数理解析研究所講究録 1263 巻 2002 年 117-130

117

Discrete final-0ffer arbitration
model
Vladimir V. Mazalov
Institute of Applied Mathematical Research
Karelian Research Center of Russian Academy of Sciences
Pushkinakaya str. 11, Petrozavodsk 185610, Russia
e-mail : vmazalov@krc.karelia.ru

Anatoliy A. Zabelin
Chita State Pedagogical University
Babushkin str. 121, Chita 672090, Russia
Abstract
Abargaining problem with two players Labor (player L) and Management (player M) is considered.
The players must decide the monthly wage payed to by M. At the begining players andthere is an agreement at $(s_{1}+s_{2})/2$ . If not, the arbitrator
and . If
submit their offers
is called in and he chooses the offer which is nearest for his solution . We suppose that asolution
is concentrated in two points , $1-a$ at the interval $[0, 1]$ with probabilities , $q=1-p$ . The
equilibrium in the arbitration game among pure and mixed strategies is derived.
Key words: bargaining problem, arbitration, equilibrium strategy.
.
AMS Subject Classification:$\mathrm{M}$$\mathrm{L}$$\mathrm{L}$$s_{1}\leq s_{2}$$s_{2}$$s_{1}$$\alpha$$p$$a$$\alpha$$91\mathrm{A}05,91\mathrm{A}80,91\mathrm{B}26$

1Introduction
We consider azer0-sum game related with amodel of the labor-management negotiations
using an arbitration procedure. Imagine that two players: Labor (player ) and Management(player $M$ ) bargain on awage bill which has to be in the range $[0, 1]$ where the current wage
bill is normalised at zero, and the known maximum management ability to pay is at 1.
Player is interested to maximize awage bill as much as possible and the player $M$ has the
opposite goal.
respectively, ,andAt the begining the players and $M$ submit their offers$[0,1]$ . If
there is an agreement at $(s_{1}+s_{2})/2$ . If not, the arbitrator is called inand he has to choose one of the decisions.
There are different approaches in analyzing the arbitration models [1-6]. We consider
here the final-0ffer arbitration procedure [3] which allows the arbitrator only to choose oneof the two final offers made by the players. We suppose here that the arbitrator imposes asolution which is random variable being concentrated in two points and $b=1-a$ withdifferent probabilities and $q=1-p$ , $0\leq a,p\leq 1$ . The arbitrator chooses the offer which
is nearest for his solution . The solution of this game with equal $p=q=1/2$ was obtainedin [6]. In this paper we obtain the solution of this game where and can be non-equal
$L$

$L$

$L$

$s_{1}$$s_{2}$$s_{1}$$A$$s_{1}\leq s_{2}$$a$$\alpha$$p$$\alpha$$p$$q$$s_{2}\in$118
So, we have azer0-sum game determined in the unit square where the strategies ofplayers L and M are the real numbers , $s_{2}\in[0,$ 1] and payoff function in this game hasform $H(s_{1}, s_{2})=EH_{\alpha}(s_{1},s_{2})$ , where$s_{1}$$(s_{1}+\mathit{8}_{2})/2$$s_{1}$$H_{\alpha}(s_{1},s_{2})=\{$$s_{2}$$\alpha$

,

if
if
if
if

,
,

,

$s_{1}\leq s_{2}$$s_{1}>s_{2}$, $|s_{1}-\alpha|<|s_{2}-\alpha|$, $|s_{1}-\alpha|>|s_{2}-\alpha|$$s_{1}>s_{2}$ , $|s_{1}-\alpha|=|s_{2}-\alpha|$$s_{1}>s_{2}$(1)

Below we show that the equilibrium in this game in dependence on value a can be among
pure (section 2) and mixed (sections 3-4) strategies.

2Solution of the game. Pure strategies
Theorem 1. Let $p\in(0,0.5]$ and $a\in[0,p/2]$ . Equilibrium consists of pure strategies andhas form,. The value of the game $v=q$ .
Proof. Let player II uses $s_{2}=0$ . The payoff of player Iis equal to:for $s_{1}\in[0,2a)H(s_{1},0)=ps_{1}+qs_{1}=s_{1}<2a\leq p\leq q$ ,for $s_{1}=2aH(2a,0)=pa+(1-p)2a=(2-p)a<2a\leq p\leq q$ ,for$+qs_{1}=qs_{1}$ .
The maximum of the function is reached for $s_{1}=1$ and equals to . Now, suppose that
player Iuses $s_{1}=1$ . For $s_{2}\in[0,1-2a)H(1,s2)=ps_{2}+q$ . Minimum of this function lies in$s_{2}=0$ and equal to . For$s_{2}=1-2aH(1,1-2a)=p(1-2a)+q(1-a)=1-a-ap$ . Because$p/(1+p)>p/2\geq a$ , it follows $p>a+ap$ and $1-a-ap>1-p=q$ . For $s_{2}\in(1-2a, 1]$$H(s_{1},s_{2})=ps_{2}+qs_{2}=s_{2}$ . According to condition $p\geq 2a$
we have $s_{2}\geq 1-2a>1-p=q$ .
So, for all $s_{2}H(1, s_{2})\geq q$ andfor aU . Hence, $\{s_{1}=1,s_{2}=0\}$ ia anequiliblium in the game and $=q$ .
Analogous arguments leads to$s_{1}^{*}=1$$s_{2}^{*}=0$$s_{1}\in(2a, 1]\#(\mathrm{s}\mathrm{i},0)=fl$

$q$

$q$

$\#(\mathrm{s}\mathrm{i},0)\leq q$$s_{1}$$v$

Theorem 2. Let $p\in(0.5,1)$ and $a\in[0,q/2]$ . Equilibrium consists of pure strategies andhas form,, and value of the game $v=q$ .
$s_{1}^{*}=1$$s_{2}^{*}=0$3Method for obtaining the equilibrium among mixed
startegies
In case $a> \min\{p/2,q/2\}$ equilibrium consists of mixed strategies, i.e. randomised strategiesof players and M. Denote (si) anddistribution functions of the strategies for$\mathrm{L}$and$\mathrm{M}$$F_{1}$$F_{2}(s_{2})$, respectively. Suppose, that Fi(si)of two intervals ($\alpha_{1};\alpha_{2}]$and ($\alpha_{3};\alpha_{4}][(\beta_{1};$$\mathrm{L}$$[F_{2}(s_{2})]$&], (a3;$\mathrm{a}4$

is continuous and its support consists] at the [0; 1] with$]$$\alpha_{2}\leq\alpha_{3}[\beta_{2}\leq\beta_{3}]$.

