Existence of a unique Nash equilibrium for an asymmetric lottery Blotto game with weighted majority

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J. Math. Anal. Appl. ••• (••••) •••–•••

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Existence of a unique Nash equilibrium for an asymmetric lottery Blotto game with weighted majority Bara Kim a , Jeongsim Kim b,∗ a

Department of Mathematics, Korea University, 145 Anam-ro, Seongbuk-gu, Seoul, 02841, Republic of Korea b Department of Mathematics Education, Chungbuk National University, 1 Chungdae-ro, Seowon-gu, Cheongju, Chungbuk, 28644, Republic of Korea

a r t i c l e

i n f o

Article history: Received 10 September 2018 Available online xxxx Submitted by J.A. Filar Keywords: Lottery Blotto game Nash equilibrium Weighted majority

a b s t r a c t We consider an asymmetric lottery Blotto game with two agents and n items, where both agents wish to maximize their probability of winning a majority value of all n items. Duffy and Matros [2] showed that if there exists a Nash equilibrium, then the equilibrium is unique, and it is found in an explicit expression. They also provided sufficient conditions for the existence of a Nash equilibrium in the cases of n = 3 and n = 4. In this paper, we prove that the lottery Blotto game always has a unique Nash equilibrium for any value of n. © 2019 Elsevier Inc. All rights reserved.

1. Introduction The classic Colonel Blotto game is a two-player constant-sum game [1]. Each player (or agent) simultaneously allocates a fixed and symmetrical amount of resources (or budgets) over a finite number of items (or battlefields). The player who allocates the most resources wins the battlefield, and each player’s payoff is equal to the total number of battlefields won. Roberson [10] characterized the unique equilibrium payoffs for both symmetric and asymmetric resources in the classic Blotto game. In this paper, we consider a two-player lottery Blotto game, where the winner of each item is determined by a lottery in which the chance of winning is proportional to each player’s allocation of resources for the item, i.e., a standard Tullock’s [12] contest success function. The players’ objective is either to maximize their total expected item values (plurality rule), or to maximize the probability of winning a majority of all items’ values (majority rule). Duffy and Matros [3] provides a theoretical framework for comparing equilibria under the two different rules for the asymmetric lottery Blotto game. * Corresponding author. E-mail addresses: [email protected] (B. Kim), [email protected] (J. Kim). https://doi.org/10.1016/j.jmaa.2019.07.004 0022-247X/© 2019 Elsevier Inc. All rights reserved.

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Friedman [4], Osorio [9], Duffy and Matros [2], Kim and Kim [5], and Kim et al. [6] considered an asymmetric lottery Blotto game with plurality rule. Friedman [4] showed that a two-player lottery Blotto game has a unique Nash equilibrium. Duffy and Matros [2] generalized Friedman’s results to a multi-player lottery Blotto game. Kim and Kim [5] investigated a multi-player lottery Blotto game with a possible budget surplus. Kim et al. [6] investigated a Nash equilibrium for a multi-player lottery Blotto game, where item valuations are asymmetric across the players, and also across the items. An asymmetric lottery Blotto game with majority rule was first studied by Lake [7]. A motivating example of the majority rule version of the asymmetric lottery Blotto game is the electoral college system for electing the U.S. president. Lake [7] showed that a two-player lottery Blotto game has a unique Nash equilibrium, when both players have equal budget constraints. Duffy and Matros [2] generalized Lake’s [7] theoretical findings to asymmetric lottery Blotto game with asymmetric budget constraints. More specifically, Duffy and Matros [2] considered an asymmetric lottery Blotto game with majority rule, where two players compete for n items and where the budget constraints and item valuations are both asymmetric. They showed that if there exists a Nash equilibrium, then the equilibrium is unique, and it is found in an explicit expression. They also provided sufficient conditions for the existence of a Nash equilibrium in the cases of n = 3 and n = 4. In this paper, we prove that Duffy and Matros’ [2] lottery Blotto game with majority rule always has a unique Nash equilibrium for any value of n. The paper is organized as follows. In Section 2, we describe our model in detail. In Section 3, we review the related literature. In Section 4, we prove our main result on the existence of a Nash equilibrium. 2. The model We consider an asymmetric lottery Blotto game with weighted majority. There are two agents (agent 1 and agent 2) and n ≥ 2 items. The value of item i is Wi > 0, i = 1, . . . , n, and the total value of all the n n items is W = i=1 Wi . Agent 1 has a budget X > 0 and has to allocate that budget across a set of items. Agent 2 has a budget Y > 0 and has to allocate that budget across a set of items. Both agents allocate their budgets simultaneously. If xi is the budget allocated to item i by agent 1, and if yi is the budget allocated i to item i by agent 2, then agent 1 obtains item i with probability xix+y , and agent 2 obtains item i with i yi probability xi +yi . If both agents do not allocate any budget to item i (i.e., if xi = yi = 0), then agent 1 obtains item i with probability qi , and agent 2 obtains item i with probability 1 − qi . If an agent obtains items i1 , i2 , . . . , il , and if l 

