Product Design Competition Under Different Degrees of Demand Ambiguity

Abstract

Product differentiation decisions are frequently made under imperfect probabilistic information about consumer tastes (demand ambiguity). We investigate a Hotelling duopoly game of product-design-then-price choices that incorporates demand ambiguity. Our model allows for different levels of demand ambiguity. We find that the impact of ambiguity on product differentiation depends on firms’ ambiguity attitudes. Furthermore, our model generalizes the probabilistic model of Meagher and Zauner (J Econ Theory 117:201–216, 2004) and the non-probabilistic model of Krol (Int J Ind Org 30:593–604, 2012).

Appendix

In many of the following proofs, we use the following upper and lower boundaries:

• Support sizes: $L\in (0,\frac{1}{2}]$ and $\underline{t}\in (0,1]$ and $\overline{t}\in [1, \infty )$,

• Ambiguity (attitude) and variance: $\delta \in [0,1]$ and $\alpha \in [0,1]$ and $\sigma ^2\in [0,\frac{1}{4}]$.

Proof of Lemma 1

Lemma 1 states that, at any pure strategy SPNE, there exists no ${\hat{M}}$ so that one of the firms can monopolize the market ($[-L, L]\subset [{\bar{x}}-\frac{3}{2}, {\bar{x}}+\frac{3}{2}]$). In the following, we will show that, under Assumptions 1, 2, and 3, there exists indeed no such ${\hat{M}}$. We have to consider three cases:

Case 1 :

One of the two firms can monopolize the market for every ${\hat{M}}$. Suppose that firm j is a monopolist. Then, firm i will deviate from its original product design to obtain a positive market share and to make a strictly positive profit. Consequently, this situation cannot be an equilibrium.

Case 2 :

For each of the firms, there exist materializations of the midpoint M such that the firm can monopolize the market. If firm 1 can monopolize the market for certain realizations of M, we can conclude that firm 1 will monopolize the market if ${\hat{M}}=-L$ (since w.lo.g. firm 1 is the firm left of firm 2). Similarly, firm 2 can monopolize the market for ${\hat{M}}=L$. By Lemma 2, we have then $\frac{3}{2}-L<{\bar{x}}<L-\frac{3}{2}$ which implies $\left| {\bar{x}} \right| <L-\frac{3}{2}$—a contradiction since $L\in (0,\frac{1}{2}]$.

Case 3 :

For only one of the firms, there exist materializations of the midpoint M such that the firm can monopolize the market. W.l.o.g. assume that firm 1 can monopolize the market for some ${\hat{M}}$. Then, firm 1 can monopolize the market for ${\hat{M}}=-L$, and, by Lemma 2, ${\bar{x}}>\frac{3}{2}-L$. For the remaining realizations, in particular for ${\hat{M}}=L$, there exists a competitive equilibrium. If it is beneficial for firm 2 to relocate to the left of its original product design given that ${\hat{M}}=L$ materializes, then relocating to the left is also beneficial given any other ${\hat{M}}$. Therefore, firm 2 will deviate from $x_2$ whenever it is beneficial to relocate to the left for ${\hat{M}}=L$. It thus suffices to show that firm 2 wants to relocate to the left for ${\hat{M}}=L$. The derivative of firm 2’s second-stage profit w.r.t. $x_2$ is

$$\begin{aligned} \frac{\partial \varPi _2 (x_1,x_2,{\hat{t}},L)}{\partial x_2}=\frac{\hat{t}}{18}\big [(3+2L+x_1-3x_2)(3+2L-x_1-x_2)\big ] \end{aligned}$$

(4)

From ${\bar{x}}>\frac{3}{2}-L$ it follows that the second term of the right-hand side of (4) is positive. Furthermore, it holds that $x_2>3-2L-x_1$. Therefore, for the first term of the right-hand side of (4), we have $(3+2L+x_1-3x_2)<8L-6+4x_1<8L-6<0$, where the last two inequalities follow from the fact that $L\le \frac{1}{2}$ and $x_1<0$.¹⁷ Hence,

$$\begin{aligned} \frac{\partial \varPi _2 (x_1,x_2,{\hat{t}},L)}{\partial x_2}<0, \end{aligned}$$

which means that firm 2 has an incentive to deviate from its design. Since we consider a symmetric model, a similar argument holds for a scenario where firm 2 is a monopolist for some ${\hat{M}}$.

