In this post, I present a generalization of Nash equilibria to non-CDT agents. I will use this formulation to model mutual cooperation in a twin prisoners' dilemma, caused by the belief that the other player is similar to you, and not by mutual prediction. (This post came mostly out of a conversation with Sam Eisenstat, as well as contributions from Tsvi Benson-Tilsen and Jessica Taylor)

This post is sketchy. If someone would like to go through the work of making it more formal/correct, let me know. Also, let me know if this concept already exists.

Let $A_1,\dots ,A_n$ be a finite collection of players. Let $M_i$ be a finite collection of moves available to $A_i$. Let $U_i:{\prod }_i{M}_i\to R$ be a utility function for player $A_i$.

Let ${\Delta }_i$ be the simplex of all probability distributions over $M_i$. Let $V_i$ denote the vector space of functions $v:M_i\to R$, with ${\sum }_{m\in M_i}v\left(m\right)=0$.

A vector given a distribution in $p\in {\Delta }_i$ and a vector $v\in V_i$ we say that $v$ is unblocked from $p$ if there exists an $\epsilon >0$, such that $p+\epsilon v\in {\Delta }_i$. i.e., it is possible to move in the direction of $v$ from $p$, and stay in ${\Delta }_i$.

Given a strategy profile $P=(p_1,\dots ,p_n)$ in ${\prod }_{j\le n}{\Delta }_j$, and a vector $V=(v_1,\dots ,v_n)$ in ${\prod }_{j\le n}{V}_j$, we say that $V$ improves $P$ for $A_i$ if

$$\underset{\epsilon \to 0}{lim}\frac{{\sum }_{m_1\in M_1}\dots {\sum }_{m_n\in M_n}{U}_i(m_1,\dots ,m_n)\left({\prod }_{j\le n}{p}_j\right(m_j)+\epsilon v_j(m_j)-{\prod }_{j\le n}{p}_j(m_j\left)\right)}{\epsilon }>0.$$

We call this limit the utility differential for $A_i$ at $P$ in the direction of $V$.

i.e., $V$ improves $P$ for $A_i$ if $U_i$ is increased when $P$ is moved an infinitesimal amount in the direction of $V$. (not that this is defined even when the vectors are not unblocked from the distributions)

A "counterfactual gradient" for player $A_i$ is a linear function from $V_i$ to ${\prod }_{j\le n}{V}_j$, such that the function into the $V_i$ component is the identity. This represents how much player $A_i$ expects the probabilities of the other players to move when she moves her probabilities.

A "counterfactual system" for $A_i$ is a continuous function which takes is a strategy profile $P$ in ${\prod }_{j\le n}{\Delta }_j$, and outputs a counterfactual gradient. This represents the fact that a player's counterfactual gradient could be different depending on the strategy profile the game ends up in. We fix a counterfactual system $C_i$ for each player.

Claim: There exists a strategy profile $P=(p_1,\dots ,p_n)$ in ${\prod }_{j\le n}{\Delta }_j$ such that for all players $A_i$, if $v_i\in V_i$ is unblocked from $p_i$, then $C_i\left(P\right)\left(v_i\right)$ does not improve $P$.

I will call such a point an entangled equilibrium.

Proof Sketch: (Very sketchy, and I have not verified this. There is probably a better way. It might not be true.)

We will construct a continuous function from ${\prod }_{j\le n}{\Delta }_j$ to itself. We do this by moving $p_i$ by adding the gradient of function from $p_i$ to to $U_i$, assuming all other players probabilities change according to the linear function $C_i\left(P\right)$. If adding this gradient would take the point out of the simplex ${\Delta }_i$, you hit a boundary after moving some proportion $\alpha$ of the gradient. Then, now that you are on the boundary, you use move by $1-\alpha$ times the gradient of the same function restricted to the boundary, and repeat.

This function has a fixed point, by Brouwer. For each player, this fixed point must have 0 gradient, or be on the boundary, pointing outward. If it is on a boundary, it has the same property when restricted to that boundary.

If there was an unblocked direction a player could move that would improve its utility, then the gradient would be nonzero. If the player is on a boundary, the gradient is pointing outward, and there is an unblocked direction that would be an improvement, there will be such a direction that stays on that boundary, and the gradient restricted to that boundary would be nonzero. $\square$

Example 1

Consider a twin prisoners' dilemma game. Two players can either cooperate or defect. They get 0 utility for being exploited, 3 for exploiting, 2 for mutual cooperation, and 1 for mutual defection.

Both players believe that the other is using a decision procedure that is entangled with their own, but they are not completely sure.

Formally, both players think that when they increase their probability by $\epsilon$, the other player increases by $2\epsilon /3$, regardless of where the probabilities start.

Thus $C_1(p,q)\left(v_1\right)=(v_1,2v_1)/3$, where $v_1$ is the vector ($A_1$ cooperates)-($A_1$ defects) in $V_1$, and $v_2$ is the analogous vector for player 2. Similarly, $C_2(p,q)\left(v_2\right)=(2v_2/3,v_2)$.

Since both players always think that increasing their probability of cooperation increases utility, the only equilibrium is when both players cooperate with probability 1.

Example 2

Now let look at a case where the gradients are a function of the distributions. Again, both players believe that the other is using a decision procedure that is entangled with their own, but they are not completely sure.

Formally, player 1 thinks that when they increase (or decrease) their probably of cooperation by $\epsilon$, the other player will increase (or decrease) their probability by $\delta \epsilon$, where $\delta$ is one minus the square of the difference between their two probabilities.

Player 2 on the other hand, thinks the other player will increase (or decrease) their probability by $\delta \epsilon$, where $\delta$ is one minus the absolute value of the difference between their two probabilities.

Thus $C_1(p,q)\left(v_1\right)=(v_1,(1-(p-q{)}^2)\cdot v_2)$, where $v_1$ is the vector ($A_1$ cooperates)-($A_1$ defects) in $V_1$, and $v_2$ is the analogous vector for player 2. Similarly, $C_2(p,q)\left(v_2\right)=\left(\right(1-|p-q|)\cdot v_1,v_2)$.

Thus, if player 1 cooperates with probability $p$, and player 2 cooperates with probability $q$, then player 1 expects to lose gain $\left(2\right(1-(p-q{)}^2)-1)\epsilon$ utility by increasing his probability by $\epsilon$. This function is only 0 if $|p-q|=1/\sqrt{2}$. Thus the only entangled equilibria with mixed strategies for player 1 have $(p-q)=1/\sqrt{2}$. Similarly, the only entangled equilibria with mixed strategies for player 2 have $|p-q|=1/2$. Thus, there are no mixed strategies for both players in equilibria.

The only pure equilibrium is $p=q=1$

If $p=0$, then $\left(2\right(1-(p-q{)}^2)-1)$ must be nonpositive, so $q$ must be at least $1/\sqrt{2}$. None of these points are equilibria for player 2.

If $p=1$, then $\left(2\right(1-(p-q{)}^2)-1)$ must be nonnegative, so $q$ must be at least $1-1/\sqrt{2}$. This is in equilibrium if $q=1/2$.

If $q=0$, then $p$ must be at least $1/2$, This in in equilibrium if $p=1/\sqrt{2}$.

If q=1, then $p$ must be at least $1/2$, and there is no equilibrium.

Thus, there are three entangled equilibria (1,1), (1,1/2), (1/$\sqrt{2}$,0).