An experimental study on the effect of ambiguity in a coordination game

Abstract

We report an experimental test of the influence of ambiguity on behaviour in a coordination game. We study the behaviour of subjects in the presence of ambiguity and attempt to determine whether they prefer to choose an ambiguity-safe option. We find that this strategy, which is not played in either Nash equilibrium or iterated dominance equilibrium, is indeed chosen quite frequently. This provides evidence that ambiguity-aversion influences behaviour in games. While the behaviour of the Row Player is consistent with randomising between her strategies, the Column Player shows a marked preference for avoiding ambiguity and choosing his ambiguity-safe strategy.

Appendix

Proof of Proposition 3.1:

Part (1) $\ 0\leqslant x<75$: By inspection, $(T,M)$ and $(B,L)$ are pure strategy Nash equilibria. Let the probability of Player 1 choosing $T$ (resp. $B)$ be $q$ (resp. $(1-q))$, and the probability of Player 2 choosing $L$ (resp. $M)$ be $p$ (resp. $(1-p))$. For $x$ in this range, $R$ is dominated by $\frac{1}{4}\cdot L+\frac{3}{4}\cdot M$, which yields an expected payoff of $75$. Thus, $R$ cannot be played in Nash equilibrium. There are 3 equilibria: $(T,M), (B,L)$ and $\left( \frac{3}{4}\cdot T+\frac{1}{4} \cdot B,\frac{1}{4}\cdot L+\frac{3}{4}\cdot M\right)$.

Part (2) $75<x\leqslant 100:$ For $x$ in this range, $(T,M)$ and $(B,L)$ remain pure strategy Nash equilibria. Player 2 is indifferent between $M$ and $R$ when $100q=x$ or $q=x/100$. Player 1 is indifferent between $T$ and $B$ when $300p\,+\,50(1-p)=55(1-p)$, or $p=1/61$. There are 3 equilibria: $(T,M), (B,L)$ and $\left( \frac{x}{100}\cdot T+\frac{ 100-x}{100}\cdot B,\frac{1}{61}\cdot M+\frac{60}{61}\cdot R\right)$.²⁵

Part (3) $100<x<300:$ For this range, $M$ is dominated for Player 2 by $R$. Once $M$ is eliminated, Player 1 will never play $T$, which is now a dominated strategy. He thus plays $B$. The best response for Player 2 is to play $L$. In this case, there is a unique Nash equilibrium: $(B,L)$, which is also an iterated dominance equilibrium. $\square$

Proof of Proposition 3.2:

Part 1. $0\leqslant x\leqslant \left( 1-\delta _{2}\right) 75$: In this range, there are two EUA in pure strategies and one in mixed strategies. In the pure equilibria, the supports of the equilibrium beliefs are given by $(T,M)$ and $(B,L)$. Consider the first of these. Define a neo-additive capacity, $\nu _{1}$, by $\nu _{1}=\left( 1-\delta _{1}\right) \pi _{M}\left( A\right)$, where $\pi _{M}$ is the additive probability on $S_{2}$ defined by $\pi _{M}\left( A\right) =1$ if $M\in A$, $\pi _{M}\left( A\right) =0$ otherwise. Similarly, define Player $2$’s beliefs $\nu _{2}$ by $\nu _{2}=\left( 1-\delta _{2}\right) \pi _{T}\left( A\right) .$ By definition, $\hbox {supp}\nu _{1}=M$ and $\hbox {supp}\nu _{2}=T.$ With these beliefs the (Choquet) expected payoffs of Players 1 and $2$ are, respectively, $V_{1}\left( T\right) =\left( 1-\delta _{1}\right) 300,V_{1}\left( B\right) =0,V_{2}\left( L\right) =0,V_{2}\left( M\right) =\left( 1-\delta _{2}\right) 100,V_{2}\left( R\right) =75.$ It can be been that the support of both players’ beliefs consists only of best responses. Denote this equilibrium by $\left\langle T,M\right\rangle$ . By similar reasoning, we may show that there exists a pure equilibrium where $\hbox { supp}\nu _{1}=L$ and $\hbox {supp}\nu _{2}=B,$ which we denote by $\left\langle B,L\right\rangle$ .

