Price competition with differentiated goods and incomplete product awareness

Abstract

We investigate the effects of increased transparency on prices in the Bertrand duopoly model. Market transparency is defined as the proportion of consumers that are fully informed about the market and thus not captive to one firm. We consider two main cases of strategic interaction, prices as strategic complements and as strategic substitutes. For the former class of games, conventional wisdom concerning prices is confirmed, in that they decrease with market transparency. Consumer welfare always increases with higher transparency but changes in firms’ profits are ambiguous. For the latter class of games, an increase in market transparency may lead to an increase in one of the prices, which implies ambiguous effects on both consumer welfare and firms’ profits. An example with linear demand for differentiated products is also investigated. The results of the paper shed light on the mixed evidence concerning the effects of the Internet on retail markets and may illuminate some of the ongoing related public policy debates.

Appendix

This section contains all the proofs of Sect. 4.

Proof of Proposition 5

The demand elasticity (in absolute value) of informed consumers is

$$\begin{aligned} \left| \varepsilon _{D^{i}}\right|= & {} \left| \frac{p_{i}}{D^{i}} D_{i}^{i}\right| =\left| -\left( \frac{2a_{i}b_{j}-\gamma a_{j}}{ 4b_{i}b_{j}-\gamma ^{2}}-\frac{2b_{j}}{4b_{i}b_{j}-\gamma ^{2}}p_{i}+\frac{ \gamma }{4b_{i}b_{j}-\gamma ^{2}}p_{j}\right) ^{-1}\right. \\&\times \left. p_{i}\frac{2b_{j}}{ 4b_{i}b_{j}-\gamma ^{2}}\right| \end{aligned}$$

The absolute value of the elasticity of the uniformed consumers is given by

$$\begin{aligned} \left| \varepsilon _{d_{i}}\right| =\left| \frac{p_{i}}{d_{i}} d_{i}^{\prime }\right| =\left| -\left( \frac{a_{i}}{2b_{i}}-\frac{ p_{i}}{2b_{i}}\right) ^{-1}\frac{p_{i}}{2b_{i}}\right| \end{aligned}$$

Setting $\left| \varepsilon _{D^{i}}\right| \ge \left| \varepsilon _{d_{i}}\right|$ leads (after computations) to the condition $0\le \gamma \left( a_{j}-p_{j}\right)$, or $\gamma \ge 0$ .$\square$

Proof of Proposition 6

When $\gamma <0\,$ the imperfect transparency game is of strategic substitutes. Propositions 1 and 5 follow that an increase in the transparency level makes both reaction curves shift up; hence, the equilibrium prices cannot both decrease.

A necessary condition for one of the equilibrium prices, say i, to decrease, is

$$\begin{aligned}&\frac{1}{2b_{i}}\left[ a_{i}-2p_{i}^{*}+c_{i}\right] \left[ -2\left( \phi \frac{2b_{i}}{4b_{i}b_{j}-\gamma ^{2}}+\frac{1-\phi }{2}\frac{1}{2b_{j}} \right) \right] \\&\quad -\frac{\phi }{2b_{j}}\left[ a_{j}-2p_{j}^{*}+c_{j}\right] \frac{ \gamma }{4b_{i}b_{j}-\gamma ^{2}}>0 \end{aligned}$$

Reordering yields after simplification $-\left[ a_{i}-2p_{i}^{*}+c_{i} \right] A_{1}>2b_{i}\phi \gamma [a_{j}-2p_{j}^{*}+c_{j}\quad ].$ Using the equilibrium prices (14), we obtain condition (16) after some computations.$\square$

