Written during the SERI MATS program under the joint mentorship of John Wentworth, Nicholas Kees, and Janus.

Motivation

I am surrounded, incessantly, by other people, possessing a range of different capabilities, striving for goals both common and opposing — and, what is more, they are incessantly surrounded by me. Fangs pierce the cartesian boundary. Outsideness leaks in. Insideness leaks out. It is the source of all joy and misery in my life.

Anyway. It's the interaction of $n\ge 2$ agents that characterises game theory.^[1] Classical game theory says that when two utility-maximisers interact in a shared environment, they will choose options in nash equilibrium, where a profile of options is in nash equilibrium if no player can increase their utility function by unilaterally changing their option.

What does higher-order game theory (which dispenses with utility functions and maximisation) say about the interaction between agents? Would a satisficer cooperate with a quantiliser in prisoner's dilemma?

Let's find out.

Preliminaries

All you need to know is Definition 3 from [Part 1], and some basic classical game theory.

Definition 3: Let $X$ be any set of options and $R$ be any set of payoffs. An optimiser is any functional $\psi :(X\to R)\to P\left(X\right)$. A $\psi$-task is any function $u:X\to R$. An option $x\in X$ is $\psi$-optimal for a task $u:X\to R$ if and only if $x\in \psi \left(u\right)$.

Let's review the textbook definition of classical simultaneous games.

Definition 5.  

1. Let $\{X_i{\}}_{i\in I}$ be a family of sets indexed by a set $I$ of players.
   
   A classical simultaneous game over $\{X_i{\}}_{i\in I}$ is a function $g:{\prod }_{i\in I}{X}_i\to R^I$.
    
2. The option-profiles of $g$ are the elements of the product $X=\prod _{i\in I}{X}_i$.
   
3. The $i$-th deviation map $U_i:X\to (X_i\to X)$ is given by $U_i\left(x\right)(\alpha {)}_j=\{\begin{array}{cc}\alpha & \text{if}i=j\\ x_j& \text{otherwise}\end{array}$ for all $x\in X$, $i\in I$, $\alpha \in X_i$, and $j\in I$.
   
   In the finite case, where $I=\{1,\dots ,n\}$, then $U_i(x_1,\dots ,x_n):X_i\to X,\alpha \mapsto (x_1,\dots ,x_{i-1},\alpha ,x_i,\dots ,x_n)$.
   
   The deviation map $U_i:X\to (X_i\to X)$ describes how the $i$th player can change an option-profile $x$ by unilaterally changing their own option. Note that the deviation maps only depend on the option spaces, not the game itself.
    
4. The best-response function of $g$ is the the function $B:X\to P\left(X\right)$ defined by$B:x\mapsto \prod _{i\in I}{\text{argmax}}_{X_i}({\pi }_i\circ g\circ U_i(x\left)\right)$ where $U_i:X\to (X_i\to X)$ is the $i$th deviation map.
    
5. Finally, the nash equilibria of $g$ is the set of option-profiles $x\in X$ such that $x\in B\left(x\right)$.

Note that the fact that the option-profiles are in nash equilibrium follows logically from the assumption each player's choice is utility-maximising for the task they face, plus the assumption that, if the option-profile is $x\in X$, then the $i$th player faces the task ${\pi }_i\circ g\circ U_i\left(x\right):X_i\to R$.

It doesn't require the assumption that the player's have "common knowledge" of each other's rationality, or that the players have any kinds of beliefs whatsoever. That being said, an economist might explain the optimality of the choices by assuming that each player is rational and knows all the relevant facts (i.e. the rules of the game, the rationality of their peers, and the beliefs of their peers). However, an evolutionary biologist might appeal to natural selection to explain optimality. An ML engineer might appeal to SGD, etc.

Higher-Order Nash Equilibrium

By inspection, we can see that at no point does the definition of the classical nash equilibrium appeal to any special properties of $R$ or ${\text{argmax}}_X$. This suggests that we can effortless generalise the definition of nash equilibrium to any family of optimisers, rather than only utility-maximisers.

Definition 6. 