119can have agap. Let also
andIn extreme points of the interval [0; 1] functions
.
and
,
;
at the intervals ( ; ] and (denote the rm of
and
Let
.
at (
and
-for the function
andand, respectively,
Firtsly, consider the case $p\leq 0.5$ . Admit, that the intervals ( ; ] and ( ; ] areare symmetric inand (symmetric in respect on the point and the intervals (respect on 6. Otherwords,$F_{2}(s_{2})$$F_{1}(s_{1})$$\beta_{4}\leq\alpha_{1}$$F_{2}(\beta_{4})=1$$F_{1}(\alpha_{1})=0$$F_{1,34}(s_{1})$$F_{1,12}(s_{1})$$F_{2,12}(s_{2})$$F_{1}(s_{1})$$\mathrm{f}\mathrm{o}$$F_{2}(s_{2})$$F_{2,34}(s_{2})$$\alpha_{1}=2a-\beta_{2}$Suppose, that player,$\beta_{1}=2a-\alpha_{2}$$\mathrm{L}(\mathrm{M})$,$\alpha_{4}=2b-\beta_{3}$,$\beta_{1}$$\alpha_{2}$$\beta_{2}$$\beta_{3};\beta_{4}]$$\alpha_{3};\alpha_{4}]$$a$$(\beta_{3};\beta_{4}]$$\beta_{1};\beta_{2}]$$\alpha_{1}$$\alpha_{3};\alpha_{4}]$$\alpha_{2}$$\alpha_{1}$$\beta_{4}=2b-\alpha_{3}$.

(2)uses amixed strategy $F_{1}(s_{1})(F_{2}(s_{2}))$ and consider the

payoffs of the players.
; ],For$s_{1}\in(\alpha_{1}$$\alpha_{2}$$H(s_{1}, F_{2}(s_{2}))=p \{s_{1}F_{2,12}(2a-s_{1})+\int_{2a-s_{1}}^{\beta_{2}}s_{2}dF_{2,12}(s_{2})+\int_{\beta_{3}}^{2b-\alpha \mathrm{s}}s_{2}dF_{2,34}(s_{2})\}+qs_{1}$For$s_{1}\in(\alpha_{3};\alpha_{4}]$$H(s_{1}$$s_{2}\in(\beta_{1;}\beta_{2}]$(3),, F2(s2 ) $=p\{$ 0$\cdot F_{2}(0)+\int_{2a-\alpha_{2}}^{\beta_{2}}s_{2}dF_{2,12}(s_{2})+\int_{\beta_{3}}^{2b-\alpha_{3}}s_{2}dF_{2,34}(s_{2})\}$$+q \{s_{1}F_{2,34}(2b-s_{1})+\int_{2b-s_{1}}^{2b-\alpha_{3}}s_{2}dF_{2,34}(s_{2})\}$

For

.

.

(4)

,

$H(F_{1}(s_{1}), s_{2})=p[ \int_{a-\beta_{2}}^{2a-s_{2}}s_{1}dF_{1,12}(s_{1})+s_{2}(1-F_{1,12}(2a-s_{2}))\}$$+q \{\int_{2a-\beta_{2}}^{\alpha_{2}}s_{1}dF_{1,12}(s_{1})+\int_{\alpha_{3}}^{2b-\beta_{3}}s_{1}dF_{1,34}(s_{1})+1\cdot(1-F_{1}(1))\}$

For$s_{2}\in(\beta_{3;}\beta_{4}]$

.

(5)

,
$H(F_{1}(s_{1}),s_{2})=ps_{2}+q[ \int_{a-\beta_{2}}^{\alpha_{2}}s_{1}dF_{1,12}(s_{1})+$.

$+ \int_{\alpha_{3}}^{2b-s_{2}}s_{1}dF_{1,34}(s_{1})+s_{2}(1-F_{1,34}(2b-s_{2}))\}$,
are optimal then the equations $H(s_{1}, F_{2}^{*}(s_{2}))=v$ andIf, must be satisfied in the support-intervals where -value of the game. Hence,$F_{1}^{*}(s_{1})$

$v$

(6)
$H(F_{1}^{*}(s_{1}), s_{2})=$$F_{2}^{*}(s_{2})$$v$$H(s_{1}, F_{2}^{*}(s_{2}))=v$,$s_{1}\in(\alpha_{1};\alpha_{2}]\cup(\alpha_{3};\alpha_{4}]$,120$H(F_{1}^{*}(s_{1}), s_{2})=v$,$s_{2}\in(\beta_{1}$;$\beta_{2}$]$\cup(\beta_{3;}\beta_{4}]$.

From here,$\frac{\partial H(s_{1},F_{2}^{*}(s_{2}))}{\partial s_{1}}=0$,$\frac{\partial H(F_{1}^{*}(s_{1}),s_{2})}{\partial s_{2}}=0$$s_{1}\in(\alpha_{1}$,;$\alpha_{2}$]$\cup(\alpha_{3;}\alpha_{4}]$$s_{2}\in(\beta_{1;}\beta_{2}]\cup(\beta_{3;}\beta_{4}]$,.

