Downstream new product development andupstream process innovation
Akio Kawasaki
Tomomichi Mizuno
Kazuhiro Takauchi
June 30, 2021
Discussion Paper No.2118
Revised No.2022

GRADUATE SCHOOL OF ECONOMICS
KOBE UNIVERSITYROKKO, KOBE, JAPANDownstream new product development andupstream process innovation
Akio Kawasaki∗

Tomomichi Mizuno†

Kazuhiro Takauchi‡§First version: December 8, 2020
This version: June 30, 2021
Abstract
It is well known that when a rival introduces a new product, a firm’s response is aﬀectedby conflicting factors. For example, a certain factor stimulates firms to introduce their new
products in a quick and retaliatory manner if their rivals introduce new products. Based on
this fact, we build a simple vertical relation model: two downstream firms decide whether tointroduce a horizontally diﬀerentiated new product, whereas a single upstream supplier invests incost-reducing research and development (R&D). We show that the equilibrium of downstream
innovation depends on upstream eﬃciency. If upstream R&D eﬃciency is high, downstream
innovation is a strategic complement; this corresponds to the scenario in which downstream
firms act in a retaliatory manner against their rivals introducing new products. Conversely, if
upstream eﬃciency is low, downstream innovation is a strategic substitute: this implies that
downstream firms behave passively when their rivals introduce new products. We also find that
upstream R&D eﬃciency works similarly to the R&D spillover parameter in the d’Aspremont andJacquemin’s (1988) model. When R&D spillover is high (low), the firm’s innovation behavior isa strategic complement (substitute). Hence, we oﬀer a new insight into the innovation literature.
Keywords: Upstream cost-reducing R&D; New product introduction; Strategic complement
JEL classification: D43; L13; O31∗

Faculty of Economics, Oita University, 200, Dannoharu, Oita, Japan. E-mail: a-kawasaki@oita-u.ac.jp
Graduate School of Economics, Kobe University, 2-1 Rokkodai-cho, Nada-ku, Kobe-City, Hyogo 657-8501,
Japan. E-mail: mizuno@econ.kobe-u.ac.jp.
‡
Corresponding author: Kazuhiro Takauchi, Faculty of Business and Commerce, Kansai University, 3-3-35
Yamate-cho, Suita, Osaka 564-8680, Japan. E-mail: kazh.takauch@gmail.com; Tel.: +81-6-6368-1817.
§
Research Fellow, Graduate School of Economics, Kobe University.
†

1

Introduction

When rivals introduce or develop new products, how do firms react?1 Reaction time is an important indicator that is knowing their responses. In their investigation, Bowman and Gatignon
reported that in response to rival firms introducing new products, 31.4% of firms introduced
their new products in less than one year, whereas 28.7% of firms took a year or more to introduce
their new products (Bowman and Gatignon, 1995, Table 1 on p.46.)
It is well known that a firm’s decision regarding whether to introduce its new product withina shorter time and aggressively depends on multiple factors. For example, according to empirical analysis by Bowman and Gatignon (1995) and Kuester et al. (1999), the factor “market
growth” increases a firm’s market size and profitability, and hence makes its market more attractive; that is, these studies empirically showed that the factor “market growth” motivates firms
to aggressively and quickly introduce their new products in response to their rivals introducing
new products.2 It has also been empirically shown that the factor “firm market share (market
share of the introducing firm)” makes firms adopt a wait-and-see attitude regarding their rivals
introducing new products. This is because, if firms aggressively and quickly introduce new products, the market share of their existing products may be partly diminished by such retaliatory
action. Hence, in this case, firms do not aggressively countervail the introduction of their rivals’
new products.
If firms aggressively and quickly introduce their new products in response to their rivals
introducing new products, the behavior of firms implies strategic complements.3 Conversely, if
1Studies that investigated a firm’s response to a rival introducing a new product were conducted in the 1980s.

For example, MacMillan et al. (1985) conducted a representative study.
2

Debruyne et al. (2002) also empirically showed that the larger the degree of market growth, the higher the

possibility of a firm’s retaliatory act against its rivals introducing new products.
3

Several researchers have used strategic complements and substitutes to assess the reaction time of decision-

making by firms. For example, in the analysis of “merger waves,” the strategic complement is often used tocapture the quickness of mergers among firms; see Fauli-Oller (2000), Qiu and Zhou (2007), and Yao and Zhou

1

firms have wait-and-see attitude and take much longer to introduce new products in response to
their rivals introducing new products by, the behavior of firms is passive and it equals to strategic
substitutes. Whether the outcome of firm behavior that results from strategic interaction amongfirms becomes a strategic complement is important from the viewpoint of industrial development(or to create innovation waves); that is, if strategic complementarity works well, a firm’s new
product introduction causes another firm’s new product introduction, and, as a result, innovationin the whole society rapidly improves.4 Furthermore, it is also important from the viewpointof consumer welfare because product varieties increase and consumer benefit can be enhanced
because of strategic complementarity. However, whether the outcome of strategic interaction
among firms is a strategic complement or strategic substitute tended to be overlooked in previous
studies.
In this paper, we show that a firm’s strategic behavior against its rival introducing a new
product depends on upstream production eﬃciency. We oﬀer a vertical oligopoly model inwhich a firm’s new product introduction behavior can become both a strategic complement andsubstitute because of the degree of upstream eﬃciency. Whereas an upstream monopoly supplierinvests in cost-reducing research and development (R&D),5 downstream duopoly firms introduce
or develop new horizontally diﬀerentiated products. The production eﬃciency of the upstream
monopolist that makes an eﬀort to reduce costs is equivalent to the eﬃciency of its R&D.
In the above scenario, suppose a downstream firm introduces a new product. This behavior(2015).
4

Strategic complementarity represents player’s self-fulfilling expectations (e.g., the reason why I invest is that

I expected others to invest). Similarly, in studies on industrialization and agglomeration, great importance is
attached to the role of self-fulfilling expectations; see, for example, Krugman (1991) and Matsuyama (1991).
5

Fontana and Guerzoni (2008) empirically found that firms tend to attach the most importance to cost-reducing

R&D when their market size is large. In our setting, as downstream new product introduction increases productvarieties and its market size, the demand for inputs increases and the market size of inputs also increases. Hence,
we can consider that assuming cost-reducing R&D upstream is consistent with the empirical findings.

2increases demand for the inputs, so the upstream supplier actively invests to reduce its production cost, and hence, the input price decreases. Because the price-cost margin of downstream
firms widens as the input price lowers and results in a scenario similar to market expansion,
the incentive of the rival firm to introduce a new product becomes strong. Introducing a new
product steals a part of the rival’s market share, so the rival’s incentive to introduce a new
product weakens. At this time, if upstream R&D eﬃciency is high, the input price drops rapidly
because of the introduction of a new product. The fall of the input price facilitates the rival
to introduce a new product, so the race to introduce a new product starts to have the natureof a strategic complement. Conversely, if upstream R&D eﬃciency is low, because the eﬀectof stealing the rival’s market share is dominant, the race to introduce a new product starts tohave the nature of a strategic substitute. Bowman and Gatignon (1995) and Kuester et al.
(1999) empirically found that when the (i) market size is large, or (ii) the firm has less of a
possibility to lose its market share because of the introduction of a new product, retaliatory
action (new product introduction) often occurs. Hence, our model is consistent with the resultsof the empirical study.
Our study also contributes to innovation literature. In most cases, d’Aspremont and Jacquemin(1988)-type process R&D models have an exogenous knowledge spillover among firms. If spilloverfrom the rival’s R&D is small, the investment behavior is a strategic substitute. If R&D spilloveris large, the investment behavior is a strategic complement. Conversely, there is no R&D spilloverin our model. We show that through the input price, upstream eﬃciency works similarly to thespillover parameter in a representative process R&D model; that is, if upstream eﬃciency is low(high), downstream investment behavior is a strategic substitute (complement). In our model,
downstream R&D is the introduction of a new product and it diﬀers from cost-reducing R&D.
However, we newly demonstrate the theoretical fact that an eﬃciency parameter upstream ofa vertical structure model works similarly to the spillover rate in a horizontal process R&D

