Location choice and costly product differentiation in a mixed duopoly

Abstract

This paper introduces costly product differentiation into a mixed duopoly with strategic location choice in the first stage and price competition in the second stage. Initially, both firms locate at the center with no product differentiation. We demonstrate that the location choices critically depend on the effectiveness of investments in creating product differentiation, and the nationality of the private firm. Firstly, firms choose to move toward the edge only when the investments are sufficiently effective, regardless of whether the private firm is domestic or foreign. Secondly, a profit-maximizing (domestic or foreign) private firm invests more than the public firm which maximizes social welfare. Thirdly, a mixed duopoly with a foreign firm generates a lower degree of product differentiation in comparison to that with a domestic private firm.

Appendix

1.1 Proof of Proposition 1

(i). At $d_1=1/2$ and $d_2=1/2$, the profit of firm 2 is $\pi _2({1}/{2}, {1}/{2})=0$ and the social welfare is ${\text {SW}}({1}/{2}, {1}/{2})=s-{t}/{12}$. We first show that firm 2 obtains no incentive to choose $d_2<1/2$ given $d_1=1/2$, which holds when

$$\begin{aligned} \pi _2 \left( \frac{1}{2},d_2 \right) =\frac{t}{2} \left( \frac{1}{2}+d_2 \right) ^2 \left( \frac{1}{2}-d_2 \right) +\frac{1}{\beta }\ln (2d_2)\le 0 \end{aligned}$$

(22)

for any $d_2<1/2$. Rearranging (22) leads to

$$\begin{aligned} \beta t\le -\frac{2\ln (2d_2)}{(\frac{1}{2}+d_2)^2(\frac{1}{2}-d_2)}\triangleq f(d_2). \end{aligned}$$

Apparently, $f(d_2)$ is a decreasing function of $d_2$, and

$$\begin{aligned} \min _{d_2}f(d_2)=\lim _{d_2\rightarrow \frac{1}{2}}-\frac{2\ln (2d_2)}{(\frac{1}{2}+d_2)^2(\frac{1}{2}-d_2)}=4. \end{aligned}$$

As a result, firm 2 has no motivation to move away from 1/2 if $\beta t\le 4$.

We next prove that firm 1 obtains no incentive to choose $d_1<1/2$ given $d_2=1/2$. That is, ${\text {SW}}(d_1, 1/2)\le {\text {SW}}(1/2, 1/2)$, which can be obtained as the following after simple calculations

$$\begin{aligned} s-c(d_1)-\frac{t}{12}\left( \frac{1}{2}-d_1 \right) ^3-\frac{t}{3}d_1^3-\frac{t}{24}<s-{t}/{12}. \end{aligned}$$

(23)

Rearranging (23) leads to

$$\begin{aligned} \beta t\le -\frac{\ln (2d_1)}{\frac{1}{24}-\frac{1}{3}d_1^3-\frac{1}{12}(\frac{1}{2}-d_1)^3}\triangleq g(d_1). \end{aligned}$$

Apparently, $g(d_1)$ is a decreasing function of $d_1$, and

$$\begin{aligned} \min _{d_1}g(d_1)=\lim _{d_1\rightarrow \frac{1}{2}}-\frac{\ln (2d_1)}{\frac{1}{24}-\frac{1}{3}d_1^3-\frac{1}{12}(\frac{1}{2}-d_1)^3}=8. \end{aligned}$$

As a result, firm 1 has no motivation to move away from 1/2 if $\beta t\le 4$.