Wik >

k=1

W , 2

(1)

then the agent wins the game. A coalition of items i1 , i2 , . . . , il is called a winning coalition if it satisfies (1). Both agents wish to maximize their probability of winning the game. Throughout the paper, we make the following technical assumption that ties (or draws) are impossible in the game: l  k=1

Wik =

W 2

(2)

for any items i1 , i2 , . . . , il . A coalition of items i1 , i2 , . . . , il can be represented by an n-dimensional vector t = (t1 , . . . , tn ), where  ti =

1 if i = ik for some k = 1, . . . , l, 0 otherwise.

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We will call a ‘vector representing a coalition’ a ‘coalition’, for simplicity. Note that a coalition t is a winning n  coalition if and only if it satisfies ti Wi > W 2 . Denote by V the set of all vectors t that represent winning i=1

coalitions, i.e., n   W . V = t ∈ {0, 1}n : ti Wi > 2 i=1

Note that (2) can be written as n 

ti Wi =

i=1

W for any t ∈ {0, 1}n . 2

(3)

A strategy of agent 1 is a nonnegative n-dimensional vector x = (x1 , . . . , xn ) such that a strategy of agent 2 is a nonnegative n-dimensional vector y = (y1 , . . . , yn ) such that

n 

n 

xj = X. Also,

j=1

yj = Y . Therefore,

j=1

the sets of strategies for agent 1 and agent 2 are, respectively,

S1 = {x ∈ Rn : xj ≥ 0 for j = 1, . . . , n and

n 

xj = X},

j=1

S2 = {y ∈ Rn : yj ≥ 0 for j = 1, . . . , n and

n 

yj = Y }.

j=1

For a strategy profile (x, y) ∈ S1 × S2 , the payoff of agent 1, which is the probability of agent 1 to get a winning coalition, is given by

u1 (x, y) =

n    tj pj (xj , yj ) + (1 − tj )(1 − pj (xj , yj )) ,

(4)

t∈V j=1

where  pj (xj , yj ) =

xj xj +yj

qj

if xj + yj > 0, if xj = yj = 0,

j = 1, . . . , n.

Similarly, the payoff of agent 2 is given by

u2 (x, y) =

n    tj (1 − pj (xj , yj )) + (1 − tj )pj (xj , yj ) .

(5)

t∈V j=1

A strategy profile (x∗ , y∗ ) ∈ S1 × S2 is a Nash equilibrium of the lottery Blotto game if and only if u1 (x∗ , y∗ ) ≥ u1 (x, y∗ ) for all x ∈ S1 ,

(6)

u2 (x∗ , y∗ ) ≥ u2 (x∗ , y) for all y ∈ S2 .

(7)

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3. A review and discussion for the existence of a Nash equilibrium We will denote the lottery Blotto game (as described in Section 2) by G. In this section, we will briefly review Lake [7] and Duffy and Matros [2] results on the existence of a Nash equilibrium for the game G. We define the sets Vi , i = 1, . . . , n, as follows: Vi = {t ∈ V : ti = 1, t − ei ∈ / V },

i = 1, . . . , n,

where ei is the n-dimensional column vector whose ith element is one, and all the other elements are zero. Define the Banzhaf power index for item i as |Vi | BPI(i) =  , n |Vj |

i = 1, . . . , n,

j=1

where |Vi | denotes the number of elements in the set Vi . Lake [7] showed that if condition (3), and X = Y hold, then the game G has a unique Nash equilibrium (x∗ , y∗ ), where x∗ = (x∗1 , . . . , x∗n ) and y∗ = (y1∗ , . . . , yn∗ ) are given by x∗i = BPI(i)X

and yi∗ = BPI(i)Y,

i = 1, . . . , n.

(8)

We introduce the Duffy-Matros power index for item i as follows: DMPI(i) =

Ki , K 1 + · · · + Kn

i = 1, . . . , n,

where Ki =





X

j=i

tj



n−1−

Y

tj

j=i

.