3framework.
This paper is related to two strands in the literature. One is studies on new product introduction in an oligopoly (Basak and Mukherjee, 2018)6 and the other is studies that focus on
upstream innovation in vertically related markets (e.g., Chen and Sapppington, 2010; Hu et al.,2020; Pinopoulos, 2020; Stefanadis, 1997). Basak and Mukherjee (2018) considered new product introduction in a unionized duopoly. They showed that in their many diﬀerent settings, thestrategic substitute equilibrium such that “one firm only introduces a new product” appears,7
whereas the strategic complimentary equilibrium appears if and only if labor unions are firm
specific (i.e., decentralized labor unions) and product diﬀerentiation is asymmetric. Our modelshows that, when an upstream supplier engages in cost-reducing investment, both the strategic complementary equilibrium and strategic substitute equilibrium appear. This is in sharp
contrast to the results of Basak and Mukherjee (2018).
Although some researchers have also focused on upstream process innovation, their models
and purposes diﬀer substantially from ours. Chen and Sappington (2010) considered the eﬀects
of vertical integration and separation on upstream innovation. Hu et al. (2020) considered
upstream R&D, but their purpose was to examine the relationship between upstream costreducing investment and cross-holdings among downstream firms. Pinopoulos (2020) consideredsome types of input pricing behaviors, for example, two-part tariﬀs, by an upstream firm thatengages in cost-reducing R&D. Stefanadis (1997) considered R&D competition between two
6

Additionally, Dawid et al. (2010) considered new product development in a duopoly setting. However, theirmodel has no upstream sector, and the R&D types diﬀer between the two firms: one firm engages in a projectof new product development and the other firm engages in cost-reducing R&D. Although Dobson and Waterson(1996) and Grossman (2007) also considered a similar scenario in which firms choose their number of diﬀerentiatedgoods, there is no upstream market in these models.
7

This case is the same as the scenario in which “one is a multi-product firm and the other is a single productfirm.” Hence, the strategic substitute type of equilibrium in our new product introduction model includes that
scenario. Inomata (2018) and Kawasaki et al. (2014) also considered the coexistence of multi-product andsingle-product firms.

4

upstream suppliers, and also examined the possibility that the conclusion of exclusive supply
contracts with a downstream firm discourages upstream innovation.
This paper is structured as follows: In Section 2, we provide the basic model, and in Section
3, we present the analysis of the model. In Section 4, we perform welfare analysis, and in Section
5, we present downstream price competition. Finally, in Section 6, we draw conclusions.

2

Model

We consider a vertically related market with an upstream firm (U ) and two symmetric downstream firms (Di, i = 1, 2). Di uses one unit of input to produce one unit of the final product,and it competes in Cournot fashion.8 For simplicity, we omit other production costs for Di.
U decides the input price w and makes a take-it-or-leave-it oﬀer. For example, in the UnitedStates, the Robinson–Patman Act is enforced, so U charges a uniform input price for Di.9
U engages in R&D to reduce the constant marginal cost c ∈ (0, 1). To create demand by
introducing a new product, Di chooses whether to conduct R&D paying a fixed cost f (> 0).
Let Di’s existing product be qe,i and its new product be qn,i . When D1 and D2 introduce new
products, inverse demand is10
pe,i = 1 − qe,i − γ(qn,i + qe,j + qn,j ),pn,i = 1 − qn,i − γ(qe,i + qe,j + qn,j ),(1)where pe,i (pe,j ) is the price of the existing product of Di (Dj) and pn,i (pn,j ) is the price ofthe new product of Di (Dj), i ̸= j and i, j = 1, 2. The parameter γ (0 ≤ γ < 1) measuresthe degree of product substitutability among final products. Final products are independent if
8

Our main results do not alter in Bertrand competition. For more details, see Section 4.

9

Even if U conducts price discrimination, our results do not alter.

10

The other possible setting is that the existing and new products are diﬀerentiated. The formula in such acase is pe,i = 1 − (qe,i + qe,j ) − γ(qn,i + qn,j ), i, j = 1, 2; i ̸= j. However, our main results do not alter, so we use
a simpler form (1).

5γ = 0, whereas they are homogeneous if γ = 1.
The gross profit of Di isπDi (qe,i , qn,i ) ≡ (pe,i − w)qe,i + (pn,i − w)qn,i .

(2)

If Di innovates, its profit is πDi (qe,i , qn,i ) − f ; otherwise, its profit is πDi (qe,i , 0).
The profit of U isπU ≡ (w − (c − x))Q − kx2 ,(3)where x is the investment level and kx2 is the R&D cost. k (> 0) denotes R&D eﬃciency. Q isthe demand for input. Q =Q=∑i qe,i +∑i qe,i if no one innovates. Q =∑i qe,i + qn,j if only Dj innovates.

∑

i qn,i if everyone innovates.

We consider the following four-stage game. In the first stage, D1 and D2 independently andsimultaneously choose whether to conduct R&D by paying the fixed cost (I) or not (N ). In
the second stage, U decides the investment level. At the third stage, U charges the input price.
Finally, downstream competes à la Cournot.
This timing structure corresponds to the diﬃculty in R&D. Generally, product development
requires a sunk cost, such as a long-term contract with researchers, and it takes a much longer
time. Hence, downstream R&D is at the first stage. Eﬀort that produces a prototype andrepeatedly tests its safety is not required, so upstream R&D is at the second stage. Downstream
can flexibly adjust its production, so the quantity of final products is decided in the final stage.
The solution concept is the subgame perfect Nash equilibrium.

3

Results

Depending on the downstream investment decisions, four regimes can arise: II, IN , N I, andN N . Because downstream firms are symmetric, IN and N I are the same. We call II the all
product-developers regime, IN and N I the mixed regime, and N N the no one invests regime.

6

Using (1)–(3), we obtain the equilibrium solutions for each regime.

No one invests regime: N N .

Each Di only produces the existing product; hence, qn,i = 0.

Thus, we obtain the following equilibrium outcomes:(1 + c)(γ + 2)k − 1
1−c(1 − c)2 k; xN N =; πUN N =,2(γ + 2)k − 1
2(γ + 2)k − 1
2(γ + 2)k − 1()(1 − c)kNN =N N = q N N 2 for i = 1, 2.
qe,i
; πDi
e,i
2(γ + 2)k − 1
wN N =

Mixed regime: IN or N I.

(4)When Di invests and Dj does not, because Di produces both

existing and new products but Dj only produces its existing product, qe,i > 0, qn,i > 0, qe,j > 0,
and qn,j = 0 (i, j = 1, 2 and i ̸= j). From this we obtain the following:(1 − c)(3 − γ),
4(2 + 2γ − γ 2 )k − (3 − γ)
(1 − c)2 (3 − γ)k2(1 + c)(2 + 2γ − γ 2 )k − (3 − γ)
IN =
;
π
,
wIN = wN I =U
4(2 + 2γ − γ 2 )k − (3 − γ)
4(2 + 2γ − γ 2 )k − (3 − γ)
(1 − c)(2 − γ)k2(1 − c)k
IN = q IN =
IN =
qe,1
; qe,2,n,1
24(2 + 2γ − γ )k − (3 − γ)
4(2 + 2γ − γ 2 )k − (3 − γ)
2
2
2
(
)
IN = 2(1 − c) (2 − γ) (1 + γ)k ; π IN = q IN 2 .
πD1
e,2
D2
2
[4(2 + 2γ − γ 2 )k − (3 − γ)]xIN = xN I =(5)

N I = q N I = q IN = q IN , q IN = q N I , π N I = π IN , and π N I = π IN .
Note that qe,2
n,2
e,1
n,1
e,2
e,1
D2
D1
D1
D2

All product-developers regime: II.

Four diﬀerentiated products are produced in thisregime, so we obtain the following equilibrium outcomes:(1 + c)(2γ + 1)k − 1
1−c(1 − c)2 k; xII =; πUII =,(4γ + 2)k − 1(4γ + 2)k − 1(4γ + 2)k − 12
2(1 − c)k
II = q II =II = (1 − c) (γ + 1)kqe,i
for i = 1, 2.
;
π
n,i
Di
2 [(4γ + 2)k − 1]
2 [(4γ + 2)k − 1]2

wII =To ensure the positive marginal cost of U after investment, we need Assumption 1.
Assumption 1. k > k0 ≡1.
2c(1 + 2γ)

We establish Proposition 1 from Equations (4)–(6).

