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I've got a question to ask about the Nash Demand game from my assignment.

Sarah and Ruth find \$100 on the ground and decide to split it between them in the following manner. Each individual simultaneously and independently selects how much of the \$100 she wants to keep. Denote this value by $xi$. If $๐‘ฅ๐‘†+๐‘ฅ๐‘…โ‰ค100$, then each player receives the amount she desires. If $๐‘ฅ๐‘†+๐‘ฅ๐‘…>100$, they forfeit the \$100, and each player receives \$0. Find all of the Nash equilibria. Explain how you come to your conclusion (i.e. how you know that your answer is exhaustive.)

I think that I've solved the Pure Strategy Nash Equilibrium for this game which is basically $๐‘ฅ๐‘†+๐‘ฅ๐‘…=100$. But I'm wondering whether there are any Mixed Strategy Nash Equilibrium so if anyone could help me out that I would be very appreciative! :)

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  • $\begingroup$ That is not an equilibrium, as it does not give how much each player chooses. Hint: start from a point (a strategy for each). Then see if anyone has an incentive to deviate, given that the other sticks to that strategy. $\endgroup$ – BB King Oct 24 '16 at 18:32
  • $\begingroup$ Oh I meant {Ruth: Xr; Sarah: 100-Xr} and {Ruth: 100-Xs; Sarah: Xs} so I put that equation down for simplicity. Thanks anw! :) $\endgroup$ – Russljd Oct 25 '16 at 5:17
  • $\begingroup$ Do either of them have a 'reputation?' If Ruth is known to be greedy and want more than half, then Sarah might ask for less than half... unless Sarah has a reputation for punishing greedy people, then she could ask for more than half so both get nothing... but if Ruth knows this then she might ask for less than half. $\endgroup$ – Daniel Apr 22 at 0:06
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Nash equilibrium is a pair of strategies. So your answer needs to give two strategies. The informal definition is "pair of strategies that are best responses to each other".

So take an arbitrary strategy; let's say each player chooses 50. Can anyone gain by playing 51 or 49? No, so (50, 50) is a NE. Like your equation says, any two strategies that add up to 100 make up a PSNE. Because there can be no profitable deviation for any of the two player (the proof is trivial; show that playing less gives less, and that playing more, gives less for any pair of strategies adding up to 100)


Now about MSNE. A good way to start thinking about it is to take an arbitrary strategy and play around with it. You will find no MSNE this way.

The other way to look for them is to find a point where two strategies make players indifferent between their options and are also best responses. I could only find this in the symmetric case.

There are some MSNE if you play strategies $( a_1, a_2 )$ where $a_1+a_2 = 100$ and your mixing probabilities are $P(a_1) = a_1/a_2$ (assuming $a_1 < a_2)$. Example here would be playing $(10, 90)$ with probabilities $(11.111, 88.888)$. This makes the opponent indifferent between $10$ and $90$, and as such he could play the mirror strategy, with both being indifferent to any mixing and both being a best response to each other.

I did not find any other case of MSNE, even after some work.

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  • $\begingroup$ How about playing uniformly with support {0,100}? And what about non-uniform distributions? $\endgroup$ – Giskard Oct 24 '16 at 21:35
  • $\begingroup$ It's pretty clear for the first case you mention that the best response is playing "pure" 50. For non uniform distributions, we need something where both players are indifferent between their options. For example, {99,98} uniform makes one indifferent between playing any mix of 1 and 2, but {99, 98} won't be a best response to any of those strategies so it won't create a MSNE. $\endgroup$ – Matt Oct 25 '16 at 2:12
  • $\begingroup$ Thank you so much. I understand it now! The part on a1/a2 is very enlightening haha. $\endgroup$ – Russljd Oct 25 '16 at 5:18
  • $\begingroup$ Oh actually shouldn't the probability be a1/(a1+a2) instead for the players to be indifferent between 10 and 90? For instance, the payoff of 10 times the probability of 0.9 should be equal to the payoff of 90 times the probability of 0.1 thus both payoffs are the same. $\endgroup$ – Russljd Oct 25 '16 at 5:32
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    $\begingroup$ My initial answer was wrong. I edited it; I found MSNE in non-uniform cases where strategies are symmetric and a player's actions sum to 100. I don't know if there are more. Thanks for the input. $\endgroup$ – Matt Oct 25 '16 at 13:58
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There is a complete characterization of equilibria in Malueg (2010) [1].

The structure of these equilibria is in a sense a generalization of the property mentioned in the question - that demands should sum up to 100. Instead for every demand $x_s$ that Sarah makes with positive probability, there is some probability that Ruth would demand exactly the remaining $100 - x_s$.

Formally, suppose one player mixes over some set A, and the other - over set B. There is a unique equilibrium for any nonempty closed A and B such that $1 \notin A\cup B$ iff these sets are balanced, i.e:

$$ s \in A \iff (100-s) \in B$$

Sets A and B do not need to be identical or finite. Moreover, these are the only equilibria (see Proposition 1 in the paper).

The equilibria indeed contain an atom. For a concrete example (example 3 from the paper) take any $a \in (0,1/2]$ and $A = B = [a,1-a]$. Each player completely mixes over this set with CDF:

$$F(s)=\begin{cases} 0 & \text{if } s<a \\ \frac{a}{1-s} & \text{if } a \le s \le 1-a \\ 1 & \text{if } 1-a<s \le 1 \end{cases}$$

That is, with probability $a/(1-a)$ the player bids $x_i=a$, the lowest demand in $A$, and otherwise demands more, say $s$. If she does demand more there is a risk of running over a 100 if the other player demands more than $1-s$, which occurs with probability exactly $1-F(1-s)$ by symmetry. Then we can check that all actions in the support do bring equal payoffs as necessary for the Nash equilibrium, namely $s \times F(1-s) = a$.

[1]: Malueg, David A. "Mixed-strategy equilibria in the Nash Demand Game." Economic Theory 44.2 (2010): 243-270. It is in open access - https://core.ac.uk/download/pdf/81838757.pdf

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