Finding the derivative of (3-4) in and putting it equal to 0, and using the admission thatandis continuous at, consequently,, we obtainthe system of differential equations with boundary conditions:$s_{1}$$F_{2}^{*}(\beta_{4})=1$$F_{2}^{*}(s_{2})$$[\beta_{2};\beta_{3}]$$F_{2}^{*}(\beta_{2})=F_{2}^{*}(\beta_{3})$$p\{2(s_{1}-a)F_{2,12}^{*’}(2a-s_{1})-F_{2,12}^{*}(2a-s_{1})\}-q=0$$q\{2(b-s_{1})F_{2,34}^{*’}(2b-s_{1})+F_{2,34}^{*}(2b-s_{1})\}=0$$F_{2,34}^{*}(\beta_{4})=1$,$F_{2,12}^{*}(\ )$$=F_{2\beta 4}^{*}(\ )$

.

,

,

$s_{1}\in(\alpha_{1};\alpha_{2}]$$s_{1}\in(\alpha_{3;}\alpha_{4}]$,,Changing the arguments $t_{1}=2a-s_{1}$ ,] in the first equation and $t_{2}=2b-s_{1}$ ,] in the second one we obtain the system:$t_{1}\in(\beta_{1};\beta_{2}$$t_{2}\in(\beta_{3};\beta_{4}$,$\frac{dt_{1}}{2(a-t_{1})}=\frac{dF_{2,12}^{*}}{F_{2,12}^{*}+p/q}$$\frac{dt_{2}}{2(b-t_{2})}=\frac{dF_{2,34}^{*}}{F_{2,34}^{*}}$.

The solution which satisfies the boundary conditions has the following form

(7)

$F_{2}^{*}(s_{2})=\{\begin{array}{l}\mathrm{I})pa\not\in a---s_{2}p\mathrm{I},\mathrm{i}\mathrm{f}2a-\alpha_{2}<s_{2}\leq\beta_{2}\mathrm{i}\mathrm{f}s_{2}\leq 2a-\alpha_{2}\mathrm{i}\mathrm{f}\oe<\mathit{8}_{2}\leq\ \mathrm{i}\mathrm{f}h<\mathit{8}_{2}\leq 2b-\alpha_{3}\mathrm{i}\mathrm{f}2b-\alpha_{3}<s_{2}\end{array}$Finding the derivative of (5-6) in$F_{1}^{*}(\alpha_{1})=0$

and$F_{1}^{*}(\alpha_{2})=F_{1}^{*}(\alpha_{3})$and putting it equal to 0, and using the admission, we obtain the system:$s_{2}$$p\{1-F_{1,12}^{*}(2a-s_{2})-2(a-s_{2})F_{1,12}^{*’}(2a-s_{2})\}=0$,$p+q\{1-F_{1,34}^{*}(2b-s_{2})-2(b-s_{2})F_{1,34}^{*’}(2b-s_{2})=0\}$$F_{1,12}^{*}(\alpha_{1})=0$,$F_{1,12}^{*}(\alpha_{2})=F_{1,34}^{*}(\alpha_{3})$Let change the arguments $t_{1}=2a-s_{2}$ ,] in the second equation:.

$t_{1}\in(\alpha_{1}$;$\alpha_{2}$$s_{2}\in(\beta_{1;}\beta_{2}]$,,$s_{2}\in(\beta_{3;}\beta_{4}]$,] in the first equation, and $t_{2}=2b-s_{2}$ ,$t_{2}\in(\alpha_{3};\alpha_{4}$$\frac{dt_{1}}{2(t_{1}-a)}=\frac{dF_{1,12}^{*}}{1-F_{1,12}^{*}}$,.

$\frac{dt_{2}}{2(t_{2}-b)}=\frac{dF_{1,34}^{*}}{1+p/q+F_{2\beta 4}^{*}}$The solution of the system:0,

if,
if $2a-\sqrt 2<s_{1}\leq\alpha_{2}$ ,if,$s_{1}\leq 2a-\beta_{2}$$1-a\not\in 1-a\mathrm{F}_{-a}S\alpha_{2}-a--,$,$F_{1}^{*}(s_{1})=\{$$1+\mathrm{g}-q(_{\alpha-aq}^{a-\mathrm{z}}\mathrm{F}_{2}+)\alpha \mathrm{F}s_{1-}-$1,$\alpha_{2}<s_{1}\leq 0_{3}$,if
if $2b-\ <s_{1}$ .

$\alpha_{3}<s_{1}\leq 2b-\beta_{3}$

,

(8)

121
Now let us substitute the functions (7) $-(8)$ to (3) $-(6)$ . For$\mathit{8}1\in(\alpha_{1}$;$\alpha_{2}$],$H_{1}=H(s_{1}, F_{2}^{*}(s_{2}))=p \frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}((2a-\beta_{2})-(2b-\beta_{3}))+p\alpha_{3}+q(2a-\beta_{2})$For$s_{1}\in(\alpha_{3};\alpha_{4}]$.

,$H_{2}=H(s_{1}, F_{2}^{*}(s_{2}))=p \frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}((2a-\beta_{2})-(2b-\beta_{3}))+p\alpha_{3}+q(2a-\beta_{2})-$$-p \alpha_{2}\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}\cdot\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}-q\alpha_{2}\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}+q\alpha_{3}$For$s_{2}\in(\beta_{1}$;$\beta_{2}$.

],$H_{3}=H(F_{1}^{*}(s_{1}), s_{2})=q \frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}((\alpha_{2}-2a)-(\alpha_{3}-2b))+q\beta_{2}-p(\alpha_{3}-2b)-$$-q \beta_{3}\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}\cdot\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}-p\beta_{3}\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}+p\beta_{2}+q\theta$For$s_{2}\in(\beta_{3};\beta_{4}]$,,$H_{4}=H(F_{1}^{*}(s_{1}), s_{2})=q \frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}((\alpha_{2}-2a)-(\alpha_{3}-2b))+q\beta_{2}-p(\alpha_{3}-2b\rangle$

where

0,
$\theta=\{$$-q\mathrm{E}+(_{\alpha_{2}-aq}^{a-e}\ovalbox{\tt\small REJECT}+)\alpha \mathrm{F}_{a}^{-}$

So, take place,$H_{2}=H_{1}+\chi_{1}$,$H_{3}=H_{4}+\chi_{2}$,

if $F_{1}^{*}(1)=1$ ,if $F_{1}^{*}(1)<1$ .

where$\chi_{1}=-p\alpha_{2}\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}\cdot\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}-q\alpha_{2}\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}+q\alpha_{3}$,.

