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To Commit or Not to Commit in Product-Innovation Timing Games

  • Chia-Hung Sun EMAIL logo
Published/Copyright: February 5, 2025

Abstract

This research analyzes a firm’s timing of bringing a new product to the market, based on a product-innovation timing game in which the quality of a new product increases over time. We explore the equilibrium outcomes when both firms can precommit to their timing of market entry (the so-called precommitment game), when they cannot credibly precommit to their timing of market entry (the so-called preemption game), and when only one firm can precommit to its timing of market entry (the so-called mixed precommitment game). Sequential entry is the unique equilibrium outcome in all three games. We finally discuss two extended timing games with endogenous commitment choices. The equilibrium involves a precommitment subgame in the endogenous precommitment game with observable delay, while mixed precommitment subgames appear to be the equilibria in undominated strategies in the endogenous precommitment game with action commitment.

JEL Classification: L13; D21; O30

Corresponding author: Chia-Hung Sun, Department of Economics, Soochow University, No. 56, Kueiyang Street, Section 1, Taipei 100, Taiwan, E-mail: 

Financial support by the Ministry of Science and Technology (MOST 107-2410-H-031-014) is deeply appreciated.


Award Identifier / Grant number: MOST 107-2410-H-031-014

Appendix

Proof of Lemma 1.

Firm 1 chooses t 1 to maximize its discounted sums of profits, taking firm 2’s choice as given. The first-order condition for profit maximization is calculated as:

(A1) L ( t ̄ 1 , t 2 ) t 1 = e t ̄ 1 π 1 M + e t ̄ 1 e t 2 π 1 M t 1 + e t 2 π 1 * t 1 c ( t ̄ 1 ) = 0 .

Substituting the first-order condition for profit maximization into the second-order derivative of the leader’s profit yields:

(A2) 2 L ( t ̄ 1 , t 2 ) t 1 2 = e t ̄ 1 π 1 M t 1 + e t 2 π 1 * t 1 π 1 M t 1 + 2 π 1 * t 1 2 c ( t ̄ 1 ) + c ( t ̄ 1 ) < 0 ,

where π 1 M / t 1 = 1 / 4 > 0 , π 1 * / t 1 π 1 M / t 1 = ( 20 t ̄ 1 t 2 2 12 ( t ̄ 1 ) 2 t 2 + ( t ̄ 1 ) 3 48 t 2 3 ) / 4 ( 4 t 2 t ̄ 1 ) 3 < 0 , and 2 π 1 * / t 1 2 = 2 t 2 2 ( 7 t ̄ 1 + 8 t 2 ) / ( 4 t 2 t ̄ 1 ) 4 < 0 . We conclude that L(t 1, t 2) is strictly quasiconcave in t 1. We calculate the first-order condition for the follower’s profit maximization as:

(A3) F ( t ̄ 2 , t 1 ) t 2 = e t ̄ 2 π 2 * + e t ̄ 2 π 2 * t 2 c ( t ̄ 2 ) = 0 .

Substituting the first-order condition for profit maximization into the second-order derivative of the follower’s profit yields:

(A4) 2 F ( t ̄ 2 , t 1 ) t 2 2 = e t ̄ 2 2 π 2 * t 2 2 π 2 * t 2 c ( t ̄ 2 ) + c ( t ̄ 2 ) < 0 ,

where 2 π 2 * / t 2 2 = 8 t 1 2 ( t 1 + 5 t ̄ 2 ) / ( 4 t ̄ 2 t 1 ) 4 < 0 and π 2 * / t 2 = 4 t ̄ 2 ( 4 ( t ̄ 2 ) 2 3 t 1 t ̄ 2 + 2 t 1 2 ) / ( 4 t ̄ 2 t 1 ) 3 > 0 . Therefore, F(t 2, t 1) is strictly quasiconcave in t 2. From equations (A1) and (A3), we have:

(A5) L 1 ( v , v ) = L ( v , v ) t 1 = e v v 4 c ( v ) + e v π 1 * ( v , v ) t 1 < 0 , F 1 ( v + , v ) = F ( v + , v ) t 2 = c ( v ) + e v π 2 * ( v + , v ) t 2 = e v 1 s ( v ) 9 > 0 if  v < s 1 ( 1 ) ,

where π 1 * ( v , v ) / t 1 = 1 / 9 , and π 2 * ( v + , v ) / t 2 = 1 / 9 . We conclude that L 1(v , v) < F 1(v +, v) and F 1(0+, 0) = 1/9 > 0. From equations (A1) and (A3), we have:

(A6) L 1 ( 0 , t j ) = L ( 0 , t 2 ) t 1 = 4 3 e t 2 16 > 0 , lim t i F 1 ( t i , t j ) = lim t 2 F ( t 2 , t 1 ) t 2 = lim t 2 c ( t 2 ) < 0 .  Q.E.D.

