The Hotelling bi-matrix game

Abstract

We study the pure equilibrium set for a specific symmetric finite game in strategic form, referred to as the Hotelling bi-matrix game. General results that guarantee non-emptiness of this set (for all parametric values) do not seem to exist. We prove non-emptiness by determining the pure equilibrium set. In this proof so-called demi-modality properties of the conditional payoff functions play an important role.

Calculations

Lemma 6

• 1. Suppose $w \ne 1$.

  • (a) If $x \ge 1$, then $f_1^{(x)}(x-1) = \frac{1-w^x}{1-w}$ and $f_1^{(x-1)}(x) = \frac{1- w^{n-x+1} }{1-w}$. If $x \ge 0$, then $f_1^{(x)}(x) = \frac{1 +w - w^{x+1} - w^{n-x+1} }{2(1-w)}$.

  • (b) If $x \ge 1$, then $\Delta f_1^{(x)}(x) = \frac{-1 + w + 2 w^x - w^{x+1} - w^{n-x+1} }{2(1-w)}$.

• 2.

  • (a) If $w \ne 1$ and $\frac{n}{2} < x \le n$, then $\Delta f_1^{(x)}(x) < 0$.

  • (b) If $w =1$, then $\Delta f_1^{(x)}(x) = \frac{n+1}{2} - x$.

  • (c) If $w \ne 1$ and $x \ge 1$, then $\Delta f_1^{(x-1)}(x) = \frac{ 1 - w + w^x + w^{n-x+2} - 2 w^{n-x+1} }{ 2(1-w) }$.

• 3. If $\frac{n}{2} < x <n$, then $f_1^{(x)}(x+s) - f_1^{(x)}(x-s) < 0 \; (1 \le s \le n-x)$.

Proof

1a. Apply the formula for $f_1$ in Sect. 2.

• 1b. By part 1a.

• 2a. Use part 1b and, using $w \in ] {0},{1} \, [$ and $x > \frac{n}{2}$, note that $1- w - 2 w^x + w^{x+1} + w^{n-x+1} > 1-w - 2 w^x + w^{x+1} + w^{\frac{n}{2}+1} = (w^{\frac{n}{2}+1} - w^x) + (1-w^x) (1-w) > 0 \,+\, 0 = 0$.

• 2b. As by the formula for $f_1$ we have $\Delta f_1^{(x)}(x) = f_1(x,x) - f_1(x-1,x) = \frac{n+1}{2} -x$.

• 2c. A direct consequence of part 1a.

• 3. The locations that contribute to $f_1^{(x)}(x+s)$ are those in $V_+ := \{ x + \lfloor \frac{s+1}{2} \rfloor , \ldots , x+s-1, x+s, x+s+1, \ldots , n \}$. The locations that contribute to $f_2^{(x)}(x-s)$ are those in $V_- := V_{-+} \cup V_{--}$ where $V_{-+} = \{ x - \lfloor \frac{s+1}{2} \rfloor , \ldots , x-s+1, x-s, x-s-1, \ldots , n - 2\lfloor \frac{s+1}{2}\rfloor \}$ and $V_{--} = \{ n - 2 \lfloor \frac{s+1}{2} \rfloor -1, \ldots , 1, 0 \}$. The contribution of $V_+$ to $f_1^{(x)}(x+s)$ is the same as that of $V_{-+}$ to $f_1^{(x)}(x-s)$. As $V_{--} \ne \emptyset$ the desired result follows.$\square$

Lemma 7

• 1. If $x_1, x_2 \in H$ with $x_2 \ge 1$, then $\Delta f_1^{(x_1)}(x_2) = - \Delta f_1^{(n-x_1)}(n-x_2+1)$.

• 2. If $x_1,x_2 \in H$ with $x_2-x_1 \ge 2$, then $\Delta f_1^{(x_1)}(x_2) = - q_{n-x_2+1;n-x_1}(w)$.

Proof

• 1. $\Delta f_1^{(x_1)}(x_2) = f_1(x_2,x_1) - f_1(x_2-1,x_1) = f_1(n-x_2,n-x_1) - f_1(n-x_2+1,n-x_1) = - \Delta f_1^{(n-x_1)}(n-x_2+1)$.