(ii). We first prove that if $4<\beta t\le 8$, firm 1 has no incentive to move away from 1/2 when firm 2 chooses $d_2\in [{2}/{5}, {1}/{2})$. As before, we need to show that ${\text {SW}}(d_1, d_2)<{\text {SW}}(1/2, d_2)$ for $d_2\in [{2}/{5}, {1}/{2})$, where

$$\begin{aligned} {\text {SW}}(d_1, d_2) & = s+\frac{1}{\beta }\ln (4d_1d_2)+\frac{t}{12}\left( 3d_1(1-d_2)(1-d_1-d_2)+3d_2(1-d_2-d_2^2)-1-3d_1^3\right) ,\\ {\text {SW}}\left( \frac{1}{2}, d_2 \right) &=s+\frac{1}{\beta }\ln (2d_2)+\frac{t}{96}(-5+6d_2-12d_2^2-24d_2^3). \end{aligned}$$

After simple calculations, ${\text {SW}}(d_1, d_2)<{\text {SW}}(1/2, d_2)$ reduces to

$$\begin{aligned} \beta t\le \frac{32\ln (2d_1)}{(2d_1-1)\left( 4d_1^2+6d_1-4d_1d_2+6d_2-4d_2^2-1\right) } \triangleq h(d_1,d_2). \end{aligned}$$

It is easy to obtain that $h(d_1, d_2)$ decreases with $d_1$ and $d_2$ for all $d_1<{1}/{2}$ and $d_2< {1}/{2}$. Furthermore,

$$\begin{aligned} \min _{d_1,d_2}h(d_1,d_2)=\lim _{d_1\rightarrow \frac{1}{2}, d_2\rightarrow \frac{1}{2}} \frac{32\ln (2d_1)}{(2d_1-1)\left( 4d_1^2+6d_1-4d_1d_2+6d_2-4d_2^2-1\right) }=8. \end{aligned}$$

Thus, firm 1 will stay at the center if $4<\beta t\le 8$.

Next, we show that firm 2 chooses $d_2\in [{2}/{5}, {1}/{2})$ when $d_1={1}/{2}$. We have that

$$\begin{aligned} \pi _2 \left( \frac{1}{2},d_2 \right) =\frac{t}{2}\left( \frac{1}{2}+d_2 \right) ^2\left( \frac{1}{2}-d_2\right) +\frac{1}{\beta }\ln (2d_2). \end{aligned}$$

(24)

Differentiating (24) with respect to $d_2$ leads to

$$\begin{aligned} \frac{d_2}{2}\left( \frac{1}{2}+d_2\right) \left( \frac{1}{2}-3d_2\right) =-\frac{1}{\beta t}, \end{aligned}$$

(25)

which implies that $d_2$ decreases with $\beta t$ when $\beta t\in (4, 8]$. We thus obtain that $d_2\in [{2}/{5}, {1}/{2})$.

(iii). By (9) and (10), we obtain that $d_1=d_2\rightarrow 1/4$ when $\beta t \rightarrow +\infty$. After the usual straightforward calculations mirroring those in the proof of part (ii), we obtain that firm 1 will choose $d_1\in ({1}/{4}, {1}/{2})$. Following (9) and (10), further calculations yield that

$$\begin{aligned} 2d_2(1-d_1+d_2)(1-d_1-3d_2)=d_1(1+d_1-d_2)(1-3d_1-d_2). \end{aligned}$$

We then obtain that firm 2 will choose $d_2\in ({1}/{4}, {2}/{5})$.

1.2 Proof of Proposition 3

The calculations mirror those in the previous proof for Proposition 1.

(i). We first show that firm 2 obtains no incentive to choose $d_2<1/2$ given $d_1=1/2$, which holds when

$$\begin{aligned} \pi _2\left( \frac{1}{2}, d_2 \right) =\frac{t}{8}\left( \frac{1}{2}+d_2\right) ^2\left( \frac{1}{2}-d_2\right) +\frac{1}{\beta }\ln (2d_2)\le \pi _2\left( \frac{1}{2}, \frac{1}{2}\right) =0. \end{aligned}$$

(26)

After simple calculations, the above inequality reduces to

$$\begin{aligned} \beta t\le -\frac{8\ln (2d_2)}{(\frac{1}{2}+d_2)^2(\frac{1}{2}-d_2)}\triangleq f(d_2). \end{aligned}$$

Apparently, $f(d_2)$ is a decreasing function of $d_2$, and

$$\begin{aligned} \min _{d_2}f(d_2)=\lim _{d_2\rightarrow \frac{1}{2}}-\frac{8\ln (2d_2)}{(\frac{1}{2}+d_2)^2(\frac{1}{2}-d_2)}=16. \end{aligned}$$

As a result, firm 2 has no motivation to move away from 1/2 if $\beta t\le 16$.

