Entropy and Negentropy: Applications in Game Theory

Abstract

The concept of entropy has been applied to such different fields as thermodynamics, cosmology, biology, chemistry, information theory and economics. An interesting application of entropy in the latter field is the existence of a complete ordering of information structures represented by the decrease in entropy, computed à la Shannon, of the agent’s beliefs. In this paper we will apply this entropy ordering to information structures used in experiments assessing the role of communication in coordination games.

Appendix

In this section we will find some of the equilibria of the general game shown in Table 1.

1.1 Warm Glow Payoff

The warm glow δ that altruists players add to the payoffs shown in Table 1 when playing the cooperative strategy (3) makes this strategy neither dominated nor dominant for altruist players and strategy (3) is the best response to both (3) and (1) for this type of players. The latter condition is met when, respectively

$$\displaystyle\begin{array}{rcl} b+\delta & \geq & a{}\end{array}$$

(5)

$$\displaystyle\begin{array}{rcl} f+\delta & \geq & d{}\end{array}$$

(6)

while the former is met whenever some of these conditions (7), (8) or (9) hold but not all at the same time (if it were the case, (3) would be a dominant strategy).

$$\displaystyle\begin{array}{rcl} f+\delta & \geq & d{}\end{array}$$

(7)

$$\displaystyle\begin{array}{rcl} f+\delta & \geq & c{}\end{array}$$

(8)

$$\displaystyle\begin{array}{rcl} b+\delta & \geq & a{}\end{array}$$

(9)

Since if inequality (8) holds, then inequality (7) will also hold and since inequality (9) is required for (3) to be the best response to (3) and inequality (7) is required for (3) to be the best response to (1), we choose inequality (8) to be false. Thus, we have the condition that c − f ≥ δ ≥ a − b.

1.2 One-Way Communication

Proposition 1

If $\rho >\max \left (\frac{c-d} {a-d}, \frac{c-f-\delta } {b-f} \right )$ , there is a nonrevealing equilibrium in which all players announce (3), egoists play (1) and altruists play (3). Any other announcement leads to the play of (2) for both types of player.

Proof

For egoist players, the strategy of announcing (3) and playing (1) should result in a higher payoff than announcing and thus playing (2). Then, a ρ + d(1 −ρ) > c or equivalently $\rho > \frac{c-d} {a-d}$. For altruist players, announcing and playing (3) should give a higher payoff than announcing and playing (2). Thus, ρ(b +δ) + (1 −ρ)( f +δ) > c or equivalently $\rho > \frac{c-f-\delta } {b-f}$.

Proposition 2

If $\frac{c-d} {a-d} >\rho > \frac{c-f-\delta } {b-f}$ , there exists a totally revealing equilibrium in which altruists announce and play (3) and egoists announce and play (2). Egoists and altruists play (2) in response to an announcement of (2). When (3) is announced, egoists play (1) and altruists play (3).

Proof

As shown above, when $\frac{c-d} {a-d} \geq \rho$ egoists prefer to announce and play (2). When (3) is announced, egoists will play (1) as it is the best response to an altruist announcing and playing (3). Furthermore, when (2) is announced, playing (2) is the best response. As $\rho > \frac{c-f-\delta } {b-f}$ announcing and playing (3) for altruists dominates announcing and playing (2). For altruists, the best response to an altruist announcing (3) is playing (3) and to an egoists announcing (2) is playing (2).

Proposition 3

If $\frac{c-f-\delta } {b-f} \geq \rho$ , there is a nonrevealing equilibrium in which all players announce and play (2).

Proof

As shown above, when $\frac{c-f-\delta } {b-f} \geq \rho$ altruists will prefer to announce and play (2) rather than announcing and playing (3). As egoists cannot induce altruists to play (3) due to the low proportion of altruist players, they also announce and play (2). This is true even if $\rho > \frac{c-d} {a-d}$, as altruists will never play (3).

1.3 Two-Way Communication

Proposition 4

If $\rho >\max \left (\frac{c-d} {a-d}, \frac{c-f-\delta } {b-f} \right )$ , there is a nonrevealing equilibrium in which all players announce (3), egoists play (1) and altruists play (3). Any other pair of announcements will lead to the play of (2).

Proof

As in one-way communication, the payoff for egoists players when announcing (3) is a ρ + d(1 −ρ) which is greater than c if $\rho > \frac{c-d} {a-d}$. The same reasoning applies for altruists, being their payoff when they announce (3) ρ(b +δ) + (1 −ρ)( f +δ) greater than the payoff of announcing (2) when $\rho > \frac{c-f-\delta } {b-f}$.