(9)

t∈Vi

Here, the sum over an empty set is defined as 0. Note that it follows from (9) that Ki = 0 when Vi = φ, and Ki > 0 otherwise. Furthermore, if X = Y holds, then from (9) Ki = X n−1 |Vi |

and DMPI(i) = BPI(i), i = 1, . . . , n.

We now provide the result from Duffy and Matros [2], in the following proposition. Proposition 1. [2] If there exists a Nash equilibrium (x∗ , y∗ ) in the game G, then it is unique and has the following form: x∗i = DMPI(i)X

and

yi∗ = DMPI(i)Y,

i = 1, . . . , n.

(10)

Duffy and Matros also showed that if n ≤ 4 and condition (3) holds, then DMPI(i) = BPI(i). But, when n ≥ 5, the Duffy-Matros power index is not necessarily the same as the Banzhaf power index. In addition, Y 9 Y they ascertained that if n = 2, or if n = 3 with 13 ≤ X ≤ 3, or if n = 4 with 11 ≤X ≤ 11 9 , then there exists ∗ ∗ a unique Nash equilibrium in which the Nash equilibrium (x , y ) is given by (10) (which is equal to (8) because n ≤ 4).

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Remark 1. In the next section we will show that the game G has a Nash equilibrium, which completes Proposition 1. This implies that the restrictions on X and Y in the cases of n = 3 and n = 4 as mentioned above, can be removed. Furthermore, the form of the unique Nash equilibrium in both of these cases can be derived from (10). Without loss of generality, we can assume that W1 ≥ W2 ≥ W3 in the case of n = 3, and that W1 ≥ W2 ≥ W3 ≥ W4 in the case of n = 4. Then it is not difficult to show that for the cases of n = 3 and n = 4, formula (10) gives the following two forms of the unique Nash equilibrium (x∗ , y∗ ) in the game G,  ∗

∗

(x , y ) =

((X, 0, 0), (Y, 0, 0))

X X X  Y Y Y  3, 3, 3 , 3, 3, 3

if W1 > W2 + W3 , if W1 < W2 + W3 ,

and ⎧ ((X, 0, 0, 0), (Y, 0, 0, 0)) ⎪ ⎨



 ∗ ∗ X X X X (x , y ) = , 6 , 6 , 6 , Y2 , Y6 , Y6 , Y6 2 ⎪ ⎩

X X X  Y Y Y  3 , 3 , 3 ,0 , 3 , 3 , 3 ,0

if W1 > W2 + W3 + W4 , if W1 + W4 > W2 + W3 , if W1 + W4 < W2 + W3 ,

respectively. 4. Existence of a Nash equilibrium In this section we present our main result (Theorem 1), which completes the result of Duffy and Matros (Proposition 1). Theorem 1. The lottery Blotto game G has the unique Nash equilibrium. The Nash equilibrium (x∗ , y∗ ), where x∗ = (x∗1 , . . . , x∗n ) and y∗ = (y1∗ , . . . , yn∗ ) are given by (10). To prove Theorem 1, we need the following seven lemmas. But, before the first lemma, we need to introduce several notions. Let {(k) : k = 1, 2, . . .} be a sequence of real numbers such that 0 < (k) < 1 (k) = 0. For each k = 1, 2, . . ., we consider a lottery Blotto game G(k) with n min{X, Y } and limk→∞  (k) weighted majority. In game G , the sets of strategies for agent 1 and agent 2 are, respectively, (k)

S1

= {x ∈ Rn : xj ≥ (k) for j = 1, . . . , n and

n 

xj = X},

j=1 (k)

S2

= {y ∈ Rn : yj ≥ (k) for j = 1, . . . , n and

n 

yj = Y }.

j=1 (k)

(k)

For a strategy profile (x, y) ∈ S1 × S2 , the payoffs of agent 1 and agent 2 are given by (4) and (5), (k) (k) respectively. A strategy profile (x(k) , y(k) ) ∈ S1 × S2 is a Nash equilibrium for the game G(k) if (k)

u1 (x(k) , y(k) ) ≥ u1 (x, y(k) ) for all x ∈ S1 , (k)

u2 (x(k) , y(k) ) ≥ u2 (x(k) , y) for all y ∈ S2 . Lemma 1. For k = 1, 2, . . ., the game G(k) has a Nash equilibrium. Proof. Define a correspondence (k)

(k)

F (k) : S1 × S2

(k)

(k)

⇒ S1 × S2

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by (k)

(k)

(k)

F (k) (x, y) = B1 (y) × B2 (x)

(k)

for (x, y) ∈ S1 × S2 ,

where (k)

(k)

˜ ∈ S1 }, : u1 (x, y) ≥ u1 (˜ x, y) for all x

(k)

