# Timing Games with Irrational Types: Leverage-Driven Bubbles and Crash-Contingent Claims

Hitoshi Matsushima

## Abstract

This study investigates strategic aspect of leverage-driven bubbles from the viewpoint of game theory and behavioral finance. Even if a company is unproductive, its stock price grows up according to an exogenous reinforcement pattern. During the bubble, this company raises huge funds by issuing new shares. Multiple arbitrageurs strategically decide whether to ride the bubble by continuing to purchase shares through leveraged finance.

We demonstrate two models that are distinguished by whether crash-contingent claim, i. e. contractual agreement such that the purchaser of this claim receives a promised monetary amount if and only if the bubble crashes, is available. We show that the availability of this claim deters the bubble; without crash-contingent claim, the bubble emerges and persists long even if the degree of reinforcement is insufficient. Without crash-contingent claim, high leverage ratio fosters the bubble, while with crash-contingent claim, it rather deters the bubble.

We formulate these models as specifications of timing game with irrational types; each player selects a time in a fixed time interval, and the player who selects the earliest time wins the game. We assume that each player is irrational with a small but positive probability. We then prove that there exists the unique Nash equilibrium; according to it, every player never selects the initial time. By regarding arbitrageurs as players, we give careful conceptualizations that are necessary to interpret timing games as models of leverage-driven bubbles.

JEL Classification: C720; C730; D820; G140

Funding statement: This research was supported by a grant-in-aid for scientific research (KAKENHI 21330043, 25285059) from the Japan Society for the Promotion of Science (JSPS) and the Ministry of Education, Culture, Sports, Science and Technology (MEXT) of the Japanese government. I am grateful to the editor-in-chief, an anonymous referee, and Professor Takashi Ui for their valuable comments. All errors are mine.

## Appendix

### A Proof of Theorem 1

Suppose that the inequality (4) holds. Then, it is clear that q~ is a Nash equilibrium: it satisfies the first-order condition (3) during the interval [τ~,1], and any player has no incentive to select before τ~, because any other player selects before τ~.

Suppose that the inequality (4) does not holds, implying τ~=0 and q~1(0)>0. Then, it is clear that q~ is not a Nash equilibrium: it does satisfy the first-order condition (3) during the interval (0,1], but any player prefers the selection of the initial time 0 to any later time because of q~1(0)>0.

Suppose that the strict inequality of (4) holds:

ε < exp [ 1 n 1 0 1 R ( τ ) d τ ] .

We prove that q~ is the unique Nash equilibrium in the following manner. Consider an arbitrary symmetric Nash equilibrium q. From the strict inequality, it is clear that q1(0)<1.

We show that q1 is continuous. Suppose that q1 is not continuous; there exists τ>0 such that limττq1(τ)<q1(τ). Then, by selecting any time slightly earlier than τ, any player can dramatically increase his winning probability. This implies that no player selects τ. This is a contradiction.

Let

τ ^ = max { τ [ 0 , 1 ] : q 1 ( τ ) = q 1 ( 0 ) } .

Note that τ^<1. We show that q1 is increasing in [τ^,1]. Suppose that q1 is not increasing in [τ^,1]; there exist τ[τ^,1] and τ[τ^,1] such that τ<τ, q1(τ)=q1(τ), and the selection of τ is a best response. Since no player selects any τ in (τ,τ), it follows from the continuity of q1 that by selecting τ instead of τ, a player can increase his winner’s payoff without decreasing his winning probability. This is a contradiction.

Since q1 is increasing in [τ^,1], any selection τ[τ^,1] must satisfy the first-order condition (3). This implies

q 1 ( t ) = q ~ 1 ( t )  for all  t [ τ ^ , 1 ] .

Since any player has no incentive to select before τ^, it follows that τ^=τ~, that is, q~ is the unique symmetric Nash equilibrium.

Next, we prove that q~ is a unique Nash equilibrium, even if we consider all asymmetric strategy profiles. We set any Nash equilibrium qQ arbitrarily.

We show that qi must be continuous. Suppose that qi is not continuous; there exists τ>0 such that limττqi(τ)<qi(τ). Then, any other player can drastically increase his winning probability by selecting any time slightly earlier than τ. Hence, no other player selects any time that is either the same as or slightly later than τ; player i can postpone the time without decreasing his winning probability. This is a contradiction.

Let

τ 1 = max { τ [ 0 , 1 ] : q i ( τ ) = q i ( 0 ) f o r a l l i N } .

