When less information is good enough: experiments with global stag hunt games

Abstract

There is mixed evidence on whether subjects coordinate on the efficient equilibrium in experimental stag hunt games under complete information. A design that generates an anomalously high level of coordination, Rankin et al. (Games Econo Behav 32(2):315–337, 2000), varies payoffs each period in repeated play rather than holding them constant. These payoff “perturbations” are eerily similar to those used to motivate the theory of global games, except the theory operates under incomplete information. Interestingly, that equilibrium selection concept is known to coincide with risk dominance, rather than payoff dominance. Thus, in theory, a small change in experimental design should produce a different equilibrium outcome. We examine this prediction in two treatments. In one, we use public signals to match Rankin et al. (2000)’s design; in the other, we use private signals to match the canonical example of global games theory. We find little difference between treatments, in both cases, subject play approaches payoff dominance. Our literature review reveals this result may have more to do with the idiosyncrasies of our complete information framework than the superiority of payoff dominance as an equilibrium selection principle.

Mathematical appendix

1.1 Proof of Proposition 1

According to Harsanyi and Selton, the players’ prior belief $S_{i}$ about player i’s strategy should coincide with the prediction of an outside observer who reasons in the following way about the game:

• (i) Player i believes that his opponents will either all choose A or that they all choose B; he assigns a subjective probability $z_{i}$ to the first event and $1-z_{i}$ to the second.

• (ii) Whatever the value of $z_{i}$, player i will choose a best response to his beliefs.

• (iii) The beliefs (i.e. the $z_{i}$) of different players are independent and they are all uniformly distributed on [0,1].

From (i) and (ii), the outside observer concludes that player i chooses A if $z_{i}>Q$, and that he chooses B if $z_{i}<Q$. Hence, using (iii), the outside observer forecasts player i’s strategy as $S_{i}=(1-Q)A+QB$, with different $S_{i}$ being independent. Harsanyi and Selten assume that the mixed strategy vector $S=(S_{1},\ldots ,S_{n})$ describes the players’ prior expectations in the game. Since S is not a Nash equilibrium, this expectation is not self-fulfilling, and, thus, has to be adapted. In a stag hunt game, the situation is symmetric, either all players will have A as the unique best response against S in which case all-A is the distinguished equilibrium, or they will all have B as the unique best response against S and in which case all-B is the distinguished equilibrium.

We can write player i’s expected payoff associated with A when each player chooses A with probability t as:

$$\begin{aligned} A_{n}^{\pi }(t)=\sum _{k=1}^{n}\frac{\left( n-1\right) !}{(k-1)!(n-k)!}t^{k-1}(1-t)^{n-k}\pi \left( k,n\right) , \end{aligned}$$

where $\pi \left( k,n\right)$ is the payoff of selecting A when k players including player i select A from a total of n players.

If the players’ prior S is $(1-Q)A+QB$, then the expected payoff associated with A is $A_{n}^{\pi }(1-Q)$. Each player’s best response against S is A if $A_{n}^{\pi }(1-Q)>Q$ and B if $A_{n}^{\pi }(1-Q)<Q$. When $\pi \left( k,n\right) =\frac{k-1}{n-1}$, $Q=A_{n}^{\pi }(1-Q)$ when $Q=0.5$, we can conclude that A risk dominates B when $Q<0.5$ and B risk dominates A when $Q>0.5$.

1.2 Proof of Proposition 2

Consider a 2X2 stag hunt game with incomplete information as shown in Table 1 where q is uniform on some interval [a, b] where $a<0$ and $b>1$. Each player receives a signal $Q_{i}=Q+\epsilon _{i}$ that provides an unbiased estimate of Q. The $\epsilon _{i}$ is uniformly distributed within $[-E,E]$ where $E \le -\frac{a}{2}$ and $E\le \frac{b-1}{2}$. The signals $Q_{1}$ and $Q_{2}$ are independent. After observing the signals, the players choose actions simultaneously and get payoffs corresponding to the game in Table 1 with the actual value of Q. It is understood that the structure of the class of games and the joint distribution of Q, $Q_{1}$ and $Q_{2}$ are common knowledge.