Now consider the mixed equilibria. Denote the equilibrium beliefs of Players $1$ and $2$, respectively, by $\tilde{\nu }_{1}=\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1}$ and $\tilde{\nu }_{2}=\left( 1-\tilde{\delta } _{2}\right) \tilde{\pi }_{2}$. Player $2$’s Choquet expected payoffs are given by, $V_{2}\left( L\right) =300\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{ \delta }_{2}\right) \tilde{\pi }_{2}\left( T\right)$ and $V_{2}\left( R\right) =x$. If $V_{2}\left( L\right) <x\leqslant (1-\tilde{\delta }_{2})75$, then $\tilde{\pi }_{2}\left( B\right) <\frac{1}{4}$, which implies $\tilde{\pi }_{2}\left( T\right) >\frac{3}{4}$. Hence $V_{2}\left( M\right) =100\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) >(1-\tilde{\delta } _{2})75\geqslant x$. Thus, $R$ cannot be a best response for Player 2 and hence $\tilde{\pi }_{1}\left( R\right) =0$. Consequently, in any mixed equilibrium, $2$ ’s strategies are $L$ and $M$.

In a mixed equilibrium, Player $2$ must be indifferent between $L$ and $M$ and hence

$$\begin{aligned} V_{2}\left( L\right)&= V_{2}\left( M\right) \Leftrightarrow 300\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) =100\left( 1-\tilde{ \delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) \\&\Leftrightarrow \tilde{\pi }_{2}\left( T\right) =\frac{3}{4}. \end{aligned}$$

In this equilibrium, $V_{2}\left( L\right) =V_{2}\left( M\right) =75\left( 1- \tilde{\delta }_{2}\right) .$ Similarly, we may show that for Player $1$ to be indifferent between $T$ and $B$, we must have $\tilde{\pi }_{1}\left( L\right) =\frac{1}{4}$ and $\tilde{\pi }_{1}\left( M\right) =\frac{3}{4}$.

Thus, in the mixed equilibrium, $\tilde{\nu }_{1}=\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1}$ with $\tilde{\pi }_{1}\left( L\right) =\frac{1}{4 }$ and $\tilde{\pi }_{1}\left( M\right) =\frac{3}{4}$ and $\hbox {supp} \tilde{\nu }_{1}=\left\{ L,M\right\}$, while $\tilde{\nu }_{2}=\left( 1-\tilde{ \delta }_{2}\right) \tilde{\pi }_{2}$ with $\tilde{\pi }_{2}\left( T\right) = \frac{3}{4}$ and $\tilde{\pi }_{2}\left( B\right) =\frac{1}{4}$, with support $\left\{ T,B\right\}$ . In this equilibrium, $V_{2}\left( L\right) =V_{2}\left( M\right) =75\left( 1-\tilde{\delta }_{2}\right)$ . We shall denote this equilibrium by $\left\langle \frac{3}{4}\cdot T+\frac{1}{4}\cdot B,\frac{1}{4}\cdot L+\frac{3}{4}\cdot M\right\rangle$ .

Part 2. $(1-\delta _{2})75<x<(1-\delta _{2})100:$       In this range, there are two EUA in pure strategies: $(T,M)$ and $(B,L)$. The reasoning is similar to that used in Part 1 above.

In addition, there is a mixed strategy equilibrium. Denote the equilibrium beliefs of Players $1$ and $2$, respectively, by $\tilde{\nu }_{1}=\left( 1- \tilde{\delta }_{1}\right) \tilde{\pi }_{1}$ and $\tilde{\nu }_{2}=\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}$. Player $2$’s Choquet expected payoffs are given by, $V_{2}\left( L\right) =300\left( 1-\tilde{\delta } _{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right)$ and $V_{2}\left( R\right) =x$.

First note that there is no longer equilibrium where Player $2$ mixes between $L$ and $M$. Such an equilibrium would require Player $1$ to be indifferent between $T$ and $B$. As in part 1, this would imply $\tilde{\pi }_{2}\left( T\right) =\frac{3}{4}$ and $\tilde{\pi } _{2}\left( B\right) =\frac{1}{4}$. Hence, $V_{2}\left( M\right) =75\left( 1- \tilde{\delta }_{2}\right) <V_{2}\left( R\right)$, which implies that $M$ is not a best response. A similar argument demonstrates that there is no equilibrium where player $2$ randomises between all three of his strategies. We can also eliminate the possibility that Player 2 mixes between $L$ and $R$, as follows. If $2$ never plays $M$, then $B$ is a best response for Player $1$. However, $B$ is not compatible with Player $2$ being indifferent between $L$ and $R$.