Hence, the scenario in each case cannot be an equilibrium. $\square$

Lemma 2

(Price equilibrium) If $x_1\le x_2$ and $({\hat{M}}-{\bar{x}})\in \left( -\frac{3}{2},\frac{3}{2}\right)$, firms charge the subsequent equilibrium prices:

$$\begin{aligned} p_1^{*}=\frac{2}{3}{\hat{t}}\varDelta _x\left( {\bar{x}}-{\hat{M}}+\frac{3}{2}\right) \quad \text{and}\quad p_2^{*}=\frac{2}{3}{\hat{t}}\varDelta _x\left( -{\bar{x}}+{\hat{M}}+\frac{3}{2}\right) . \end{aligned}$$

If $x_1\le x_2$ and $({\hat{M}}-{\bar{x}})\notin \left( -\frac{3}{2},\frac{3}{2}\right)$, firms charge the subsequent equilibrium prices:

$$\begin{aligned}&\quad p_1^{*}=2{\hat{t}}\varDelta _x\left( {\bar{x}}-\hat{M}-\frac{1}{2}\right) \quad \text{and} \quad p_2^{*}=0\quad \text{if}\quad ({\hat{M}}-{\bar{x}})\le -\frac{3}{2}\\&or\\&\quad p_2^{*}=2{\hat{t}}\varDelta _x\left( {\hat{M}}-{\bar{x}}-\frac{1}{2} \right) \quad \text{and}\quad p_1^*=0\quad \text{if}\quad ({\hat{M}}-{\bar{x}})\ge \frac{3}{2}. \end{aligned}$$

Proof

See Anderson et al. (1997, p. 107) and Meagher and Zauner (2004, p. 203). $\square$

Lemma 3

If $x_1\le x_2$ w.l.o.g., then, at any pure strategy SPNE for the Hotelling game with demand uncertainty, firms choose product designs, $(x_1^*,x_2^*)$, such that firm i’s expected profit w.r.t. the reference prior $\pi$ is

$$\begin{aligned} {\text{E}}_\pi [\varPi _i(x_i^*,x_j^*,t,M)]=\frac{(-1)^{j}}{18}\,(x_j^*-x_i^*) \bigg [ (2{\bar{x}}^*-3(-1)^{i})^2 +4\sigma ^2\bigg ], \end{aligned}$$

(5)

where ${\bar{x}}^*=\frac{x_1^*+x_2^*}{2}$ and $\sigma ^2$ is the variance of M.

Proof

The first part of firms’ Choquet-expected profit is

$$\begin{aligned} {\text{E}}_\pi [\varPi _i(x_1,x_2,t,M)] =\int \limits _{\underline{t}}^{\overline{t}}\int \limits _{-L}^{L} (-1)^{j} \,\frac{2}{9}\, t\,\varDelta _x \left( {\bar{x}} -\left( M+\frac{3}{2}(-1)^{i} \right) \right) ^2 \pi (t,M)\,dM\,dt. \end{aligned}$$

This expectation is of the form ${\text{E}}_\pi [g_i(t)\,h_i(M)]$ with real-valued Borel-measurable functions $g_i$ and $h_i$ for $i=1,2$. We define

$$\begin{aligned} g_i(t)=t \text{ and } h_i(M)=(-1)^{j}\, \frac{2}{9}\, t\,\varDelta _x \left( {\bar{x}} -\left( M+\frac{3}{2}(-1)^{i} \right) \right) ^2. \end{aligned}$$

By (R7), t and M are uncorrelated. Then, by Lemma 5.20 in Meintrup and Schäffler (2005, p. 131), $g_i(t)$ and $h_i(M)$ are uncorrelated too. Hence, it holds that ${\text{E}}_\pi [g_i(t)\,h_i(M)]={\text{E}}_\pi [g_i(t)]\cdot {\text{E}}_\pi [h_i(M)]=\mu _t \cdot {\text{E}}_\pi [h_i(M)]={\text{E}}_\pi [h_i(M)].$