Proof of Lemma 2.

Differentiating L(t 1, t 2) and F(t 2, t 1) with respect to t 2 and t 1 yields:

(A7) L 2 ( t i , t j ) = L ( t 1 , t 2 ) t 2 = e t 2 π 1 M π 1 * + e t 2 π 1 * t 2 > 0 , F 2 ( t i , t j ) = F ( t 2 , t 1 ) t 1 = e t 2 π 2 * t 1 < 0 .

The cross-partial derivatives of L(t 1, t 2) and F(t 2, t 1) are calculated as:

(A8) L 12 ( t i , t j ) = 2 L ( t 1 , t 2 ) t 1 t 2 = e t 2 π 1 M t 1 π 1 * t 1 + e t 2 2 π 1 * t 1 t 2 > 0 , F 12 ( t i , t j ) = 2 F ( t 2 , t 1 ) t 1 t 2 = e t 2 π 2 * t 1 + e t 2 2 π 2 * t 1 t 2 > 0 ,

where π 1 M / t 1 π 1 * / t 1 = 20 t 1 t 2 2 + 12 t 1 2 t 2 t 1 3 + 48 t 2 3 / 4 ( 4 t 2 t 1 ) 3 > 0 , 2 π 1 * / t 1 t 2 = 2 t 1 t 2 ( 7 t 1 + 8 t 2 ) / ( 4 t 2 t 1 ) 4 > 0 , π 2 * / t 1 = 4 t 2 2 ( t 1 + 2 t 2 ) / ( 4 t 2 t 1 ) 3 < 0 , and 2 π 2 * / t 1 t 2 = 8 t 1 t 2 ( t 1 + 5 t 2 ) / ( 4 t 2 t 1 ) 4 > 0 . Q.E.D.

Proof of Lemma 4.

When t 1t 2, the slope of the leader’s best response on the plane (t 1t 2) is:

(A9) d t 2 d t 1 = 1 g l ( t 2 ) = 2 L ( t 1 , t 2 ) t 1 2 2 L ( t 1 , t 2 ) t 1 t 2 = π 1 * t 1 + π 1 M t 1 2 π 1 * t 1 2 + e ( t 2 t 1 ) π 1 M t 1 + e ( t 2 t 1 ) s ( t 1 ) ( π 1 M π 1 * ) t 1 + 2 π 1 * t 1 t 2 > π 1 * t 1 + π 1 M t 1 2 π 1 * t 1 2 + π 1 M t 1 ( π 1 M π 1 * ) t 1 + 2 π 1 * t 1 t 2 = 2 28 t 1 t 2 2 192 t 1 t 2 3 + 82 t 1 2 t 2 2 16 t 1 3 t 2 + t 1 4 + 32 t 2 3 + 224 t 2 4 64 t 1 t 2 2 128 t 1 t 2 3 + 56 t 1 2 t 2 + 68 t 1 2 t 2 2 16 t 1 3 t 2 + t 1 4 + 192 t 2 4 α ( t 1 , t 2 ) .

The slope of the follower’s best response is:

(A10) d t 2 d t 1 = g f ( t 1 ) = 2 F ( t 2 , t 1 ) t 1 t 2 2 F ( t 2 , t 1 ) t 2 2 = 2 π 2 * t 1 t 2 π 2 * t 1 π 2 * t 2 2 π 2 * t 2 2 + s ( t i ) < 2 π 2 * t 1 t 2 π 2 * t 1 π 2 * t 2 2 π 2 * t 2 2 = t 2 10 t 1 t 2 2 t 1 t 2 2 2 t 1 2 + t 1 2 t 2 8 t 2 3 16 t 1 t 2 3 10 t 1 2 t 2 11 t 1 2 t 2 2 2 t 1 3 + 2 t 1 3 t 2 16 t 2 4 β ( t 1 , t 2 ) .