7(6)

Proposition 1. (i) The equilibrium level of the R&D investment of U is largest in the all
product-developers regime, intermediate size in the mixed regime, and smallest in the no one
invests regime. More precisely, xII > xIN = xN I > xN N . (ii) In all regimes, higher R&D
eﬃciency of U increases its investment level. Higher product substitutability decreases the level of
R&D investment of U . More precisely, ∂xr /∂k < 0 and ∂xr /∂γ < 0, where r = II, IN, N I, N N .

Proof. (i) xII − xIN = 2k(1−c)(1−γ)> 0 and xIN − xN N = 2k(1−c)(1−γ)(2−γ)> 0, where LN ≡LN
LI[(4γ + 2)k − 1][4(2 + 2γ − γ 2 )k − (3 − γ)] and LI ≡ [2(γ + 2)k − 1][4(2 + 2γ − γ 2 )k − (3 − γ)]. (ii)
The partial derivative of xr with respect to k yields ∂xII /∂k = − 2(1−c)(2γ+1)< 0, ∂xIN /∂k =[(4γ+2)k−1]22)2(1−c)(γ+2)
r− 4(1−c)(3−γ)(2+2γ−γ< 0, and ∂xN N /∂k = − [2(γ+2)k−1]
2 < 0. The partial derivative of x with
[4(2+2γ−γ 2 )k−(3−γ)]24(1−c)kIN /∂γ = − 4(1−c)(γ −6γ+8)k< 0, and
respect to γ yields ∂xII /∂γ = − [(4γ+2)k−1]
2 < 0, ∂x[4(2+2γ−γ 2 )k−(3−γ)]222(1−c)k
∂xN N /∂γ = − [2(γ+2)k−1]
2 < 0.

Proposition 1 immediately yields Corollary 1.

Corollary 1. (i) wN N > wIN = wN I > wII . (ii) ∂wr /∂k > 0 and ∂wr /∂γ > 0, wherer = II, IN, N I, N N .

Proof. (i) wN N − wIN = k(1−c)(2−γ)(1−γ)> 0 and wIN − wII = k(1−c)(1−γ)
> 0.
LI
LN

(ii) The(1−c)(2γ+1)IN /∂k =partial derivative of wr with respect to k yields ∂wII /∂k = [(4γ+2)k−1]
2 > 0, ∂w
2(1−c)(3−γ)(2+2γ−γ 2 )(1−c)(γ+2)> 0, and ∂wN N /∂k = [2(γ+2)k−1]2[4(2+2γ−γ 2 )k−(3−γ)]2> 0. The partial derivative of wr with

2(1−c)k2(1−c)k(4−γ)(2−γ)
IN /∂γ =respect to γ yields ∂wII /∂γ = [(4γ+2)k−1]> 0, and2 > 0, ∂w[4(2+2γ−γ 2 )k−(3−γ)]2(1−c)k∂wN N /∂γ = [2(γ+2)k−1]
2 > 0.

The logic behind part (i) of Proposition 1 is as follows: As U engages in cost-reducing
investment, it invests a large amount if it can sell a large amount of input. Innovation by Di
increases the number of product varieties, so the demand for the input also expands. If D1
and D2 innovate, because the input demand is the largest among all regimes and the sales
8

opportunity for inputs is similarly the largest, the investment level becomes the largest. Hence,
when no one innovates, the investment size becomes the least among all regimes. If only Di
innovates, the investment level becomes intermediate.
Part (ii) is intuitive. The first result is natural. Although larger γ makes competition
tougher, in our model, it reduces downstream market size. The latter eﬀect is dominant, so the
input demand shrinks. This impedes upstream investment.
Proposition 1 immediately yields Corollary 1. Since a larger investment corresponds to alower input price, we obtain the ranking of the input price. Part (ii) is a natural one. The eﬀects
of γ are similar to those in part (ii) of Proposition 1.
Our model has a similar timing structure to those of Banerjee and Lin (2003), Gilbert andCvsa (2003), and Haucap and Wey (2004); that is, in their models, the downstream firm (in ourmodel, Di) first executes cost-reducing R&D investment, and after observing it, the upstream
agent (in our model, U ) charges its price. Banerjee and Lin (2003), Gilbert and Cvsa (2003), andHaucap and Wey (2004) emphasized that the increasing upstream price extracts the benefits ofdownstream innovation.11 Conversely, in our study, because market expansion that results fromdownstream R&D promotes upstream investment, the input price falls. Corollary 1 implies that
upstream R&D is very influential in vertical structures.
To derive the equilibrium of the game, we introduce two threshold functions ΦI and ΦN :IN − π N N = π N I − π N N and Φ
IINIII
IN 12 These thresholds areΦI ≡ πD1
N ≡ πD1 − πD1 = πD2 − πD2 .
D1
D2
D2

related to the gain (or loss) that results from the deviation from equilibrium regimes N N andII, respectively.
ΦI and ΦN are given by
11

Banerjee and Lin (2003) showed that a fixed-price contract that uses the input price resolves this hold-up

problem. Conversely, Takauchi and Mizuno (2019) demonstrated that a fixed-price contract can harm upstream
and downstream.
12

Chowdhury (2005) defined ΦI as a non-strategic benefit of R&D and ΦN as a strategic benefit of R&D.

9[(1−c)2 (1−γ)k 2
ΦN ≡

ΦI ≡

1+4γ−γ 2 + 16(2+6γ+6γ 2 +2γ 3 −γ 4 )k 2
− 8(2+4γ+3γ 2 −γ 3 )k2[(4γ + 2)k − 1]2 [4(2 + 2γ − γ 2 )k − (3 − γ)]2

]
>0

and

(1−c)2 (1−γ)k 2 [8(8+8γ−4γ 4 )k 2 − 8(2+2γ+γ 2 −γ 3 )k − 2γ 2 +5γ−1][2(γ + 2)k − 1]2 [4(2 + 2γ − γ 2 )k − (3 − γ)]2

> 0.

Di innovates if f < ΦI and does not innovate if ΦN < f . Hence, the mixed regime, IN &N I,
can appear if ΦN < ΦI ; and the complementary equilibrium, N N &II, can appear if ΦI < ΦN .
These arguments yield Proposition 2.
Proposition 2.
1. Suppose that k ∈ (k0 , 1/(4γ)). Then, ΦI < ΦN . (i) If f < ΦI , II appears; (ii) if f > ΦN ,N N appears; and (iii) if ΦI ≤ f ≤ ΦN , N N &II can appear.
2. Suppose k > 1/(4γ) or 1/(4γ) ≤ k0 . Then, ΦN < ΦI .

(i) If f < ΦN , II appears; (ii) if

f > ΦI , N N appears; and (iii) if ΦN ≤ f ≤ ΦI , IN &N I can appear.
Proof. See Appendix A.