(k)

˜ ) for all y ˜ ∈ S2 }. : u2 (x, y) ≥ u2 (x, y

B1 (y) = {x ∈ S1 B2 (x) = {y ∈ S2

(k) (k)

Note that a strategy profile (x(k) , y(k) ) is a Nash equilibrium of the game G(k) if and only if (x(k) , y(k) ) is a fixed point of F (k) , i.e., (x(k) , y(k) ) ∈ F (k) (x(k) , y(k) ). Therefore, for the proof of the lemma, it suffices to show that F (k) has a fixed point. By Kakutani’s fixed point theorem (see, for example, p. 331 of Ok [8]), we can prove that F (k) has a fixed point. This can be proved by following the same arguments as in the proof of Lemma 1 of Kim et al. [6]. 2 (k)

Lemma 2. Let (x(k) , y(k) ) be a Nash equilibrium of the game G(k) . For k = 1, 2, . . ., there exists λj j = 1, 2 such that 

∂ (k) , y(k) ) ∂xi uj (x ∂ (k) , y(k) ) ∂xi uj (x

(k)

∈ R,

(k)

if xi > (k) , (k) if xi = (k) ,

= λj (k) ≤ λj

i = 1, . . . , n.

(11)

Proof. We only prove (11) for j = 1, because the case of j = 2 can be proved in the same manner. Since (x(k) , y(k) ) is a Nash equilibrium of the game G(k) , u1 (x(k) , y(k) ) = max u1 (˜ x, y(k) ). (k)

˜ ∈S1 x

From this, we can see that x(k) solves the following maximization problem with constraints: maximize

u1 (x, y(k) )

subject to

x1 ≥ (k) , x2 ≥ (k) , . . . , xn ≥ (k) , x1 + · · · + xn = X.

Define the Lagrangian L(x, μ1 , . . . , μn , λ1 ) by L(x, μ1 , . . . , μn , λ1 ) = u1 (x, y(k) ) +

n 

μi (xi − (k) ) − λ1 (x1 + · · · + xn − X).

i=1 (k)

(k)

(k)

Then, by Theorem 18.5 of Simon and Blume [11], there exist Lagrange multipliers μ1 , . . . , μn and λ1 such that (a) (b) (c) (d) (e)

(k) (k) (k) ∂ (k) , μ1 , . . . , μn , λ1 ) = 0, ∂xi L(x (k) (k) μj (xj − (k) ) = 0, j = 1, . . . , n; (k) (k) x1 + · · · + xn = X; (k) (k) μ1 ≥ 0, . . . , μn ≥ 0; (k) (k) x1 ≥ (k) , . . . , xn ≥ (k) .

i = 1, . . . , n;

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Note that • condition (a) is equivalent to

∂ (k) , y(k) ) ∂xi u1 (x

(k)

+ μi

(k)

− λ1 = 0, i = 1, . . . , n; (k)

• condition (c) together with condition (e) is equivalent to x(k) ∈ S1 ; • under condition (e), condition (b) together with condition (d) is equivalent to the following: for i = (k) (k) (k) (k) 1, . . . , n, μi = 0 if xi > (k) and μi ≥ 0 if xi = (k) . (k)

Since we already know x(k) ∈ S1 , all the conditions together (a) through to (e), is equivalent to the following conditions: For i = 1, . . . , n, (i) (ii)

(k) (k) ∂ u (x(k) , y(k) ) = λ1 − μi ; ∂x i 1 (k) (k) μi = 0 if xi > (k) , (k) (k) μi ≥ 0 if xi = (k) . (k)

Therefore, there exists λ1

such that ⎧ ⎨ ⎩

(k)

if xi

(k)

if xi

∂ (k) , y(k) ) ∂xi u1 (x

= λ1

∂ (k) , y(k) ) ∂xi u1 (x

≤ λ1

which completes the proof of (11) for j = 1.

(k)

> (k) ,

(k)

= (k) ,

2

Before the next lemma, we introduce the following notations. Let m be the number of items i with |Vi | > 0. Without loss of generality, we assume that |Vi | > 0 if i = 1, . . . , m and |Vi | = 0 if i = m + 1, . . . , n. Let m   W , V 0 = t0 ∈ {0, 1}m : t0i Wi > 2 i=1

Vi0 = {t0 ∈ V 0 : t0i = 1, t0 − e0i ∈ / V 0 }, where t0 = (t01 , . . . , t0m ) and e0i is the m-dimensional column vector whose ith element is one, and all the other elements are zero. Then m  