It is easily seen that player i’s posterior of q will be uniform on $[Q_{i}-E,Q_{i}+E]$ if he observes $Q_{i}\in [a+E,b-E]$, so his conditionally expected payoff from choosing B will simply be $Q_{i}$. Moreover, for $Q_{i}\in [a+E,b-E]$, the conditional distribution of the opponent’s observation $Q_{j}$ will be symmetric around $Q_{i}$ and have support $[Q_{i}-2E,Q_{i}+2E]$. Hence, the probability that $Q_{j}<Q_{i}$ is equal to the probability that $Q_{j}>Q_{i}$, which is 0.5.

Now suppose player i observes $Q_{i}<0$, his conditionally expected payoff from choosing B is negative and smaller than the minimum payoff from choosing A, which is 0. Hence A is conditionally strictly dominate B for player i when he observes $Q_{i}<0$. It should be clear that iterated dominance arguments allow us to get further. For instance, if player j is restricted to playing A when observing $Q_{j}<0$, then player i, observing $Q_{i}<0$, must assign at least probability of 0.5 that player j would select A. Consequently, player i’s conditionally expected payoff from choosing A will be at least 0.5, so choosing B (which yields 0) can be excluded by iterated dominance for $Q_{i}=0$.

Let $Q_{i}^{*}$ be the smallest observation for which A cannot be established by iterated dominance. By symmetry, obviously, $Q_{1}^{*}=Q_{2}^{*}=Q^{*}$. Iterated dominance requires player i to play A for any $Q_{i}<Q^{*}$, so if player j observes $Q^{*}$ he will assign at least probability 0.5 to player i’s choosing A and, thus, player j’s expected payoff from choosing A will be at least 0.5. Since player j’s expected payoff from choosing B equals $Q^{*}$, we must have $Q^{*}\ge 0.5$, for otherwise iterated dominance would require player j to play A when he observes $Q^{*}$.

We can proceed in the same way for large values of Q. B is dominant for each player if $Q_{i}>1$ because the expected payoff from choosing B is larger than the maximum payoff of choosing A. Let $Q^{**}$ be the largest value for which B cannot be established by iterated dominance. If player j observes $Q_{j}=Q^{**}$, his expected payoff from choosing A will be at most 0.5 given that player i conforms to iterated dominance. Since $Q^{**}$ equals player j’s expected payoff from choosing B, we can conclude that $Q^{**}\le 0.5$. Combine these two findings with the fact that $Q^{*}\le Q^{**}$, we get $Q^{*}=Q^{**}=0.5$.

When $n>2$, player i’s payoff is the average payoff from the matches with the other $n-1$ players or $\pi (k,n)=\frac{k-1}{n-1}$. Since the game is symmetric, each player has the same $Q^{*}$. The probability that for each player $j\ne i$, $Q_{j}<Q_{i}$ is equal to the probability that $Q_{j}>Q_{i}$, which is 0.5. We can use the same argument as the case where $n=2$ to conclude that A strictly dominates B at high values of $Q_{i}$ and B strictly dominates A at low values of $Q_{i}$. We need to find a critical value of $Q^{*}$ in which the expected payoff from selecting A and B are the same when player i observe $Q^{*}$. When subject i observes $Q^{*}$, his expected payoff is $\sum _{k=1}^{n}\frac{\pi \left( k,n\right) \times \frac{\left( n-1\right) !}{(k-1)!(n-k)!}}{2^{(n-1)}}$. This expression is equal to 0.5 which is the same to the expected payoff in a standard 2X2 game. That is $Q^{*}$ does not change and equals to 0.5. Using the mean matching protocol preserves the expected payoff which results in the same $Q^{*}$ of 0.5.