1. Let $\{X_i{\}}_{i\in I}$ be a family of sets indexed by a set $I$ of players. Let $R$ be a set of payoffs, and let $\Psi =\{{\psi }_i{\}}_{i\in I}$ be a family of optimisers with ${\psi }_i\in J^P(X_i,R)$.^[2]
   
   A higher-order simultaneous $\Psi$-game is a function $g:{\prod }_{i\in I}{X}_i\to R$.
    
2. Like before, the option-profiles of $g$ are the elements of the product $X=\prod _{i\in I}{X}_i$.
   
3. Like before, the $i$-th deviation map $U_i:X\to (X_i\to X)$ is given by $U_i\left(x\right)(\alpha {)}_j=\{\begin{array}{cc}\alpha & \text{if}i=j\\ x_j& \text{otherwise}\end{array}$ for all $x\in X$, $i\in I$, $\alpha \in X_i$, and $j\in I$.
    
4. The $\Psi$-best-response function of $g$ is the the function $B:X\to P\left(X\right)$ defined by $B:x\mapsto \prod _{i\in I}{\psi }_i(g\circ U_i(x\left)\right)$ where $U_i:X\to (X_i\to X)$ is the $i$th deviation map
    
5. Like before, the $\Psi$-nash equilibria of $g$ is the set of option-profiles $x\in X$ such that $x\in B\left(x\right)$.

When clear from context, I'll just say game, best-response, and nash equilibrium.

The intuition is this —

• Suppose that $x=(x_1,\dots ,x_n)\in X_1\times \cdots \times X_n$ is the option-profile that's collectively chosen.
• Then player $i\in I$ can achieve a payoff $[g\circ U_i(x\left)\right]\left(\alpha \right)\in R$ by unilaterally changing their option to $\alpha \in X_i$ because $[g\circ U_i(x\left)\right]\left(\alpha \right)=g(x_1,\dots ,x_{i-1},\alpha ,x_{i+1},\dots ,x_n)$.
• So player $i\in I$ is faced with the task $g\circ U_i\left(x\right):X_i\to R$.
• Their choice $x_i\in X_i$ must be ${\psi }_i$-optimal for this task, so $x_i\in {\varphi }_i(g\circ U_i(x\left)\right)\subseteq X_i$.
• If we assume ${\psi }_i$-optimality for every player conjunctively, then $x\in B\left(x\right)$.
• If we know that $x\in B\left(x\right)$, and we know nothing else, then the possible option profiles are $\{x\in X:x\in B(x\left)\right\}\in P\left(X\right)$.

This definition applies even to unusual cases —

• Zero-player games? When $|I|=0$, then game $g$ is just an element of the payoff space $R$. There's only one option-profile, the empty sequence, and it's vacuously nash.
• One-player games? When $|I|=1$, then a game $g:X\to R$ is just a task for the solitary player. An option-profile for the game is specified by an option for the player, and this option-profile is nash whenever the option is optimal for the player.
• When $I$ is countably infinite, or indeed uncountably infinite, then the definition still makes sense and gives the right answer.^[3]

We obtain the classical simultaneous game when $R=R^n$ and ${\psi }_i={\text{argmax}}_X({\pi }_i\circ -)\in J^P(X,R^n)$, and in this case the higher-order nash equilibrium coincides with the classical nash equilibrium.

The higher-order nash equilibrium allows us to answer somewhat esoteric questions —

An argmaxer, a satisficer, and a quantiliser walk into a bar. They can order from menus $A$, $S$, and $Q$ respectively, and the barman will mix their three orders $(a,s,q)\in A\times S\times Q$ into a single cocktail $f(a,s,q)\in C$. The argmaxer's preferences over the cocktails are given by $v_a:C\to R$, and likewise $v_s:C\to R$ for the satisfier and $v_q:C\to R$ for the quantiliser.

What cocktail might they be served?

Well, the possible cocktails are $f(a,s,q)\in C$ such that $a\in {\text{argmax}}_A({\lambda }^{a^{\prime }\in A}.v_a\circ f(a^{\prime },s,q\left)\right)$, $s\in {\text{satisfice}}_{s_0}({\lambda }^{s^{\prime }\in S}.v_s\circ f(a,s^{\prime },q\left)\right)$, and $q\in {\text{quant}}_{ϵ}({\lambda }^{q^{\prime }\in Q}.v_q\circ f(a,s,q^{\prime }\left)\right)$.