In the following, we can rely on the results in Meagher and Zauner (2004, p. 205), since ${\text{E}}_\pi [h_i(M)]$ is equal to firm i’s expected profit function in the risk case:

$$\begin{aligned} {\text{E}}_\pi [\varPi _i(x_1,x_2,t,M)]=\frac{(-1)^{j}}{18}\,(x_j-x_i) \bigg [(2{\bar{x}}-3(-1)^{i})^2+4\sigma ^2 \bigg ]. \end{aligned}$$

$\square$

Lemma 4

If the condition $({\hat{M}}-{\bar{x}})\in (-\frac{3}{2},\frac{3}{2})$ is met, firm i’s profit function $\varPi _i(x_i,x_j,{\hat{t}},{\hat{M}})$ is strictly increasing in ${\hat{t}}$, strictly decreasing in ${\hat{M}}$ for firm 1, and strictly increasing in ${\hat{M}}$ for firm 2, provided that $x_1<x_2$.

Proof

Note that $({\hat{M}}-{\bar{x}})\in \left( -\frac{3}{2},\frac{3}{2}\right)$ and $\varDelta _x>0$. Lemma 1 implies that firms’ second-stage profits at the realization $({\hat{t}},{\hat{M}})$ equal the second piece of firms’ second-stage profits [see Eq. (1)]. Therefore, both profit functions are continuously differentiable w.r.t. ${\hat{t}}$ and ${\hat{M}}$. Differentiating ${{\varPi }_i}$ w.r.t. ${\hat{t}}$ yields

$$\begin{aligned} \frac{\partial \varPi _i}{\partial {\hat{t}}}=\frac{1}{18}{\varDelta }_x\left[ 3(-1)^i+2({\hat{M}}-{\bar{x}})\right] ^2>0 \end{aligned}$$

and the derivatives of ${{\varPi }_1}$ and ${{\varPi }_2}$ with respect to ${\hat{M}}$ are

$$\begin{aligned} \frac{\partial {\varPi }_1}{\partial {\hat{M}}}=\underbrace{\frac{2}{9}\hat{t}\varDelta _x}_{>0}\underbrace{\left[ -3+2({\hat{M}}-{\bar{x}})\right] }_{<0}<0 \text{ and } \frac{\partial \varPi _2}{\partial {\hat{M}}}=\underbrace{\frac{2}{9}{\hat{t}}\varDelta _x}_{>0}\underbrace{\left[ 3+2({\hat{M}}-{\bar{x}})\right] }_{>0}>0. \end{aligned}$$

$\square$

Proof of Proposition 1

The proof of the proposition consists of two parts. In the first part, we prove that neither firm has an incentive to “jump over” its rival. The second part shows that the pair $(x_1^*,x_2^*)$ is the only pair that simultaneously solves the first- and the second-order conditions of firms’ optimization problems. Taken together, there exists a unique SPNE where firms’ equilibrium designs are $(x_1^*,x_2^*)$.

Part 1

In the following, we show that firm 1 has no incentive to choose a product design to the right of that from firm 2. Similarly, firm 2 does not want to be on the left of firm 1. Our proof follows the proof of Anderson et al. (1997, pp. 113–114). Given any location $x_2$, if firm 1 chooses its best reply, $R_1^*(x_2)$, under the restriction $x_1\le x_2$, its profit equals $\widehat{{\text{CEU}}}_1={\text{CEU}}_1[R_1^*(x_2),x_2, t,M]$. At first, we show that under global competition (cf. Lemma 1) firm 1’s optimal profit is increasing in $x_2$. By using the envelope theorem on firm 1’s profit function, we obtain:

$$\begin{aligned} \frac{\widehat{{\text{CEU}}}_1}{\partial x_2}= & \, \frac{2\overline{t}\,\left( 1-\alpha \right) \,\delta \,\left( {\left( L+\frac{x_2+R_1^*(x_2)}{2}+\frac{3}{2}\right) }^{2}+\left( x_2-R_1^*(x_2)\right) \,\left( L+\frac{x_2+R_1^*(x_2)}{2}+\frac{3}{2}\right) \right) }{9}\\&+\frac{2\,\alpha \,\underline{t}\,\delta \left( \left( x_2-R_1^*(x_2)\right) \,\left( -L+\frac{x_2+R_1^*(x_2)}{2}+\frac{3}{2}\right) +{\left( -L+\frac{x_2+R_1^*(x_2)}{2}+\frac{3}{2}\right) }^{2}\right) }{9}\\&+(1-\delta )\left( \frac{\left( 4\,{\sigma }^{2}+{\left( x_2+R_1^*(x_2)+3\right) }^{2}\right) }{18}+\frac{\left( x_2-R_1^*(x_2)\right) \,\left( x_2+R_1^*(x_2)+3\right) }{9}\right) \\&> 0. \end{aligned}$$

To see that the sign of this derivative is positive, note that the quadratic terms are positive and the distance $(x_2-R_1^*(x_2))$ is non-negative. By assumption, the parameters $\underline{t},\overline{t}, \alpha$ and $\delta$ are all non-negative. Furthermore, by the global competition lemma, the terms

$$\begin{aligned} \left( L+\frac{x_2+R_1^*(x_2)}{2}+\frac{3}{2}\right) , \left( -L+\frac{x_2+R_1^*(x_2)}{2}+\frac{3}{2}\right) , \text{ and } \left( x_2+R_1^*(x_2)+3\right) \end{aligned}$$

are positive. Hence, all terms are positive or non-negative.

Now, suppose firm 1 locates to the other side of its rival. Then, given any location $x_2$, if firm 1 chooses its best reply, $R_1^*(x_2)$, now, restricting its location to $x_1\ge x_2$, its profit is decreasing in $x_2$. The proof is analogous to the one above.

To sum up, we have that $\frac{\partial \widehat{{\text{CEU}}}_1}{\partial x_2}>0$ for $x_1\le x_2$ and $\frac{\partial \widehat{{\text{CEU}}}_1}{\partial x_2}<0$ for $x_1\ge x_2$. This means that there is a unique “straddle” point, $\tilde{x}$, such that if firm 2 chooses a design $x_2>\tilde{x}$, firm 1 will optimally locate to the left of firm 2. Otherwise, if $x_2<\tilde{x}$, firm 1 chooses an optimal design to the right of firm 2. Similar arguments apply to firm 2. Hence, in equilibrium, it holds that $x_1^*<\tilde{x}<x_2^*$ and neither firm has an incentive to jump over its opponent.

Part 2

Solving the first-order conditions of the profit maximization problem with respect to $x_1$ and $x_2$ gives three solution pairs:

$$\begin{aligned} x_1^{*}= & \, \frac{\delta \left[ (1-\alpha ) (2 L+3)^2 \overline{t}+\alpha (3-2 L)^2 \underline{t}-4 \sigma ^2-9\right] +4 \sigma ^2+9}{4 (\delta ((\alpha -1) (2 L+3) \overline{t}+\alpha (2 L-3) \underline{t}+3)-3)}\\ x_2^{*}= & \, -x_1^{*}=\frac{\delta \left[ (\alpha -1) (2 L+3)^2 \overline{t}-\alpha (3-2 L)^2 \underline{t}+4 \sigma ^2+9\right] -4 \sigma ^2-9}{4 (\delta ((\alpha -1) (2 L+3) \overline{t}+\alpha (2 L-3) \underline{t}+3)-3)}, \end{aligned}$$

and¹⁸

$$\begin{aligned} x_1^{**}= & \, \frac{3+\delta [-3+\alpha \underline{t}(3-2L)+(1-\alpha )\overline{t}(3+2L)]-\sqrt{z}}{2[-1+ \delta (1+(\alpha -1)\overline{t}-\alpha \underline{t})]}\\ x_2^{**}= & \, \frac{-3-\delta [-3+\alpha \underline{t}(3-2L)+(1-\alpha )\overline{t}(3+2L)]-\sqrt{z}}{2[-1+ \delta (1+(\alpha -1)\overline{t}-\alpha \underline{t})]},\\ x_1^{***}= & \, \frac{3+\delta [-3+\alpha \underline{t}(3-2L)+(1-\alpha )\overline{t}(3+2L)]+\sqrt{z}}{2[-1+ \delta (1+(\alpha -1)\overline{t}-\alpha \underline{t})]}\\ x_2^{***}= & \, \frac{-3-\delta [-3+\alpha \underline{t}(3-2L)+(1-\alpha )\overline{t}(3+2L)]+\sqrt{z}}{2[-1+ \delta (1+(\alpha -1)\overline{t}-\alpha \underline{t})]}, \end{aligned}$$

