Loving What You Get: The Price Effects of Consumer Self-Persuasion

Abstract

The paper considers how consumers’ cognitive efforts at preference adjustment at the time of decision affect prices in competitive markets with differentiated products. Greater ease of self-persuasion implies higher prices when self-persuasion reinforces first impressions and lower prices when the best opportunities to persuade oneself exist for consumers with weak initial impressions. Exogenous interventions to ease decision-complementing cognition—e.g., advertising—predictably increase or reduce prices, depending upon how they are targeted. While facilitation of consumers’ adjustment always improves welfare in a covered market, firms’ appropriation of surplus may make consumers worse off even as they learn better to love what they get.

Appendix

1.1 Applicability of Log Concavity of $F\left( \cdot \right)$ in $p_{j}$

In this section, I show that log concavity of $F\left( \cdot \right)$ in $p_{0}$ – a critical condition for the existence of an interior equilibrium in prices—may be met (1) for the general class of pointwise-symmetric adjustment map pairs for any symmetric distribution—f, and (2) for an example of a non-pointwise-symmetric adjustment map pair when f is Beta distributed with shape parameters $\left( \alpha ,\beta \right) =\left( 3,3\right)$.

The main issue in the case of non-pointwise-symmetric map pairs is that, approaching the extreme locations $x=0$ and $x=1$, consumers’ marginal adjustment costs approach infinity for the nearby product. Thus, unless marginal adjustment costs for the distant product similarly grow without limit, sensitivity of demand to price rises precipitously at the extremes, making it potentially profitable for firms to attempt to drop price from any candidate interior maximum to a low enough level to take the whole market.

This situation is avoided if the density of consumers at the extremes is sufficiently low, as with some log-concave distributions such as the Beta. So, to summarize, an interior price equilibrium will result whenever the incentive to de-stabilize such an equilibrium is mitigated by pointwise adjustment symmetry; or when there are not enough consumers with extreme tastes for firms to want to de-stabilize an interior price equilibrium despite non-symmetry.

Let us define the log concavity of $F\left( \cdot \right)$ in $p_{0}$ as $\frac {f\left( x_{E}^{*}\right) \frac{\partial x_{E}^{*}}{\partial p_{0}}}{F\left( x_{E}^{*}\right) }$ being decreasing in $p_{0}$. This gives rise to the following necessary and sufficient condition:

$$\begin{aligned} \frac{\frac{\partial ^{2}x_{E}^{*}}{\partial p_{0}^{2}}}{\left( \frac{\partial x_{E}^{*}}{\partial p_{0}}\right) ^{2}}<\frac{f\left( x_{E}^{*}\right) }{F\left( x_{E}^{*}\right) }-\frac{f'\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) } \end{aligned}$$

(11)

Using (8),

$$\begin{aligned} \frac{\partial ^{2}x_{E}^{*}}{\partial p_{0}^{2}}= & {} -\left[ -2t+\intop _{0}^{i^{*1}\left( x_{E}^{*}\right) }\frac{\partial g^{1}}{\partial x_{E}^{*}}di-\intop _{0}^{i^{*0}\left( x_{E}^{*}\right) }\frac{\partial g^{0}}{\partial x_{E}^{*}}di\right] ^{-2}\cdot \\&\left[ \intop _{0}^{i^{*1}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*2}}\frac{\partial x_{E}^{*}}{\partial p_{0}}di-\intop _{0}^{i^{*0}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*2}}\frac{\partial x_{E}^{*}}{\partial p_{0}}di+\frac{\partial g^{1}}{\partial x_{E}^{*}}\frac{\partial i^{*1}}{\partial x}\frac{\partial x_{E}^{*}}{\partial p_{0}}-\frac{\partial g^{0}}{\partial x_{E}^{*}}\frac{\partial i^{*0}}{\partial x}\frac{\partial x_{E}^{*}}{\partial p_{0}}\right] \\= & {} -\left( \frac{\partial x_{E}^{*}}{\partial p_{0}}\right) ^{2}\left[ \intop _{0}^{i^{*1}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*2}}\frac{\partial x_{E}^{*}}{\partial p_{0}}di\right. \\&\left. \quad -\intop _{0}^{i^{*0}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*2}}\frac{\partial x_{E}^{*}}{\partial p_{0}}di+\frac{\partial g^{1}}{\partial x_{E}^{*}}\frac{\partial i^{*1}}{\partial x}\frac{\partial x_{E}^{*}}{\partial p_{0}}-\frac{\partial g^{0}}{\partial x_{E}^{*}}\frac{\partial i^{*0}}{\partial x}\frac{\partial x_{E}^{*}}{\partial p_{0}}\right] \end{aligned}$$