$\chi_{2}=-q\beta_{3}\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}\cdot\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}-p\beta_{3}\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}+p\beta_{2}+q\theta$

But must be $H_{1}=H_{2}=v$ and $H_{3}=H_{4}=v$ , hence$\chi_{1}=0$,$\chi_{2}=0$,$H_{1}=H_{4}$.

Below we will find asolution of the system (9) in different cases. The value of the}equal
$v=p \frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}\cdot((2a-\beta_{2})-(2b-\beta_{3}))+p\alpha_{3}+q(2a-\beta_{2})$.

122
Denote. After symplifications (9) can be rewritten:$\frac{\sqrt{a-}}{\sqrt{\alpha_{2}-a}}=\frac{1}{x},\alpha\ovalbox{\tt\small REJECT}_{-3}^{-}=y$$- \frac{\alpha_{2}}{x}(py+q)+q\alpha_{3}=0$,$-\beta_{3}y(q/x+p)+p\beta_{2}+q\theta=0$

,

(11)

$p(y(2a-2b- \beta_{2}+\beta_{3})+2\alpha_{3}-2b)=q(\frac{1}{x}(\alpha_{2}-\alpha_{3}-2a+2b)+2\beta_{2}-2a)$.

If $F_{1}^{*}(1)=1$ (or, $F_{1,34}^{*}(2b-\beta_{3})=1$ , or $\theta=0$ ), then $y(q/x+p)=p$ . Substituting itto (11) we receive$\beta_{2}=\beta_{3}$. If$F_{1}^{*}(1)<1$$q\theta=-p+(q/x+p)y$ .

(2b-&=l), then $\beta_{3}=2b-1$ ,$y=\not\in_{a}\alpha-$

andAnalogously, if $F_{2}^{*}(0+)=.0(F_{2,12}^{*}(2a-\alpha_{2})=0)$ , then
. Substituting to(11), we receive. If $F_{2}^{*}(0+)>0(2a-\alpha_{2}=0)$ , thenand.
Thus, take place $F_{1}^{*}(1)=1\Rightarrow\beta_{2}=\$ and.
Varying different collections of the values
andand demanding that the
support of optimal strategies belongs to [0; 1], we will obtain the form of optimal
strategies
depending on values of and (see Fig. 1).
$1/\mathrm{x}(\mathrm{p}\mathrm{y}+q)=q$$\alpha_{2}=\alpha_{3}$$\alpha_{2}=2a$$1/x=\mathrm{F}_{a}^{a-}$$F_{2}^{*}(0+)=0\Rightarrow\alpha_{2}=\alpha_{3}$$F_{1}^{*}(1)$$a$$F_{2}^{*}(0+)$$p$4Solution of the game. Mixed Strategies
4.1

Equilibrium for (p,$a)\in D_{1}$Suppose that $F_{1}^{*}(1)=1$ and $F_{2}^{*}(0+)=0$ (i.e.
,it follows$\frac{1}{x}=\mathrm{F}aA-a-$$\alpha_{2}=\alpha_{3}=A$,$\beta_{2}=\ =\mathrm{B}$). From the equations$y= \frac{\sqrt{A-}}{\sqrt{b-B}}$$\alpha_{2}=\alpha_{3}=A=\frac{bx^{2}(1+y^{2})-ay^{2}(1+x^{2})}{x^{2}-y^{2}}$,$\beta_{2}=\beta_{3}=B=\frac{a(1+x^{2})-b(1+y^{2})}{x^{2}-y^{2}}$. (12)

The first two equations in (11) give$\{$$qx=py+q$ ,$y(_{x}^{q}+p)=p$,

which positive solution is
x

(13)

$=$

$y= \frac{p^{2}-pq-q^{2}+\sqrt{p^{4}+2p^{3}q-p^{2}q^{2}+2pq^{3}+q^{4}}}{2p^{2}}$

.

(14)

It is not difficult to check that it satisfies to the third equation in (11).
The values $x,y$ and (12) give the solution of the game iff the following system of inequalities be satisfied,,$\beta_{1}\geq 0$$\alpha_{4}\leq 1$

123

0.4
0.381

Fig. 1

Theorem 3. For $(p,a)\in D_{1}$ the equilibrium isdetemined by (12-14). The value of the game:$(F_{1}^{*}, F_{2}^{*})$of the form (7-8) with parameters$v=q(2a-\beta_{2})+p\alpha_{3}-2p(2b-1)\mathrm{f}^{\alpha}\mathrm{f}^{-}-\mathrm{i}_{2}$.

Notice some properties of the solution:$\lim_{parrow 0}a_{1}(p)=0.4$

where

$z$

,

$\lim_{parrow 0.5}a_{1}(p)=z^{2}$,is the “golden section” of the interval $[0, 1]$ . It follows from$\lim_{parrow 0+}x=1$,$\lim_{parrow 0+}y=0$,$\lim_{parrow 0.5-}x=\frac{\sqrt{5}+1}{2}$,$\lim_{parrow 0.5-}y=z=\frac{\sqrt{5}-1}{2}$.

$a=a_{1}(p)$Notice also, that for fixed if decreases then 04 increases to 1and reaches it for
instead of to $\alpha_{4}=2-2a-\beta_{3}$ ) . For values $a\leq a_{1}(p)$ ,(to obtain it we can substitutethe solution of the game is different.
$p$

$a$

$a_{1}(p)$4.2

Equilibrium for (p, a)$a$$\in D_{2}$If $\mathrm{F};(1)<1$ and $F_{2}^{*}(0+)=0$ (or, equiavently,
we obtain
andfrom the equations$\alpha_{2}=\alpha_{3}=A$$\frac{1}{x}=\not\in_{-a}^{a-}$, $\beta_{2}=B$ , $\beta_{3}=2b-1$ ), then$y=\Leftrightarrow_{a}^{-b}$$\alpha_{2}=\alpha_{3}=A=ay^{2}+b$,$\beta_{2}=B=\frac{a(1+x^{2})-(ay^{2}+b)}{x^{2}}$

.