The remaining possibility is that $2$ mixes between $M$ and $R$. Player 2 is indifferent between $M$ and $R$ when

$$\begin{aligned} V_{2}\left( M\right) =V_{2}\left( R\right) \Leftrightarrow 100\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) =x\Leftrightarrow \tilde{\pi }_{2}\left( T\right) =\frac{x}{\left( 1-\tilde{\delta }_{2}\right) 100}. \end{aligned}$$

Similarly, Player 1’s Choquet expected payoff is given by $V_{1}(T){\,=\,}300\left( \!1-\tilde{\delta }_{1}\!\right) \tilde{\pi }_{1}(M) +50\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\left( R\right)$ and $V_{1}\left( B\right) =55(1-\tilde{\delta }_{1})\tilde{\pi } _{1}\left( R\right)$ . Player 1 is indifferent between $T$ and $B$ when

$$\begin{aligned} V_{1}\left( T\right) =V_{1}\left( B\right) \Leftrightarrow \tilde{\pi } _{1}\left( M\right) =\frac{1}{61}. \end{aligned}$$

Thus, in the mixed equilibrium, $\tilde{\nu }_{1}=\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1},$ with $\tilde{\pi }_{1}\left( M\right) =\frac{1}{ 61}$ and $\tilde{\pi }_{1}\left( R\right) =\frac{60}{61}$ and $\hbox {supp} \tilde{\nu }_{1}=\left\{ M,R\right\}$, while $\tilde{\nu }_{2}=\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }^{2}$ with $\tilde{\pi }_{2}\left( T\right) =\frac{x}{\left( 1-\tilde{\delta }_{2}\right) 100}$ and $\tilde{\pi } _{2}\left( B\right) =\frac{(1-\tilde{\delta }_{2})100\text { }-\text { }x}{(1- \tilde{\delta }_{2})100},$ with support $\left\{ T,B\right\} .$ In this equilibrium $V_{2}\left( M\right) =V_{2}\left( R\right) =x$. The mixed strategy equilibrium is $\left\langle \frac{x}{(1-\tilde{\delta }_{2})100} \cdot T+\frac{(1-\tilde{\delta }_{2})100\text { }-\text { }x}{(1-\tilde{\delta } _{2})100}\cdot B,\frac{1}{61}\cdot M+\frac{60}{61}\cdot R\right\rangle$.

Part 3. $\ (1-\delta _{2})100<x<(1-\delta _{2})300:$ Denote the equilibrium beliefs of Players $1$ and $2$, respectively, by $\tilde{\nu }_{1}=\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}$ and $\tilde{\nu }_{2}=\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}.$ Player $2$’s Choquet expected payoffs are given by $V_{2}\left( L\right) =300\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi } _{2}\left( T\right)$ and $V_{2}\left( R\right) =x$. For $x$ in this range, $V_{2}\left( R\right) >V_{2}\left( M\right)$ for any beliefs of Player 2 and hence $\tilde{\pi }_{1}\left( M\right) =0$. Player $1$’s Choquet expected payoffs are given by $V_{1}\left( T\right) =50\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1}\left( R\right)$ and $V_{1}\left( B\right) =100\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\left( L\right) +55\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\left( R\right)$ . Strategy $B$ yields a higher Choquet expected payoff than $T$ for any beliefs of Player 1, with support contained in $\{L,R\}$. For Player 2, $L$ is the best response to $B$. In this case, there is a unique EUA: $\left\langle B,L\right\rangle$.

Part 4. $x>(1-\delta _{2})300:$ Denote the equilibrium beliefs of Players $1$ and $2$, respectively, by $\tilde{\nu } _{1}=\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}$ and $\tilde{\nu } _{2}=\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}$. Player $2$’s Choquet expected payoffs are given by $V_{2}\left( L\right) =300\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right)$ and $V_{2}\left( R\right) =x$, where $\ x>(1-\tilde{\delta } _{2})300$.

For $x$ in this range, $R$ strictly dominates both $L$ and $M$ for any beliefs of Player 2 and hence $\tilde{\pi }_{1}\left( L\right) =\tilde{\pi } _{1}\left( M\right) =0$. Player $1$’s best response is to play $B$, with $\hbox {supp}\nu _{1}=R$. There is a unique EUA: $\left\langle B,R\right\rangle$ .  $\square$