where $\frac{d^{2}g^{j}}{dx_{E}^{*2}}$ represents the second derivative of the $g^{j}$ with respect to x, evaluated at $x_{E}^{*}$. Hence (11) may be re-written

$$\begin{aligned}&-\frac{\partial x_{E}^{*}}{\partial p_{0}}\left[ \intop _{0}^{i^{*1}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*2}}di-\intop _{0}^{i^{*0}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*2}}di+\frac{\partial g^{1}}{\partial x_{E}^{*}}\frac{\partial i^{*1}}{\partial x}-\frac{\partial g^{0}}{\partial x_{E}^{*}}\frac{\partial i^{*0}}{\partial x}\right] \nonumber \\&\quad <\frac{f\left( x_{E}^{*}\right) }{F\left( x_{E}^{*}\right) }-\frac{f'\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) } \end{aligned}$$

(12)

where $\frac{\partial x_{E}^{*}}{\partial p_{0}}$, which is not a function of i, has been pulled out of the integrals.

Consider first the set of pairs of pointwise-symmetric adjustment maps, $\left\{ {\mathscr {G}}^{0},{\mathscr {G}}^{1}:{\mathscr {G}}^{0}=-{\mathscr {G}}^{1}\right\}$. Considering the exemplar pair ${{\bar{\mathscr G}}}^{0}$ and ${{\bar{\mathscr G}}}^{1}$, for each adjustment curve in ${{\bar{\mathscr G}}}^{0}$ corresponding to a given location $x\in \left( 0,1\right)$, the corresponding curve in ${{\bar{\mathscr G}}}^{1}$ would be its mirror image about x.¹⁸ A subset of such pairs, for $j=0,1$ and $\rho =\left[ -1,1\right]$, is given by

$$\begin{aligned} g^{j}=&{\left\{ \begin{array}{ll} \frac{x}{2\left( x-i\right) } &{} for~x\in \left[ 0,\frac{1}{4}\right] \\ \frac{\rho x}{2\left( x-i\right) }+\frac{1-\rho }{2-8i} &{} for~x\in \left[ \frac{1}{4},\frac{1}{2}\right] \\ \frac{\rho \left( 1-x\right) }{2\left( 1-x-i\right) }+\frac{1-\rho }{2-8i} &{} for~x\in \left[ \frac{1}{2},\frac{3}{4}\right] \\ \frac{1-x}{2\left( 1-x-i\right) } &{} for~x\in \left[ \frac{3}{4},1\right] \end{array}\right. } \end{aligned}$$

(13)

Map pairs corresponding to the values $\rho =1$ and $\rho =-1$ are shown in Fig. 6.

Fig. 6

Pointwise-Symmetric Adjustment Map Pairs Per (13)

With pointwise-symmetric adjustment map pairs, and symmetric distribution f, the following conditions hold: (i) $\frac{\partial g^{0}}{\partial x_{E}^{*}}=\frac{\partial g^{1}}{\partial x_{E}^{*}}$, (ii) $\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*2}}=\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*2}}$, (iii) $i^{*0}=i^{*1}$, (iv) $\frac{\partial i^{*0}}{\partial x}=\frac{\partial i^{*1}}{\partial x}$, and (v) $f'\left( x_{E}^{*}\right) =0$. It may be verified, based on these, that the necessary and sufficient condition (12) above for log concavity is met, whence log concavity of $F\left( x\left( p_{0}\right) \right)$ holds for any symmetric distribution f.

Consider now an example of a non-pointwise-symmetric adjustment map pair, given by $g^{0}\left( i,x\right) \equiv \frac{x}{2\left( x-i\right) }$ and $g^{1}\left( i,x\right) \equiv \frac{1-x}{2\left( 1-x-i\right) }$ for $x\in \left[ 0,1\right]$. These functions have the property that $g^{0}\left( 0,x\right) =g^{1}\left( 0,x\right) =\frac{1}{2}$. Observe further that $i^{*0}\left( x,t\right) =\frac{2t-1}{2t}x$ is defined for $t\ge \frac{1}{2}$, whence $i^{*}<x$; similarly $i^{*1}\left( x,t\right) =\frac{2t-1}{2t}\left( 1-x\right)$, whence $i^{*}<1-x$. We also have $\frac{\partial i^{*0}}{\partial x}=\frac{2t-1}{2t}$ and $\frac{\partial i^{*1}}{\partial x}=-\left( \frac{2t-1}{2t}\right)$. We evaluate the left-hand side of (12) at $i=0$ (i.e., the position at which the indifferent consumer evaluates his decision between product options) for all $x\in \left[ 0,1\right]$, using integration by parts:

$$\begin{aligned}&\intop _{0}^{i^{*1}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*2}}di-\intop _{0}^{i^{*0}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*2}}di+\frac{\partial g^{1}}{\partial x_{E}^{*}}\frac{\partial i^{*1}}{\partial x}-\frac{\partial g^{0}}{\partial x_{E}^{*}}\frac{\partial i^{*0}}{\partial x}\nonumber \\&\quad =\intop _{0}^{\frac{2t-1}{2t}\left( 1-x\right) }\frac{i}{\left( 1-x-i\right) ^{3}}di-\intop _{0}^{\frac{2t-1}{2t}x}\frac{i}{\left( x-i\right) ^{3}}di+\frac{i}{2\left( 1-x-i\right) ^{2}}\left( -\frac{2t-1}{2t}\right) -\frac{-i}{2\left( x-i\right) ^{2}}\left( \frac{2t-1}{2t}\right) \nonumber \\&\quad =\left[ \frac{i}{2\left( 1-x-i\right) ^{2}}-\frac{1}{2\left( 1-x-i\right) }\right] _{0}^{\frac{2t-1}{2t}\left( 1-x\right) }\nonumber \\&\qquad -\left[ \frac{i}{2\left( x-i\right) ^{2}}-\frac{1}{2\left( x-i\right) }\right] _{0}^{\frac{2t-1}{2t}x}=\frac{\left( 4t^{2}-4t+1\right) \left( 2x-1\right) }{2x\left( 1-x\right) } \end{aligned}$$

(14)

where $\frac{\partial g^{0}}{\partial x}=\frac{-i}{2\left( x-i\right) ^{2}}\le 0$, $\frac{\partial ^{2}g^{0}}{\partial x^{2}}=\frac{i}{\left( x-i\right) ^{3}}\ge 0$, $\frac{\partial g^{1}}{\partial x}=\frac{i}{2\left( 1-x-i\right) ^{2}}\ge 0$, and $\frac{\partial ^{2}g^{1}}{\partial x^{2}}=\frac{i}{\left( 1-x-i\right) ^{3}}\ge 0$. Substituting into (8) for our example functions we obtain $\frac{\partial x_{E}^{*}}{\partial p_{0}}=-\frac{1}{1-\ln \frac{1}{2t}}$, whence we may re-write the left-hand side of (12) as

$$\begin{aligned}&-\frac{\partial x_{E}^{*}}{\partial p_{0}}\left[ \intop _{0}^{i^{*1}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*2}}di-\intop _{0}^{i^{*0}\left( x_{E}^{*}\right) }\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*2}}di+\frac{\partial g^{1}}{\partial x_{E}^{*}}\frac{\partial i^{*1}}{\partial x}-\frac{\partial g^{0}}{\partial x_{E}^{*}}\frac{\partial i^{*0}}{\partial x}\right] \nonumber \\&\quad =\frac{\left( 4t^{2}-4t+1\right) \left( 2x-1\right) }{2x\left( 1-x\right) \left( 1-\ln \frac{1}{2t}\right) } \end{aligned}$$

(15)

Now assume f is distributed Beta with shape parameters $\left( \alpha ,\beta \right) =\left( 3,3\right)$. We have:

$$\begin{aligned} f\left( x\right) =\frac{\left[ x\left( 1-x\right) \right] ^{2}}{\int _{0}^{1}\left[ u\left( 1-u\right) \right] ^{2}du};~F\left( x\right) =\frac{\int _{0}^{x}\left[ u\left( 1-u\right) \right] ^{2}du}{\int _{0}^{1}\left[ u\left( 1-u\right) \right] ^{2}du} \end{aligned}$$

whereby

$$\begin{aligned} f'\left( x\right) =\frac{2x\left( 1-x\right) \left( 1-2x\right) }{\int _{0}^{1}\left[ u\left( 1-u\right) \right] ^{2}du} \end{aligned}$$

Thus,

$$\begin{aligned} \frac{f'\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) }-\frac{f\left( x_{E}^{*}\right) }{F\left( x_{E}^{*}\right) }= & {} \frac{2x\left( 1-x\right) \left( 1-2x\right) }{\left[ x\left( 1-x\right) \right] ^{2}}-\frac{\left[ x\left( 1-x\right) \right] ^{2}}{\int _{0}^{x}\left[ u\left( 1-u\right) \right] ^{2}du}\nonumber \\= & {} \frac{2\left( 1-2x\right) }{x\left( 1-x\right) }-\frac{30\left[ 1-x\right] ^{2}}{x\left[ 6x^{2}-15x+10\right] } \end{aligned}$$

(16)

One may verify using (15) and (16) that (12) holds for all $x\in \left( 0,1\right)$, $t>\frac{1}{2}$.