Define the means

$$\begin{aligned} \overline{x}^{**}=\frac{x_1^{**}+x_2^{**}}{2} \qquad \text{and} \qquad \overline{x}^{***}=\frac{x_1^{***}+x_2^{***}}{2}. \end{aligned}$$

Using numerical optimization techniques, we obtain that the range of $\overline{x}^{**}$ is [1, 2]. Similarly, the range of $\overline{x}^{***}$ is $[-2,-1]$. This implies that $\overline{x}^{**}\ge 1$ and $\overline{x}^{***} \le -1$. Moreover, $\overline{x}^{**}$ attains its minimum value 1 for $L=\frac{1}{2}$ and $\overline{x}^{***}$ attains its maximum value $-1$ for $L=\frac{1}{2}$. However, the global competition condition would require that $\overline{x}^{**}<-\frac{1}{2}+\frac{3}{2}=1$ and $\overline{x}_3 > \frac{1}{2}-\frac{3}{2}=-1$. Consequently, the solution pairs $(x_1^{**}, x_2^{**})$ and $(x_1^{***}, x_2^{***})$ do not fulfill the global competition condition. This excludes these pairs as equilibrium designs.

The first pair of solutions $(x_1^{*},x_2^{*})$ satisfies the global competition condition in Lemma 1 and satisfies the second-order condition of the profit maximization problem. We have checked this by using Wolfram Mathematica version 10.0.0.0. The Mathematica code will be provided by the authors on request. $\square$

Corollary 3

(Ex-ante equilibrium prices and profits) Before uncertainty is resolved, in the unique SPNE, firm i’s Choquet-expected equilibrium price is

$$\begin{aligned} {\text{CEU}}(p_i^*)= \frac{2}{3}\varDelta _x^*\left(\left(1-\delta \right)\frac{3}{2}+\delta \left[\alpha \underline{t}\left(L(-1)^j+\frac{3}{2}\right)+(1-\alpha )\overline{t}\left(L(-1)^i+\frac{3}{2}\right)\right]\right), \end{aligned}$$

and firm i’s Choquet-expected profit is

$$\begin{aligned} {\text{CEU}}_i[x_1^*,x_2^*,t,M]=-\frac{\left( \delta \left[ -(\alpha -1) (2 L+3)^2 \overline{t}+\alpha (3-2 L)^2 \underline{t}-4 \sigma ^2-9\right] +4 \sigma ^2+9\right) ^2}{36 (\delta [(\alpha -1) (2 L+3) \overline{t}+\alpha (2 L-3) \underline{t}+3]-3)}. \end{aligned}$$

Proof

The corollary follows from Lemma 2 and Proposition 1. $\square$

Proof of Proposition 2

Result (i). The derivative of $x_1^*$ with respect to $\delta$ is

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial \delta }=\frac{6(\sigma ^2-L^2)(\overline{t}(1-\alpha )+\underline{t}\alpha )-L(\overline{t}(\alpha -1)+\underline{t}\alpha )(-9+4\sigma ^2)}{2(-3+\delta (3+(3+2L)\overline{t}(\alpha -1)+(-3+2L)\underline{t}\alpha ))^2} \end{aligned}$$

The denominator is positive. Therefore, ${\text{sign}}\big (\frac{\partial x_1^{*}}{\partial \delta }\big )={\text{sign}}(\text{numerator})$. By determining the sign of the numerator depending on $\alpha$, we obtain:

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial \delta }={\left\{ \begin{array}{ll} >0,\quad \text{for}\,\, \alpha \in (\alpha ^*,1]\\ =0,\,\, \text{for}\,\, \alpha =\alpha ^*,\qquad \text{where}\,\, \alpha ^{*}=\frac{\overline{t}[6(\sigma ^2-L^2)+L\left( -9+4\,{\sigma }^{2}\right) ]}{6\,\left( \sigma ^2-L^2\right) \left( \overline{t}-\underline{t}\right) +\left( -9+4\,{\sigma }^{2}\right) \,L\,\left( \overline{t}+\underline{t}\right) }\\ <0,\quad \text{for}\,\, \alpha \in [0,\alpha ^*) \end{array}\right. } \end{aligned}$$