• 2. By part 1, $\Delta f_1^{(x_1)}(x_2) = -\Delta f_1^{(n-x_1)}(n-x_2+1)$. As $n-x_2+1 < n -x_1$, we obtain $\Delta f_1^{(x_1)}(x_2) = - q_{n-x_2+1;n-x_1}(w)$. $\square$

Lemma 8

Consider $(x_1,x_2) \in H \times H$ with $0< x_1 < x_2 < n$ and $x_2-x_1 \ge 2$.

• 1. Suppose $x_2-x_1$ is even.

  • (a) $0 \ge \Delta f_1^{(x_2)}(x_1+1) \; \Leftrightarrow \; w^{\frac{3x_1-x_2+2}{2}} \le \frac{1}{2}$;

  • (b) $\Delta f_1^{(x_2)}(x_1) \ge 0 \; \Leftrightarrow \; w^{\frac{3x_1-x_2}{2}} \ge \frac{1}{2}$;

  • (c) $0 \ge \Delta f_1^{(x_1)}(x_2+1) \Leftrightarrow w^{\frac{2n-3x_2+x_1}{2}} \ge \frac{1}{2}$;

  • (d) $\Delta f_1^{(x_1)}(x_2) \ge 0 \Leftrightarrow w^{\frac{2n-3x_2+x_1+2}{2}} \le \frac{1}{2}$.

• 2. Suppose $x_2-x_1$ is odd.

  • (a) $0 \ge \Delta f_1^{(x_2)}(x_1+1) \; \Leftrightarrow \; w^{\frac{3x_1-x_2+3}{2}} \le \frac{1}{2}$;

  • (b) $\Delta f_1^{(x_2)}(x_1) \ge 0 \; \Leftrightarrow \; w^{\frac{3x_1-x_2-1}{2}} \ge \frac{1}{2}$;

  • (c) $0 \ge \Delta f_1^{(x_1)}(x_2+1) \Leftrightarrow w^{\frac{2n-3x_2+x_1-1}{2}} \ge \frac{1}{2}$;

  • (d) $\Delta f_1^{(x_1)}(x_2) \ge 0 \Leftrightarrow w^{\frac{2n-3x_2+x_1+3}{2}} \le \frac{1}{2}$.

Proof

Proposition 2, (4) and Lemma 7(2) imply $\Delta f_1^{(x_2)}(x_1+1) \le 0 \; \Leftrightarrow \; w^{x_1+1} \le \frac{1}{2} w^{ \lfloor \frac{x_2-x_1}{2} \rfloor }; 0 \le \Delta f_1^{(x_2)}(x_1) \; \Leftrightarrow \; w^{x_1} \ge \frac{1}{2} w^{ \lfloor \frac{x_2-x_1+1}{2} \rfloor }; \Delta f_1^{(x_1)}(x_2+1) \le 0 \; \Leftrightarrow \; w^{n-x_2} \ge \frac{1}{2} w^{ \lfloor \frac{x_2-x_1+1}{2} \rfloor }; 0 \le \Delta f_1^{(x_1)}(x_2) \; \Leftrightarrow \; w^{n-x_2+1} \le \frac{1}{2} w^{ \lfloor \frac{x_2-x_1}{2} \rfloor }$. This leads to the desired results. $\square$

Lemma 9

• 1. Suppose n is even.

  • (a) $\Delta f_1^{(p)} (p) = - \Delta f_1^{( p )} ( p + 1 ) = w^p -\frac{1}{2}$.

  • (b) If $p \ge 2$, then $\Delta f_1^{(p)} (p-1) = - \Delta f_1^{(p )} ( p + 2 ) = w ( w^{p-2} - \frac{1}{2} )$.

  • (c) $\Delta f_1^{(p+1)} (p) = - \Delta f_1^{( p-1 )} ( p + 1 ) = w^p - \frac{w}{2}$.

  • (d) $\Delta f_1^{(p-1)} (p) = - \Delta f_1^{(p+1)}(p+1) = - w^{p+1} + \frac{1}{2}$.

• 2. Suppose n is odd.

  • (a) If $p \ge 1$, then $\Delta f_1^{(p+1)} (p) = - \Delta f_1^{( p )} ( p + 2 ) = w( w^{p-1} - \frac{1}{2} )$.

  • (b) $\Delta f_1^{(p+1)} (p+1) = - \Delta f_1^{( p )} ( p + 1 ) = \frac{w^{p+1}-1}{2}$.

Proof

It is sufficient to prove the statements for $w \ne 1$. Well, use Proposition 2 and Lemmas 6(1b) and 7. $\square$