1.2 Proof of Lemma 1

Beginning with the expression $g^{j}\left( i^{*j}\left( x,t,\theta \right) ,x,\theta \right) =t$ which implicitly defines $i^{*j}$, and totally differentiating (here, for $j=0$),

$$\begin{aligned} \frac{\partial g^{0}}{\partial i^{*0}}di^{*0}=-\frac{\partial g^{0}}{\partial \theta }d\theta \end{aligned}$$

Using Cramer’s rule, one obtains, given $g^{j}\left( i,x,\theta \right) ={\bar{g}}^{j}\left( i,x\right) -\theta$,

$$\begin{aligned} \frac{\partial i^{*0}}{\partial \theta }=-\frac{\frac{\partial g^{0}}{\partial \theta }}{\frac{\partial g^{0}}{\partial i^{*0}}}=-\frac{-1}{\frac{\partial g^{0}}{\partial i^{*0}}}>0 \end{aligned}$$

Corresponding results can be derived along the same lines for $j=1$.

1.3 Proof of Lemma 2

The proof is an extension of the proof of Bloch & Manceau’s (1999) Lemma 1. Suppose that the market is not covered, that is, at equilibrium prices $\left( p_{0}^{*},p_{1}^{*}\right)$ there exists a consumer x for whom

$$\begin{aligned} V-p_{0}^{*}-t\left[ x-i^{*0}\left( x,t,\theta \right) \right] -\intop _{0}^{i^{*0}\left( x,t,\theta \right) }g^{0}\left( i,x,\theta \right) di<0 \end{aligned}$$

and

$$\begin{aligned} V-p_{1}^{*}-t\left[ 1-x-i^{*1}\left( x,t,\theta \right) \right] -\intop _{0}^{i^{*1}\left( x,t,\theta \right) }g^{1}\left( i,x,\theta \right) di<0 \end{aligned}$$

One can show these prices do not constitute a Nash equilibrium, in that firm 0 can increase its profit by lowering its price $p_{0}$ without altering the profit, hence strategy, of firm 1. Begin by noting that, under $\left( p_{0}^{*},p_{1}^{*}\right)$, because there is a consumer for whom neither good provides nonnegative utility somewhere between the firms, the profit of firm 0 can be written

$$\begin{aligned} \Pi _{0}=p_{0}^{*}\left( x_{0}\right) F\left( x_{0}\right) \equiv \left\{ V-t\left[ x_{0}-i^{*0}\left( x_{0},t,\theta \right) \right] -\intop _{0}^{i^{*0}\left( x_{0},t,\theta \right) }g^{0}\left( i,x_{0},\theta \right) di\right\} F\left( x_{0}\right) \end{aligned}$$

(17)

where $x_{0}$ is the position of the consumer who, at prices $\left( p_{0}^{*},p_{1}^{*}\right)$, is just indifferent between buying product 0 and buying nothing. By assumption, $\frac {\partial \Pi _{0}}{\partial x_{0}}>0$. Now note that

$$\begin{aligned} \frac{\partial p_{0}}{\partial x_{0}}= & {} -t+t\frac{\partial i^{*0}}{\partial x}-g^{0}\left( i^{*0},x_{0},\theta \right) \frac{\partial i^{*0}}{\partial x}-\intop _{0}^{i^{*0}\left( x_{0},t,\theta \right) }\frac{\partial g^{0}}{\partial x_{0}}di\\= & {} -t-\intop _{0}^{i^{*0}\left( x_{0},t,\theta \right) }\frac{\partial g^{0}}{\partial x_{0}}di<-t+\intop _{0}^{i^{*0}\left( x_{0},t,\theta \right) }\frac{\partial g^{0}}{\partial i}di\\= & {} -t+g\left( i^{*0}\left( x_{0}\right) ,x_{0},\theta \right) -g\left( 0,x_{0},\theta \right) =-g\left( 0,x_{0},\theta \right) <0 \end{aligned}$$

