J.Appl. Prob. 30, 627-638(1993)
Printed in Israel
© Applied Probability Trust 1993
EQUILIBRIUM POINTS FOR THREE GAMES BASED
ON THE POISSON PROCESS
M. T. DIXON,* University of Cambridge
Abstract
An arbitrary number of competitors are presented with independent Poisson
streams of offers consisting of independent and identically distributed random
variables having the uniform distribution on [0, 1]. The players each wish to accept a
single offer before a known time limit is reached and each aim to maximize the
expected value of their offer. Rejected offers may not be recalled, but they are passed
on to the other players according to a known transition matrix. This paper finds
equilibrium points for two such games, and demonstrates a two-player game with an
equilibrium point under which the player with the faster stream of offers has a lower
expected reward than his opponent.
OPTIMAL STOPPING; SEQUENTIAL GAMES
AMS 1991 SUBJECT CLASSIFICATION: PRIMARY 90006
1. Introduction
In [2] Enns and Ravindran study the following two-player game:
Each player receives a Poisson stream with rate qj of offers consisting of independent
and identically distributed random variables having the uniform distribution on [0, 1],
denoted by U[O, 1]. The streams are assumed to be independent. If a player accepts an
offer, he drops out of the game and all subsequent offers on his stream go directly to his
opponent. If instead he rejects the offer, it is passed to the other player for his consideration. Offers rejected by both competitors are lost and may not be recalled. This
procedure continues until either both players have accepted or a predetermined time
limit expires, when a reward of one unit is paid to whichever of the two players (if either)
holds the highest offer. Both players seek to maximize their expected reward Of,
equivalently, to maximize the probability of taking a larger offer than their opponent.
The methods used to analyze the above game are of general application, in particular
they adapt easily to incorporate more than two players. However, an attempt to do this
with the above reward structure fails to provide even asymptotic results because of the
awkwardness of the expressions involved. This paper investigates three games of the
above type but with different reward structures which allow, in many cases, a complete
Received 26 November 1991; revision received 4 August 1992.
* Present address: 18 Orwell Close, St. Ives, Huntingdon, Cambs PE 17 6FP, UK.
627
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628
M. T. DIXON
solution. All games have a given time limit and a matrix of transition probabilities for
passing on rejected offers. In the first game, play terminates when the first offer is
accepted and the accepting player receives the value of the offer. The second game also
allows at most one player to accept, but this time he receives a reward in payments from
the other players, making this a zero-sum game. The third game allows every player to
accept at most one offer. When an offer is accepted, the accepting player drops out of the
game and the arrival rates and transition probabilities are allowed to change. At the end
of the game, all players who have accepted receive the value of their offer. Section 2
explains the methods used and gives some basic results which will be employed later.
Sections 3-5 give the analysis for three possible reward structures. Section 6 looks at the
game of Section 5 in more detail for two players.
Sakaguchi [6] has studied the third game here for two players; he also gives results for a
general discrete-time formulation. His work differs from ours in that he imposes a
constraint on the players' behaviour to eliminate all but one of the possible equilibrium
points, including the most interesting one. To the author's knowledge, all other work
considers maximizing the probability of the accepted offer satisfying a given condition
rather than maximizing the value of the offer.
Majumdar [3] solves the two-player, discrete-time game where the number of offers is
fixed and the offer values are uniformly distributed on [0, 1] - a problem posed by
Sakaguchi who deals with the case where the distribution of the offer values is unknown
in a series of papers ([5], and references therein).
A general solution to a two-player, zero-sum, discrete-time optimal stopping game was
given by Dynkin (see [4]) but, in general, the games considered here do not fit into his
formulation.
2. Preliminaries
The solution concept employed by Enns and Ravindran is that of an equilibrium
point, which is defined as a set of strategies for each player such that no player has any
incentive to deviate unilaterally. For competitive games with a unique equilibrium point
or one point which all players prefer to all others (a dominating point), equilibrium
points provide a natural solution. However, a game need not have a unique equilibrum
point, and if different players prefer different points there is no reason for basing
strategies on equilibrium points. In such situations we might consider the alternative
concept of a minimax-type solution in which each player bases his strategy on the
assumption that opponents will play purely to minimize his expected winnings. Playing
this minimax strategy provides a guaranteed minimum expected income with no
requirements that other players toe the line.
