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I am trying to solve a certain problem that I wrote in great detail here. The following lemma I thought of may be useful somewhere. I'm posing it as another question:

Suppose sets $I$ and $J$, each with cardinality $\geq 2$, are given. If we are also given the values of $\alpha_{ij} \in \mathbb{R}$ for all $(i,j) \in I \times J$ such that there exists $(i_1,j_1)$ and $(i_2, j_2)$ satisfying $\alpha_{i_1j_1} > 0 > \alpha_{i_2j_2}$, when does $$\sum_{i} \sum_{j} p_i q_j \alpha_{ij} = 0$$ have a solution in $p_i$s and $q_j$s such that each $p_i, q_j \geq 0$ and $\sum_{i \in I} p_i = \sum_{j \in J} q_j = 1$?

Just to clarify, the notation $[m]$ denotes $\{1,2,\cdots, m\}$.

breakfasttheorist
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  • I don't understand your mention of $ \alpha_{ij} < 0$ when you state $\alpha_{ij} \in [0, \infty] $. It also isn't clear how you plan to handle $\infty - \infty$. – user8675309 Apr 02 '25 at 15:54
  • @user8675309 Thanks for pointing it out. The $\alpha_{ij}$s are just finite real numbers. The question is basically to determine whether there exists two probability vectors $p = [p_1, \cdots, p_{|I|}]$ and $q = [q_1, \cdots, q_{|J|}]$ such that the given sum is $0$. The sets $I$ and $J$ are also given (you can interpret that as the desired length of each probability vector). I suppose there's some linear-algebra way to do this, although I might be wrong. – breakfasttheorist Apr 02 '25 at 18:00
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    Cann you do the case $|I|=|J|=2$ ? – GEdgar Apr 02 '25 at 18:04
  • where does it say that the $p_i$ and $q_j$ are non-negative? I don't see this in any iteration of the original post. – user8675309 Apr 02 '25 at 18:05
  • If $I$ and $J$ both have only one element, the statement is false. Then $p = q = 1$ and the only solution is $\alpha = 0$. – Tzimmo Apr 02 '25 at 18:06
  • @user8675309 I hope there's no other mistake. – breakfasttheorist Apr 02 '25 at 18:33
  • @Tzimmo The cardinality must be at least $2$, since otherwise (if both strategies are not mixed) the Nash equilibrium can not be calculated simply by equating the two pay-offs which is being done here (if you see the linked post). I'm sorry for taking it too long to mention this in the post. – breakfasttheorist Apr 02 '25 at 18:40
  • @TedShifrin I can see how staying awake for ~50 hours can really take a toll. So many errors that it is now embarrassing, but I genuinely thank you all so much for having the patience to deal with this. – breakfasttheorist Apr 02 '25 at 18:44

1 Answers1

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Translating to the language of vectors and matrices: you want ${\bf p}^\top A {\bf q} = 0$ where ${\bf p} \in [0,\infty)^m$ and ${\bf q} \in [0,\infty)^n$ are column vectors and $A \in \mathbb R^{m\times n}$ a matrix with at least one positive and at least one negative entry, and $m,n\ge 2$. If there is a solution with ${\bf p} \ne {\bf 0}$ and ${\bf q} \ne {\bf 0}$, you can normalize so $\sum_{i} p_i = 1$ and $\sum_{i} q_i = 1$.

Hint: reduce to the case $m=n=2$, which is easy, by considering ${\bf p}$ and ${\bf q}$ which each have only two nonzero elements.

EDIT: There are three cases to consider.

  1. If $i_1 \ne i_2$ and $j_1 \ne j_2$, we can take all $p_i = 0$ except for $i=i_1$ and $i=i_2$, and all $q_j = 0$ except $j=j_1$ and $j=j_2$. For simplicity of notation, I'll assume $i_1 = j_1 = 1$, $i_2 = j_2 = 2$, so we want $f(p,q) = p_1 \alpha_{11} q_1 + p_1 \alpha_{12} q_2 + p_2 \alpha_{21} q_1 + p_2 \alpha_{22} q_2 = 0$. We have $f([1,0],[1,0]) = \alpha_{11} > 0$ and $f([0,1],[0,1]) = \alpha_{22} < 0$. By Intermediate Value Theorem there is some $t \in (0,1)$ such that $f([t,1-t],[t,1-t]) = 0$.
  2. If $i_1 = i_2$ and $j_1 \ne j_2$, I'll assume $i_1 = i_2 = 1$, $j_1 = 1$ and $j_2 = 2$. Take $p_1 = 1$, $p_i = 0$ otherwise, $q_j = 0$ for $j > 2$. We have $f(p,q) = \alpha_{11} q_1 + \alpha_{12} q_2$, with $f([1,0],[1,0]) = \alpha_{11} > 0$ and $f([1,0],[0,1]) = \alpha_{12} < 0$. By Intermediate Value Theorem there is $t \in (0,1)$ such that $f([1,0],[t,1-t]) = 0$.
  3. If $i_1 \ne i_2$ and $j_1 = j_2$, this is like case (2) with $p$ and $q$ interchanged.
Robert Israel
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