which follows from Assumption 4. Since $\frac {\partial \Pi _{0}}{\partial x_{0}}=\left( \frac {\partial \Pi _{0}}{\partial p_{0}}\right) \left( \frac {\partial p_{0}}{\partial x_{0}}\right)$, it follows that $\frac {\partial \Pi _{0}}{\partial p_{0}}<0$. Therefore a small downward deviation in the price $p_{0}$ from $p_{0}^{*}$ increases firm 0’s profits while not affecting firm 1’s profits. This contradicts the assertion that $\left( p_{0}^{*},p_{1}^{*}\right)$ constitutes an equilibrium.

1.4 Proof of Proposition 2

Differentiating (9) with respect to $\theta$ yields

$$\begin{aligned} \frac{\partial ^{2}D_{0}}{\partial p_{0}\partial \theta }= & {} -\frac{f\left( x_{E}^{*}\right) }{\left[ -2t+\int _{0}^{i^{*1}(x_{E}^{*}t,\theta )}\frac{\partial g^{1}}{\partial x_{E}^{*}}di-\int _{0}^{i^{*0}(x_{E}^{*}t,\theta )}\frac{\partial g^{0}}{\partial x_{E}^{*}}di\right] ^{2}}\\&\cdot \left[ \frac{\partial g^{1}}{\partial x_{E}^{*}}\left( \frac{\partial i^{*1}}{\partial \theta }+\frac{\partial i^{*1}}{\partial x}\frac{\partial x_{E}^{*}}{\partial \theta }\right) +\int _{0}^{i^{*1}(x_{E}^{*}t,\theta )}\left( \frac{\partial ^{2}g^{1}}{\partial x_{E}^{*}\partial \theta }+\frac{\partial ^{2}g^{1}}{\partial x^{2}}\frac{\partial x_{E}^{*}}{\partial \theta }\right) di\ \right. \\&\left. -\frac{\partial g^{0}}{\partial x_{E}^{*}}\left( \frac{\partial i^{*0}}{\partial \theta }+\frac{\partial i^{*0}}{\partial x}\frac{\partial x_{E}^{*}}{\partial \theta }\right) -\int _{0}^{i^{*0}(x_{E}^{*}t,\theta )}\left( \frac{\partial ^{2}g^{0}}{\partial x_{E}^{*}\partial \theta }+\frac{\partial ^{2}g^{0}}{\partial x^{2}}\frac{\partial x_{E}^{*}}{\partial \theta }\right) di\right] \end{aligned}$$

Inspection of (2) reveals $\frac{\partial ^{2}g^{0}}{\partial x_{E}^{*}\partial \theta }=\frac{\partial ^{2}g^{1}}{\partial x_{E}^{*}\partial \theta }=0$. Given symmetry, $\frac{\partial x_{E}^{*}}{\partial \theta }=0$. Therefore, simplifying,

$$\begin{aligned} \frac{\partial ^{2}D_{0}}{\partial p_{0}\partial \theta }=\frac{f\left( x_{E}^{*}\right) \left( \frac{\partial g^{0}}{\partial x_{E}^{*}}-\frac{\partial g^{1}}{\partial x_{E}^{*}}\right) \frac{\partial i^{*0}}{\partial \theta }}{\left[ -2t+\int _{0}^{i^{*1}(x_{E}^{*}t,\theta )}\frac{\partial g^{1}}{\partial x_{E}^{*}}di-\int _{0}^{i^{*0}(x_{E}^{*}t,\theta )}\frac{\partial g^{0}}{\partial x_{E}^{*}}di\right] ^{2}} \end{aligned}$$

Per Lemma 1, $\frac{\partial i^{*0}}{\partial \theta }>0$, whence $\frac{\partial ^{2}D_{0}}{\partial p_{0}\partial \theta }$ takes the sign of $\frac{\partial g^{0}}{\partial x_{E}^{*}}-\frac{\partial g^{1}}{\partial x_{E}^{*}}$. Given $\frac{\partial D_{0}}{\partial p_{0}}<0$, the effect of ease of adjustment on price sensitivity of demand takes the sign of $\frac{\partial g^{1}}{\partial x_{E}^{*}}-\frac{\partial g^{0}}{\partial x_{E}^{*}}$.