Many well-studied variations of the classical nash equilibrium can be seen as the higher-order nash equilibrium between non-classical optimisers. For example, epsilon-equilibrium is precisely the higher-order nash equilibrium between ${\text{argmax-}ϵ\text{-slack}}_X({\pi }_i\circ -)$.^[4]

Optimisation is nash-sum closed

There's a nice upshot of higher-order game theory — the nash equilibria between optimisers is itself an optimiser!

Suppose that two agents are modelled by ${\psi }_A\in J^P(A,R)$ and ${\psi }_B\in J^P(A,R)$. We can combine them into a single function which assigns, to each game $g:A\times B\to R$, the set of $\{{\psi }_A,{\psi }_B\}$-nash equilibria of the game $g$. This function has type-signature $(A\times B\to R)\to P(A\times B)$, so it corresponds to a single optimiser with option space $(A\times B)$ with payoff space $R$.

This gives us a binary operator on optimisers,$\oplus :J^P(A,R)\times J^P(B,R)\to J^P(A\times B,R)$, given by $(a,b)\in ({\psi }_A\oplus {\psi }_B)\left(g\right)⟺a\in {\psi }_A(g\circ U_1(a,b\left)\right),b\in {\psi }_B(g\circ U_2(a,b\left)\right)$.

This operator is both associative and commutative, justifying the name "nash sum".

For example, the nash sum $({\text{argmax}}_A\oplus {\text{argmax}}_B):(A\times B\to R\times R)\to P(A\times B)$ will calculate the classical nash equilibria of a generic payoff matrix $g:A\times B\to R^n$.

The operator can be extended to any family of optimisers, i.e. $\underset{i\in I}{⨁}:\prod _{i\in I}{J}^P(X_i,R)\to J^P(\prod _{i\in I}{X}_i,R)$ where $x\in \left(\underset{i\in I}{⨁}{\psi }_i\right)\left(g\right)⟺x\in \prod _{i=1}^n{\psi }_i(g\circ U_i(x\left)\right)$.

Terminology:

• ${⨁}_{i\in I}{\psi }_i$ is a single optimiser, while $\{{\psi }_i{\}}_{i\in I}$ is a family of optimisers.
• ${⨁}_{i\in I}{\psi }_i$ is the nash-sum of $\{{\psi }_i{\}}_{i\in I}$, and $\{{\psi }_i{\}}_{i\in I}$ are subagents of ${⨁}_{i\in I}{\psi }_i$.
• An option for ${⨁}_{i\in I}{\psi }_i$ is an option-profile for $\{{\psi }_i{\}}_{i\in I}$.
• A task for the optimiser ${⨁}_{i\in I}{\psi }_i$ is a game for the family of optimisers $\{{\psi }_i{\}}_{i\in I}$.
• An option $x\in {\prod }_{i\in I}{X}_i$ is optimal for ${⨁}_{i\in I}{\psi }_i$ in the task $g:{\prod }_{i\in I}{X}_i\to R$ if and only if the option-profile $x\in {\prod }_{i\in I}{X}_i$ is a nash equilibrium for $\{{\psi }_i{\}}_{i\in I}$ in the game $g:{\prod }_{i\in I}{X}_i\to R$.

Totality is not nash-sum closed

Just because two optimisers ${\psi }_a\in J^P(A,R)$ and ${\psi }_b\in J^P(B,R)$ are total does not imply that their nash sum $({\psi }_a\oplus {\psi }_b)\in J^P(A\times B,R)$ is total.^[5] In other words, we may have a model of an agent which works perfectly in all solitary situations, but breaks down when many such agents interact in a shared environment.

The textbook example is Matching Pennies. Each player must choose either heads or tails — player 1 wins if the coins match and player 2 wins if they don't. If both players are utility-maximisers then there's no nash equilibria, even though utility-maximisers are total optimisers.

