Generic finiteness of equilibrium distributions for bimatrix outcome game forms

Abstract

We provide sufficient and necessary conditions for the generic finiteness of the number of distributions on outcomes, induced by the completely mixed Nash equilibria associated to a bimatrix outcome game form. These equivalent conditions are stated in terms of the ranks of two matrices constructed from the original game form.

A The proof of Theorem 2.8

Lemma A.1

Suppose that Assumption 2.3 holds. Let $u^1, u^2 \in U$. The set of QE of the game $\left( u^1( \phi ),u^2( \phi )\right)$ induce finitely many quasi-distributions on outcomes if and only if for every $\omega \in \Omega$ and every QE $\left( x\left( u^2\right) ,y\left( u^1\right) \right)$ of that game the following two conditions hold.

• (a) $x \left( u^2\right) \phi ^{\omega }$ is in the image of $u^1( \phi )^t$

• (b) $\phi ^{\omega } y\left( u^1\right)$ is in the image of $u^2( \phi )$.

Proof

Let $u^1, u^2 \in U$. By Proposition 2.6, the set of QE of the game $\left( u^1( \phi ),u^2( \phi )\right)$ induce finitely many quasi-distributions on outcomes if and only if for every $\omega \in \Omega$, every solution $x\left( u^2\right)$ of the system of Eq. (2) and every solution $y\left( u^1\right)$ of the system of Eq. (1) we have that $x\left( u^2\right) \phi ^{\omega } K_1\left( u^1 \right) =\{0\}$ and $K_2(u^2) \phi ^{\omega } y\left( u^1\right) =\{0\}$. This is equivalent to the statement that $x\left( u^2\right) \phi ^{\omega }$ is orthogonal to $K_1\left( u^1 \right)$ and $\phi ^{\omega } y\left( u^1\right)$ is orthogonal to $K_2(u^2)$, which occurs if and only if $x \left( u^2\right) \phi ^{\omega }$ is in the image of $u^1( \phi )^t$ and $\phi ^{\omega }\, y\left( u^1\right)$ is in the image of $u^2( \phi )$. $\square$

Lemma A.2

Let $\omega \in \Omega$ and $u^1, u^2 \in U$. Let Assumption 2.3 hold and suppose that $K_2(u^2) \phi ^{\omega } K_1(u^1) =\{0\}$. Then,

$$\begin{aligned} \mathrm{rank}\left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi ) \\ \end{array}\right) =2k \end{aligned}$$

Proof

We use the notation

$$\begin{aligned} \phi ^{\omega } = \left( \begin{array}{cc} B^{\omega } &{} C^{\omega } \\ D^{\omega } &{} E^{\omega } \end{array}\right) \end{aligned}$$

to denote the decomposition of the matrix $u( \phi )$ in (3) applied to the matrix $\phi ^{\omega }$. Let

$$\begin{aligned} F= \left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi ) \\ \end{array}\right) \end{aligned}$$

We can write now

$$\begin{aligned} F=\left( \begin{array}{cccc} B \left( u^1 \right) &{} C \left( u^1 \right) &{} 0 &{} 0 \\ D\left( u^1 \right) &{} E\left( u^1 \right) &{} 0 &{} 0 \\ B^{\omega } &{} C^{\omega } &{} B \left( u^2 \right) &{} C \left( u^2 \right) \\ D^{\omega } &{} E^{\omega } &{} D \left( u^2 \right) &{} E \left( u^2 \right) \end{array}\right) \end{aligned}$$

By elementary row and column operations,

$$\begin{aligned} \mathrm{rank}F = \mathrm{rank}\left( \begin{array}{ccc} B \left( u^1 \right) &{} C \left( u^1 \right) &{} 0 \\ B^{\omega } &{} C^{\omega } &{} B \left( u^2 \right) \\ D^{\omega } &{} E^{\omega } &{} D \left( u^2 \right) \end{array}\right) = \mathrm{rank}\left( \begin{array}{ccc} B \left( u^1 \right) &{} C \left( u^1 \right) &{} 0 \\ B^{\omega } &{} C^{\omega } &{} B \left( u^2 \right) \\ D^{\omega }_1 &{} E^{\omega }_1 &{} 0 \end{array}\right) \end{aligned}$$

where

$$\begin{aligned} D^{\omega }_1= & {} D^{\omega } - D \left( u^2 \right) B^{-1}\left( u^2 \right) B^{\omega } \end{aligned}$$

(8)

$$\begin{aligned} E^{\omega }_1= & {} E^{\omega } - D \left( u^2 \right) B^{-1}\left( u^2 \right) C^{\omega } \end{aligned}$$

(9)