Therefore, 1 / g l ( t 2 ) g f ( t 1 ) > α ( t 1 , t 2 ) β ( t 1 , t 2 ) . Solving for α(t 1, t 2) − β(t 1, t 2) yields:

(A11) α ( t 1 , t 2 ) β ( t 1 , t 2 ) = ( 4 t 2 t 1 ) h ( t 1 , t 2 ) f ( t 1 , t 2 ) ,

where

(A12) f ( t 1 , t 2 ) = 576 t 1 t 2 5 2816 t 1 t 2 6 + 1008 t 1 2 t 2 4 + 2696 t 1 2 t 2 5 480 t 1 3 t 2 3 1424 t 1 3 t 2 4 + 90 t 1 4 t 2 2 + 446 t 1 4 t 2 3 + 30 t 1 5 t 2 71 t 1 5 t 2 2 4 t 1 6 + 4 t 1 6 t 2 + 256 t 2 6 + 1408 t 2 7 , h ( t 1 , t 2 ) = t 2 3 ( 192 t 2 128 t 1 ) + t 1 2 t 2 ( 68 t 2 16 t 1 ) + 64 t 1 t 2 2 + 56 t 1 2 t 2 + t 1 4 t 2 3 ( 16 t 2 16 t 1 ) + t 1 2 t 2 ( 11 t 2 2 t 1 ) + 10 t 1 2 t 2 + 2 t 1 3 > 0 .

If t 2 > t 1, then h(t 1, t 2) > 0, implying sign α ( t 1 , t 2 ) β ( t 1 , t 2 ) = sign f ( t 1 , t 2 ) . Because there is no real root for 2 f ( t 1 , t 2 ) / t 2 2 = 0 and 2 f ( t 1 , t 2 ) / t 2 2 t 2 = t 1 > 0 , we conclude 2 f ( t 1 , t 2 ) / t 2 2 > 0 , ∀t 2t 1 and f(t 1, t 2) is strictly convex in t 2. Since f ( t 1 , t 2 ) t 2 = t 1 = 81 t 1 6 ( 4 + 3 t 1 ) > 0 and f ( t 1 , t 2 ) / t 2 t 2 = t 1 = 486 t 1 5 ( 3 + 4 t 1 ) > 0 , we conclude f(t 1, t 2) > 0 and α(t 1, t 2) − β(t 1, t 2) > 0, ∀t 2 > t 1. This implies 1 / g l ( t 2 ) g f ( t 1 ) > 0 , ∀t 2 > t 1. Q.E.D.

Proof of Corollary 1.

Suppose that t 1 ≤ t 2. If the R&D cost is s(t) = t, then:

(A13) g f ( t 1 ) = 1 12 μ ( t 1 ) 1 / 3 12 1 9 t 1 36 13 t 1 2 144 μ ( t 1 ) 1 / 3 + 5 t 1 12 + 1 3 ,

where μ ( t 1 ) = 64 + 24 t 1 + 390 t 1 2 + 35 t 1 3 + 6 1104 t 1 2 + 504 t 1 3 + 4029 t 1 4 + 702 t 1 5 27 t 1 6 . Substituting t 2 = g f (t 1) into ∂L(t 1, t 2)/∂t 1 = 0, we obtain t ̂ 1 = 0.0568 . Substituting t ̂ 1 = 0.0568 into t 2 = g f (t 1) yields t ̂ 2 = g f ( t ̂ 1 ) = 0.2123 . It can then be checked that F ( t ̂ 2 , t ̂ 1 ) = 0.0165 > 0.00279 = L ( t ̂ 1 , t ̂ 2 ) , g f ( t ̂ 2 ) = 0.2951 , and L ( t ̂ 1 , t ̂ 2 ) = 0.00279 > 0.0129 = F ( g f ( t ̂ 2 ) , t ̂ 2 ) . Therefore, ( t ̂ 1 = 0.0568 , t ̂ 2 = 0.2123 ) is the unique Nash equilibrium and there is a second-mover advantage. Q.E.D.

Proof of Lemma 5.