Formally speaking, $A=\{H,T\}$, $B=\{H,T\}$, and $u:A\times B\to R\times R,(x,y)\mapsto ({\delta }_{xy},1-{\delta }_{xy})$.^[6] Then $\left({\text{argmax}}_A\right({\pi }_1\circ -)\oplus {\text{argmax}}_B({\pi }_2\circ -\left)\right)\left(u\right)=\varnothing$, although ${\text{argmax}}_A({\pi }_1\circ -)\left(w\right)\ne \varnothing$ and ${\text{argmax}}_B({\pi }_1\circ -)\left(w\right)\ne \varnothing$ for all $w:\{H,T\}\to R\times R$.

This shows why we couldn't have defined optimisers as functions with type-signature $(X\to R)\to P^{\ast }\left(X\right)$ where $P^{\ast }\left(X\right)$ denotes the set of non-empty subsets of $X$. In order to ensure nash-sum closure of optimisers, we needed to include non-total optimisers.

Utility-maximisation is not nash-sum closed.

It is well-known that the nash sum of $n$ utility-maximisers isn't a utility-maximiser for any $n\ge 2$. For many games, none of the nash equilibrium are even pareto-optimal!

The textbook case is the prisoner's dilemma. Let $B=\{C,D\}$ and consider two players${\text{argmax}}_B({\pi }_1\circ -)$ and ${\text{argmax}}_B({\pi }_2\circ -)$ which respectively value the first and second coordinate of the payoff. Their nash sum is the optimiser in $J^P(B^2,R^2)$ which calculates the classical nash equilibria of 2-by-2 payoff matrices $u:B^2\to R^2$. In particular, when $u:B^2\to R^2$ is the payoff matrix defined below, then $\left({\text{argmax}}_B\right({\pi }_1\circ -)\oplus {\text{argmax}}_B({\pi }_2\circ -\left)\right)\left(u\right)=\left\{\right(D,D\left)\right\}$. However, both players prefer the payoff $u(C,C)=(2,2)$ to $u(D,D)=(1,1)$.

Consequentialism is not nash-sum closed.

In fact, the nash sum of two utility-maximisers isn't even consequentialist!^[7]

Suppose Angelica is babysitting her cousin Tommy, and each child has been given a cookie by their grandmother. If Angelica is awake while Tommy sleeps, then she can steal both cookies for herself. If Angelica sleeps while Tommy is awake then he'll cause mayhem getting both children in trouble. If Angelica and Tommy both stay sleep, or if they both wake up, then they'll each keep one cookie.

The payoffs $u:\{W_A,S_A\}\times \{W_T,S_T\}\to R\times R$ are summarised in the matrix below.

When Angelica and Tommy are both awake, this is nash — she doesn't want to sleep because then he'll cause mayhem, and he doesn't want to sleep because then she'll steal his cookie. In contrast, when they're both sleeping, this isn't nash — she can wake up and steal his cookie. But the same outcome will result whether Angelica and Tommy are both sleeping or both awake. So the nash sum of Angelica and Tommy isn't a consequentialist optimiser!^[8] 

This shows why we couldn't have defined optimisers as functions with type-signature $(X\to R)\to P\left(R\right)$, which bakes-in the assumption that all optimisers are consequentialist. In order to ensure the nash-sum closure of optimisers, we needed to include nonconsequentialist optimisers.^[9]

Beyond CDT?

Warning: This section is a speculative extension of the existing literature, it's merely a proof-of-concept of how higher-order game theory would extend to non-CDT agents.

Let $u:\{C,D\}\times \{C,D\}\to R\times R$ be the game defined by the payoff matrix below. We may readily check that $({\text{argmax}}_{\{C,D\}}\oplus {\text{argmax}}_{\{C,D\}})\left(u\right)=\left\{\right(1,1\left)\right\}$, which corresponds to the well-known result that two utility-maximisers will defect in the prisoners' dilemma.