Proposition 5

If $\frac{c-d} {a-d} >\rho$ , there exists an equilibrium in which all altruists announce (3) and egoists announce (3) with probability $\epsilon = \frac{\rho } {1-\rho }\frac{a-c} {c-d}$ and (2) with probability 1 −ε. When both announcements are (3) altruists will play (3) and egoists will play (1). All other pair of announcements lead to the play of (2).

Proof

For the egoists to be indifferent with respect to the announcement of (2) and (3) we have

$$\displaystyle\begin{array}{rcl} \rho a + (1-\rho )\left [\epsilon d + (1-\epsilon )c\right ] = c& &{}\end{array}$$

(10)

Right side of Eq. (10) is the payoff of announcing (2). If the egoist player announces (3), receives an announcement of (3) and plays (1), he will win a with probability ρ (he was confronted to an altruist) or d with a probability (1 −ρ)ε (the opponent announcing (3) was an egoist). If he receives an announcement of (2), he is confronted with an egoist, plays (2) and gains c. The probability of this event is (1 −ρ)(1 −ε). In order to be indifferent, we have that $\epsilon = \frac{\rho } {1-\rho }\frac{a-c} {c-d}$.

For altruists, announcing (3) is better than announcing (2) if, following the same reasoning as above, we have

$$\displaystyle\begin{array}{rcl} \rho (b+\delta ) + (1-\rho )\left [\epsilon (\,f+\delta ) + (1-\epsilon )c\right ] \geq c& &{}\end{array}$$

(11)

Substituting the value of ε in Eq. (11) we obtain, after some algebra

$$\displaystyle\begin{array}{rcl} \rho \left [(b +\delta -c)(c - d) + (\,f +\delta -c)(a - c)\right ] \geq 0& &{}\end{array}$$

(12)

The first term inside brackets is positive while the second is negative. Indeed, ( f +δ − c) < 0. Thus, we can rewrite inequality (12) as (b +δ − c)(c − d) ≥ (c −δ − f)(a − c) where all of its terms are positive. Since b +δ − c ≥ a − c we just have that c − d ≥ c − ( f +δ). As d ≤ f +δ, which is required for the warm glow conditions, we have that inequality (12) holds for any value of its parameters.

Proposition 6

If $\frac{c-f-\delta } {b-f} >\rho$ , there exists an equilibrium in which all altruists announce (3) and egoists announce (3) with probability $\epsilon = \frac{\rho } {1-\rho }\frac{a-c} {c-d}$ and (2) with probability 1 −ε. When both announcements are (3) altruists will play (3) and egoists will play (1). All other pair of announcements lead to the play of (2). In addition to this, there exists another equilibrium in which altruists and egoists both announce and play (2).

Proof

Since $\frac{c-f-\delta } {b-f} >\rho$, altruists have no incentive to announce and play (3) unless egoists reveal themselves announcing (2) with probability 1 −ε. For egoists being indifferent between announcing (2) and (3) we have

$$\displaystyle\begin{array}{rcl} \rho \left [\epsilon a + (1-\epsilon )c\right ] + (1-\rho )\left [\epsilon ^{2}d + 2\epsilon (1-\epsilon )c + (1-\epsilon )^{2}c\right ] = c& &{}\end{array}$$

(13)

The first term in Eq. (13) is the payoff when facing an altruist that announces (3). With probability ε the egoist announces (3), receives (3) and plays (1) and with probability 1 −ε announces (2), receives (3) and plays (2). The second term is the payoff when facing an egoist. With probability ε ² both announce (3) and thus play (1). The other possible announcements lead to the play of (2). After some algebra it can be shown that either ε = 0 or $\epsilon = \frac{\rho } {1-\rho }\frac{a-c} {c-d}$. For ε = 0, egoists always announce and play (2) and thus altruists always announce and play (2).

When $\epsilon = \frac{\rho } {1-\rho }\frac{a-c} {c-d}$, altruist’s payoff should be greater or equal than c

$$\displaystyle\begin{array}{rcl} \rho (b+\delta ) + (1-\rho )\left [\epsilon (\,f+\delta ) + (1-\epsilon )c\right ] \geq c& &{}\end{array}$$

(14)

Since Eqs. (11) and (14) are similar, we can apply the same analysis obtaining the conclusion that inequality (14) identically holds.