When the fixed cost f is small (large) because I (N ) is the dominant strategy, II (N N )appears. If f is an intermediate size, Di’s strategy depends on upstream R&D eﬃciency k: (i)if k is small, N N &II can appear; and (ii) if k is large, IN &N I can appear (see Figure 1).
The intuition behind this is as follows: (i) When k is small, upstream R&D is eﬃcient.
In this case, if Di deviates from II, its market halves. Furthermore, the input price jumps,
so Di does not deviate from II. If Di deviates from N N , the number of product markets
increases. As upstream R&D eﬃciency is high and the range of the drop in input price islarger, downstream production costs largely fall. However, this promotes the rival’s production
and makes competition in the existing product market tougher, so the benefit of R&D can becanceled. Di does not deviate.
10f /(1 − c)2

N N &IINN
ΦN /(1 − c)2
ΦI /(1 − c)2

IN &N III0k0k1/(4γ)f
Figure 1: Equilibrium of the new product introduction game in (k, (1−c)
2 )-space (k0 < 1/(4γ))

Basak and Mukherjee (2018) found that, in a unionized duopoly, the emergence of thecomplementary equilibrium needs asymmetric product diﬀerentiation and decentralized unions.
Conversely, we show that the complementary equilibrium appears even if there is no asymmetry in product diﬀerentiation. This implies that upstream R&D plays an important role indownstream innovation, and therefore, gives a new insight into the literature.
(ii) When k is large (or 1/(4γ) ≤ k0 ), because upstream R&D is ineﬃcient, the eﬀects of
upstream investment on the input price weakens. If Di deviates from II, its product market
halves and input price rises. However, in this case, the input price is relatively high because of
the large k (see Corollary 1). Thus, the production cost is also high, and the profit from thenew product market becomes relatively small; that is, the R&D benefit is relatively small. Asa deviation from II raises the input price, it also increases the rival’s production cost. This
results in a lessening of competition in the existing product market. Because the R&D benefit issmall and competition in the existing product market loosens, Di has an incentive to choose N

11

if the rival chooses I. The deviation from N N increases the sales market of products and lowers
the input price. Then, although the R&D benefit is small, the input price is high because k islarge. Hence, a fall in production cost through the decrease in input price becomes attractive.
Di chooses I when the rival chooses N .
We believe that our model makes two types of contributions to the literature. The firstis that it oﬀers results that are partially consistent with the results of an empirical study inthe marketing area. As Bowman and Gatignon (1995) and Kuester et al. (1999) empiricallyshowed, the factor “market growth” strengthens a firm’s incentive to introduce a new product ina retaliatory manner; hence, it produces the equilibrium of strategic complementarity. Simultaneously, the factor “firm market share” strengthens the firm’s incentive to wait and see; hence,
it generates the equilibrium of strategic substitutability. In our model, when the parameter k issmall, and, thus, upstream is eﬃcient, the market expanding eﬀect caused by introducing a new
product is strong. This corresponds to the scenario in which the eﬀects of the factor “market
growth” are stronger. Conversely, when k is large, and, thus, the upstream is less eﬃcient, thesnatching eﬀect of the existing-product market caused by introducing a new product becomesstronger. This is related to the scenario in which the eﬀect of the factor “firm market share” isstrong. Hence, our model undoubtedly has empirical relevance to marketing studies.
The second contribution is that our model oﬀers a new finding with respect to the worksof variables between the traditional R&D model and simple vertical structure model. To thebest of our knowledge, one of the most popular models for cost-reducing R&D investment is thed’Aspremont and Jacquemin (1988) model, in which the R&D spillover rate β is an exogenous
variable. More specifically, in that model, if A (> 0) is defined as a constant marginal cost of
R&D firms, the cost function of firm i is defined as Ci (qi , xi , xj ) = [A − xi − βxj ]qi , i, j = 1, 2,i ̸= j with 0 < β < 1. Then, the best response function is a strategic substitute (complement)
if the spillover rate β is small (large).13
13

Henriques (1990) showed that in the d’Aspremont and Jacquemin (1988) model, each firm’s best response to12

Conversely, in our model, there is no term related to R&D spillover “−βxj .” However, by
way of the work in upstream input price w, if k is small (or upstream is eﬃcient), strategic
complementarity appears in downstream innovation. This corresponds to the case in which thespillover rate β is large in the d’Aspremont and Jacquemin (1988) model. Conversely, if k is large(or upstream is less eﬃcient), strategic substitutability appears in downstream innovation. This
corresponds to the case in which the spillover rate β is small in the d’Aspremont and Jacquemin
model.
While d’Aspremont and Jacquemin (1988) considered process R&D, the downstream firmin our model conducts product R&D (new product introduction), so these two models consider
diﬀerent types of R&D. However, from the viewpoint of promotion or the restriction factor ofinnovation, the variable k in our model works similarly to the spillover rate β in the d’Aspremontand Jacquemin model. Because we newly found such an essential fact, our main result oﬀers a
new insight into the study of innovation activities.

4

Welfare analysis

Even if a downstream firm innovates, it will not capture all the surplus it generates. Therefore,
if the innovation cost f is large, downstream firms give up on introducing new products, andunderinvestment (i.e., welfare loss) occurs. In this section, we provide the conditions that cancause underinvestment.
To avoid unnecessary algebraic complexity and facilitate our welfare analysis, we must set alower limiting value for k. Therefore, we make Assumption 2.

Assumption 2.

1
k≥ .
2

First, we discuss underinvestment in terms of consumer surplus. We define consumer surplus
the R&D level is a strategic complement (substitute) if β > 1/2 (β < 1/2).

13

and gross total surplus (excluding downstream R&D cost f ) as follows:CS =2 + q2 + q2 + q2
qe,1
e,2
n,1
n,2
+ γ [qe,1 (qe,2 + qn,1 + qn,2 ) + qe,2 (qn,1 + qn,2 ) + qn,1 qn,2 ]2and T S = CS + πU +∑i πDi .

We have the following equilibrium surpluses: in the all product-developers regime,
CS II =

(1 − c)2 (3γ + 1)k 2(1 − c)2 k[(13γ + 7)k − 2]
II
;
T
S
=
.
2[(4γ + 2)k − 1]22[(4γ + 2)k − 1]2

In the mixed regime,
CS IN =

(1 − c)2 (γ 3 − 7γ 2 + 8γ + 6)k 2,[γ + (−4γ 2 + 8γ + 8)k − 3]2

T S IN =(1 − c)2 k[(7γ 3 − 33γ 2 + 24γ + 42)k − (γ − 3)2 ].
[γ + (−4γ 2 + 8γ + 8)k − 3]2

Note that CS IN = CS N I and T S IN = T S N I .
In the no one invests regime,
CS N N =(1 − c)2 (γ + 1)k 2(1 − c)2 k[(3γ + 7)k − 1]
NN
;
T
S
=
.
[1 − 2(γ + 2)k]2[1 − 2(γ + 2)k]2

By comparing the consumer surpluses, we obtain Result 1.

Result 1. (i) Assume that “II” appears if the equilibrium regime is II and N N . Then, fromthe viewpoint of consumer surplus, underinvestment in downstream occurs if f > ΦN . (ii)Assume that “N N ” appears if the equilibrium regime is II and N N . Then, from the viewpointof consumer surplus, underinvestment in downstream occurs if f > min{ΦN , ΦI }.
Proof. See Appendix B.
Consumers always welcome an increase in the variety of goods they consume. Moreover, from
Proposition 2, equilibrium regime II is not realized when the R&D cost f is large. Therefore,
the consumer surplus is not maximized if f is large.
Next, we discuss underinvestment in terms of total surplus. We assume k > max{1/2, k0 }.
To consider the best regime that maximizes the total surplus, we define the gross benefits of14

an increase in the number of downstream firms conducting R&D: ΨT21S ≡ T S II − T S IN =T S II − T S N I , ΨT10S ≡ T S IN − T S N N = T S N I − T S N N , and ΨT20S ≡ (T S II − T S N N )/2. Moreprecisely, a rise in the number of downstream firms conducting R&D increases the total surplus
if the following conditions are satisfied: ΨT21S > f , ΨT10S > f , or ΨT20S > f .
To provide the result for the total surplus, we need to compare the gross benefits of anincrease in the number of downstream innovating firms. We define g T S (γ, k), which has thesame sign as ΨT21S − ΨT10S . Additionally, we implicitly define k T S (> max{1/2, k0 }), which is thelargest root of the following equation:g T S (γ, k) ≡ 64γ(9γ 4 +γ 3 −88γ 2 −106γ−32)k 4 + 32(−15γ 4 +20γ 3 +142γ 2 +112γ+20)k 3+4(29γ 3 −97γ 2 −298γ−126)k 2 − 4(γ 2 −17γ−26)k − 3 − γ = 0.
Note that all roots of g T S (γ, k) = 0 are depicted in Figure 2. The blue curves represent the setof pair (γ, k), which satisfies g T S (γ, k) = 0. The dashed line represents k = 1/2. In the shadow
area, g T S (γ, k) > 0.
1.0

0.8

TS
ΨTS
21 > Ψ10TS
ΨTS
21 < Ψ100.6

k = kTS0.4

k = 1/2
0.2

0.0

-0.2
0.0

0.2

0.4

0.6

0.8

1.0

Figure 2: Region for g T S (k, γ) > 0

Then, by comparing the total surpluses, we can show the condition under which underinvestment occurs.