Now, some decision theorists think that it's not generally true that two rational utility-maximisers will defect in the prisoners' dilemma — e.g. if they're both perfect replicas of one another then they'll both cooperate! The reasoning is this — cooperation from the first prisoner would count as evidence (to the first player) of high utility, and defection from the first prisoner would count as evidence of low utility. This corresponds to the observation that ${\pi }_1\circ u(C,C)>{\pi }_1\circ u(D,D)$. So the first player should cooperate. Ditto the second player.

So what went wrong?

It all comes down to an ambiguity with the task $u:X\to R$. What relationship is the task $u$ supposed to capture exactly? The formalism itself is agnostic.

• According to causal decision theory, the task $u:X\to R$ is supposed to track which payoff $u\left(x\right)\in R$ is caused by the agent choosing $x\in X$.
• According to evidential decision theory, the task $u:X\to R$ is supposed to track which payoff $u\left(x\right)\in R$ is evidenced by the agent choosing $x\in X$.

So let's explain our answer $({\text{argmax}}_{\{C,D\}}\oplus {\text{argmax}}_{\{C,D\}})\left(u\right)=\left\{\right(1,1\left)\right\}$, which agrees with CDT. We can trace the problem to the deviation maps $U_i:X\to (X_i\to X)$ in the definition of $\oplus$.

The $i$-th deviation map $U_i:X\to (X_i\to X)$ is given by $U_i\left(x\right)(\alpha {)}_j=\{\begin{array}{cc}\alpha & \text{if}i=j\\ x_j& \text{otherwise}\end{array}$ for all $x\in X$, $i\in I$, $\alpha \in X_i$, and $j\in I$.

The map $U_i\left(x\right)$ is the causal dependence of the overall option-profile on the $i$th player's choice. Hence, when the game $u:X\to R$ is the causal dependence of the overall payoff on the option-profile, it follows that the task $u\circ U_i\left(x\right):X_i\to R$ is the causal dependence of the payoff on the $i$th player's choice. This is why we got the CDT answer — causation composed with causation is causation.

According to EDT, however, the task is supposed to capture the evidential dependency of the player's choice on the payoff. And when the game $u:X\to R$ is the evidential dependence of the overall payoff on the option-profile, it isn't true that that the task $u\circ U_i\left(x\right):X_i\to R$ is the evidential dependence of the payoff on the $i$-th player's choice. Evidence composed with causation isn't evidence.

To achieve the EDT answer, we need is a function $U_i^E\left(x\right):X_i\to X$ which captures the evidential dependence of the option-profile on the $i$th player's choice. In the case that all the players are replicas, $U_i^E\left(x\right):X_i\to X,\alpha \mapsto (\alpha ,\dots ,\alpha )$. Replacing $U_i:X\to (X_i\to X)$ in the definition of $⨁$ with $U_i^E:X\to (X_i\to X)$ then we'd get a different binary operator ${⨁}^E$ which yields the EDT prediction, namely that $\left({\text{argmax}}_A{\oplus }^E{\text{argmax}}_B\right)\left(u\right)=\left\{\right(C,C\left)\right\}$.

It seems that we have another free parameter in our model! We can the replace the family of deviation maps $U_i:X\to (X_i\to X)$ with an arbitrary family of non-casual deviation maps $D_i:X\to (X_i\to X)$, and thereby encode information about each player's idiosyncratic decision theory. If the $i$th player is CDT then $D_i=U_i$ and if the $i$th player is EDT then $D_i=U_i^E$. 

Suppose we have a payoff space $R$, a family of sets $\{X_i{\}}_{i\in I}$ with $X={\prod }_{i\in I}{X}_i$, a family of optimisers $\{{\psi }_i{\}}_{i\in I}$ where ${\psi }_i\in J^P(X_i,R)$, and a family of decision theories $\{D_i{\}}_{i\in I}$ where $D_i:X\to (X_i\to X)$, and a game is a map $g:X\to R$. Then we can define the non-CDT best-response function of $g$ to be the function $B:X\to P\left(X\right),x\mapsto \prod _{i=1}^n{\psi }_i(u\circ D_i(x\left)\right)$, and the non-CDT nash equilibria of $g$ to be the set $\{x\in X:x\in B(x\left)\right\}$. Or something like that.