Proof of Lemma 3

The sum of the demand effects $\sum \nolimits _{i=1}^{2}\big ( D^{i}-\frac{1}{2} d_{i}\big )$ is $\frac{K}{8b_{i}b_{j}\big ( 4b_{i}b_{j}-\gamma ^{2}\big ) }\times \big [ \big ( a_{i}-c_{i}\big ) b_{j}\big ( A_{2}-2b_{i}\gamma (A_{1}+\big ( 4b_{i}b_{j}-\gamma ^{2}\big ) )\big ) -\big ( a_{j}-c_{j}\big ) b_{i}\big ( 2b_{j}\gamma (A_{1}+\big ( 4b_{i}b_{j}-\gamma ^{2}\big ) )-A_{2}\big ) \big ] ,$ where $A_{1}$ is as defined in Sect. 4, $A_{2}=\left( 4b_{i}b_{j}-z^{2}\right) \left( 4b_{i}b_{j}+z^{2}\right) +\phi \left( z^{4}+16b_{i}^{2}b_{j}^{2}\right) >0$, and

$$\begin{aligned} K= & {} \frac{A_{1}}{A_{1}^{2}-4b_{i}b_{j}z^{2}\phi ^{2}}>0\hbox { as } A_{1}^{2}-4b_{i}b_{j}z^{2}\phi ^{2}=\left( 4b_{i}b_{j}-z^{2}\right) ^{2}\\&+2\phi \left( 4b_{i}b_{j}+z^{2}\right) \left( 4b_{i}b_{j}-z^{2}+2\phi b_{i}b_{j}\right) +z^{4}\phi ^{2}>0. \end{aligned}$$

We wish to show that

$$\begin{aligned}&\left( a_{i}-c_{i}\right) b_{j}\left( A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )\right) -\left( a_{j}-c_{j}\right) \nonumber \\&\quad \times b_{i}\left( 2b_{j}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )-A_{2}\right) >0. \end{aligned}$$

(17)

There are four possible cases to be considered:

• 1. $A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )>0$ and $\left( 2b_{j}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )-A_{2}\right) >0$

• 2. $A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )<0$ and $\left( 2b_{j}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )-A_{2}\right) >0$

• 3. $A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )>0$ and $\left( 2b_{j}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )-A_{2}\right) <0$

• 4. $A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )<0$ and $\left( 2b_{j}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )-A_{2}\right) <0$

Note that in case of $\gamma <0$ the only possibility is case 3, and in this case (17) is naturally satisfied. Hence, we consider $\gamma >0$ and we would like to show that (17) is satisfied also in cases 1 and 4 and case 2 is not possible, given our assumptions.

• Case 1 Using (15) for firm i  we can replace $\left( a_{j}-c_{j}\right)$ in the left-hand side (henceforth LHS) of (17) to obtain a smaller expression. After rearranging this expression, we obtain $\left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{2}+2\left( 4b_{i}b_{j}-\gamma ^{2}\right) \left( 4b_{i}b_{j}+\gamma ^{2}\right) \phi +\left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}-4\gamma ^{2}b_{i}b_{j}\right) \phi ^{2}$ which is positive, given our initial assumptions. Hence, we conclude that (17) holds.

• Case 4 As before we use (15), but this time for firm j and we can replace $\left( a_{i}-c_{i}\right)$ in the LHS of (17) to obtain a smaller expression. After rearranging this expression we obtain, the same as in case 1, $\left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{2}+2\left( 4b_{i}b_{j}-\gamma ^{2}\right) \left( 4b_{i}b_{j}+\gamma ^{2}\right) \phi +\left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}-4\gamma ^{2}b_{i}b_{j}\right) \phi ^{2}$ which is positive, given our initial assumptions. Hence, we conclude that ( 17) holds.

• Case 2 Suppose $A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )<0$ and $\big ( 2b_{j}\gamma (A_{1}+\big ( 4b_{i}b_{j}-\gamma ^{2}\big ) )-A_{2}\big ) >0,$ then $\gamma \left( b_{i}+b_{j}\right) \left( A_{1}+4b_{i}b_{j}-\gamma ^{2}\right) >A_{2}.$ Using definitions of $A_{1}$ and $A_{2}$, this can be rewritten as

  $$\begin{aligned}&\gamma ^{4}-\gamma ^{3}\left( b_{i}+b_{j}\right) -4\gamma b_{i}b_{j}\left( b_{i}+b\right) +16b_{i}^{2}b_{j}^{2}\nonumber \\&\quad <-\frac{1}{\phi }\left( 2b_{j}-\gamma \right) \left( 2b_{i}-\gamma \right) \left( 4b_{i}b_{j}-\gamma ^{2}\right) . \end{aligned}$$

  (18)

  The right-hand side (henceforth RHS) is negative. We want to show that the LHS is positive; hence, there is a contradiction.

