{\partial s_{i}^{\,j}} {\partial X^{k}}\} =\mathop{ \mathrm{sgn}}\nolimits \{ \frac{\partial ^{2}u_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X^{k}}\}$. As such, since group members are qualitatively symmetric, it follows that $\mathop{\mathrm{sgn}}\nolimits \{ \frac{\partial \tilde{X}^{j}(\mathbf{X}^{-j})} {\partial X^{k}} \} =\mathop{ \mathrm{sgn}}\nolimits \{ \frac{\partial ^{2}u_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X^{k}}\}$, as stated.

$\left \vert \frac{\partial \tilde{X}^{j}(\mathbf{X}^{-j})} {\partial X^{k}} \right \vert < 1$ if the numerator in (10) is less than the denominator, a sufficient (but by no means necessary) condition for which is $\left \vert \frac{\partial s_{i}^{\,j}} {\partial X^{k}}\right \vert < \left \vert \frac{\partial s_{i}^{\,j}} {\partial X^{j}}\right \vert$ for all i ∈ I ^j, which is implied by the inequality in the proposition. □ 

Proof of Proposition 6

If x ^∗ is a Nash equilibrium, then by definition of share functions $x_{i}^{\,j{\ast}} = X^{j{\ast}}\hat{s}_{i}^{\,j}(X^{j{\ast}};X^{{\ast}})$ for all i ∈ I ^j, j ∈ J. As such, $\sum _{i\in I^{j}}\hat{s}_{i}^{\,j}(X^{j{\ast}};X^{{\ast}}) = 1$ and therefore $X^{j{\ast}} =\hat{ X}^{j}(X^{{\ast}})$ for all j ∈ J, implying $X^{{\ast}} =\sum _{j\in J}\hat{X}^{j}(X^{{\ast}})$. For necessity of the condition, define a player’s best response in a nested aggregative game as b _i ^j(X _−i ^j; X ^−j). Consider the strategy $x_{i}^{\,j{\ast}} =\hat{ X}^{j}(X^{{\ast}})\hat{s}_{i}^{\,j}(X^{j{\ast}};X^{{\ast}})$. By definition of share functions and the consistency of $\hat{X}^{j}(X^{{\ast}})$ within group j (which implies $\hat{X}^{j}(X^{{\ast}}) - x_{i}^{\,j{\ast}} = X_{-i}^{j{\ast}}$), $x_{i}^{\,j{\ast}} = b_{i}^{\,j}(X_{-i}^{j{\ast}};X^{{\ast}}-\hat{ X}^{j}(X^{{\ast}}))$. When $X^{{\ast}} =\sum _{j\in J}\hat{X}^{j}(X^{{\ast}})$, it follows that $X^{{\ast}}-\hat{ X}^{j}(X^{{\ast}}) = X^{-j{\ast}}$, and therefore x _i ^j∗ = b _i ^j(X _−i ^j∗; X ^−j∗) for all i ∈ I ^j, j ∈ J, giving the conclusion that x ^∗ is a Nash equilibrium. □ 

Proof of Proposition 7

The properties of individual share functions in $\hat{\mathcal{G}}^{j}(X)$ are first deduced. The conditions stated on preferences are equivalent to assuming

$$\displaystyle\begin{array}{rcl} \frac{\partial \hat{l}_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} & =& X^{j}\left [ \frac{\partial ^{2}\hat{u}_{ i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial x_{i}^{\,j}\partial X^{j}} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial x_{i}^{\,j}\partial X}\right ] < 0, {}\\ \frac{\partial \hat{l}_{i}^{\,j}} {\partial X^{j}}& =& \sigma _{i}^{\,j} \frac{\partial ^{2}\hat{u}_{ i}^{\,j}} {\partial (x_{i}^{\,j})^{2}} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial x_{i}^{\,j}\partial X^{j}} +\sigma _{ i}^{\,j} \frac{\partial ^{2}\hat{u}_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X^{j}} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial (X^{j})^{2}} +\sigma _{ i}^{\,j} \frac{\partial ^{2}\hat{u}_{ i}^{\,j}} {\partial x_{i}^{\,j}\partial X} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial X^{j}\partial X} < 0,\text{ and} {}\\ \frac{\partial \hat{l}_{i}^{\,j}} {\partial X} & =& \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial x_{i}^{\,j}\partial X} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial X^{j}\partial X} + \frac{\partial ^{2}\hat{u}_{i}^{\,j}} {\partial (X)^{2}} < 0. {}\\ \end{array}$$