Firm 1 obtains no incentive to choose $d_1<1/2$ given $d_2=1/2$ if ${\text {SW}}(d_1, 1/2)\le {\text {SW}}(1/2, 1/2)$, which can be reduced to

$$\begin{aligned} \beta t (-23+98d_1-100d_1^2-8d_1^3)\le -128\ln (2d_1). \end{aligned}$$

If $-23+98d_1-100d_1^2-8d_1^3\le 0$, the inequality automatically holds; on the other hand, if $-23+98d_1-100d_1^2-8d_1^3>0$, we need

$$\begin{aligned} \beta t\le \frac{-128\ln (2d_1)}{-23+98d_1-100d_1^2-8d_1^3}\triangleq g(d_1). \end{aligned}$$

Notice that $g(d_1)$ is a decreasing function of $d_1$. Hence, we have

$$\begin{aligned} \min _{d_1}g(d_1)=\lim _{d_1\rightarrow \frac{1}{2}} \frac{-128\ln (2d_1)}{-23+98d_1-100d_1^2-8d_1^3}=32. \end{aligned}$$

As a result, firm 1 has no motivation to move away from 1/2 if $\beta t\le 16$.

(ii). We first prove that if $16<\beta t\le 32$, firm 1 has no incentive to move away from 1/2 when firm 2 chooses $d_2\in [{2}/{5}, {1}/{2})$. As before, we need to show that ${\text {SW}}(d_1, d_2)<{\text {SW}}(1/2, d_2)$ for $d_2\in [{2}/{5}, {1}/{2})$, where

$$\begin{aligned} {\text {SW}}(d_1, d_2)=s+\frac{1}{\beta }\ln (2d_1)-\frac{t}{48}\left( 13+3d_1^3+3d_1^2(13-d_2)-3d_2+3d_2^2+3d_2^3-3d_1(13-2d_2+d_2^2)\right) , \end{aligned}$$

$$\begin{aligned} {\text {SW}} \left( \frac{1}{2}, d_2 \right) =s-\frac{t}{384}(29-6d_2+12d_2^2+24d_2^3). \end{aligned}$$

After simple calculations, ${\text {SW}}(d_1, d_2)<{\text {SW}}(1/2, d_2)$ reduces to

$$\begin{aligned} \beta t>\frac{128\ln (2d_1)}{(1-2d_1)\left( 25-4d_1^2-54d_1+4d_1d_2+6d_2-4d_2^2\right) } \triangleq h(d_1,d_2). \end{aligned}$$

It is easy to obtain that $h(d_1, d_2)<0$ for all $d_1\in (0,1/2)$ and $d_2\in [2/5,1/2)$. Thus, firm 1 will never move to $d_1<1/2$ given $d_2\in [2/5,1/2)$.

Next, we show that firm 2 chooses $d_2\in [{2}/{5}, {1}/{2})$ when $d_1={1}/{2}$. We have that

$$\begin{aligned} \pi _2\left( \frac{1}{2},d_2\right) =\frac{t}{64}(1-2d_2)(1+2d_2)^2+\frac{1}{\beta }\ln (2d_2). \end{aligned}$$

(27)

Differentiating (27) with respect to $d_2$ leads to

$$\begin{aligned} \frac{d_2}{32}(1+2d_2)(1-6d_2)=-\frac{1}{\beta t}, \end{aligned}$$

(28)

which implies that $d_2$ decreases with $\beta t$ when $\beta t\in (16, 32]$. We thus obtain that $d_2\in [{2}/{5}, {1}/{2})$.