(15)

124
The first two equations of (11) take form
$qx=py+q$ ,

(11)

$\{$

$2ay(_{x}\mathrm{A}+p)=p(1-B)$

From the first equation it follows
receive after symplification.

. Substituting it to the second equation we$x=\ovalbox{\tt\small REJECT}+_{\mathrm{A}}q$(2 $y^{3}ap^{3}+(-p^{3}+4ap^{2}q+paq^{2}+ap^{3})y^{2}+$$+(2aq^{3}-2p^{2}q+2paq^{2}+2ap^{2}q)y+3paq^{2}-2pq^{2})/(py+q)^{2}=0$

.

(17)

Substituting it to the third equation in (11) we obtain$y(2y^{3}ap^{3}+(-p^{3}+4ap^{2}q+paq^{2}+ap^{3})y^{2}+$$+(2aq^{3}-2p^{2}q+2paq^{2}+2ap^{2}q)y+3paq^{2}-2pq^{2})/(py+q)^{2}=0$ .

It is sufficient to find only positive roots of (17).
Denoting A $=p/q$ we have
$2a\lambda^{3}y^{3}+\lambda(a+4a\lambda-\lambda^{2}+a\lambda^{2})y^{2}+2(a+a\lambda-\lambda^{2}+a\lambda^{2},)y+\lambda(3a-2)=0$

.

(18)

Denote the cubic polynomial at the left side of (18) as, $\nu(0)=\lambda(3a-2)<0$,$a\in[0;0.5$ ). The coefficient in higher degree ofin (18) is positive, hence, at least one
postive root exists. From here also follows that the maximum lies before minimum. The
functionha two extreme points
and. With$\nu(0)<0$ it gives the uniqueness of the positive root of (18).
The solution takes place in case of, or $a(3-y^{2})\geq 1$ . It determines the lower
borderof the region
on the plane $(p,a)$ .
$\nu(y)$

$y$

$\nu$

$=\mathrm{v}(\mathrm{y})$$y_{1}= \frac{1}{3}(\frac{1}{a}-\frac{1+\lambda+\lambda^{2}}{\lambda^{2}})$$y_{2}=- \frac{1}{\lambda}<0$$\beta_{1}\geq 0$$a_{2}(p)$$D_{2}$

Theorem 4. For $(p,a)\in D_{2}$ the equilibrium isof the form (7-8) with parameters
determined by (15-17). The value of the game: $v=q(2a-\beta_{2})+p\alpha_{3}-p(2b-1+\beta_{2})^{\alpha}\not\in_{a}^{-}$ .
$(F_{1}^{*}, F_{2}^{*})$

In case $a<a_{2}(p)$ the following solution

4.3$\mathrm{w}\mathrm{i}\mathrm{U}$take place.

Equilibrium for (p,$a)\in D_{3}$If $F_{1}^{*}(1)<1$ and $F_{2}^{*}(0+)>0$ (or, equivalently, $\alpha_{2}=2a$ , $\beta_{3}=2b-1$ , $\alpha_{3}=A$ , $\beta_{2}=B$ ),the first two equations in (11) with
and(or, $\beta_{2}=B=a-a/x^{2}$ and$\alpha_{3}=A=ay^{2}+b)$ take the form
$1/x=\Leftrightarrow_{a}^{a-B}$$y=\Leftrightarrow_{a}^{A-b}$$2a(py+q)=q(ay^{2}+b)x$ ,$\{$$2ay( \frac{q}{x}+p)=p(b+\frac{a}{x^{2}})$

From the first equation in (19) it follows.

$x= \frac{2a(py+q)}{q(ay^{2}+b)}$(19). Substituting it to the second

equation in (19) and the third equation in (11) we obtain(3 $a^{2}y^{4}q^{2}p+(8a^{2}p^{3}+4a^{2}q^{3})y^{3}+(-2pa^{2}q^{2}-4p^{3}a+16a^{2}p^{2}q+4a^{2}p^{3}+2paq^{2})y^{2}+$125$+(8a^{2}p^{2}q+8pa^{2}q^{2}-4a^{2}q^{3}-8p^{2}aq+4q^{3}a)y+3pa^{2}q^{2}-pq^{2}-2paq^{2})/(4a(py+q)^{2})=0$

.

(20)

and
$y(3a^{2}y^{4}q^{2}p+(8a^{2}p^{3}+4a^{2}q^{3})y^{3}+(-2pa^{2}q^{2}-4p^{3}a+16a^{2}p^{2}q+4a^{2}p^{3}+2paq^{2})y^{2}+$$+(8a^{2}p^{2}q+8pa^{2}q^{2}-4a^{2}q^{3}-8p^{2}aq+4q^{3}a)y+3pa^{2}q^{2}-pq^{2}-2paq^{2})/(4a(py+q)^{2})=0$.