15

Result 2. Suppose k > max {1/2, k0 }. (i) Assume that “N N ” is realized if the equilibrium
regimes are “II” and “N N .” Then, from the viewpoint of the total surplus, the condition underwhich underinvestment for downstream development occurs is given as follows:ΦI < f < ΨT20S if max{1/2, k0 } < k ≤ 1/(4γ),ΦN < f < ΨT20S if max{1/2, k0 , 1/(4γ)} < k ≤ k T S ,ΦN < f < ΨT10S if max{1/2, k0 , k T S , 1/(4γ)} < k.
(ii) Assume that “II” is realized if the equilibrium regimes are “II” and “N N .” Then, fromthe viewpoint of the total surplus, underinvestment for downstream development occurs if thefollowing condition is satisfied:ΦN < f < ΨT20S if max{1/2, k0 } < k ≤ max{1/(4γ), k T S },
ΦN < f < ΨT10S if max{1/(4γ), k T S } < k.

Proof. See Appendix C.

0.22
TSII -2f

TSII -2f

0.35

0.20
0.30
0.18
TSIN -f (=TSNI -f )

0.25
0.16

TSIN -f (=TSNI -f )
0.20

ΦI
0.02

ΨTS
10

ΦN
0.04

TSNN

0.14

TSNN

0.06

0.08

0.10

ΦN

ΨTS
20

f

0.12

Figure 3: Total welfare comparison: k = 1

0.01

ΨTS
20

ΦI
0.02

0.03

0.04

ΨTS
10
0.05

0.06

Figure 4: Total welfare comparison: k = 5

Note: In both Figures 3 and 4, c = 2/5 and γ = 1/5.

We depict the results in Figures 3 and 4, in which the horizontal axis represents the fixed
cost f of introducing a new product. For Figure 3 (or Figure 4), we assume c = 2/5, k = 1, and
γ = 1/5 (or 2/5, k = 5, and γ = 1/5). Each figure has three lines. The bold gray, dashed black,16fand solid black lines represent T S II − 2f , T S IN − f (= T S N I − f ), and T S N N , respectively.
Each figure has four vertical dotted lines: ΦI , ΦN , ΨT10S , and ΨT20S . From Proposition 2, ΦI andΦN determine the investment decision for downstream firms. Additionally, at f = ΨT10S andf = ΨT20S , T S IN − f = T S N I − f = T S N N and T S II − 2f = T S N N , respectively.
We explain the intuition behind Result 2. First, we consider a case with small f . At f = 0,
downstream firms always invest. Additionally, the total surplus increases with the number ofinvesting downstream firms because downstream investment increases final demand. From thecontinuity of the total surplus function, both D1 and D2 invest, and this investment decision issocially optimal if f is small.
Second, we consider a case with large f . When no downstream firm invests, the total surplus
is independent of f ; when at least one downstream firm invests, the total surplus decreases withf . Hence, we obtain two threshold values: ΨT10S and ΨT20S . Because from Proposition 2, nodownstream firm invests if f is large, the equilibrium number of investing downstream firms issocially optimal.
Finally, we consider a case in which f takes an intermediate value. In our model, max{ΦI , ΦN } <max{ΨT10S , ΨT20S }. Hence, for any max{ΦI , ΦN } < f < max{ΨT10S , ΨT20S }, no downstream firm
invests while it is socially desirable for both downstream firms to invest. In the case withΦI < f < ΦN in Figure 3, equilibrium investment decisions are II&N N . Hence, whether theunderinvestment of downstream firms occurs depends on the manner in which the equilibrium
is refined. If N N (or II) is realized under ΦI < f < ΦN , downstream investment is insuﬃcient(or socially optimal).

5

Downstream price competition

In this section, we discuss the case in which downstream firms compete on price to show therobustness of the main result in the previous sections. In diﬀerentiated Bertrand competition,17

the demand function depends on whether downstream firms introduce a new product.

No one invests regime: N N . In this regime, the demand function isqe,i =(1 − γ) − pe,i + γpe,j(1 − γ) − pe,j + γpe,i; qe,j =, i ̸= j.
21−γ
1 − γ2Mixed regime: IN or N I.

(7)

As only Di introduces a new product, the demand function is(1 − γ) − (γ + 1)pe,i + γ(pn,i + pe,j ),(1 − γ)(2γ + 1)
(1 − γ) − (γ + 1)pe,j + γ(pe,i + pn,i )
qe,j =,(1 − γ)(2γ + 1)
(1 − γ) − (γ + 1)pn,i + γ(pe,i + pe,j )qn,i =.
(1 − γ)(2γ + 1)
qe,i =All product-developers regime: II.

(8)

In this regime, the demand function is(1 − γ) − (2γ + 1)pe,i + γ(pn,i + pn,j + pe,j ),(1 − γ)(3γ + 1)
(1 − γ) − (2γ + 1)pn,i + γ(pn,j + pe,i + pe,j )qn,i =.
(1 − γ)(3γ + 1)
qe,i =(9)

The timing of the game is similar to that of the Cournot competition case. By applying thesame procedure as that in the previous setting, we obtain each regime’s equilibrium outcomes,
which are shown in Appendix D. Using their equilibrium outcomes, we can derive the best
response of downstream firms.
Suppose that Dj introduces a new product. If f < Φ̂I , Di introduces a new product;otherwise, it does not, andΦ̂I ≡ (1 − c)2 (γ − 1)k 2[

2(2γ+1)(3γ+2)2
γ+1− (γ(γ+5)+4(2γ+1)((γ−2)γ−2)k+3)2(2(γ−2)(γ+1)k+1)2].

Suppose that Dj does not introduce a new product. If f < Φ̂N , Di introduces a new product;otherwise, it does not, and
Φ̂N ≡

[
]
1
8(γ+1)2 (2γ+1)
3γ+1(1 − c)2 (γ − 1)(γ + 1)k 2 (γ(γ+5)+4(2γ+1)((γ−2)γ−2)k+3)−.
2(γ−2(3γ+1)k+1)22

18

These two thresholds, Φ̂I and Φ̂N , have similar properties to those in the Cournot competition
case; that is, even if downstream has a diﬀerent competition form, we obtain similar results.14
The logic behind the results in the Bertrand case is very similar to that of Proposition 2.
Hence, we conclude that the complementary equilibrium appears without asymmetry in productdiﬀerentiation and verify that Proposition 2 has a certain robustness.

6

Conclusion

In a standard vertically related market comprising an upstream monopoly supplier and downstream diﬀerentiated duopoly firms, we showed that whether the behavior of a downstream firm
introducing a new product becomes a strategic complement or substitute depends on upstream
eﬃciency; that is, the eﬃciency level of cost-reducing R&D by the upstream supplier. When
upstream is eﬃcient, the result of a new product introduction game is a strategic complement.
A strategic complement in the behavior of introducing a new product implies that each firm
quickly introduces or develops a new product if its rival introduces a new product, so the varieties of products expand and product innovation rapidly progresses in such a case. Thus, if therival coincidentally introduces a new product in the equilibrium of strategic complementarities,
consumer welfare can increase and further industrial development can be realized, and therefore,
it may be desirable for the whole society.
Although we considered a simple vertical structure in this paper, its analysis is limited to adomestic or single region’s market. When downstream firms have two options, that is, “produce
the product for domestic consumers” and “introduce a new product for foreign consumers,”
it would be interesting to consider what equilibrium patterns arise. Additionally, we did notconsider the eﬀect of vertical integration/separation on upstream and downstream innovation.
14

As the figure for equilibrium in the diﬀerentiated Bertrand competition is almost the same as that in the

Cournot case, we omit it.

19

When upstream and a downstream firm are integrated, because their organizational forms alter,
we expect that innovation activities would be aﬀected. However, this consideration is beyond
the scope of this study. Hence, we leave it to future research.