To begin setting up a model of the game, let ({Jij(x, t) denote the probability that an
offer of value x that was first offered to player j will be accepted by player i with time t to
go (given that he has not yet accepted an offer, ofcourse). We can then describe strategies
in terms of these ({Jij. The ({Jij take values in [0, 1], so we are permitting the use of
randomized strategies by each player. Using t to represent residual time rather than time
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629
Equilibrium points for three games based on the Poisson process
elapsed will give us slightly simpler expressions. Some continuity condition is needed on
these qJ - we restrict attention to strategies satisfying:
for all to > 0, qJ;j(x, t) is continuous in t at to for almost all x.
At to = 0 we require the above condition to hold with 'right continuous' replacing
'continuous'.
Next, we need a way of expressing v;(t) - player i's expected winnings with time t to
go given that no offers have yet been accepted - in terms of the strategy. We will refer to
the function v;(t) as the value of the game to player i at (residual) time t corresponding to
the strategies denoted by the qJij. The notation suppresses the dependence of the value on
each player's choice of strategy. By considering the possible events in a small time
interval (t, t + lit), we get a differential equation for v in terms of the 'P;j' In all our
games, at least one offer must be accepted for any rewards to be earned and so we always
have the boundary condition v(O) = O. To see how this works, consider a single player
allowed to choose just one offer from a rate 1 Poisson stream with a reward equal to the
value of that offer. We have
v(t
+ Llt) =
11
+ 11
(l - Llt)v(t) + M
V'(t)
= -
v(t)
(qJX
(qJX
+ (1 -
+ (l -
qJ)v)dx
+ o(Llt),
qJ)v)dx.
In expressions such as this we frequently suppress the arguments to v; and the qJij
for clarity.
Clearly, choosinge to maximize the derivative at all times will maximize the resulting
v(t), so the optimal strategy for the above player will accept only those offers which
exceed v; that is
qJij(X, t)
=
{~
if x
~
v(t)
if x < v(t)
Substituting this qJ into the above expression yields
v'(t) =
! (1 -
V)2.
This we can solve by separation of variables to get v(t) = t/(t + 2).
One problem that arises when we have two or more players is that to obtain explicit
differential equations for the values it is necessary to know their relative magnitudes at
all times. We conclude this section with some simple results which are useful in the
resolution of this difficulty and indeed throughout the study of these games. In both
cases, the proofs are elementary analysis.
Lemma 2.1. Let f be a differentiable function on [0,(0) with f'(t) > 0 whenever
f(/) = 0 andf(O)~ O. Thenf(t) > ofor all t >0.
Lemma 2.2. Let w be a differentiable function on [0, (0) with w(O) ~ 0 and w'(t)
whenever w(t) > O. Then w(t) ~ 0 for all t.
~O
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630
M. T. DIXON
3. Game 1
Let P = {I, 2,· · ., n} be the set of players. Player i receives a Poisson stream rate qj of
independent identically distributed (i.i.d.) U[O, 1] offers, where L~ qi = 1 and (without
loss of generality) ql = min {qi}. Rejected offers are repeatedly passed on according to an
irreducible transition matrix P until either someone accepts the offer or everyone has
seen it, in which case it is discarded. We assume that this process is instantaneous so that
there is no delay between the arrival of an offer and its acceptance by some player or
rejection by all. We also insist that once a player has rejected an offer, he may not then
accept it if it is presented to him again while it is doing the rounds. This continues until
either (i) the time limit expires or (ii) some player accepts an offer. In case (i) no player
earns any reward, otherwise the player who accepted collects the value of his offer.
Clearly one equilibrium point is for all players to take the first offer to come their way.