(iii) By (14) and (15), we obtain that $d_1\rightarrow 0.47$ and $d_2\rightarrow 0.18$ when $\beta t \rightarrow +\infty$. After the usual straightforward calculations mirroring those in the proof of part (ii), we obtain that firm 1 will choose $d_1\in (0.47, 0.5)$. Following (14) and (15), further calculations yield that

$$\begin{aligned} 2d_2(1-2d_1-2d_2+2d_1d_2+d_1^2-3d_2^2)=d_1(13-26d_1-3d_1^2+2d_1d_2-2d_2+d_2^2). \end{aligned}$$

We then obtain that firm 2 will choose $d_2\in (0.18, 0.4)$.

1.3 Proof of Proposition 4

It is obvious from (21) that $d_i^*>0$ and it decreases with $\beta$. By setting $({\sqrt{1+\frac{96}{\beta t}}-1})/{8}=1/2$, we obtain that $\beta =4/t$. As a results, $d_i^*=1/2$ if $\beta \le 4/t$; otherwise, $0<d_i^*<1/2$.

1.4 Proof of Proposition 5

Firstly, we show that $(d_1^*+d_2^*)|_{\mathrm {Case\, I}}\le (d_1^*+d_2^*)|_{\mathrm{Case\,II}}$. To this end, we rewritten the equilibrium locations under private duopoly from (21) as the following:

• (i). if $\beta t \le 4,$ we have that $d_1^*=d_2^*=1/2$;

• (ii). if $4<\beta t \le 8,$ we have that $0.33\le d_1^*=d_2^*<1/2$;

• (iii). if $8<\beta t\le 12,$ we have that $1/4\le d_1^* =d_2^*<0.33$;

• (iv). if $\beta t >12$ we have that $0< d_1^* =d_2^*<1/4.$

By looking at Proposition 2, it is obvious that (a) when $\beta t \le 4$, $(d_1^*+d_2^*)|_{\mathrm{Case\,I}}=(d_1^*+d_2^*)|_{\mathrm{Case\,II}}$, and (b) when $\beta t > 12$, $(d_1^*+d_2^*)|_{\mathrm{Case\,I}}<(d_1^*+d_2^*)|_{\mathrm{Case\,II}}$. Thus, we only need to prove that $(d_1^*+d_2^*)|_{\mathrm{Case\,I}}<(d_1^*+d_2^*)|_{\mathrm{Case\,II}}$ when $4<\beta t <12$.

For $4<\beta t \le 8$, we observe that $d_1^*|_{\mathrm{Case\,I}}<d_1^*|_{\mathrm{Case\,II}}$. Furthermore, by (21) and (25), we obtain that $d_2^*|_{\mathrm{Case\,I}}<d_2^*|_{\mathrm{Case\,II}}$ after standard calculations. As a result, $(d_1^*+d_2^*)|_{\mathrm{Case\,I}}<(d_1^*+d_2^*)|_{\mathrm{Case\,II}}$ holds for $4<\beta t \le 8$.

For $8<\beta t \le 12$, we have $0.25\le d_1^*=d_2^*<0.33$ in Case I, and $0.44 \le d_1^* <0.5$, and $0.36\le d_2^* <0.4$ in Case II. It is straightforward to see that $d_1^*|_{\mathrm{Case\,I}}<d_1^*|_{\mathrm{Case\,II}}$ and $d_2^*|_{\mathrm{Case\,I}}<d_2^*|_{\mathrm{Case\,II}}$. It thus follows that $(d_1^*+d_2^*)|_{\mathrm{Case\,I}}<(d_1^*+d_2^*)|_{\mathrm{Case\,II}}$ holds for $8<\beta t \le 12$.