It is sufficient to find only positive solutions of (20).
Denoting $\lambda=p/q$ we rewrite (20) in the form
$3a^{2}\lambda y^{4}+4a^{2}(1+2\lambda^{3})y^{3}+2a\lambda(1-a+8a\lambda-2\lambda^{2}+2a\lambda^{2})y^{2}+$$+4\mathrm{a}(1-a+2a\lambda-2\lambda^{2}+2a\lambda^{2})y-(1-a)(1+3a)\lambda=0$.

polynomials at the left side of the equation. Then $\nu(0)=-(1-a)(1+3a)\lambda<$
Denote, and because the coefficient in higher degree of is positive then there exists at least one
positive root of the equation. Let us show that it is unique. It follows from the fact that the
points where $\nu’(y)=0$ are negative.
$\nu(y)$$0$$y$$\nu’(y)=36a^{2}\lambda y^{2}+24a^{2}(1+2\lambda^{3})y+4a\lambda((1-a)(1-2\lambda^{2})+8a\lambda)$

If this parabola has no roots thenthere are two roots$\nu(y)$.

is concave and the positive root is unique. Let$y_{1,2}=$is negative. Coefficient in higher degree of of $\nu’(y)$ is positive, hence, theThe rootis negative, iff the coefficient in lower degree of $\nu(y)$ is positive. It is equallargest rootis
to $\xi(a,\lambda)=(1-a)(1-2\lambda^{2})+8\mathrm{a}\mathrm{A}$ . We have: $\xi(a,\mathrm{O})=1-a>0$ , the function$\nu’(y)$, then $\xi(a, \lambda)>0$ , coefficient in lower degree inconvex in , $\xi(a, 1)=9a-1$ . If
is positive, is negative, hence, the positive root of the equation is inique.
. this enequality determines the loweror
The solution takes place, iffisthe inequality
on the plane $(p, a)$ . Notice, that inof the region
bordersatisfied automatically.
$y$$y_{1}$$y_{2}$$\xi(a, \lambda)$$a> \frac{1}{9}$$\lambda$$y_{2}$$\beta_{2}\geq 0$$a_{3}(p)$$\frac{ay^{2}+b}{2a(1+\lambda y)}\leq 1$$D_{3}$$D_{3}$

Theorem 5. For $(p, a)\in D_{3}$ the equilibrium isdetemined by (19-20). The value of the game:, thento
For fixed , if decreases fromthe case $a<a_{3}(p)$ .

$(F_{1}^{*}, F_{2}^{*})$$a< \frac{1}{9}$of the form (7-8) with parameters.

$v=q(2a- \beta_{2})+p\alpha_{3}-p(2b-1+\beta_{2})\frac{\sqrt{\alpha_{3}-}}{\sqrt{a}}$

$p$

4.4
For

$a_{2}(p)$$a$$a_{3}(p)$$\beta$decreases to zero. Finally, consider

Equilibrium for (p,$a)\in D_{4}$$\alpha_{1}=\alpha_{2}=2a$,$\alpha_{4}=1$,$\beta_{1}=\beta_{2}=0$, $\beta_{3}=2b-1$ the optimal strategies are0,$F_{1}^{*}(s_{1})=\{$ $\frac{1}{q}(1-\alpha Fs_{1}-b-b)$

1,

,

,

if

$s_{1}\leq\alpha_{3}$if$\alpha_{3}<s_{1}\leq 1$

if $1<s_{1}$ ,

,

(20)

126
0,

if

$s_{2}\leq 0$,’$F_{2}^{*}(s_{2})=\{$$\frac{\sqrt{\alpha_{3}-}}{\mathrm{F}_{-2}^{\alpha-}\sqrt{a}}.,$(22)$\mathrm{i}\mathrm{f}2b-1<s_{2}\leq 2b-\mathrm{a}_{3}\mathrm{i}\mathrm{f}0<s_{2}\leq 2b-1,$,
if $2b-a_{3}<s_{2}$ .

1,Then, for$s_{1}\in(\mathrm{a}_{3};1]$$H_{2}=H(s_{1}, F_{2}^{*}(s_{2}))=p \{0\cdot F_{2}^{*}(0)+\int_{2b-1}^{2b-\alpha_{3}}s_{2}dF_{2}^{*}(s_{2})\}+$.

$+q \{s_{1}F_{2}^{*}(2b-s_{1})+\int_{2b-s_{1}}^{2b-\alpha_{3}}s_{2}dF_{2}^{*}(s_{2})\}=\alpha_{3}-\frac{p\sqrt{\alpha_{3}-b}}{\sqrt{a}}$If $s_{2}=0$ , then.

$H_{3}=H(F_{1}^{*}(s_{1}),s_{2})=q \{\int_{\alpha \mathrm{s}}^{1}s_{1}dF_{1}^{*}(s_{1})+1\cdot(1-F_{1}^{*}(1))\}=2\sqrt{a}\sqrt{\alpha_{3}-b}+2b-\alpha_{3}-p$

If $s_{2}\in(2b-1;2b=\alpha$ ], then$H_{4}=H(F_{1}^{*}(s_{1}),s_{2}) \cdot=ps_{2}+q\{\int_{\alpha \mathrm{s}}^{2b-s_{2}}s_{1}dF_{1}^{*}(s_{1})+s_{2}(1-F_{1}^{*}(2b-s_{2}))\}=2b-\alpha_{3}$$F_{1}^{*}(s_{1})$,$F_{2}^{*}(s_{2})$.

be optimal iff$2b-a_{3}$$\{$$2b-\alpha_{3}$Solution of this system:.
This form fortakes place, iff
the region
on the plane $(p, a)$ .
$\alpha_{3}=b+\not\simeq 4a$$H_{2}.-H_{4}$,$\alpha_{3}$$\leq 1$or, equivalently, $a>p/2$ . That determines$D_{4}$

Theorem 6. For $(p,a)\in D_{4}$ the equilibrium isthe game: $v=b-4L^{2}a$ .

$(F_{1}^{*}, F_{2}^{*})$of the form (21-22). The value of

The case $a<p/2$ was analysed in section 2.

5

Solution for p $>0.5$

At the begining we assumed p. In case p $>0.5$ the solution follows from the following
theorem.
Theorem 7. Let for some fixed values of and we found the optimal strategies
and F2(s2,) in the game with$\leq 0.5$$a$$\mathrm{p},$$F_{1}^{*}(s_{1},p,a)$

$p$

$a$

$P\{\alpha=a\}=p$ ,$P\{\alpha=b\}=q$,$a+b=1$ ,$p+q=1$ ,$a<b$,$p\leq q$.