However, we show in this section that there is a dominating equilibrium point (i.e. one
that all players prefer to all others) which is characterized by all players refusing offers
below some (time-dependent) threshold and accepting any higher offers.
Notation 3.1. Denote by {ca}~~ll)! the (n - I)! cycles of the n elements {I, 2" · " n}
(indexed by a). Then ca ( } ) is the image of} under c; and c;(}) is the image of} under r
applications of c..
Let Ati .], a) = {c~U): 0 ~ r < m, m = inf{s: c~ (}) = i}}. Thus if an offer is given
first to player} and passed on effectively according to the cycle Ca , A ti.], a) denotes the
players who must reject the offer before player i sees it. In particular,} EA ti.], a) for all
a and i =1=}.
Let (Zn) be a Markov chain with transition matrix P and let Tj be the usual first-passage
time to state} for each}. Define Pa,j = Pj(TcaU) < Tc;U) < · · · < TCa"-IU) < (0), that is, the
probability that offers are effectively passed on according to the cycle c., By irreducibility, La Pa,j = 1 for each}.
The differential equations for this game are, for i = 1, 2,· · " n,
vi(t)
= -
v;(t)
+ L qj
j+i
+ q;
i
r
Jo
+
1 [rp;;x
l [(n-I)!
L
0
ft (1 -
rpki )Vi]
dx
k-I
Pa,j
Il
+ L qj
j
v
i
rift (l -
Jo
f/Jkj w.dx
k=1
]
(1 - ({Jkj )({Jij x dx.
kEA(i,j,a)
a-I
The equilibrium strategies are thus the solutions to
rp;; =
1
if x·~
n (1 -
k:;.i
(3.1)
=0
otherwise,
(n-I)!
if
L
Pa,j
a-I
(3.2)
=0
rpkj)Vi ,
Il
kEA(i,j,a)
(1 -
rpkj)X
~
Il (1 -
rpkj)V,.,
k:;:i
otherwise.
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Equilibrium points for three games based on the Poisson process
631
As a matter of convention, we set ({Jij = 1 whenever it is unspecified by the above
relations. This is included in the above equations.
From (3.1), we see that ({J;j == ({Jjj for all i,) and hence that no rejected offers are ever
accepted by other players - offers are either accepted immediately or rejected by all.
Given this, it is straightforward to deduce from (3.1) and (3.2) and the differential
equation that the dominating equilibrium point is
'Pij(X, t)
Set w = (1 -
Ql)V 1 -
(1
W' =
{~
if x < min{vk(t)}
Ql(V2
n
~ min{vk(t)}
if x
For all i and} we have q;vj
Theorem 3.2.
Proof,
Then
=
qjv;.
=
+ ··· + v
n).
n
W + W Jo j~l qj k~l (1- 'Pkj)dx
-
+
.C
['Pllql(q2
+ · · · + qn) -
'P22Q2ql - • • • - 'Pnnqnqdx dx.
The second integral vanishes because ({J;j is independent of i and). Further, the first
integral lies in [0, 1] and so applying Lemma 2.2 to wand - w we deduce that w == O.
Expressing this in terms of the v's and q's gives VI = ql(V I + · · · + vn ) . By symmetry,
V; = q;(v 1 + · · · + vn ) and the proportionality result follows.
Theorem 3.3.
v;(t)
For each i
= _q.
,_
2 - ql
=
1, 2, · · ., n.
(1 - exp( - (1 - ql)t)) / ( 1 - -ql- exp( - (1 - ql)t) )
2 - ql
and hence
as t
t 00.
If ql > 0 then by Theorem 3.2, V; ~ VI for all i and we can explicitly write down
and solve the differential equation for VI to give the result quoted when i = 1. The
general result then follows by Theorem 3.2.
If ql = 0 then by Theorem 3.2 VI == 0 and so (from (3.1) and (3.2)) ({JII = · · · = ({Jnn = 1
for all x and t. The differential equations reduce to v[(t) = q;/2 - V; which have the
solutions quoted.
Proof.