0  u1 (x, y) = ti pi (xi , yi ) + (1 − t0i )(1 − pi (xi , yi )) , t0 ∈V 0 i=1

u2 (x, y) =

m  

0  ti (1 − pi (xi , yi )) + (1 − t0i )pi (xi , yi ) . t0 ∈V 0 i=1 (k)

Lemma 3. There exist positive real numbers Λi , i = 1, 2 such that 0 ≤ λi i = 1, 2. (k)

Proof. Note that if x ∈ S1

(k)

and y ∈ S2 , then u1 (x, y) =

m 0   tj xj + (1 − t0j )yj . xj + yj 0 0 j=1

t ∈V

A direct calculation gives

≤ Λi for all k = 1, 2, . . . and

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m m ∂ ∂   t0j xj + (1 − t0j )yj ∂   t0j xj + (1 − t0j )yj u1 (x, y) = + ∂xi ∂xi 0 0 j=1 xj + yj ∂xi 0 0 j=1 xj + yj t ∈V t0i =1

yi = (xi + yi )2 yi = (xi + yi )2

=

t ∈V t0i =0





m 0 m 0     tj xj + (1 − t0j )yj tj xj + (1 − t0j )yj − xj + yj xj + yj 0 0 j=1 0 0 j=1

t ∈V t0i =1 j=i



t ∈V t0i =0 j=i

m 0   tj xj + (1 − t0j )yj − xj + yj 0 0 j=1

t ∈V t0i =1 j=i

m 0  tj xj + (1 − t0j )yj xj + yj j=1





t0 ∈V 0 j=i t0i =1 t0 −e0i ∈V 0

m   t0j xj + (1 − t0j )yj yi . (xi + yi )2 0 0 j=1 xj + yj t ∈Vi

j=i

Therefore, (11) can be written as ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

(k) yi (k) (k) (xi +yi )2



(k)

(k) 2 )

+yi

(k)

(k)

(k)

1≤i≤n

(k)

(k) 2

xik + yik

(k)

(k)

(k) x1 (k) (k) x1 +y1

,...,

xj

t0 ∈Vi0 j=1 k j=i k

Therefore, there exists Λ1 such that 0 ≤ λ1 (k) 0 ≤ λ2 ≤ Λ2 for all k = 1, 2, . . .. 2  Since x(k) , y(k) ,

min{X,Y } n

(k)

if xi

(k)

if xi

= λ1 ≤ λ1

x(k) n (k) (k) xn +yn

(k)

> (k) ,

(k)

= (k) .

(12)

(k)

< xik , we have

(k) (k) m   t0j xj + (1 − t0j )yj

(k)

yik

(k)

xj +yj

t0 ∈Vi0 j=1 j=i

Choose ik such that xik = max xi . Since (k) <



(k)

j=1 j=i m t0 x(k) +(1−t0 )y (k)  j j j j



(k)

(k)

t0j xj +(1−t0j )yj (k) (k) xj +yj

t0 ∈Vi0

yi (xi

m 

(k)

+ yj

(k)

= λ1 .

≤ Λ1 for all k = 1, 2, . . .. Similarly, there exists Λ2 such that

(k)

(k)



, λ1 , λ2

(k)

(k)

is a sequence in the compact set S1 × S2 × [0, 1]n ×

[0, Λ1 ] × [0, Λ2 ], we have the following lemma. Lemma 4. There exists an increasing sequence {kl } of positive integers such that (x(kl ) , y(kl ) ), i = 1, . . . , n, and

(k ) (k ) (λ1 1 , λ2 l )

(k ) xi l (kl ) (k ) +yi l

xi

,

converge as l → ∞.

We denote the above limits as follows: (x∗ , y∗ ) = lim (x(kl ) , y(kl ) ), l→∞

(kl )

αi = lim

l→∞

xi (kl )

xi

λ∗j = lim λj

(kl )

l→∞

(kl )

,

i = 1, . . . , n,

+ yi ,

j = 1, 2.

We will investigate the properties of these limits in the following three lemmas.