127Then the optimal strategies in the game for the same values ,$a$$P\{\alpha=a\}=q$,$P\{\alpha=b\}=p$,$p$

and for$p+q=1$ ,$a+b=1$ ,$a<b$ ,

$p\leq q$are$G_{1}^{*}(s_{1}, q,a)=1-F_{2}^{*}(1-s_{1},p, a)$,$G_{2}^{*}(s_{2},q,a)=1-F_{1}^{*}(1-s_{2},p, a)$,.

Proof. We have0,
$1-f_{-}^{\alpha-}1-\alpha \mathrm{F}sf\overline{-}i_{3}a’$,$G_{1}^{*}(s_{1}, q, a)=\{$$1+lp-( \frac{\sqrt{\alpha_{3}-}}{\sqrt{-\mathrm{s}}}+\mathrm{I})ps_{1}\ovalbox{\tt\small REJECT}_{-}a-$1,$1-2a+\alpha_{2}<\mathit{8}_{1}$0,’$G_{2}^{*}(s_{2}, q, a)=\{$,

if $s_{1}\leq 1-2b+\alpha_{3}$ ,if $1-2b+\alpha_{3}<s_{1}\leq 1-\sqrt 3$ ,if $1-\sqrt 3<s_{1}\leq 1-\beta_{2}$ ,if $1-\beta_{2}<s_{1}\leq 1-2a+\alpha_{2}$ ,,if$\mathrm{F}_{2}^{\alpha_{2}aq}\alpha\not\in_{-s_{2}}(_{a-}^{a-2\mathrm{g}}\mathrm{F}_{2},’+)\frac{\sqrt{\alpha_{3}-}}{\sqrt{a-s_{2}}}-q\mathrm{E}a--a$1,

if $s_{2}\leq 1-2b+\beta_{3}$ ,,if $1-2b$ %if $1-\alpha_{3}<s_{2}\leq 1-\alpha_{2}$ ,if $1-\alpha_{2}<s_{2}\leq 1-2a+\beta_{2}$ ,if $1-2a+\beta_{2}<s_{2}$ .
$\sqrt 3<s_{2}\leq 1-\alpha_{3}$These functions will represent the optimal strategies, iff$\mathrm{H}\{\mathrm{s}\mathrm{u}G_{2}^{*}(s_{2},q,a))=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}$$H(G_{1}^{*}(s_{1}, q, a),s_{2})=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}$for $s_{1}\in(1-2b+\alpha_{3};1-\ ]\cup(1-\beta_{2};1-2a+\alpha_{2}]$,
for $s_{2}\in(1-2b+\beta_{3};1$ -Q3]$\cup(1$-a2; $1-2a+\beta_{2}$ ].

at the intervals (1as the form of function
at the,for theand ( $1-\beta_{2};1-2a+\alpha_{2}]$ andintervals ( $1-2b+\beta_{3};1$ -a3], ( $1-\alpha_{2};1-2a+\beta_{2}]$ , respectively.
We obtain for $s_{1}\in(1-2b+\alpha_{3};1-\beta_{3}]$Denote$G_{1,12}^{*}(s_{1})$and$G_{1}^{*}(s_{1},q, a)$$G_{1,34}^{*}(s_{1})$$2b+\alpha_{3};1-\beta_{3}]$$G_{2,12}^{*}(s_{1})$$G_{2}^{*}(s_{1}, q, a)$$G_{2,34}^{*}(s_{1})$$H_{1}’=H(s_{1}, G_{2}^{*}(s_{1}, q,a))=q \{s_{1}G_{2,12}^{*}(2a-s_{1})+\int_{2a-s_{1}}^{1-\alpha_{3}}s_{2}dG_{2,12}^{*}(s_{2})+\int_{1-\alpha_{2}}^{1-2a+\beta_{2}}s_{2}dG_{2,34}^{*}(s_{2})\}+$$+ps_{1}=q \frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}((\alpha_{3}-2b)-(\alpha_{2}-2a))+p(\alpha_{3}+2a-1)+q(1-\beta_{2})$.

If $s_{1}\in(1-\beta_{2};1-2a+\alpha_{2}$ ], then$H_{2}’=H(s_{1}, G_{2}^{*}(s_{1}, q,a))=q \{0\cdot G_{2}^{*}(0, q, a)+\int_{1-2b+\beta_{3}}^{1\alpha_{3}}s_{2}dG_{2,12}^{*}(s_{2})$%$1-1- \int_{\alpha_{2}}^{2a+\beta_{2}}s_{2}dG_{2,34}^{*}(s_{2})\}+$$+p \{s_{1}G_{2,34}^{*}(2b-s_{1})+\int_{2b-*1}^{1-2a+\beta_{2}}s_{2}dG_{2,34}^{*}(s_{2})\}=$

128$=q \frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}((\alpha_{3}-2b)-(\alpha_{2}-2a))+p(\alpha_{3}+2a-1)+q(1-\beta_{2})-$$-q(1- \beta_{3})\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}\cdot\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}+p(1-\beta_{2})-p(1-\beta_{3})\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\ }}$.

If $s_{2}\in(1-2b+\beta_{3};1-\alpha_{3}$ ], then$H_{3}’=H(G_{1}^{*}(_{\mathit{8}_{1}},q, a), s_{2})=q( \int_{-2b+\alpha_{3}}^{2a-s_{2}}s_{1}dG_{1,12}^{*}(s_{1})+s_{2}(1-G_{1,12}^{*}(2a-s_{2}))\}+$$+p \{\int_{1-2b+\alpha_{3}}^{1-\beta_{3}}\mathit{8}_{1}dG_{1,12}^{*}(s_{1})+\int_{1-h}^{1-2a+\alpha_{2}}s_{1}dG_{1,34}^{*}(s_{1})+1\cdot(1-G_{1}^{*}(1))\}=$$=p \frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\ }}((2b-\beta_{3})-(2a-\beta_{2}))+p(1-\alpha_{3})-q(1-2b-\ )-$.