1.5 Proof of Corollary 2

Using $p_{0}^{*}=-\frac{F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) \frac{\partial x_{E}^{*}}{\partial p_{0}}}$, and noting that symmetry makes $\frac {F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) }$ constant in $\theta$:

$$\begin{aligned} \frac{\partial p_{0}^{*}}{\partial \theta }= & {} \frac{F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) \frac{\partial x_{E}^{*}}{\partial p_{0}}^{2}}\frac{\partial ^{2}x_{E}^{*}}{\partial p_{0}\partial \theta }=\frac{F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) \frac{\partial x_{E}^{*}}{\partial p_{0}}^{2}}\cdot \frac{\left( \frac{\partial g^{0}}{\partial x_{E}^{*}}-\frac{\partial g^{1}}{\partial x_{E}^{*}}\right) \frac{\partial i^{*0}}{\partial \theta }}{\left[ -2t+\int _{0}^{i^{*1}(x_{E}^{*}t,\theta )}\frac{\partial g^{1}}{\partial x_{E}^{*}}di-\int _{0}^{i^{*0}(x_{E}^{*}t,\theta )}\frac{\partial g^{0}}{\partial x_{E}^{*}}di\right] ^{2}} \end{aligned}$$

which takes the sign of $-\left( \frac{\partial g^{1}}{\partial x_{E}^{*}}-\frac{\partial g^{0}}{\partial x_{E}^{*}}\right)$.

1.6 Proof of Proposition 3

Begin by deriving adjustment productivity for ${\tilde{g}}^{j}$ for generic x:

$$\begin{aligned} {\tilde{g}}^{0}\left( i^{*0}\left( x,t\right) ,x\right) =t\Leftrightarrow & {} \frac{t-\theta }{1-\frac{i^{*0}}{2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}}}=t \end{aligned}$$

(18)

$$\begin{aligned} \Rightarrow \frac{t-\theta }{t}\left[ 2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}\right]= & {} 2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}-i^{*0}\nonumber \\ \Rightarrow i^{*0}= & {} \frac{\theta }{t}\left( 2\sigma x+\frac{1}{4}-\frac{1}{2}\sigma \right) \end{aligned}$$

(19)

and by symmetry,

$$\begin{aligned} i^{*1}= & {} \frac{\theta }{t}\left( 2\sigma \left( 1-x\right) +\frac{1}{4}-\frac{1}{2}\sigma \right) \nonumber \\= & {} \frac{\theta }{t}\left( -2\sigma x+\frac{1}{4}+\frac{3}{2}\sigma \right) \end{aligned}$$

(20)

whence one can obtain the expression $i^{*1}\left( x,t,\theta \right) -i^{*0}\left( x,t,\theta \right) =\frac{2\sigma \theta }{t}\left( 1-2x\right)$. Using (5) and (10), one may then derive $x_{E}^{*}$ explicitly. Substitution of our expressions for $i^{*1}\left( x,t,\theta \right) -i^{*0}\left( x,t,\theta \right)$ and ${\tilde{g}}^{j}$ into (5) yield

$$\begin{aligned} p_{1}-p_{0}+\left( t-2\sigma \theta \right) \left( 1-2x_{E}^{*}\right)+ & {} \intop _{0}^{i^{*1}\left( x_{E}^{*},t,\theta \right) }\frac{t-\theta }{1-\frac{i}{2\sigma \left( 1-x_{E}^{*}\right) +\left( 1-2\sigma \right) \cdot \frac{1}{4}}}di\\&-\intop _{0}^{i^{*0}\left( x_{E}^{*},t,\theta \right) }\frac{t-\theta }{1-\frac{i}{2\sigma x_{E}^{*}+\left( 1-2\sigma \right) \cdot \frac{1}{4}}}di=0 \end{aligned}$$

Integration, followed by substitution of (18) and (20), leads to

$$\begin{aligned}&p_{1}-p_{0}+\left( t-2\sigma \theta \right) \left( 1-2x_{E}^{*}\right) \\&\quad +\left( t-\theta \right) \left( -\frac{1}{4}+2\sigma x_{E}^{*}-\frac{3}{2}\sigma \right) \ln \frac{t-\theta }{t}+\left( t-\theta \right) \left( 2\sigma x_{E}^{*}+\frac{1}{4}-\frac{1}{2}\sigma \right) \ln \frac{t-\theta }{t}=0\\&\quad \Leftrightarrow p_{1}-p_{0}+\left( t-2\sigma \theta \right) \left( 1-2x_{E}^{*}\right) -2\sigma \left( t-\theta \right) \left( 1-2x_{E}^{*}\right) \ln \frac{t-\theta }{t}=0 \end{aligned}$$