(13)

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Lemma 5. We have λ∗1 > 0 and λ∗2 > 0. Proof. We will show λ∗1 > 0. With αi given in (13), we define J0 = {i ∈ M : αi = 0}, J1 = {i ∈ M : αi = 1}, J2 = {i ∈ M : 0 < αi < 1}, where M = {1, 2, . . . , m}. Also, we define the set J as J = {{i : t0i = 1} : t0 ∈ V 0 }. We will show that J0 ∈ / J and J1 ∈ / J . To prove J1 ∈ / J , suppose on the contrary that J1 ∈ J . Then u1 (x(kl ) , y(kl ) ) =



(kl )

J∈J j∈J

≥

(kl )

xj

 

+ yj

(kl )

 

(kl )

+ yj

(kl )

xj

(kl )

(kl )

(kl )

+ yj

(kl )

xj



(kl )

(kl )

+ yj

yj

j∈M \J

xj

J⊂M j∈J J⊃J1

(kl )

xj



(kl )

xj

(kl )

yj

j∈M \J

xj

J∈J j∈J J⊃J1

=



(kl )

xj

+ yj (kl )

yj (kl )

(kl )

xj

j∈M \J

.

+ yj

By changing the set J to J1 ∪ J in the last summation above, we have u1 (x

(kl )

(kl )

,y





)≥

(kl )





J  ⊂M \J1 j∈J1

=



=



(kl )

+ yj

→1

(kl )

(kl )

xj

+ yj

j∈J 





J  ⊂M \J1 j∈J 



(kl )

xj (kl )

xj

(kl )

+ yj

j∈(M \J1 )\J 

(kl )

(kl )

+ yj

j∈(M \J1 )\J 

(kl )

xj

(kl )

+ yj

as l → ∞,

where the last equality follows from 



J  ⊂M \J1 j∈J 



(kl )

xj (kl )

xj

(kl )

+ yj

j∈(M \J1 )\J 

(kl )

yj (kl )

xj

(kl )

= 1.

+ yj

Since u1 (x(kl ) , y(kl ) ) ≤ 1, we have lim u1 (x(kl ) , y(kl ) ) = 1. Therefore, l→∞

lim u2 (x(kl ) , y(kl ) ) = lim (1 − u1 (x(kl ) , y(kl ) )) = 0.

l→∞

l→∞

(kl )

yj (kl )

xj



(kl )

xj xj

(kl )

+ yj

xj (kl )

j∈J1

(kl )

xj

(kl )

yj

j∈(M \J1 )\J 



(kl )

(kl )

xj

+ yj

xj

xj (kl )

j∈J1

(kl )

xj

J  ⊂M \J1 j∈J1 ∪J 

=



(kl )

xj

(kl )

+ yj (kl )

yj (kl )

xj

(kl )

+ yj

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B. Kim, J. Kim / J. Math. Anal. Appl. ••• (••••) •••–•••

10

On the other hand,  

Y

Y Y  Y  ,..., = u2 x∗ , ,..., > 0, lim u2 x(kl ) , l→∞ n n n n which is a contradiction. Therefore, we have proved that J1 ∈ / J . Similarly, we can prove J0 ∈ / J. Since J0 ∈ / J , we have M \ J0 ∈ J , i.e., J1 ∪ J2 ∈ J . Since J1 ∈ / J and J1 ∪ J2 ∈ J , there exists a ˜ Define ˜t0 by nonempty set J˜ ⊂ J2 such that J1 ∪ J˜ ∈ J , but J1 ∪ J ∈ / J if J is a proper subset of J.  t˜0j

˜ if j ∈ J1 ∪ J, ˜ if j ∈ M \ (J1 ∪ J).

1 0

=

0 0 ˜ then ˜t0 ∈ V and ˜t0 − e0 ∈ ˜ ˜ ˜0 Note that if i ∈ J, i / V . Hence if i ∈ J, then t ∈ Vi . Therefore, for i ∈ J,

(kl )

λ1

(k ) l)   t0j x(k + (1 − t0j )yj l j

(kl )

≥

yi (kl )

(xi

(kl ) 2 )

+ yi

yi (kl )

(xi

(kl )

+ yj

(k ) l)  t˜0j x(k + (1 − t˜0j )yj l j

(kl )

≥

(kl )

xj

t0 ∈Vi0 j∈M j=i

(kl ) 2 )

+ yi

(kl )

j∈M j=i

xj

(kl )

,

(14)

+ yj

where the first inequality follows from (12), and the second inequality follows from the fact that ˜t0 ∈ Vi0 . Letting l → ∞ in (14) gives 1

λ∗1 ≥ lim

l→∞ x(kl ) i

1

= lim

l→∞ x(kl ) i

×

+

(k ) yi l



+

(k ) yi l

(1 − αi )



(t˜0j αj + (1 − t˜0j )(1 − αj ))

j∈M j=i

(1 − αi )



