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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.

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Notes

  1. See for instance Colman and Pulford (2007), Eichberger et al. (2008), Ivanov (2011), Kelsey and le Roux (2013) or Di Mauro and Castro (2011).

  2. The theory is based on earlier research by Dow and Werlang (1994) and Eichberger and Kelsey (2000).

  3. Of course, this convention is for convenience only and bears no relation to the actual gender of subjects in our experiments.

  4. If \(v(R)>v(B)\) and \(v(R\cup Y)<v(B\cup Y),\) then these beliefs will be compatible with the Ellsberg Paradox.

  5. Note that Chateauneuf et al. (2007) write a neo-additive capacity in the form \(\mu (E)=\delta \alpha +(1-\delta )\pi (E).\) We have modified their definition to be consistent with the majority of the literature where \(\alpha \) is the weight on the minimum expected utility.

  6. This definition is justified in Eichberger and Kelsey (2014).

  7. The notation \(\frac{3}{4}\cdot T+\frac{1}{4}\cdot B\) denotes the mixed strategy where \(T\) is played with probability \(\frac{3}{4}\) and \(B\ \)is played with probability \(\frac{1}{4}.\)

  8. Proposition 3.2 confirms that \((B,L)\) is an equilibrium for a greater parameter range than \((T,M)\).

  9. Experimental protocols can be found here: http://saraleroux.weebly.com/experimental-protocols.html

  10. For the purpose of the Indian experiments, \(1ECU=Rs.1\), while for the Exeter experiments, \(100ECU=\pounds 2.\) In addition, a show-up fee of \(Rs.250\) was paid to the Indian subjects and \(\pounds 5\) to the Exeter subjects\(.\)

  11. The computer simulated urn can be found at the following link: http://saraleroux.weebly.com/experimental-protocols.html.

  12. The programme was produced by the FEELE lab programmer, Tim Miller. Even the experimenters were unaware of the processes determining the composition of the urn.

  13. Subjects remained in the same role throughout the game. It is thus possible that subjects could consider the properties of the game more fully, as the rounds progressed, even though they were not given any feedback. It may be noted that this “repeated” nature of play may have led to more Column Players choosing the option \(R\) when \(x=60,\) than is expected, as it was the last game round. These subjects may have got used to choosing \(R\) and not re-optimised for the lower value of \(x\).

  14. The binomial test was conducted for each value of \(x\), except \(x=60\), where EUA predicts that the column player can play \(L.\) It may be noted that for \( x=60,\) subjects play \(L+M\) more than 50 % of the time.

  15. The dummy for \(x\_60\) was dropped from the probit regression, in order to avoid the dummy variable trap.

  16. An initial probit regression showed that the dummy variable for location \(( Delhi/Exeter)\) was not significant. This reinforced our beliefs that the behaviour of the Indian subjects was very similar to that of the Exeter subjects. Thus, the location dummy variable was dropped and the model was re-run without it.

  17. The coefficients from a probit regression do not have the same interpretation as coefficients from an Ordinary Least Squares regression. From the probit results, we can interpret that if a subject had a quantitative degree, his z-score increases by \(0.538,\) making him more likely to pick \(R.\) If the subject is male, the z-score decreases by \(0.402\); hence males are less likely to opt for the ambiguity-safe option \(R\) than females. When \(x=120\), the z-score increases by \(1.16\); for \(x=170\), the z-score increases by \(1.08\); for \(x=200\), the z-score increases by \(1.75\); for \(x=230\), the z-score increases by \(2.27\); and for \(x=260\), the z-score increases by \(2.57;\) more than the base which is \(x=60.\)

  18. We would like to thank Peter Dursch, whose suggestions helped the design of the experiment.

  19. Only \(12\) of the \(80\) subjects that took part in our experiment always chose Red; 3 of these subjects were Row Players and thus do not have an ambiguity-safe option. As such, the Pearson correlation between choosing \( Right\) in the game round and Red in the urn round was close to zero.

  20. The data for \(y=100\) are from the classic Ellsberg paradox round. It is not completely comparable, as subjects were not given the option of choosing yellow.

  21. We ignore the subjects who switch multiple times between Red and Blue/Yellow. Such subjects may have been (incorrectly) using the notion of diversifying a portfolio.

  22. We do not consider \(x=60\), where \(R\) is a dominated strategy.

  23. We would like to thank our anonymous referee for this suggestion.

  24. These observations are as recorded by Itzhak Gilboa, in a discussion on observed ambiguity in Ellsberg experiments.

  25. There are no equilibria where Player 2 mixes between \(L\) and \(R\). Such an equilibrium would require Player 2 to be indifferent between \(L\) and \(R\) when \(300(1-q)=x\) or \(q=\frac{300-x}{300}\). Player 1 is indifferent between \(T\) and \(B\) when \(100p+55(1-p)=50(1-p)\). However, there is no solution to this equation with a positive value of \(p \ (p=-\frac{5}{95}\) is a solution).

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Acknowledgments

We would like to thank Dieter Balkenborg, Adam Dominiak, Peter Dursch, Jürgen Eichberger, Todd Kaplan, Joshua Teitelbaum, Horst Zank, participants in seminars in Bristol, Exeter, Manchester and RUD Paris and some anonymous referees for their comments and suggestions. We would like to thank Tim Miller for programming our Ellsberg Urn simulator.

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Correspondence to Sara le Roux.

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This paper is based on a chapter from Sara le Roux’s PhD thesis.

Appendix

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 \(\left( \frac{x}{100}\cdot T+\frac{ 100-x}{100}\cdot B,\frac{1}{61}\cdot M+\frac{60}{61}\cdot R\right) \).Footnote 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: \(\left\langle B,R\right\rangle \)\(\square \)

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Kelsey, D., le Roux, S. An experimental study on the effect of ambiguity in a coordination game. Theory Decis 79, 667–688 (2015). https://doi.org/10.1007/s11238-015-9483-2

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Keywords

  • Ambiguity
  • Choquet expected utility
  • Coordination game
  • Ellsberg urn
  • Experimental economics