Simplifying, one obtains $x_{E}^{*}=\frac{1}{2}-\frac{p_{0}-p_{1}}{2\left[ t-2\sigma \theta -2\sigma \left( t-\theta \right) \ln \frac{t-\theta }{t}\right] }$, whence $\frac{\partial x_{E}^{*}}{\partial p_{0}}=-\frac{1}{2\left[ t-2\sigma \theta -2\sigma \left( t-\theta \right) \ln \frac{t-\theta }{t}\right] }=-\frac{\partial x_{E}^{*}}{\partial p_{1}}$. Now substitution into the expressions provided for equilibrium price levels in Proposition 1 yields

$$\begin{aligned} p_{0}^{*}= & {} -\frac{F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) \frac{\partial x_{E}^{*}}{\partial p_{0}}}=2\left[ t-2\sigma \theta -2\sigma \left( t-\theta \right) \ln \frac{t-\theta }{t}\right] \frac {F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) }\\ p_{1}^{*}= & {} \frac{1-F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) \frac{\partial x_{E}^{*}}{\partial p_{1}}}=2\left[ t-2\sigma \theta -2\sigma \left( t-\theta \right) \ln \frac{t-\theta }{t}\right] \frac {\left[ 1-F\left( x_{E}^{*}\right) \right] }{f\left( x_{E}^{*}\right) } \end{aligned}$$

whence $\frac{\partial p_{0}}{\partial \sigma }=2\left( -2\theta -2\left( t-\theta \right) \ln \frac{t-\theta }{t}\right) \frac {F\left( x_{E}^{*}\right) }{f\left( x_{E}^{*}\right) }$ and $\frac{\partial p_{1}}{\partial \sigma }=2\left( -2\theta -2\left( t-\theta \right) \ln \frac{t-\theta }{t}\right) \frac {\left[ 1-F\left( x_{E}^{*}\right) \right] }{f\left( x_{E}^{*}\right) }$. Both of these expressions are negative for all $t>\theta \ge 0$.

1.7 Proof of Proposition 4

Firms incur no costs of production, and based on Corollary 1 we know that there is a common market price. Welfare may therefore be calculated as the integral of utility over the mass of consumers, plus the market price. Given that the market is evenly divided (again, Corollary 1), one may express welfare using (3) and (4) as

$$\begin{aligned} W= & {} V-\intop _{0}^{\frac{1}{2}}\left\{ t\left[ x-i^{*0}\left( x,t,\theta \right) \right] +\intop _{0}^{i^{*0}\left( x,t,\theta \right) }g^{0}\left( i,x,\theta \right) di\right\} dx\\&- \intop _{\frac{1}{2}}^{1}\left\{ t\left[ 1-x-i^{*1}\left( x,t,\theta \right) \right] +\intop _{0}^{i^{*1}\left( x,t,\theta \right) }g^{1}\left( i,x,\theta \right) di\right\} dx \end{aligned}$$

which can be re-written again, using symmetry,

$$\begin{aligned} W= & {} V-2\intop _{0}^{\frac{1}{2}}\left\{ t\left[ x-i^{*0}\left( x,t,\theta \right) \right] +\intop _{0}^{i^{*0}\left( x,t,\theta \right) }g^{0}\left( i,x,\theta \right) di\right\} dx \end{aligned}$$

(21)

Differentiating,

$$\begin{aligned} \frac{\partial W}{\partial \theta }= & {} -2\intop _{0}^{\frac{1}{2}}\left\{ -t\frac{\partial i^{*0}}{\partial \theta }+\intop _{0}^{i^{*0}\left( x,t,\theta \right) }\frac{\partial g^{0}}{\partial \theta }di+g^{0}\left( i^{*0},x,\theta \right) \frac{\partial i^{*0}}{\partial \theta }\right\} dx \end{aligned}$$

and, recognizing that $g^{0}\left( i^{*0},x,\theta \right) \equiv t$ and $\frac{\partial g^{0}}{\partial \theta }=-1$,

$$\begin{aligned} \frac{\partial W}{\partial \theta }= & {} -2\intop _{0}^{\frac{1}{2}}\intop _{0}^{i^{*0}\left( x,t,\theta \right) }\frac{\partial g^{0}}{\partial \theta }didx=2\intop _{0}^{\frac{1}{2}}i^{*0}\left( x,t,\theta \right) dx \end{aligned}$$

which is positive for $\theta >0$ (whence $i^{*0}>0$ $\forall x\in \left( 0,1\right)$, by Lemma 1).

