For an arbitrary normal-form game, suppose I add the same constant to the payoff of the players everywhere in the game. Will the Nash equilibria (both pure and mixed) stay the same? Intuitively I would say yes, but I was wondering if other people could confirm that. Thanks.
Do Nash equilibria stay the same when you add a constant to all payoffs?
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$\begingroup$
game-theory
nash-equilibrium
2 Answers
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Yes, this is correct. Adding a constant to each payoff for all players just 'shifts up the graph' of the respective utility functions for all players. In particular,
$$\textrm{argmax}_{\Delta(S_i)} u_i(\sigma_i, \sigma_{-i}) = \textrm{argmax}_{\Delta(S_i)} u_i(\sigma_i, \sigma_{-i}) + c$$
for all $\sigma_{-i} \in \times_{j\neq i} \Delta(S_j)$ hence all best responses remain unchanged (and thus so too their fixed points). Intuitively (as you suggest), unless you change the relative worth of one outcome versus another, nothing will change.
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I agree with @Pete Caradona conditional on assuming that the way to deal with uncertainty (mixed strategies) is the expected utility. Other ways to calculate utility of lotteries might not be invariant to adding constant to utility of all outcomes.