Secondly, we show that $(d_1^*+d_2^*)|_{\mathrm{Case\,II}}<(d_1^*+d_2^*)|_{\mathrm{Case\,III}}$. From Proposition 1 and Proposition 3, we have $(d_1^*+d_2^*)|_{\mathrm{Case\,II}}\le (d_1^*+d_2^*)|_{\mathrm{Case\,III}}$ for all $\beta t \le 32$. Next, we will prove $(d_1^*+d_2^*)|_{\mathrm{Case\,II}}<(d_1^*+d_2^*)|_{\mathrm{Case\,III}}$ for $\beta t > 32$.

Simple calculations yield that $0.25<d_1^*<0.35$ and $0.25<d_2^*<0.29$ in Case II. Furthermore, from Proposition  3, we have that $0.47<d_1^*<0.5$ and $0.18<d_2^*<0.4$ in Case III. Notice that the upper limit of $d_1^*+d_2^*$ in Case II is smaller than the lower limit of $d_1^*+d_2^*$ in Case III. As a result, $(d_1^*+d_2^*)|_{\mathrm{Case\,II}}<(d_1^*+d_2^*)|_{\mathrm{Case\,III}}$ holds.

1.5 Proof of Proposition 6

Recall that both firms charge equal prices in equilibrium in both Case I and Case II. The demand of firm 1 is thus $x^*=\frac{1+d_1-d_2}{2}$. We then write down social welfare as

$$\begin{aligned} {\text {SW}}=s-\int _{0}^{x^*}t(y-d_1)^2{\mathrm{d}}y-\int _{x^*}^1t(1-y-d_2)^2{\mathrm{d}}y-c(d_1)-c(d_2). \end{aligned}$$

The socially optimal locations solve the following first-order conditions:

$$\begin{aligned}&\frac{\partial {\text {SW}}}{\partial d_1}=\frac{t }{4}\left( 1+d_1-d_2\right) \left( 1-3 d_1-d_2\right) +\frac{1}{\beta d_1}=0, \end{aligned}$$

(29)

$$\begin{aligned}&\frac{\partial {\text {SW}}}{\partial d_2}=\frac{t }{4}\left( 1-d_1+d_2\right) \left( 1-d_1-3d_2\right) +\frac{1}{\beta d_2}=0. \end{aligned}$$

(30)

Notice that the first-order condition in (29) is the same as that in (10). We then evaluate (30) at the equilibrium location choices under a mixed duopoly in (9), and obtain that

$$\begin{aligned} \frac{\partial {\text {SW}}}{\partial d_2}&=\frac{t }{4}\left( 1-d_1+d_2\right) \left( 1-d_1-3d_2\right) -\frac{t }{2}\left( 1-d_1+d_2\right) \left( 1-d_1-3d_2\right) \\&= -\frac{t }{4}\left( 1-d_1+d_2\right) \left( 1-d_1-3d_2\right) >0. \end{aligned}$$

Hence, a decrease in $d_2$ reduces social welfare. We next show that the private firm invests more on product differentiation in a private duopoly than that in a mixed duopoly. Evaluating $\partial \pi _2/\partial d_2$ in (9) at the equilibrium location choices under a private duopoly in (20) leads to that

$$\begin{aligned} \frac{\partial \pi _2}{\partial d_2}&= \frac{t}{2}(1-d_1+d_2)(1-d_1-3d_2)+\frac{t}{18}\left( (3-d_1+d_2)(1+d_1+3d_2)\right) \\&= \frac{t}{18}\left( 9\left( 1- d_1+ d_2\right) \left( 1-d_1-3 d_2\right) +\left( 3-d_1+d_2\right) \left( 1+d_1+3 d_2\right) \right) \\&=\frac{2t}{9}\left( 3+2 d_1^2-4 d_1 \left( 1-d_2\right) -2 d_2 \left( 1+3 d_2\right) \right) >0, \end{aligned}$$

which indicates that $d_2^*|_{\mathrm{Case\,II}}>d_2^*|_{\mathrm{Case\,I}}$. As a result, ${\text {SW}}^*|_{\mathrm{Case\,I}}<{\text {SW}}^*|_{\mathrm{Case\,II}}$.