Acknowledgements
We specially thank Noriaki Matsushima for his highly constructive comments and suggestions.
We thank Keizo Mizuno and Keiichi Kishi, in addition to participants at the JEA Spring Meeting(Kwansei Gakuin University), for their useful comments. This work was supported by JSPS
KAKENHI [grant numbers 18K01613, 20K01646]. We thank Maxine Garcia, PhD, from Edanz(https://jp.edanz.com/ac) for editing a draft of this manuscript. All errors are our own.

Appendix A. Proof of Proposition 2.
By comparing ΦN with ΦI , we obtain
ΦN − ΦI =(1 − c)2 (1 − γ)2 k 2 (1 − 4γk) g(k, γ)
2[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [γ + (−4γ 2 + 8γ + 8) k − 3]2

and g(k, γ) ≡ 16(3γ 4 +5γ 3 +16γ 2 +22γ+8)k 3 −12(5γ 3 +5γ 2 +8γ+6)k 2 +24γ 2 k−3γ+3.
We show that g(k, γ) > 0 and sign{ΦN − ΦI } depend only on 1 − 4γk. To prove g(k, γ) > 0, it
is suﬃcient to show that g(k, γ) has its minimum value at k = k0 and c = 1, and that value ispositive.
First, we show that g(k, γ) is an increasing function of k; that is, g(k, γ) is smallest at k = k0 .
/The first derivative g(k, γ) with respect to k is ∂g(k, γ) ∂k = 24[2(3γ 4 + 5γ 3 + 16γ 2 + 22γ +8)k 2 − (5γ 3 + 5γ 2 + 8γ + 6)k + γ 2 ]. ∂g(k, γ)/∂k is a quadratic function of k and the coeﬃcientof k 2 is positive. Hence, by solving ∂g(k, γ)/∂k > 0 for k, we obtain k < k1 and k > k2 , wherek1 and k2 are roots of g(k, γ) = 0 for k and k1 < k2 .
As k0 = 1/[2c(2γ + 1)] decreases with c, k0 has its minimum value at c = 1. We illustrate

20

k
0.5
0.4
k1
0.3

k2
k0 at c = 1

0.2
0.1
0.2

0.4

0.6

0.8

1.0

γ

Figure 5: k0 at c = 1 and the two roots of g(k, γ) = 0

k1 , k2 , and k0 at c = 1 in Figure 5. Using numerical calculation, we find that for γ ∈ [0, 1], theunique root of k2 − k0 |c=1 = 0 is γ = 1. Hence, ∂g(k, γ)/∂k > 0 for any k > k0 . Therefore,g(k, γ) has its minimum value at k = k0 .
Second, we show ∂g(k0 , γ)/∂c < 0. The derivation yields[]∂g(k0 , γ)−1∂g(k0 , γ) ∂k0∂g(k0 , γ)
< 0.
=
×
=
∂c
∂k
∂c
∂k
2c2 (2γ + 1)
The last inequality is satisfied because ∂g(k, γ)/∂k > 0. Hence, g(k0 , γ) is a decreasing function
for c and it has its minimum value when c = 1.
From the above discussion, g(k0 , γ) has the following minimum value at k = k0 and c = 1:2

(γ+1)
g(k0 , γ) c=1 = (1−γ)> 0. Because g(k0 , γ) c=1 is positive, ∀k > k0 , g(k, γ) > 0. This result(2γ+1)3

implies that sign{ΦN − ΦI } depends only on “1 − 4γk.” Hence, ΦN > ΦI iﬀ k < 1/(4γ).

Appendix B. Proof of Result 1.
Case (i). From Proposition 2, the equilibrium regime is either “IN &N I” or “N N ” if f > ΦN ;and the equilibrium regime is “II” if f < ϕN . Hence, to prove the first part, we should showthat CS II > CS IN (= CS N I ) > CS N N . This is because underinvestment occurs only if f > ΦN .
Case (ii). Applying a similar discussion to that in case (i), we find that underinvestment
occurs only if f > min{ΦN , ΦI }, where the equilibrium regime is also either “IN &N I” or21

“N N .” Hence, in both cases, if we show CS II > CS IN > CS N N , the proof is complete.
First, we consider sign{CS II − CS IN }:CS II − CS IN =(1 − c)2 (1 − γ)k 2 ψ1CS
2[(4γ + 2)k − 1)]2 [γ + (−4γ 2 + 8γ + 8) k − 3]2,

where ψ1CS ≡ 8(2 + 10γ + 9γ 2 − 4γ 3 − 2γ 4 )k 2 − 8γ(2 − γ 2 )k − γ 2 + 2γ − 3.
sign{CS II − CS IN } depends only on ψ1CS . Because ψ1CS is a quadratic function of k andthe coeﬃcient of k 2 is positive, ψ1CS = 0 has two roots: k1CS and k2CS . By solving ψ1CS > 0 fork, we obtain k < k1CS or k > k2CS , where
k1CS ≡√2γ(2−γ 2 )− 6γ 4 −40γ 3 +34γ 2 +52γ+12;4(2+10γ+9γ 2 −4γ 3 −2γ 4 )k2CS ≡√2γ(2−γ 2 )+ 6γ 4 −40γ 3 +34γ 2 +52γ+12.
4(2+10γ+9γ 2 −4γ 3 −2γ 4 )We compare k2CS with k0 . We consider the case c = 1. Using numerical calculation, we find∀γ ∈ [0, 1), k0 |c=1 > k2CS . Because k0 takes the minimum value at c = 1, k0 > k2CS > k1CS for
any c > 0. Hence, CS II − CS IN > 0.
Next, we consider CS IN − CS N N and apply a proof similar to that above:
CS IN − CS N N = −(c − 1)2 (γ − 1)k 2 ψ2CS
(1 − 2(γ + 2)k)2 [γ + (−4γ 2 + 8γ + 8) k − 3]2,

where ψ2CS ≡ 4(3γ 4 − 6γ 3 − 6γ 2 + 16γ + 8)k 2 − 4γ(γ 2 − 2γ + 2)k − 3 + 2γ.
sign{CS IN − CS N N } depends only on ψ2CS . Because the coeﬃcient of k 2 in ψ2CS is positive,
by solving ψ2CS > 0 for k, we obtain k < k3CS or k > k4CS , where
k3CS ≡√γ(γ 2 −2γ+2)−(2−γ) γ 4 −6γ 3 +γ 2 +14γ+6;2(3γ 4 −6γ 3 −6γ 2 +16γ+8)

k4CS ≡√γ(γ 2 −2γ+2)+(2−γ) γ 4 −6γ 3 +γ 2 +14γ+6.
2(3γ 4 −6γ 3 −6γ 2 +16γ+8)

We show k0 > k4CS (> k3CS ). At c = 1, using numerical calculation, we find that, ∀γ ∈ [0, 1),
k0 |c=1 > k4CS . Because k0 has its minimum value at c = 1, for any c > 0, k0 > k4CS > k3CS holds.
Therefore, CS IN − CS N N > 0.

22

Appendix C. Proof of Result 2.
In both regimes, we need to determine, given the downstream R&D cost f , which regime maximizes the total surplus.
We consider case (i) and compare the gross benefits. First, we consider ΨT21S − ΨT10S :ΨT21S − ΨT10S =(1 − c)2 (1 − γ)2 k 2 g T S (k, γ)
2[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [(−4γ 2 + 8γ + 8) k − 3γ]2.

sign{ΨT21S − ΨT10S } depends only on g T S (γ, k). From Figure 2, we obtain g T S (γ, k) > 0 if
max{1/2, k0 } < k < k T S .
Next, we consider ΨT21S − ΨT20S :ΨT21S − ΨT20S =(1 − c)2 (1 − γ)2 k 2 g T S (γ, k)4[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [(−4γ 2 + 8γ + 8) k − 3 + γ]2.