We note L(t 1, g f (t 1)) is continuous in t 1 ∈ [0, d] and that F(t 1, g l (t 1)) is continuous in t 1 [ d , t ̄ ] . (1) If t 1 t ̂ 1 , then t 1 < g l (g f (t 1)). Since L 2(t 1, t 2) > 0 and g f ( t 1 ) > 0 , we conclude d L ( t 1 , g f ( t 1 ) ) / d t 1 = L 1 ( t 1 , g f ( t 1 ) ) + L 2 ( t 1 , g f ( t 1 ) ) g f ( t 1 ) > 0 . This implies if t 2 m = t l * , then t 2 m = t l * > t ̂ 1 , resulting in t 1 m = g f ( t l * ) > t ̂ 2 from strategic complements. (2) If t 1 t ̂ 2 , then t 1 > g f (g l (t 1)) and F 1(t 1, g l (t 1)) < 0. Since F 2(t 1, t 2) < 0 and g l ( t 1 ) > 0 , we conclude d F ( t 1 , g l ( t 1 ) ) / d t 1 = F 1 ( t 1 , g l ( t 1 ) ) + F 2 ( t 1 , g l ( t 1 ) ) g l ( t 1 ) < 0 . This implies if t 2 m = t f * , then t 2 m = t f * < t ̂ 2 , resulting in t 1 m = g l ( t f * ) < t ̂ 1 from strategic complements. Q.E.D.

Proof of Proposition 2.

(1) If the commitment firm is the leader t 2 m t 1 m , then Π 2 t 2 m , t 1 m = L ( t l * , g f ( t l * ) ) and Π 1 t 1 m , t 2 m = F ( g f ( t l * ) , t l * ) . We claim that L ( t l * , g f ( t l * ) ) F ( t ̂ 2 , t ̂ 1 ) > F ( g f ( t l * ) , t l * ) . The first inequality follows since the commitment firm always has the option of enforcing the Nash equilibrium profit in the precommitment game; and the second inequality follows from F 2(t 2, t 1) < 0 and t l * > t ̂ 1 .

(2) If the commitment firm is the follower t 2 m t 1 m , then Π 2 t 2 m , t 1 m = F ( t f * , g l ( t f * ) ) and Π 1 t 1 m , t 2 m = L ( g l ( t f * ) , t f * ) . We claim that F ( t f * , g l ( t f * ) ) L ( t ̂ 1 , t ̂ 2 ) > L ( g l ( t f * ) , t f * ) . The first inequality follows since the commitment firm always has the option of enforcing the Nash equilibrium profit in the precommitment game; and the second inequality follows from L 2(t 2, t 1) > 0 and t ̂ 2 > t f * . Therefore, regardless of whether the commitment firm is the leader or the follower in equilibrium, there is a commitment advantage. Q.E.D.

Proof of Corollary 2.

Because t 1 e = arg max L t 1 , g f ( t 1 ) = 0.0594 and L ( t l * , g f ( t l * ) ) L t 1 e , g f t 1 e = 0.00280 < F ( t ̂ 2 , t ̂ 1 ) = 0.0165 F ( t f * , g l ( t f * ) ) , we conclude the commitment firm is the follower in equilibrium, and its timing of market entry is earlier than that of the follower in the precommitment game ( t 2 m < t ̂ 2 = 0.0568 ). The strategic complements imply that the leader’s timing of entry is also earlier than that of the leader in the precommitment game t 1 m < t ̂ 1 = 0.2123 . This implies that there is a commitment advantage. Q.E.D.

Proof of Proposition 5.

We first consider the equilibrium strategy in the precommitment game t 1 c , t 2 c . Since t 2 c is a best response to t 1 c , neither waiting nor any other committed timing of entry can raise firm 2’s profit, and this goes similarly for firm 1. We next analyze the situation when firm 1 chooses t 1 m and firm 2 waits and then plays its best response g t 1 m = t 2 m . When firm 2 waits, firm 1 knows it is the only commitment firm with an optimal strategy of t 1 m . We note g t 1 m = t 2 m is firm 2’s best response to t 1 m , which can be committed or after observation. However, if it were committed, then t 1 m would not be the best response to it. By similar reasoning, firm 2 choosing t 2 m and firm 1 waiting and then playing its best response g t 2 m = t 1 m is an equilibrium. Q.E.D.

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Received: 2024-04-30
Accepted: 2025-01-13
Published Online: 2025-02-05

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