Finally,

$$\begin{aligned} \mathrm{rank}F = \mathrm{rank}\left( \begin{array}{ccc} B \left( u^1 \right) &{} 0 &{} 0 \\ B^{\omega } &{} C^{\omega }_2 &{} B \left( u^2 \right) \\ D^{\omega }_1 &{} E^{\omega }_2 &{} 0 \end{array}\right) \end{aligned}$$

with $C^{\omega }_2 = C^{\omega } - B^{\omega } B^{-1}\left( u^1 \right) C \left( u^1 \right)$ and

$$\begin{aligned} E^{\omega }_2= & {} E^{\omega }_1 - D^{\omega }_1 B^{-1}\left( u^1 \right) C \left( u^1 \right) \\= & {} E^{\omega } - D \left( u^2 \right) B^{-1}\left( u^2 \right) C^{\omega } - D^{\omega } B^{-1}\left( u^1 \right) C \left( u^1 \right) + D \left( u^2 \right) B^{-1}\left( u^2 \right) B^{\omega } B^{-1}\left( u^1 \right) C \left( u^1 \right) \\= & {} \left( \begin{array}{cc} - D\left( u^2\right) B^{-1} \left( u^2 \right)&I_{m-k} \end{array}\right) \left( \begin{array}{cc} B^{\omega } &{} C^{\omega }\\ D^{\omega } &{} E^{\omega } \end{array}\right) \left( \begin{array}{c} -B^{-1} \left( u^1 \right) C \left( u^1\right) \\ I_{n-k} \end{array}\right) \\= & {} \left( \begin{array}{cc} - D\left( u^2\right) B^{-1} \left( u^2 \right)&I_{m-k} \end{array}\right) \phi ^{\omega } \left( \begin{array}{c} -B^{-1} \left( u^1 \right) C \left( u^1\right) \\ I_{n-k} \end{array}\right) \end{aligned}$$

It follows from Lemma 2.4 that

$$\begin{aligned} K_1(u^1) =\left\{ y^h(u^1,v) : v \in \mathbb {R}^{n-k} \right\} = \left\{ \left( \begin{array}{c} -B^{-1} \left( u^1 \right) C \left( u^1\right) \\ I_{n-k} \end{array}\right) v : v \in \mathbb {R}^{n-k} \right\} \end{aligned}$$

and

$$\begin{aligned} K_2(u^2) = \left\{ x^h(u^2,w) : w \in \mathbb {R}^{m-k} \right\} = \left\{ w \left( \begin{array}{cc} - D\left( u^2\right) B^{-1} \left( u^2 \right)&I_{m-k} \end{array}\right) : w \in \mathbb {R}^{m-k} \right\} \end{aligned}$$

Since $K_2(u^2) \phi ^{\omega } K_1(u^1) =\{0\}$, for any $v\in \mathbb {R}^{n-k}$ and $w\in \mathbb {R}^{m-k}$ we have that

$$\begin{aligned} w\, E^{\omega }_2 \, v= w \left( \begin{array}{cc} - D\left( u^2\right) B^{-1} \left( u^2 \right)&I_{m-k} \end{array}\right) \phi ^{\omega } \left( \begin{array}{c} -B^{-1} \left( u^1 \right) C \left( u^1\right) \\ I_{n-k} \end{array}\right) v = 0 \end{aligned}$$

Therefore, $E^{\omega }_2=0$ and

$$\begin{aligned} \mathrm{rank}F = \mathrm{rank}\left( \begin{array}{ccc} B \left( u^1 \right) &{} 0 &{} 0 \\ B^{\omega } &{} C^{\omega }_2 &{} B \left( u^2 \right) \\ D^{\omega }_1 &{} 0 &{} 0 \end{array}\right) = \mathrm{rank}\left( \begin{array}{ccc} B \left( u^1 \right) &{} 0 &{} 0 \\ B^{\omega } &{} C^{\omega }_2 &{} B \left( u^2 \right) \end{array}\right) = 2k \end{aligned}$$

because, since $\mathrm{rank}B \left( u^1 \right) = \mathrm{rank}B \left( u^2 \right) = k$, the rows of $D^{\omega }_1$ are a linear combination of the rows of $B \left( u^1 \right)$. $\square$

Lemma A.3

Suppose that Assumption 2.3 holds. Let $u^1, u^2 \in U$, $\omega \in \Omega$. Then,

• (a) $x\left( u^2 \right) \phi ^{\omega } K_1\left( u^1 \right) =\{0\}$ for every solution $x\left( u^2 \right)$ of the system of Eq. (2) if and only if

  $$\begin{aligned} \mathrm{rank}\left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi ) \\ 0 &{} d_n \end{array}\right) =2k \end{aligned}$$

• (b) $K_2(u^2) \phi ^{\omega } y \left( u^1 \right) =0$ for every solution $y \left( u^1 \right)$ of the system of Eq. (1) if and only if

  $$\begin{aligned} \mathrm{rank}\left( \begin{array}{ccc} u^2( \phi ) &{} \phi ^{\omega } &{} 0\\ 0 &{} u^1( \phi )&{} d_m \end{array}\right) =2k \end{aligned}$$