(t˜0j αj + (1 − t˜0j )(1 − αj ))

j∈J0



(t˜0j αj + (1 − t˜0j )(1 − αj ))

j∈J1

(t˜0j αj + (1 − t˜0j )(1 − αj ))

j∈J2 j=i

> 0. Therefore, λ∗1 > 0 is proved. Similarly, we can prove λ∗2 > 0. 2 Lemma 6. For all i = 1, . . . , m, min{λ∗1 , λ∗2 } max{λ∗1 , λ∗2 } ≤ αi ≤ . ∗ ∗ λ1 + λ2 λ∗1 + λ∗2 Proof. We will show that αi ≤

max{λ∗1 , λ∗2 } for all i = 1, . . . , m, λ∗1 + λ∗2

(15)

αi ≥

min{λ∗1 , λ∗2 } for all i = 1, . . . , m. λ∗1 + λ∗2

(16)

To show (15), suppose on the contrary that αi > (k)

lim

xi (k)

k→∞ xi

(k)

+yi

>

1 2

and

(k) yi

≥ (k) , we have

(k) xi

∗ max{λ∗ 1 ,λ2 } ∗ λ∗ 1 +λ2

for some i = 1, . . . , m. Since αi =

> (k) except for finitely many k. Therefore,

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[m3L; v1.260; Prn:5/07/2019; 15:13] P.11 (1-13)

B. Kim, J. Kim / J. Math. Anal. Appl. ••• (••••) •••–••• (k) m ˜ (k)   tj xj + (1 − t˜j )yj

(k)

yi (k)

(xi

(k)

+ yi )2

(k)

(k)

xj

˜ t∈Vi0 j=1

+ yj

11

(k)

except for finitely many k.

(17)

(k)

except for finitely many k.

(18)

= λ1

Also, (k) (k) m   (1 − t˜j )xj + t˜j yj

(k)

xi (k)

(xi

(k)

+ yi )2

(k)

(k)

xj

˜ t∈Vi0 j=1

+ yj

≤ λ2

Since (k) m ˜ (k)   tj xj + (1 − t˜j )yj (k)

˜ t∈Vi0 j=1

xj

(k)

=

+ yj

(k) (k) m   (1 − t˜j )xj + t˜j yj (k)

xj

˜ t∈Vi0 j=1

(k)

,

+ yj

(17) and (18) imply that (k)

(k)

xi (k)

xi

(k)

+ yi

≤

λ2 (k)

(k)

.

λ1 + λ2

Letting k → ∞ through the subsequence {kl }, we obtain αi ≤

λ∗1

λ∗2 , + λ∗2

which is a contradiction. Therefore, we have proved (15). In a very similar way to the derivation of (15), we can prove (16). 2 Lemma 7. We have x∗i > 0 and yi∗ > 0 for all i = 1, . . . , m. Proof. First we show that x∗i > 0 if and only if yi∗ > 0 for all i = 1, . . . , m.

(19)

Suppose x∗i > 0. Then (k)

yi (k) (xi

+

(k) m ˜ (k)   tj xj + (1 − t˜j )yj

(k) yi )2 ˜t∈V 0 j=1 i j=i

(k) xj

+

(k) yj

(k)

= λ1

except for finitely many k.

Letting k → ∞ through the subsequence {kl }, we obtain m ˜ ∗   tj xj + (1 − t˜j )yj∗ yi∗ = λ∗1 . ∗ ∗ 2 (xi + yi ) ˜ 0 j=1 x∗j + yj∗ t∈Vi

j=i

Since λ∗1 > 0 by Lemma 5, we have yi∗ > 0. Similarly, we can show that if yi∗ > 0, then x∗i > 0. Thus we have proved (19).

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B. Kim, J. Kim / J. Math. Anal. Appl. ••• (••••) •••–•••

12

Next, we show that either x∗i > 0, or yi∗ > 0 for all i = 1, . . . , m. For i = 1, . . . , m, we have (k) m ˜ (k)   tj xj + (1 − t˜j )yj

(k)

yi (k) (xi

+

(k) yi )2 ˜t∈V 0 j=1 i j=i

(k)

+

(k)

≤ λ1 ,

(k) yj

(k) (k) m   (1 − t˜j )xj + t˜j yj

(k)

xi (xi

(k) xj

(k)

+ yi )2

(k)

xj

˜ t∈Vi0 j=1 j=i

(k)

≤ λ2 .

(k)

+ yj

Since (k) m ˜ (k)   tj xj + (1 − t˜j )yj (k)

xj

˜ t∈Vi0 j=1 j=i

(k)

(k) (k) m   (1 − t˜j )xj + t˜j yj

=

+ yj

(k)

˜ t∈Vi0 j=1 j=i

xj

(k)

,

+ yj

we have (k) m ˜ (k)   tj xj + (1 − t˜j )yj

1 (k) xi

+

(k) yi ˜t∈V 0 j=1 i j=i

(k) xj

+

(k) yj

(k)

(k)

≤ λ1 + λ2 .