Hence, sign{ΨT21S − ΨT20S } is the same as sign{ΨT21S − ΨT10S }. Then, we obtain ΨT21S − ΨT10S > 0 if
max{1/2, k0 } < k < k T S .
Finally, we consider ΨT20S − ΨT10S :ΨT20S − ΨT10S =(1 − c)2 (1 − γ)2 k 2 g T S (γ, k)4[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [(−4γ 2 + 8γ + 8) k − 3 + γ]2.

sign{ΨT20S − ΨT10S } is also the same as sign{ΨT21S − ΨT10S }. Thus, ΨT20S − ΨT10S > 0 if max{1/2, k0 } <k < kT S .
From these, we have the following ranking of the thresholds:
ΨT10S < ΨT20S < ΨT21S if
max{1/2, k0 } < k < k T S , ΨT21S ≤ ΨT20S ≤ ΨT10S if k ≥ k T S .



(10)

Then, we establish Lemma 1.
Lemma 1. (i) For max{1/2, k0 } < k < k T S , the best regime for total welfare is “II” if f ≤ ΨT20S ;and “N N ” if f > ΨT20S . (ii) For k ≥ k T S , the best regime for total welfare is “II” if f ≤ ΨT21S ;“IN ” or “N I” if ΨT21S < f ≤ ΨT10S ; and “N N ” if f > ΨT10S .
By comparing ΨT10S , ΨP20S , ΨT21S , ΦI , and ΦN , we show the ranking of thresholds. First, we23show ΦI < min{ΨT10S , ΨT21S }. The diﬀerence ΨT10S − ΦI yields
ΨT10S − ΦI =TS
(1 − c)2 (1 − γ)k 2 g10,I[1 − 2(γ + 2)k]2 [(−4γ 2 + 8γ + 8) k − 3 + γ]2,

T S ≡ 4(7γ 4 − 10γ 3 − 42γ 2 + 32γ + 40)k 2 − 4(4γ 3 − 11γ 2 − 6γ + 16)k + 2γ 2 − 7γ + 4.
where g10,I
T S . By solving g T S = 0 for k, we obtain two roots k 1
sign{ΨT10S −ΦI } depends only on g10,I
10,I and
10,I
2 , where k 1
2
2
TS
TS
1
k10,I
10,I < k10,I . Because the coeﬃcient of k in g10,I is positive, g10,I > 0 if k < k10,I
2 . Additionally, using numerical calculation, we can show k 1
2
or k > k10,I
10,I < k10,I < 1/2. As we
T S > 0, which leads to ΨT S − Φ > 0.
assume k > max{1/2, k0 }, we obtain g10,II
10

Next, we consider ΨT21S − ΦI :
ΨT21S − ΦI =TS
(1 − c)2 (1 − γ)k 2 g21,I2[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [(−4γ 2 + 8γ + 8) k − 3 + γ]2,

T S ≡ 32(10γ 6 + 4γ 5 − 23γ 4 − 14γ 3 + 98γ 2 + 128γ + 40)k 4 − 96(5γ 5 − 2γ 4 − 12γ 3 + 7γ 2 +where g21,I26γ + 12)k 3 + 4(65γ 4 − 78γ 3 − 87γ 2 + 118γ + 90)k 2 + 4(−15γ 3 + 28γ 2 + γ − 14)k + 5γ 2 − 12γ + 5.
T S . To prove ΨT S − Φ > 0, we show (i) g T S > 0 at
sign{ΨT21S − ΦI } depends only on g21,I
I
21
21,I
T S /∂k > 0; and (ii) ∂g T S /∂k > 0 at k = 1/2 and ∂ 2 g T S /∂k 2 > 0.
k = 1/2 and ∂g21,I
21,I
21,I
T S /∂k > 0 at k = 1/2 and ∂ 2 g T S /∂k 2 > 0:
First, we show (ii) ∂g21,I
21,I


(
)
TS
∂ 2 g21,I
48 10γ 6 + 4γ 5 − 23γ 4 − 14γ 3 + 98γ 2 + 128γ + 40 k 2
.
= 8
(
)
∂k 2
−72 5γ 5 −2γ 4 −12γ 3 +7γ 2 +26γ+12 k + 65γ 4 −78γ 3 −87γ 2 +118γ+90/
T S ∂k 2 = 0 for k, we obtain two roots. Using numerical calculation, we can show
By solving ∂ 2 g21,I
T S /∂k 2
that all roots are smaller than 1/2. Because the coeﬃcient of k 2 in the equation ∂ 2 g21,I
T S /∂k 2 > 0. Furthermore, by substituting
is positive and we assume k > max{1/2, k0 }, ∂ 2 g21,I(/ )T S /∂k, we obtain ∂g T S ∂kk = 1/2 into ∂g21,I= 4(40γ 6 − 74γ 5 + 9γ 4 + 67γ 3 + 207γ 2 +
21,I
k=1/2
T S /∂k > 0 at k = 1/2 and ∂ 2 g T S /∂k 2 > 0, which leads
163γ + 20) > 0. Thus, we obtain (ii) ∂g21,I
21,I
T S /∂k > 0 ∀k > 1/2.
to ∂g21,I
T S /∂k > 0, to prove (i) g T S > 0 at k = 1/2 and ∂g T S /∂k > 0,
Because we already have ∂g21,I
21,I
21,I
T S > 0 at k = 1/2: g T S
6
5
4
3
2
we only show g21,I
21,I k=1/2 = 20γ − 52γ + 43γ + 8γ + 86γ + 52γ + 3 > 0.

24

T S > 0, which implies ΨT S − Φ > 0. From this and ΨT S − Φ > 0, we
Therefore, ∀k > 1/2, g21,I
I
I
21
10

obtain Lemma 2.
Lemma 2. ΦI < min{ΨT10S , ΨT21S }.
We show ΨT21S > ΦN and ΨT20S > ΦN . We consider ΨT21S − ΦN :
ΨT21S − ΦN =TS
(1 − c)2 (1 − γ)k 2 g21,N[(4γ + 2)k − 1]2 [(−4γ 2 + 8γ + 8) k + γ − 3]2,

T S ≡ 4(4γ 4 − 16γ 3 + 3γ 2 + 26γ + 10)k 2 − 4(2γ 3 − 7γ 2 + 3γ + 5)k + γ 2 − 3γ + 1.
where g21,N
T S . By solving g T S = 0 for k, we obtain two roots k 1
sign{ΨT21S −ΦN } depends only on g21,N
21,N
21,N
2
1
2
T S is positive; hence, g T S > 0 if
and k21,N, where k21,N< k21,N. The coeﬃcient of k 2 in g21,N
21,N
1
2
1
2
k < k21,Nor k > k21,N. Additionally, using numerical calculation, we can show k21,N< k21,N<
T S > 0, which leads to ΨT S − Φ > 0.
1/2. Because we assume k > max{1/2, k0 }, we obtain g21,NN
21

Next, we consider ΨT20S − ΦN :
ΨT20S − ΦN = −TS(c − 1)2 (γ − 1)k 2 g20,N

4[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [γ + (−4γ 2 + 8γ + 8)k − 3]2,

T S ≡ 64(13γ 6 −8γ 5 −134γ 4 −44γ 3 +200γ 2 +176γ+40)k 4 −32(27γ 5 −49γ 4 −214γ 3 +72γ 2 +where g20,N280γ +100)k 3 +12(27γ 4 −78γ 3 −131γ 2 +140γ +114)k 2 −4(13γ 3 −50γ 2 −17γ +54)k+3γ 2 −14γ +7.
T S . To prove ΨT S − Φ > 0, we show (i) g T S > 0 at
sign{ΨT20S − ΦN } depends only on g20,IN
N
20
20,N
T S /∂k > 0; and (ii) ∂g T S /∂k > 0 at k = 1/2 and ∂ 2 g T S /∂k 2 > 0.
k = 1/2 and ∂g20,N
20,N
20,N
T S /∂k > 0 at k = 1/2 and ∂ 2 g T S /∂k 2 > 0:
First, we show (ii) ∂g20,N
20,N


TS
∂ 2 g20,N
32(13γ 6 −8γ 5 −134γ 4 −44γ 3 + 200γ 2 +176γ+40)k 2
.
=24 
∂k 2
−8(27γ 5 −49γ 4 −214γ 3 +72γ 2 +280γ+100)k+27γ 4 −78γ 3 −131γ 2 +140γ+114
T S /∂k 2 = 0 for k, we obtain two roots. Using numerical calculation, we can show
Solving ∂ 2 g20,N
T S /∂k 2 is positivethat both roots are smaller than 1/2. Because the coeﬃcient of k 2 in ∂ 2 g20,N
T S /∂k 2 > 0. Additionally, by substituting k = 1/2 into
and we assume k > max{1/2, k0 }, ∂ 2 g20,N