4. Game 2
The set-up for the second game is as in the previous section except that we change the
rewards. Introduce a non-negative measure (l5 j ) with L~ l5j = ~ > O. If player i is the first
player to accept and the value of his offer is x, he receives (~ - l5;)x via a contribution of
l5jx from each other player j. This is therefore a zero-sum game. Precisely as in Section 3,
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632
M. T. DIXON
we deduce that all offers larger than min {Vj/(~ - l5 j)} are accepted under any equilibrium
strategy. But, since the game is zero-sum, this minimum is at most zero and consequently the unique equilibrium strategy is to accept all offers. Given this, we can solve
the differential equations easily to get
Example 4.1.
1. l5j = I/(n - 1) for all i. In this case, the first player to stop receives the value of his
offer in equal payments from all other players. The game is profitable (i.e. Vj is positive)
only for those players with qj > lin.
2. l5j = qj for all i. With this choice, Vj == 0 for every i.
5. Game 3
Again, P = {I, 2,· · ., n} is the set of players. For each subset S of P and T E IR + we
define the game r*(S, T) as follows. For each i E S, player i receives a Poisson stream of
i.i.d. U[O, 1] offers at rate qf ~ O. If a player rejects an offer then it is passed on to at most
one other player with probabilities given by the substochastic matrix
If this player
also rejects then the offer is lost. The game terminates when either (i) a player accepts an
offer or (ii) time T has elapsed with no acceptances. In case (i) the player who accepted
receives the value of his offer and all others get nothing, in case (ii) no one gets anything.
We now combine these games into a new game F(S', T) in the following natural way. The
players start by playing r*(S, T). If T" terminates due to T elapsing then r also
terminates. Otherwise some player - t.. say - has accepted with time t l left. In this
case, the other players play r*(S\ {il}' (1) until this terminates. This procedure continues until either all players have accepted or the time limit expires, and the reward to
each player is then just the sum of the rewards in those games in which he participated.
Thus players who stopped receive the value of their offer, other players get nothing. To
simplify the notation, we write q~j
for qf\{jJ and, where there is no danger of confusion,
I
qf for q~i}.
I
We impose three conditions on the way in which the arrival rates and transition
matrix are allowed to change when players drop out.
1. qfj~qf+qlPj1,foralli1=j,
2. P~k
~ Pf'J'" for all i 1= k, j 1= k,
I}
3. LjES ql > 0, for all S with IS I > 1.
Note that the expression on the right-hand side of condition 1 is just the arrival rate of
offers to player i if we assume that player j rejects all his offers (which is a reasonable
definition of player j dropping out). Condition 2 says that when player k drops out, the
other players cannot arbitrarily reallocate all their rejected offers; they can only
reallocate those offers which would have gone to player k. Condition 3 is included to
avoid degeneracy.
For each S with i E S, this game is described by the differential equation
Pv.
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633
Equilibrium points for three games based on the Poisson process
d;;S = qf
.C
+ j~i ql
{rpff(x - vf) + (1 - rpff) [j~i PJrpff(vf; - Vf)]} dx
[11 {rpE(V f; -
vf) + (l - rpE) [pffrpJ(x - vf) +
k~/;1rpt [Vfk - vfl]} dx] ·
k
vj
Equilibrium strategies are given by
rpJ= 1
(5.1)
if x
=0
vf
+ L PJrpjf(vfj - vf),
j+i
.
otherwise,
if(l - rpl)Pj~X ~ (1 - rpl)Plvl,
rpJ(i =Fj)= 1
(5.2)
~
=0
otherwise.
To see that there exists at least one solution to (5.1) and (5.2) satisfying the continuity
condition, note that defining rpJ(x, t) = 1 for x ~ vl(t) and then using (5.1) to get rp~
will work.
Lemma 5.1. Ifi ES, then qf ~ ql
Proof.
+ Lj+i qlPl.
Repeated application of Conditions 1 and 2.
Lemma 5.2.
IfiES,thenvl~vf.