Since $\varDelta _x^*=-2x_1^*$ (see Proposition 1), we have that ${\text{sign}}(\frac{\varDelta _x^*}{\partial \delta })=-{\text{sign}}\big (\frac{\partial x_1^{*}}{\partial \delta }\big )$.

Result (ii). The derivative of $x_1^{*}$ with respect to $\alpha$ is

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial \alpha }=\frac{\delta [(\overline{t}-\underline{t})(1-\delta )6(L^2-\sigma ^2)+\overline{t}\underline{t}L\delta (18-8L^2)+L(9-4\sigma ^2)(1-\delta )(\overline{t}+\underline{t})]}{2(-3+\delta (3+(3+2L)\overline{t}(\alpha -1)+(-3+2L)\underline{t}\alpha ))^2} \end{aligned}$$

The denominator is obviously positive. It can be easily verified that the numerator is also positive. Hence, by ${\text{sign}}(\frac{\varDelta _x^*}{\partial \alpha })=-{\text{sign}}\big (\frac{\partial x_1^{*}}{\partial \alpha }\big )$, we have $\frac{\varDelta _x^*}{\partial \alpha }<0$. $\square$

Proof of Proposition 3

The derivative of $x_1^{*}$ with respect to $\sigma ^2$ is

$$\begin{aligned} \frac{\partial {x_1^{*}}}{\partial \sigma ^2}=\frac{1-\delta }{-3+\delta (3+(3+2L)\overline{t}(\alpha -1)+(-3+2L)\underline{t}\alpha )}. \end{aligned}$$

The numerator is non-negative. Furthermore, the denominator is negative. For $\delta <1$, it holds that $\frac{\partial x_1^{*}}{\partial \sigma ^2}<0$ and, by ${\text{sign}}(\frac{\varDelta _x^*}{\partial \sigma ^2})=-{\text{sign}}\big (\frac{\partial x_1^{*}}{\partial \sigma ^2}\big )$, $\frac{\varDelta _x^*}{\partial \sigma ^2}>0$. $\square$

Proposition 4

At any SPNE for the Hotelling game with demand ambiguity ($\delta >0$):

• (i) A c.p. increase in the support of t ($\underline{t}\downarrow$ and/or $\overline{t}\uparrow$) expands equilibrium differentiation ($\varDelta ^{*}_x \uparrow$).

• (ii) A c.p. increase in the support of M $(L\uparrow )$, expands equilibrium differentiation ($\varDelta ^{*}_x \uparrow$) if $\alpha \in [0,{\hat{\alpha }})$), leaves equilibrium differentiation unchanged ($\varDelta _x^* \rightarrow$) if $\alpha ={\hat{\alpha }}$, and reduces equilibrium differentiation ($\varDelta ^{*}_x \uparrow$) if $\alpha \in ({\hat{\alpha }},1]$.

Proof

Result (i). The derivative of $x_1^{*}$ with respect to $\underline{t}$ is

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial \underline{t}}=\frac{\alpha \delta (-3+2L)[-\overline{t}2L(1-\alpha )\delta (3+2L)-(1-\delta )(3L+2\sigma ^2)]}{2(-3+\delta (3+(3+2L)\overline{t}(\alpha -1)+(-3+2L)\underline{t}\alpha ))^2}. \end{aligned}$$

The denominator is positive. For $\alpha ,\delta >0$, the first term in the numerator is strictly negative since $2L<3$. The second term is strictly negative. Hence, $\frac{\partial x_1^{*}}{\partial \underline{t}}>0$, which implies that $\frac{\varDelta _x^*}{\partial \underline{t}}<0$.