Proof

We prove only part (a). The proof of part (b) is similar. Fix a solution $x= x\left( u^2 \right)$ of the system of Eq. (2). Let

$$\begin{aligned} F= \left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi ) \\ 0 &{} d_n \end{array}\right) \end{aligned}$$

Since, $d_n= \frac{ 1}{\alpha \left( u^2 \right) }x u^2( \phi )$, by elementary row operations we have that

$$\begin{aligned} \mathrm{rank}F = \mathrm{rank}\left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi ) \\ \frac{ 1}{\alpha \left( u^2 \right) } x \phi ^{\omega } &{} 0 \end{array}\right) \end{aligned}$$

Assume that $x\, \phi ^{\omega } K_1\left( u^1 \right) =0$. Then, by Lemma A.1, $x\, \phi ^{\omega }$ is in the image of $u^1( \phi )^t$ and hence $x\, \phi ^{\omega }$ is a linear combination of the rows of $u^1( \phi )$. Therefore,

$$\begin{aligned} \mathrm{rank}F = \mathrm{rank}\left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi )\end{array}\right) \end{aligned}$$

and, by Lemmas 2.5 and A.2 we have that $\mathrm{rank}F = 2k$.

Conversely, suppose now that $\mathrm{rank}F = 2k$. We proceed now as in Lemma A.2 and write F as

The above argument shows that

Since $\mathrm{rank}u^2( \phi ) = \mathrm{rank}B \left( u^2 \right) = k$, we get that

where $D^{\omega }_1, E^{\omega }_1$ are defined in (8) and  (9). Since, $\mathrm{rank}u^1( \phi ) = \mathrm{rank}B \left( u^2 \right) = k$, the rows of the matrix $\left( \begin{array}{cc} D^{\omega }_1&E^{\omega }_1 \end{array}\right)$ and $x\, \phi ^{\omega }$ are a linear combination of the rows of $u^1( \phi )$. It follows that $x\, \phi ^{\omega }$ is orthogonal to $K_1\left( u^1\right)$ and (a) follows. $\square$

The following result follows now immediately from Proposition 2.6 and Lemma A.3.

Theorem A.4

Suppose that Assumption 2.3 holds. Let $u^1, u^2 \in U$. Then, the set of all the QE of the game $\left( u^1( \phi ),u^2( \phi )\right)$ induces finitely many quasi-distributions on outcomes iff for every $\omega \in \Omega$

$$\begin{aligned} \mathrm{rank}\left( \begin{array}{cc} u^1( \phi )&{} 0 \\ \phi ^{\omega } &{} u^2( \phi ) \\ 0 &{} d_n \end{array}\right) =2k \quad \text {and} \quad \mathrm{rank}\left( \begin{array}{ccc} u^2( \phi ) &{} \phi ^{\omega } &{} 0\\ 0 &{} u^1( \phi )&{} d_m \end{array}\right) =2k \end{aligned}$$

We address now the proof of Theorem 2.8. We show first the ‘if’ part. If the rank conditions in Theorem 2.8 hold, then by Theorem A.4 the set of QE induce a unique quasi-distribution on outcomes. Since the set of CMNE is a subset of the set of QE, the set of CMNE also induces, at most, a unique distribution on outcomes. Thus, the ‘if’ part of Theorem 2.8 holds.

Conversely, suppose that the set of CMNE of the game $\left( u^1( \phi ),u^2( \phi )\right)$ induces finitely many distributions on outcomes and that the game $\left( u^1( \phi ),u^2( \phi )\right)$ has, at least, a CMNE , say $\bar{x} = x^p(u^2) + x^h(u^2,w_0)\in \Delta _+(S^1)$ and $\bar{y} = y^p\left( u^1 \right) + y^h(u^1,v_0)\in \Delta _+(S^2)$ with $w_0\in \mathbb {R}^{m-k}, v_0\in \mathbb {R}^{n-k}$. By continuity, there are open sets $H_1 \subset \mathbb {R}^{n-k}$ and $H_2 \subset \mathbb {R}^{m-k}$ such that for $v\in H_1$, $w\in H_2$ we have that $x = x^p(u^2) + x^h(u^2,w)\in \Delta _+(S^1)$ and $y= y^p \left( u^1 \right) + y^h(u^1,v)\in \Delta _+(S^2)$ is a CMNE of the game $\left( u^1( \phi ),u^2( \phi )\right)$.

If the rank conditions in Theorem 2.8 do not hold, then, by Theorem A.4 for some outcome $\omega \in \Omega$ the polynomial $q_{\omega }(v,w)$ in (7) is not constant and, hence, it takes a continuum of values as the variables (v, w) vary on the open set $H_1\times H_2$. It follows that the set CMNE of the game $\left( u^1( \phi ),u^2( \phi )\right)$ induce infinitely many distributions on outcomes, which contradicts our assumption. And the ‘only if’ part of Theorem 2.8 follows.