Proof. The proof is by induction on IS I and the result is trivial for IS I = 1. Suppose
then that vlj ~ vf and set w = vl- vf. We show that w/(t) ~ 0 whenever w(t» 0 - the
result then follows from Lemma 2.2 since \v(O) = O.
If w > 0, vf > vf ~ vfj and so
vf
dd
t
~ qf
(I rpff(x - vf)dx
J0
~ [qf + .L. qlPl]
) +1
J:
+ L ql (I (l - rpE)PffrpJ(x - vf)dx
j vi
J0
(x - vf)dx
Vi
Recalling that
dvf =
-
dt
-
l'
~
q~(1
1
-
. 2
v~)
"
ifw > 0, then
~O,
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634
M. T. DIXON
using Lemma 5.1, completing the proof.
Theorem 5.3. If i ES
~
P, then
vf==O
and vf
-+
1 as t -+
00
if and only if q = 0
ifand only if q > O.
Proof. Lemma 5.2 implies that if qf = 0 then vf == 0, so assume from now on that
q f > O. The proof is again by induction on IS I and the result is clear for IS I = 1.
Suppose then that vfj -+ 1 for each} =1= i. Define t* = sup{ t : v1 ~ vl for some} =1= i}.
j
Case 1: t* = 00. Given e E (0, 1), there exists to such that vl ~ 1 - e for every} =1= i,
1 > to. Since t* = 00, we can find t, > to such that vl(t l ) ~ 1 - e. Suppose there exists
12> t l with vf(t2) < 1 - e. Then vf(t2) < vfj(t 2) and so (from the differential equation)
dvf/dl ~ 0 at 1 = 12 • From Lemma 2.2 we deduce that vl(t) ~ 1 - e for t > t.. But e was
arbitrary and thus vf -+ 1.
j
Case 2: t* < 00. Restrict attention from now on to t > t*. From the definition of t*
we have vf <vfj for all} =1= i and so dvi tdt ~ O. Thus vf(t)f u and it only remains to
. show u = 1. By Condition 3 we must have at least one of qf > 0 and ql > 0 for some
j
=1=
i.
First suppose that qf > 0 and let qJ be any solution to (5.1) and (5.2). Then
d;;S = qf
.C {9?~(X
- vf) + (l -
9?~) [j~i PJ9?X(vf
J -
vf)]} dx
+ non-negative terms not involving qJji
~
above with qJji replaced by any valid strategy 'II;;
(since qJji is chosen to maximize the integrand)
~ qf
11
'IIii(X - vf)dx.
Take
'IIii(X, t) =
{~
if x
~
vf(t)
if x < vf(t)
As in the example of Section 2, we have dvf/dt ~ !qf(l - Vf)2. Ifu < 1, then dvt tdt ~
!qf(l - U)2 > 0 which is clearly impossible.
If, instead, ql > 0 for some } =1= i then, exactly as in Section 2, we have vj =
qft/(qft + 2). By Lemma 5.2, vl ~ vf and vlk ~ vj and so vf + L.k+jPjlqJ~(vfk - vf) ~
vf. Thus any solution to (5.1) and (5.2) has qJl = 1 for x > vj. Putting this into the
differential equation gives
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635
Equilibrium points for three games based on the Poisson process
If u < 1, the third factor is bounded away from 0 and so dv f / dt
e > O. This contradicts vf remaining bounded and so u = 1.
~ e/(2
+ qjt) for some
In general there will not be a unique or dominating equilibrium point, and so a
different solution concept is required. As a possible alternative, we consider a minimax
solution. This will be the solution to the differential equation for vi(t) by first choosing
«(fJj.)j'fl=i to minimize the integrand and then choosing ({li. to maximize this new integrand.
We have the following analogue of Theorem 5.3, which is proved in a similar way.