Under the first two conditions (as previously) two thresholds $\hat{\underline{X}}_{i}^{\,j}(X)$ (which is X ^j > 0 such that $\hat{l}_{i}^{\,j}(1,X^{j};X) = 1$) and $\hat{\bar{X}}_{i}^{\,j}(X)$ (which is X ^j such that $\hat{l}_{i}^{\,j}(0,X^{j};X) = 1$ if such an X ^j exists, otherwise it is defined as + ∞) can be defined, between which the share function is defined and takes positive values, and where

$$\displaystyle{ \frac{\partial \hat{s}_{i}^{\,j}} {\partial X^{j}} = -\frac{\frac{\partial \hat{l}_{i}^{\,j}} {\partial X^{j}}} {\frac{\partial \hat{l}_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} } < 0. }$$

If $\hat{\underline{X}}_{i}^{\,j}(X)$ as defined above does not exist then the share function is defined for all X ^j > 0 with $\lim _{X^{j}\rightarrow 0}\hat{s}_{i}^{\,j}(X^{j};X) =\hat{\bar{ s}}_{i}^{\,j}(X)$.

As before, the aggregation of individual share functions is taken to be defined only for values of X ^j where all group members’ share functions are defined. Noting that share functions are either equal to zero for large enough X ^j, or are vanishing in the large X ^j limit, if either $\hat{\underline{X}}_{i}^{\,j}(X) > 0$ for any i ∈ I ^j, or $\sum _{i\in I^{j}}\hat{\bar{s}}_{i}^{\,j}(X) > 1$ then there is a single consistent aggregate action $\hat{X}^{j}(X)$ in $\hat{\mathcal{G}}^{j}$ which is such that $\sum _{i\in I^{j}}\hat{s}_{i}^{\,j}(\hat{X}^{j}(X);X) = 1$. If this is not the case then $\hat{X}^{j}(X) = 0$.

Consider now varying X to change the partial game played by group j. Group j’s share of the total aggregate is $\hat{S}^{j}(X) =\hat{ X}^{j}(X)/X$, defined by (9) if the resulting share is between 0 and 1. Note that

$$\displaystyle\begin{array}{rcl} \frac{\partial \hat{L}^{j}} {\partial \Lambda ^{j}}& =& X\sum _{i\in I^{j}} \frac{\partial \hat{s}_{i}^{\,j}} {\partial X^{j}}\text{ and} {}\\ \frac{\partial \hat{L}^{j}} {\partial X} & =& \sum _{i\in I^{j}}\Lambda ^{j} \frac{\partial \hat{s}_{i}^{\,j}} {\partial X^{j}} + \frac{\partial \hat{s}_{i}^{\,j}} {\partial X}. {}\\ \end{array}$$

Now,

$$\displaystyle{ \frac{\partial \hat{s}_{i}^{\,j}} {\partial X} = -\frac{\frac{\partial \hat{l}_{i}^{\,j}} {\partial X} } {\frac{\partial \hat{l}_{i}^{\,j}} {\partial \sigma _{i}^{\,j}} } < 0, }$$

which, combined with the monotonicity of share functions with respect to group aggregate, implies both of the expressions above are negative. Given this, the thresholds and the monotonicity of group share functions stated in the proposition can be derived analogously to the case within groups, so the details are omitted. Aggregate share functions are either equal to zero for $X >\hat{\bar{ X}}^{j}$ or, if this is not finite, vanish in the large X limit—this follows by recalling that individual share functions vanish in the large X ^j limit, so as X → ∞ (9) can hold only if $\lim _{X\rightarrow \infty }\Lambda ^{j}X < \infty$ which requires $\Lambda ^{j} \rightarrow 0$. Given this, if either $\hat{\underline{X}}^{j} > 0$ for any j ∈ J, or $\sum _{j\in J}\hat{\bar{S}}^{j} > 0$ if $\hat{\underline{X}}^{j} = 0$ for all j ∈ J, the aggregate share function will exceed one for small enough X and since it is strictly decreasing in X will be equal to one at exactly one value of X, so consequently there is a unique Nash equilibrium. □