Graham wrote:

I mean the hyperplane

Oh! I hadn’t considered that option. In math, half the time words like ‘line’, ‘plane’ and ‘hyperplane’ mean things that contain the origin, so that they’re vector spaces. And half the time they don’t. It’s about linear versus affine geometry.

Let me think about this some more now. Thanks.

]]>Here is that part of the argument more explicitly:

Let and let , that is, is a unit vector normal to the hyperplane . Let

and

Then

and expanding that out and using leads to

where does not depend on .

]]>I mean the hyperplane I want regard the unit simplex in as a hypertetrahedron in

A bit later, where I say “…B’s payoff can be improved by moving in direction from any point in …” I think it would be clearer if I added that this works for any choice by player A.

]]>This looks interesting—sorry to take so long to reply!

Now belongs to a hyperplane in […]

Which hyperplane? Any old random hyperplane? Yes, belongs to infinitely many different hyperplanes, but this makes me nervous, mainly because I’m trying to understand your proof and it’s not making sense yet.

Now belongs to a hyperplane in so we can project and the onto it, with becoming […]

If belongs to a hyperplane, projecting onto this hyperplane won’t change at all, so we get Or am I not understanding you?

]]>Lemma. Let be vectors in a finite dimensional real vector space. Then either (1) there is a vector such that for all or (2) there are nonnegative reals , not all zero, such that

Proof idea for lemma: consider the convex hull of the . If the origin is outside we can find a for (1), otherwise we can show (2).

Sketch proof of existence of Nash equilibrium. Denote the i’th row of by regarded as a vector in Player A’s payoff can be written as

Now belongs to a hyperplane in so we can project and the onto it, with becoming and becoming in Note is constrained to a hypertetrahedron Then

where does not depend on .

Now apply the lemma. If there is a such that for all i, then player B’s payoff can be improved by moving in direction from any point in , so player B will choose a point on the boundary of , which means at least one strategy is never chosen. So the game is equivalent to a smaller game and induction provides a Nash equilibrium. Otherwise there is some (not necessarily optimal) choice for player A such that so player B cannot change the payoff.

The same argument with A and B reversed either gives an equilibrium by induction, or a choice for player B such that player A cannot change the payoff. So if induction fails in both cases, we have

]]>I noticed it is used in this paper you pointed to:

Andrew McLennan and Rabee Tourky, From imitation games to Kakutani

… but they’re economists.

]]>Okay, I didn’t understand the proof either.

]]>That’s for sure. I’ve been doing math for decades and have never seen that notation. It looks useful, but I’ll need to explain it whenever I use it in front of mathematicians.

]]>I am sure that’s what Luis means. I was initially surprised that you didn’t know the notation, but I think it is more commonly used by programmers than mathematicians.

]]>I don’t understand Luis’ proof. Let

The set is like with an inequality replaced by an equality. It is the set of points in which is an optimal choice for player 2 for some choice by player 1. Similarly we can define a function and a set What we want to show is that the intersection of and is non-empty.

I don’t see how it helps to extend to and look for points in that.