1.8 Proof of Proposition 5

It follows from (10) that, over the range $x=0$ to $x=\frac{1}{4}$ (and $x=\frac{3}{4}$ to $x=1$), all the terms in (21) are invariant to $\sigma$. Differentiation of (21) with respect to $\sigma$ thus yields

$$\begin{aligned} \frac{\partial W}{\partial \sigma }= & {} -2\intop _{\frac{1}{4}}^{\frac{1}{2}}\left\{ -t\frac{\partial i^{*0}}{\partial \sigma }+\intop _{0}^{i^{*0}\left( x,t,\theta \right) }\frac{\partial g^{0}}{\partial \sigma }di+g^{0}\left( i^{*0},x,\theta \right) \frac{\partial i^{*0}}{\partial \sigma }\right\} dx \end{aligned}$$

whence $g^{0}\left( i^{*0},x,\theta \right) =t$ allows us to write

$$\begin{aligned} \frac{\partial W}{\partial \sigma }= & {} -2\intop _{\frac{1}{4}}^{\frac{1}{2}}\intop _{0}^{i^{*0}\left( x,t,\theta \right) }\frac{\partial g^{0}}{\partial \sigma }didx \end{aligned}$$

(22)

Now using the expression for ${\tilde{g}}^{0}$ in (10), for $x\in \left[ \frac{1}{4},\frac{1}{2}\right]$,

$$\begin{aligned} \frac{\partial {\tilde{g}}^{0}}{\partial \sigma }= & {} -\frac{2i\left( t-\theta \right) \left( x-\frac{1}{4}\right) }{\left[ 2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}-i\right] ^{2}} \end{aligned}$$

Using integration by parts and the expression for $i^{*0}$ in (18) ,

$$\begin{aligned} \intop _{0}^{i^{*0}\left( x,t,\theta \right) }\frac{\partial g^{0}}{\partial \sigma }di= & {} \left[ -\frac{2i\left( t-\theta \right) \left( x-\frac{1}{4}\right) }{2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}-i}\right] _{0}^{i^{*0}\left( x,t,\theta \right) }-\intop _{0}^{i^{*0}\left( x,t,\theta \right) }\frac{-2\left( t-\theta \right) \left( x-\frac{1}{4}\right) di}{2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}-i}\\= & {} \left[ -\frac{2\theta \left( \frac{t-\theta }{t}\right) \left( x-\frac{1}{4}\right) \left( 2\sigma x+\frac{1}{4}-\frac{1}{2}\sigma \right) }{\left( 2\sigma x+\frac{1}{4}-\frac{1}{2}\sigma \right) \left( \frac{t-\theta }{t}\right) }\right] -\left\{ 2\left( t-\theta \right) \left( x-\frac{1}{4}\right) \ln \left[ 2\sigma x+\left( 1-2\sigma \right) \cdot \frac{1}{4}-i\right] \right\} _{0}^{i^{*0}\left( x,t,\theta \right) }\\= & {} -2\theta \left( x-\frac{1}{4}\right) -2\left( t-\theta \right) \left( x-\frac{1}{4}\right) \ln \left( \frac{t-\theta }{t}\right) =-\left( x-\frac{1}{4}\right) \left[ 2\theta +\left( 2t-2\theta \right) \ln \left( \frac{t-\theta }{t}\right) \right] \end{aligned}$$

Substituting back into (22),

$$\begin{aligned} \frac{\partial W}{\partial \sigma }= & {} 2\left[ 2\theta +\left( 2t-2\theta \right) \ln \left( \frac{t-\theta }{t}\right) \right] \intop _{\frac{1}{4}}^{\frac{1}{2}}\left( x-\frac{1}{4}\right) dx\\= & {} 4\left[ \theta +\left( t-\theta \right) \ln \left( \frac{t-\theta }{t}\right) \right] \left[ \frac{x^{2}}{2}-\frac{x}{4}\right] _{\frac{1}{4}}^{\frac{1}{2}}\\= & {} 4\left[ \theta +\left( t-\theta \right) \ln \left( \frac{t-\theta }{t}\right) \right] \left( -\frac{1}{32}+\frac{1}{16}\right) \\= & {} \frac{1}{8}\left[ \theta +\left( t-\theta \right) \ln \left( \frac{t-\theta }{t}\right) \right] \end{aligned}$$

which is positive for $\theta >0$.