The derivative of $x_1^{*}$ with respect to $\overline{t}$ is

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial \overline{t}}=\frac{(1-\alpha ) \delta (3+2L)[-\underline{t}2L\alpha \delta (3-2L)-(1-\delta )(3L-2\sigma ^2)]}{2(-3+\delta (3+(3+2L)\overline{t}(\alpha -1)+(-3+2L)\underline{t}\alpha ))^2}. \end{aligned}$$

Again, the denominator is positive. For $\delta >0$ and $\alpha <1$, the numerator is strictly positive. Therefore, $\frac{\partial \varDelta _x^{*}}{\partial \overline{t}}>0$.

Result (ii). Let $\delta \in (0,1]$. Then, the derivative of $x_1{^*}$ with respect to L is strictly negative for $\alpha =0$:

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial L}\vert _{\alpha =0}=-\frac{\overline{t}\delta [(9-4\sigma ^2+12L)(1-\delta )+\overline{t}\delta (2L+3)^2]}{2(-3+\delta (3-(3+2L)\overline{t}))^2}<0, \end{aligned}$$

and strictly positive for $\alpha =1$:

$$\begin{aligned} \frac{\partial x_1^{*}}{\partial L}\vert _{\alpha =1}=\frac{\underline{t}\delta [(9-4\sigma ^2-12L)(1-\delta )+\underline{t}\delta (2L-3)^2]}{2(-3+\delta (3+(-3+2L)\underline{t}))^2}>0. \end{aligned}$$

Since $\frac{\partial x_1^{*}}{\partial L}$ is continuous in $\alpha$, by the intermediate value theorem, we know that there exists an ${\hat{\alpha }}\in (0,1)$ such that $\frac{\partial x_1^{*}}{\partial L}=0$ for $\alpha ={\hat{\alpha }}$.

What remains to be shown is that ${\hat{\alpha }}$ is unique. In this case, $x_1^{*}$ is strictly decreasing in L for values of $\alpha$ smaller than ${\hat{\alpha }}$, constant for $\alpha ={\hat{\alpha }}$, and increasing for $1\ge \alpha >{\hat{\alpha }}$. The numerator of the derivative $\frac{\partial x_1^{*}}{\partial L}$ is a polynomial of degree 2 in $\alpha$ (the denominator is strictly positive). Hence, there exists at most one $\hat{\hat{\alpha }}\in (0,1)$, $\hat{\hat{\alpha }}\ne \hat{\alpha }$, such that $\frac{\partial x_1^{*}}{\partial L}\vert _{\alpha =\hat{\hat{\alpha }}}=0$. Assume, to the contrary, that $\hat{\hat{\alpha }}\in (0,1)$ and, w.l.o.g., that $\hat{\alpha }<\hat{\hat{\alpha }}$. We have to consider two cases:

Case 1 :

The numerator of $\frac{\partial x_1^{*}}{\partial L}$ has a global maximum. Then, the global maximum is located in the interval $({\hat{\alpha }}, \hat{\hat{\alpha }})\subset (0,1)$. In this case, it holds that $\frac{\partial x_1^{*}}{\partial L}$ is smaller zero for $\alpha <{\hat{\alpha }}$, equal to zero for $\alpha \in \{{\hat{\alpha }}, \hat{\hat{\alpha }}\}$, and smaller zero for $\alpha \in (\hat{\hat{\alpha }},1]$—a contradiction to $\frac{\partial x_1^{*}}{\partial L}\vert _{\alpha =1}>0$, which was shown above.

Case 2 :

The numerator of $\frac{\partial x_1^{*}}{\partial L}$ has a global minimum, which has to be located in the interval $({\hat{\alpha }}, \hat{\hat{\alpha }})\subset (0,1)$. In this case, we can conclude that $\frac{\partial x_1^{*}}{\partial L}$ is larger than zero for $\alpha \in (\hat{\hat{\alpha }},1]$, equal to zero for $\alpha \in \{\hat{\alpha }, \hat{\hat{\alpha }}\}$, and again larger zero for $\alpha <\hat{\alpha }$, which contradicts that $\frac{\partial x_1^{*}}{\partial L}<0$ for $\alpha =0$.

To sum up, there exists only one ${\hat{\alpha }}\in (0,1)$ such that $\frac{\partial x_1^{*}}{\partial L}\vert _{\alpha ={\hat{\alpha }}}=0$. $\square$