(
(
)
/ )
T S /∂k, we obtain ∂g T S
6
5
4
3
2
∂g20,N
20,N ∂k k=1/2 = 4 104γ −226γ −697γ +685γ +825γ +165γ +8 >T S /∂k > 0 at k = 1/2 and ∂ 2 g T S /∂k 2 > 0, which leads to
0. Therefore, we obtain (ii) ∂g20,N
20,N

25

1.0
0.8

k

0.6
0.4

k = kTS
10,N
k = 1/2

0.2
0.0
-0.2
0.0

0.2

0.4

0.6

0.8

1.0

T S (k, γ) > 0
Figure 6: Region for g10,N

T S /∂k > 0 for any k > 1/2.
∂g20,NT S /∂k > 0, to prove (i) g T S > 0 at k = 1/2 and ∂g T S /∂k > 0,
Because we already have ∂g20,N
20,N
20,N
T S > 0 at k = 1/2: g T S
6
5
4
3
2
we only show g20,N
20,N k=1/2 = 52γ −140γ −259γ +420γ +222γ +24γ+1 >T S > 0, which implies ΨT S − Φ > 0. Hence, we obtain Lemma 3.
0. Therefore, ∀k > 1/2, g20,NN
20

Lemma 3. ΨT21S > ΦN and ΨT20S > ΦN .
Finally, we compare ΨT10S with ΦN :
ΨT10S − ΦN =TS
(1 − c)2 (1 − γ)k 2 g10,N

2[1 − 2(γ + 2)k]2 [(4γ + 2)k − 1]2 [(−4γ 2 + 8γ + 8) k + γ − 3]2,

T S ≡ 32(22γ 6 −16γ 5 −223γ 4 −62γ 3 +274γ 2 +208γ+40)k 4 −96(7γ 5 −14γ 4 −56γ 3 +17γ 2 +where g10,N62γ + 20)k 3 + 4(55γ 4 − 180γ 3 − 297γ 2 + 296γ + 234)k 2 − 4(7γ 3 − 34γ 2 − 13γ + 40)k + γ 2 − 8γ + 5.
Note that all roots for the above equation are depicted in Figure 6. The blue curves represent
T S (γ, k) = 0. The dashed line represents k = 1/2. Hence,the set of pair (γ, k), which satisfies g10,N
T S as the largest root if for k > 1/2, it exists. In thegiven γ, we can implicitly define k = k10,N
T S (γ, k) > 0.
shadow area, g10,NThis result yields Lemma 4.
T
T
Lemma 4. ΨT10S > ΦN if k > k10,N; ΨT10S ≤ ΦN if max{1/2, k0 } < k ≤ k10,N.

26

From Proposition 2 and Lemmas 1 and 4, we obtain three thresholds for k: 1/(4γ), k T S , andT S . They are depicted in Figure 7. Then, we obtain k T S < min{k T S , 1/(4, γ)}. However,
k10,N
10,N
TS
TS
we cannot conclude that k10,N< k T S because k10,N
is implicitly defined and it diverges toTS ,infinity as γ → 0. Hence, we potentially have five regions: (i) max{1/2, k0 } < k ≤ k10,N
T S } < k ≤ min{1/(4γ), k T S }, (iii) max{1/2, k , k T S } < k ≤ 1/(4γ), (iv)
(ii) max{1/2, k0 , k10,N0

max{1/2, k0 , 1/(4γ)} < k ≤ k T S , and (v) max{1/2, k0 , k T S , 1/(4γ)} < k. Each case is depicted
in Figure 8.
k

1/(4γ)

1.0

(iii)
0.9

(v)

0.8
k

kTS
k = kTS
10,N

0.7

(ii)

k=1/(4γ)
0.6

0.5
0.0

TS
k10,N
0.2

0.4

0.6

0.8

(i)

1.0

γ

kT S
(iv)

γ

O
Figure 8: Cases for threshold ranking

Figure 7: Thresholds

From Proposition 2, Lemmas 2–4, and (10), we obtain the threshold ranking:TS ,ΦI < ΨT10S ≤ ΦN < ΨT20S < ΨT21S if (i) max{1/2, k0 } < k ≤ k10,N
T S } < k ≤ min{1/(4γ), k T S },
ΦI ≤ ΦN < ΨT10S ≤ ΨT20S ≤ ΨT21S if (ii) max{1/2, k0 , k10,NΦI ≤ ΦN < ΨT21S < ΨT20S < ΨT10S if (iii) max{1/2, k0 , k T S } < k ≤ 1/(4γ),ΦN < ΦI < ΨT10S ≤ ΨT20S ≤ ΨT21S if (iv) max{1/2, k0 , 1/(4γ)} < k ≤ k T S ,ΦN < ΦI < ΨT21S < ΨT20S < ΨT10S if (v) max{1/2, k0 , k T S , 1/(4γ)} < k.
By combining this ranking with Lemma 1 and Proposition 2, we can identify the condition
for downstream underinvestment. Therefore, the proof is complete.

27

Appendix D. SPNE outcomes in downstream Bertrand.
To identify Bertrand rivalry, we attach “ˆ” to the variables of the equilibrium solutions.

No one invests regime: N N
2(1−c) k
1−c
NN =
NN =
ŵN N = 1−(c+1)(2−γ)(γ+1)k1−2(2−γ)(γ+1)k , x̂2(2−γ)(γ+1)k−1 , π̂U2(2−γ)(γ+1)k−1 ,2

2

1−(γ+1)k(c−2γ+3)N
N N = (1−c) (1−γ)(γ+1)k .
p̂N, and π̂Die,i =2(γ 2 −γ−2)k+1(2(γ 2 −γ−2)k+1)2

Mixed regime: IN or N I
ŵIN = 2(c+1)(2γ+1)((2−γ)γ+2)k−(γ(γ+5)+3)4(2γ+1)((2−γ)γ+2)k−(γ(γ+5)+3) ,2(1−c) (γ(γ+5)+3)k(1−c)(γ(γ+5)+3), π̂UIN = 4(2γ+1)((2−γ)γ+2)k−(γ(γ+5)+3),x̂IN = 4(2γ+1)((2−γ)γ+2)k−(γ(γ+5)+3)−(2γ+1)k(c(γ+1)(γ+2)+5(1−γ)γ+6)+γ(γ+5)+3−(2γ+1)k(c(γ+1)(γ+2)+5(1−γ)γ+6)+γ(γ+5)+3p̂IN, p̂IN
,
e,i =
n,i =
γ(γ+5)−4(2γ+1)((2−γ)γ+2)k+3
γ(γ+5)−4(2γ+1)((2−γ)γ+2)k+3−2(2γ+1)k(2γc+c+2(1−γ)γ+3)+γ(γ+5)+32(1−c) (1−γ)(2γ+1)(3γ+2) k
IN =
p̂IN
, π̂Di,e,j =γ(γ+5)−4(2γ+1)((2−γ)γ+2)k+3(4(2γ+1)((2−γ)γ+2)k−(γ(γ+5)+3))2
2

2

3

2 2

2

4(1−c) (1−γ)(γ+1) (2γ+1)kIN =.
and π̂Dj(4(2γ+1)((2−γ)γ+2)k−(γ(γ+5)+3))2

All product-developers regime: II2(1−c) (γ+1)k(1−c)(γ+1)(3γ+1)k(γc+c−γ+3)−2(γ+1)
II =
II
II
ŵII = (c+1)(3γ+1)k−(γ+1),(6γ+2)k−(γ+1) , x̂
(6γ+2)k−(γ+1) , π̂U = (6γ+2)k−(γ+1) , q̂e,i =4(3γ+1)k−2(γ+1)
22II = (1−c) (1−γ)(γ+1)(3γ+1)k .
and π̂Di
2[2(3γ+1)k−(γ+1)]2

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