Note that τ1<1. We show that D(τ;q) must be increasing in [τ1,1]. Suppose that D(τ;q) is not increasing in [τ1,1]; there exist τ(τ1,1] and τ(τ1,1] such that τ<τ, D(τ;q)=D(τ;q), and the selection of τ is a best response for some player. Since no player selects any time τ in (τ,τ), it follows from the continuity of q that, by selecting τ instead of τ, any player can postpone the time from τ to τ without decreasing his winning probability. This is a contradiction.

We show that q must be symmetric. Suppose that q is asymmetric. From the strict inequality, the selection of time zero is a dominated strategy. Hence, we have τ1>0, and

q i ( τ ) = 0  for all  i N  and  τ [ 0 , τ 1 ] .

Since q is asymmetric and continuous and D(τ;q) is increasing in [τ1,1], there must exist τ>0, τ>τ, and iN such that

q 1 ( t ) = q j ( t )  for all  j N  and  t [ 0 , τ ] ,

(13) D i ( τ ; q i ) 1 D i ( τ ; q i ) > min h i D h ( τ ; q h ) 1 D h ( τ ; q h )  for all  t ( τ , τ )

and

(14) D i ( τ ; q i ) 1 D i ( τ ; q i ) = min h i D h ( τ ; q h ) 1 D h ( τ ; q h ) > 0.

Since D(τ;q) is continuous and increasing in [τ1,1], any selection of time t in (τ,τ) must be a best response for any player jN satisfying

D j ( τ ; q j ) 1 D j ( τ ; q j ) = min h i D h ( τ ; q h ) 1 D h ( τ ; q h ) .

This implies qj(t)t>0. Hence, the first-order condition (2) must hold for this player during the interval (τ,τ):

D j ( t ; q j ) 1 D j ( t ; q j ) = min h i D h ( t ; q h ) 1 D h ( t ; q h ) = v ¯ ( t ) v ¯ ( t ) v _ ( t )  for all  t ( τ , τ ) .

However, from (13),

D i ( τ ; q i ) 1 D i ( τ ; q i ) > v ¯ ( t ) v ¯ ( t ) v _ ( t ) ,

implying that the first-order condition (2) does not hold for player i during the interval (τ,τ); instead ui(τ,qi)τ<0 holds for all τ(τ,τ). This implies that player i prefers τ to any time in (τ,τ+η];

D i ( τ ; q i ) = 0  for all  τ ( τ , τ + η ]

where η>0 is set to be close to zero. This is a contradiction, because the inequality in (14) implies Di(τ;qi)>0. Hence, we have proved that any Nash equilibrium must be symmetric.

From the above observations, we have proved Theorem 1.

### B Proof of Theorem 2

For every t(0,1],

u 1 ( t , q ^ 1 ) = ε n 1 v ¯ ( t ) + ( 1 ε n 1 ) v _ ( 0 )

ε n 1 v ¯ ( 1 ) + ( 1 ε n 1 ) v _ ( 0 ) = u 1 ( 1 , q ^ 1 ) ,

and

u 1 ( 0 , q ^ 1 ) = { 1 l n 1 ( n 1 ) ! l ! ( n 1 l ) ! ( 1 ε ) l ε n 1 l 1 l + 1 } v ¯ ( 0 )

+ { 1 1 l n 1 ( n 1 ) ! l ! ( n 1 l ) ! ( 1 ε ) l ε n 1 l 1 l + 1 } v _ ( 0 ) .

Hence, a necessary and sufficient condition for q^ to be a Nash equilibrium is given by

u 1 ( 0 , q ^ 1 ) = { 1 l n 1 ( n 1 ) ! l ! ( n 1 l ) ! ( 1 ε ) l ε n 1 l 1 l + 1 } v ¯ ( 0 ) + { 1 1 l n 1 ( n 1 ) ! l ! ( n 1 l ) ! ( 1 ε ) l ε n 1 l 1 l + 1 } v _ ( 0 ) ε n 1 v ¯ 1 ( 1 ) + ( 1 ε n 1 ) v _ 1 ( 0 ) u 1 ( 1 , q ^ 1 ) .

This inequality is equivalent to

1 l n 1 ( n 1 ) ! l ! ( n 1 l ) ! ( 1 ε ε ) l 1 l + 1 v ¯ 1 ( 1 ) v ¯ 1 ( 0 ) v ¯ 1 ( 0 ) v _ 1 ( 0 ) = R .

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Published Online: 2019-08-23

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