# Nash Equilibria change linearly in (some) game parameters. Already known / follows from a more general result?

EDIT: The key thing that I am wondering about is the linearity of the P2 strategy, not the constancy of P1. (The latter is straightforward.)

Question: Is the following result already known? Or is it a straightforward corollary of some more general principle?

Intuitive version (slightly wrong): In a normal form game, adding a small cost $$c$$ to a particular action of a player (independently of the action of the opponent) changes the opponent's Nash equilibrium strategy linearly (in the size of $$c$$) and leaves the player's NE strategy constant.

Formal version: Let $$G$$ be a two-player normal-form game with actions denoted $$i$$, $$j$$ and payoffs $$u_1(i,j)$$, $$u_2(i,j)$$. For a fixed action $$i_0$$ of P1, denote by $$G'_c$$ the game that is identical to $$G$$, except $$u'_1(i_0, j) := u_1(i_0, j) - c$$. Then there is a finite set of exception points $$-\infty = e_0 < \dots < e_n = \infty$$ such that for every $$c_0 \in (e_k, e_{k+1})$$ and $$\sigma^{c_0} \in \textrm{NE}(G'_{c_0})$$, there is a linear mapping $$c \in [e_k, e_{k+1}] \mapsto \sigma^c_2$$ such that $$(\sigma^{c_0}_1, \sigma^c_2) \in \textrm{NE}(G'_c)$$ (and the value for $$c = c_0$$ coincides with $$\sigma^{c_0}_2$$).

Comments: (1) I know how to prove the statement "from scratch" using the LP formulation of NE (and the notion of basic solutions). But I was wondering whether there is perhaps a more general result that this follows from. (2) It seems something similar holds if we only change a single cell in the payoff matrix. (This will cause a linear change in the relative weights of probabilities in $$\sigma_2^{(\, \cdot \, )}$$. IE, no longer linear, but still quite simple.)

• I'm not sure the result is known in this form, but the "phenomenon" is; see Section B in this paper: stat.berkeley.edu/~aldous/157/Papers/goeree.pdf Dec 5, 2022 at 23:08
• I should have clarified: The key thing I am wondering about is the linearity of P2 strategy (not the constancy of P1 strategy). (Apologies there, the mistake was in my phrasing. And thanks a lot, this result is useful to know nevertheless.) Dec 6, 2022 at 2:15
• The linearity is not that complicated either, the only messy part is dealing with corner cases in which a strategy is played with probability zero. Otherwise it follows directly from the condition that the other players must be indifferent between all pure strategies in the support of a mixed best response. Dec 6, 2022 at 9:50
• Just flaggin that I don't see this as resolved yet. (The linearity seemed pretty non-obvious to me. The basic idea is simple, but my current proof --- which involves dealing with the "messy cases" --- is 1.5 pages. Surely that isn't optimal, but I would count that as decently complicated.) Dec 8, 2022 at 19:32
• You are right, I should have chosen my words a bit more carefully. I do think the theorem is somewhat unsurprising in that this is what you get whenever you calculate equilibria in "practice." Of course, this does not give a direct handle on how to deal with all cases in a systematic fashion. Dec 9, 2022 at 1:28