Theorem 5.4. For each S with i ES
for player i,
vf
0
~
P, if all players play the minimax strategy
if and only if qi w L qlP} =0
j =1=;
and v f
--. 1 as t --. a:
ifandonlyifqf+
L qlPj~>O.
j =l=i
The essential difference between Theorem 5.3 and Theorem 5.4 is the case qf > 0,
+ Lj'fl=i qlPjf = 0; i.e. the case where player i will start receiving offers but only after at
least one other player has dropped out. In the minimax case, all other players sacrifice
their offers so that player i never gets any. However, if we assume that all players play
their own minimax strategies, this problem disappears and the result of Theorem 5.3
holds.
qf
6. A closer look at Game 3 with two players
In the notation established above set P = {I, 2}, qi = qI' qf = q2' Pi; = P£ = 1 and
2
2}
qI{1} = qi = 1 where ql + q2 = 1 and 0 < ql < q2 < 1. Write v for v {1} = vi }, i.e. the best
I
that one player can expect to do receiving a rate 1 stream of offers. As we saw in Section
2, v = t /(t + 2).
There are four equilibrium points, differing only in their treatment of offers in the
interval VI v V2 ~ X ~ v. They are given in Table 6.1. Notice the importance of not
cancelling the common (1 - ((ljJ) factor in (5.1) and (5.2) - doing so would lose three of
the four equilibrium points.
TABLE
VI /\ V2
V I/\ V2
~II
0
'fJ21
0
0
0
~22
'P12
=
VI
<
6.1
X ~ VI V V2
VI /\ V2
=
VI V V2
II
V2
<x
v<x~1
~ V
III
IV
0
0
0
1
1
1
1
0
1
0
0
1
0
1
0
0
1
1
1
0
0
1
1
0
We know that in each case, v;(t) --. 1 so it is clearly of interest to investigate the rate of
convergence. To this end, we seek constants Xi for each point such that
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636
M. T. DIXON
W· =
(t
I
+2) (V, - 1 +~)
t +2
--+
I
o.
The following paragraphs explain the methods used for the four points. We begin with a
result which will be useful: the proof is by elementary analysis.
Lemma 6.1. Suppose w satisfies w'(t)=(aw 2+bw+c)/(t+2) with a>O, b 24ac > 0 and w(O) < [- b + Jb 2 - 4ac]/2a. Then w(t)--+ [- b - Jb 2 - 4ac]/2a monotonically as t --+ 00. Further, this limit is zero ifand only if b ~ 0, c = o.
Example 6.2: Equilibrium point 1. If V2 ~
VI we
have
and similarly
dV I
dt
-
I
= 1(V 1 -
1) 2 -
I
1) 2.
1(V -
We can use the above expressions to apply Lemma 2.1 to w(t) = V2(t) - vl(t) (since this
only needs the derivatives when vl(t) = v2(t» and deduce that in fact V2(t) > vl(t) for all
t > O. Hence the above expressions for the derivatives are valid at all times.
Now define
w2(t ) = (t
+2) (V
2 -
1 +~)
t+2
wl(t) = (t
+ 2) (VI
-
1 + -y-) .
t+2
and
Then
dW2
dt
-
1
= - - [W 1W2
t
+2
+ (1 -
y)w2 + (2 - x)w 1 + (y(x - 2) + 2q2 - x)]
with w2(0) = x - 2, wl(O) = y - 2.
Lemma 6.1 shows that necessary and sufficient conditions for
and !y2 - y - 2q2 = O. Thus y = 1 +
+ 4q2 = 1 +
4ql.
The obvious candidate for x is the solution to y(x - 2) + 2q2 -
Jl
J5 -
2ql
x=2+ J 5 - 4ql
WI
--+
X =
0 are 1 - y < 0
0, i.e.
'
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637
Equilibrium points for three games based on the Poisson process
and the lemmas of Section 2 will provide a proof that this works. The trick is to apply
Lemma 2.1 to the function wi - WI to show that wi(t) < wl(t) for t > O. The result then
follows since wl(t) --+ O.
Example 6.3: Equilibrium point II. Using the method of the previous example, we
can show that v2(t) > vl(t) for t > O. Defining WI and W 2 as before, we see from their
derivatives that the obvious candidates for x and yare solutions to
Since these are quadratic equations, they may not have a unique solution. At this stage
we use the intuitive observation that y ~ x ~ 2 (since VI ~ V2 ~ V) to single out one
solution - the subsequent application of Lemma 6.1 will justify this choice.