By letting k → ∞ through the subsequence {kl }, we obtain lim

l→∞

By Lemma 6,

m  

(k ) xi l

1

m  

+

(k ) yi l ˜t∈V 0 j=1 i j=i

(t˜j αj + (1 − t˜j )(1 − αj )) ≤ λ∗1 + λ∗2 .

(20)

(t˜j αj + (1 − t˜j )(1 − αj )) > 0. Therefore, (20) implies that x∗i + yi∗ > 0. Hence, we

˜ t∈Vi0 j=1 j=i

have either x∗i > 0, or yi∗ > 0 for all i = 1, . . . , m. Therefore, the proof is complete. 2 Now, we are able to prove Theorem 1. Proof of Theorem 1. Note that (x∗ , y∗ ) is a Nash equilibrium of the game G if and only if (x∗ , y∗ ) satisfies ˜ ∈ S1 be an (6) and (7). Since the proofs of (6) and (7) are very similar, we will only prove (6). Let x arbitrary strategy of agent 1. Then u1 (x(k) , y(k) ) ≥ u1 (˜ x(k) , y(k) ),

(21)

where (k)

x ˜i

=

X − n(k) x ˜i + (k) , X

i = 1, . . . , n.

Note that (21) can be written as (k) m 0 (k)   tj xj + (1 − t0j )yj (k)

t0 ∈V 0 j=1

xj

Letting k → ∞, we have, by Lemma 7,

(k)

+ yj

≥

(k) m 0 (k)   tj x ˜j + (1 − t0j )yj (k)

t0 ∈V 0 j=1

x ˜j

(k)

+ yj

.

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[m3L; v1.260; Prn:5/07/2019; 15:13] P.13 (1-13)

B. Kim, J. Kim / J. Math. Anal. Appl. ••• (••••) •••–•••

13

m 0 ∗ m 0     tj xj + (1 − t0j )yj∗ tj x ˜j + (1 − t0j )yj∗ ≥ , ∗ ∗ xj + yj x ˜j + yj∗ 0 0 j=1 0 0 j=1

t ∈V

t ∈V

that is, ˜ ∈ S1 , u1 (x∗ , y∗ ) ≥ u1 (˜ x, y∗ ) for all x which proves (6). 2 Acknowledgments We are grateful to the reviewer for valuable comments and suggestions, which greatly improved this paper. B. Kim’s research was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (NRF-2017R1A2B4012676). J. Kim’s research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2017R1D1A1B03029542). References [1] E. Borel, La théorie du jeu les équations intégrales à noyau symétrique, C. R. Acad. 173 (1921) 1304–1308; English translation by L. Savage, The theory of play and integral equations with skew symmetric kernels, Econometrica 21 (1953) 97–100. [2] J. Duffy, A. Matros, Stochastic asymmetric Blotto games: some new results, Econom. Lett. 134 (2015) 4–8. [3] J. Duffy, A. Matros, Stochastic asymmetric Blotto games: an experimental study, J. Econ. Behav. Organ. 139 (2017) 88–105. [4] L. Friedman, Game-theory models in the allocation of advertising expenditures, Oper. Res. 6 (1958) 699–709. [5] J. Kim, B. Kim, An asymmetric lottery Blotto game with a possible budget surplus and incomplete information, Econom. Lett. 152 (2017) 31–35. [6] G.J. Kim, J. Kim, B. Kim, A lottery Blotto game with heterogeneous items of asymmetric valuations, Econom. Lett. 173 (2018) 1–5. [7] M. Lake, A new campaign resource allocation model, in: S.J. Brams, A. Schotter, G. Schwodiauer (Eds.), Applied Game Theory, Physica-Verlag, Wurzburg, Germany, 1979, pp. 118–132. [8] E.A. Ok, Real Analysis with Economic Applications, Princeton University Press, 2007. [9] A. Osorio, The lottery Blotto game, Econom. Lett. 120 (2013) 164–166. [10] B. Roberson, The Colonel Blotto game, Econom. Theory 29 (2006) 1–24. [11] C.P. Simon, L. Blume, Mathematics for Economists, W.W. Norton & Company, Inc., 1994. [12] G. Tullock, Efficient rent-seeking, in: J.M. Buchanan, R.D. Tollison, G. Tullock (Eds.), Toward a Theory of the RentSeeking Society, Texas A&M University Press, College Station, Texas, 1980, pp. 97–112.