$-p(1- \alpha_{2})\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\ }} \frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}+q(1-\alpha_{3})-q(1-\alpha_{2})\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}+p\eta$

If $s_{2}\in(1-\alpha_{2};1-2a+\beta_{2}$], then$H_{4}’=H(G_{1}^{*}(_{\mathit{8}1},q,a),s_{2})=qs_{2}+p \{\int_{1-2b+\alpha_{3}}^{1\beta_{3}}s_{1}dG_{1,12}^{*}(s_{1})+$$+ \int_{1-\beta_{2}}^{2b-s_{2}}s_{1}dG_{1,34}^{*}(\mathit{8}_{1})+s_{2}(1-G_{1,34}^{*}(2b-s_{2}))\}=$,$=p \frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\overline{\beta}_{3}}}((2b-\beta_{3})-(2a-\beta_{2}))+p(1-\alpha_{3})-q(1-2b-\beta_{2})$where $\eta=\{$0, if $G_{1}^{*}(1)=1$ ,$-1p+(\mathrm{f}_{-3p}^{\alpha-\iota}\mathrm{f}\mathrm{i}+)a\mathrm{F}_{a}-$, if $G_{1}^{*}(1)<1$ .

We have,$\psi_{1}=H_{2}’-H_{1}’=-q(1-\beta_{3})\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}\cdot\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\ }}+p(1- \sqrt 2)-p(1-\beta_{3})\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\sqrt 3}}$$\psi_{2}=H_{3}’-H_{4}’=-p(1-\alpha_{2})\frac{\sqrt{\alpha_{3}-b}}{\sqrt{b-\beta_{3}}}\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}+q(1-\alpha_{3})-q(1-\alpha_{2})\frac{\sqrt{a-\beta_{2}}}{\sqrt{\alpha_{2}-a}}+p\eta$There are only four possible forms for the functions, it gives:$F_{1}^{*}(s_{1},p, a)$and$F_{2}^{*}(s_{2},p, a)$.

. With$/\chi_{1}=\chi_{2}=0$

1. For $\alpha_{2}=\alpha_{3}=A$ ,$\psi_{1}=\psi_{2}=0$.

$\beta_{2}$$=\beta_{3}=B$

take place

2. For $\alpha_{2}=\alpha_{3}=A$ , $\beta_{3}=2b-1$ take place$\psi_{1}=\psi_{2}=0$.

$\simeq\underline{1}A=\frac{\psi_{2}}{1-A}$$\mathrm{g}\underline{\iota}A=\frac{\psi_{2}}{1-A}$

and
and

$\frac{X2}{B}=\frac{\psi_{2}}{1-B}$, consequently,$\chi_{2}=-\psi_{1}$, consequently,

129
For
0.
For

$\alpha_{2}=2a$, $\beta_{3}=1-2a$ take place$\alpha_{1}=\alpha_{2}=2a$,$\alpha_{4}=1$,$\chi_{1}=-\psi_{2}$$\beta_{1}=\beta_{2}=0$$G_{1}^{*}(s_{1}, q, a)=\{$

and$\chi_{2}=-\psi_{1}$, $\beta_{3}=2b-1$ , the form of

if

1–

if
if $2a<s_{1}\leq 1$ ,if $1<s_{1}$ ,$s_{1} \leq a+\frac{p^{2}}{4a}$$a+ \mathrm{E}\frac{2}{a}4<s_{1}\leq 2\mathrm{a}$

,

1,

0,
$G_{2}^{*}(s_{2}, q,a)=\{$ $1- \frac{1}{q}(1-$Then for$G_{1}^{*}(s_{1})$

,

if
if
if

$s_{2}\leq 0$,$\psi_{1}=\psi_{2}=$$G_{2}^{*}(s_{2})$

is:

,

0,

$1- \frac{p}{2a}$1,, consequently,

,

,

,

$0<s_{2} \leq a-\frac{p^{2}}{4a}$$a- \frac{p^{2}}{4a}<s_{2}$.

$s_{2} \in(0;a-\frac{p^{2}}{4a}]$$H(G_{1}^{*}(s_{1}, q, a), s_{2})=q \{\int_{a+_{4a}^{\mathrm{L}_{-}^{2}}}^{2a-s_{2}}s_{1}dG_{1}^{*}(s_{1}, q, a)+s_{2}(1-G_{1}^{*}(2a-s_{2}, q, a))\}+$.

$+p \{\int_{a+_{4}^{e}\frac{2}{a}}^{2a}s_{1}dG_{1}^{*}(s_{1}, q, a)+1\cdot(1-G_{1}^{*}(1, q, a))\}=a+\frac{p^{2}}{4a}$For$s_{1}\in(a+E^{2};4\overline{a}2a]$$H(s_{1}, G_{2}^{*}(s_{2}, q, a))=q \{s_{1}G_{2}^{*}(2a-s_{1}, q, a)+\int_{2a-s_{1}}^{4}s_{2}dG_{2}^{*}(s_{2}, q, a)\}a-R_{\frac{2}{a}}+ps_{1}=a+\frac{p^{2}}{4a}$.

Finally, for $s_{1}=1$$H(s_{1}, G_{2}^{*}(s_{2}, q, a))=q \int_{0}^{a-_{4}^{\mathrm{p}_{\frac{2}{a}}}}s_{2}dG_{2}^{*}(s_{2}, q, a)+p=a+\frac{p^{2}}{4a}$.

In all cases the payoff is constant, and with $H_{1}+H_{4}’=1$ , $H_{4}+H_{1}’=1$ and $H_{1}=H_{4}$ ,and;ives $H_{1}’=H_{4}’$ , and all , $i=1$ , .., 4 are equal. It proves the optimality$H_{j}’$$(s_{2}, q,a)$$G_{1}^{*}(s_{1}, q,a)$.

knowledgements.
The research was supported by the Russian Fund for Basic Research (projects 01-01was supported also by the Japan Society for the Promotion 01, 01-01-00113).
ence (grant 01530)$\mathrm{V}\mathrm{M}$$\mathrm{L}26$$\mathrm{L}$130
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