The only solution satisfying this condition is
x
=
2+
Y= I +
V
V
5 - 4ql - 2q?2- J25 - 40q( - 4q? ,
ql /2 + 2ql
5 - 3q(/2
.
+ J25 -
40ql - 4q?
q 1 /2 + 2
'
where the second term in the expression for x is to be taken as 0 when ql = O.
It remains to show that, with this choice of x and y, WI and W 2 both go to 0 as t goes to
infinity. To do this, we look for a linear change of variables such that the equations
separate into a form suitable for Lemma 6.1. Thus we seek k, I and A such that
Equating coefficients shows that we must have k = 1/2, Aa root of A2 + qlA - ql = 0 and
1= - A(X - 2) + (I - y). With these values, the hypotheses of Lemma 6.1 are verified
easily and hence WI and W2 do indeed go to zero.
Example 6.4: Equilibrium point III. This is almost the same as the previous
example, the only difference being that the values of A to use in the change of
variables are
A = ![ - 1 ± qi + 4q2]·
J
Applying the method to point IV shows that, provided the difference between the
two arrival rates is not too great, the player receiving the faster stream of offers can in
fact be weaker.
Theorem 6.5: Equilibrium Point IV.
q2 > ~ then V2(t) ~ vl(t) for all t.
If!
~
q2 <
~,
there exists t with vl(t) > V2(t). If
Proof. By calculating the derivatives of VI and V2 under the assumptions firstly that
VI ~ V2 and then that VI ~ V2, we see that
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638
M. T. DIXON
d
Ia(l--v)=(V-v2)-I(I-v)
- I
2
dt(a(l-v)-(v-V2))
(a 2 +a-qI)
whatever the values of VI and V2. To apply Lemma 2.1, we need therefore a to be such that
a 2 + a - qi > 0 and a(l - v(O)) ~ (v(O) - viO)). These conditions are satisfied by a>
! [ - l +)1 +4qI].
Now we can compare VI and V2:
By the previous result (with a = ) q2- qI), a sufficient condition for this to be strictly
positive is that
Jq2-qI>![-1 +Jl +4ql]·
Rearranging this gives ~ < q2 < 1, proving the first part of the result.
The case q2 < ~ is proved by contradiction. The idea is to start by assuming that
V2(t) ~ vI(t) for all t > o. Using the same change of variables as in Example 6.3 to find the
rate of convergence to 1 shows that if WI and W 2 are as in Example 6.2 with.x and y chosen
so that w2 -+ 0, WI -+ 0, then x > y. But limt_oo(t + 2)(v2 - VI) = Y - x < 0 giving the
contradiction.
However, these calculations fail for ~ < q2 < i (5
7) and we need a further step to
complete the proof. The intuitive result that increasing q2 increases V2 and decreases VI
would be sufficient, and this may be shown by uncoupling the differential equations with
the usual change of variables and then applying Lemma 2.1 to the solutions resulting
from two different values of q2.
J5 -
References
[1] ENNS, E. G. AND FERENSTEIN, E. Z. (1989) A competitive best choice problem with Poisson
arrivals. J. Appl. Probe 27,333-342.
[2] ENNS, E. G. AND RAVINDRAN, K. (1989) A two Poisson stream competitive decision problem.
Preprint.
[3] MAJUMDAR, A. A. K. (1987) Optimal stopping for a full-information bilateral sequential game
related to the secretary problem. Math. Japonica 32, 63-74.
[4] NEVEU, J. (1975) Discrete Parameter Martingales. North-Holland, Amsterdam.
[5] SAKAGUCHI, M. (1989) Multiperson multilateral secretary problems. Math. Japonica 34,
459-473.
[6] SAKAGUCHI, M. (1991) Sequential games with priority under expected value maximization.
Math. Japonica 36, 545-562.
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available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/S0021900200044363