  Assume w.l.o.g that $b_{j}<b_{i}$ and replace in the LHS $\gamma$ by $2b_{j}.$ This way we obtain $(2b_{j})^{4}-\left( b_{i}+b_{j}\right) (2b_{j})^{3}-4b_{i}b_{j}\left( b_{i}+b_{j}\right) 2b_{j}+16b_{j}^{2}b_{i}^{2}=8b_{j}^{2}\left( b_{i}-b_{j}\right) ^{2}>0$. Now we show that

  $$\begin{aligned}&\gamma ^{4}-\gamma ^{3}\left( b_{i}+b_{j}\right) -4\gamma b_{i}b_{j}\left( b_{i}+b\right) +16b_{i}^{2}b_{j}^{2}\nonumber \\&\quad >(2b_{j})^{4}-\left( b_{i}+b_{j}\right) (2b_{j})^{3}-4b_{i}b_{j}\left( b_{i}+b_{j}\right) 2b_{j}+16b_{j}^{2}b_{i}^{2} \end{aligned}$$

  (19)

  since this means that the LHS of (18) is positive.

  To do this, we compute the difference of the left- and the right-hand side of (19). This is positive if and only if $\left( b_{i}-b_{j}\right) \gamma ^{2}-\gamma ^{3}+2b_{j}\left( b_{i}-b_{j}\right) \gamma +4b_{j}\left( b_{i}^{2}-b_{j}^{2}+2b_{i}b_{j}\right) >0.$ To show this, we add and subtract an additional term $4b_{i}b_{j}\gamma$ and obtain $\left( b_{i}-b_{j}\right) \gamma ^{2}+4b_{i}b_{j}\gamma -\gamma ^{3}+2b_{j}\left( b_{i}-b_{j}\right) \gamma +4b_{j}\left( b_{i}^{2}-b_{j}^{2}+2b_{i}b_{j}\right) -4b_{i}b_{j}\gamma =\left( b_{i}-b_{j}\right) \gamma ^{2}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) \gamma +2b_{j}\left( b_{i}-b_{j}\right) \gamma +$ $4b_{j}\left( b_{i}\left( 2b_{j}-\gamma \right) +b_{i}^{2}-b_{j}^{2}\right)$. This is positive since all the components are positive.

Summarizing, we have shown that the LHS of (18) is positive; hence, it cannot be less than the RHS, which is negative. That is why $A_{2}-2b_{i}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )<0\$and$\ 2b_{j}\gamma (A_{1}+\left( 4b_{i}b_{j}-\gamma ^{2}\right) )-A_{2}>0$ is a contradiction.$\square$

Proof of Proposition 7

The indirect effect on firm i’s equilibrium output is given by $[A_{1}A_{3}\left( a_{j}-c_{j}\right) -16b_{j}^{2}b_{i}\phi ^{3}\gamma ^{3}\left( a_{i}-c_{i}\right) ]\gamma /4A_{4}^{2}$, where $A_{1}$ is as defined in Sect. 4,

$A_{3}=\phi ^{2}\left( \gamma ^{4}+4b_{i}b_{j}\left( 4b_{i}b_{j}-\gamma ^{2}\right) \right) +2\phi \left( 4b_{i}b_{j}-\gamma ^{2}\right) \left( 4b_{i}b_{j}{+}\gamma ^{2}\right) +\left( 4b_{i}b_{j}{-}\gamma ^{2}\right) ^{2}>0,$ and

$A_{4}=\phi ^{2}\left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}\!-4\gamma ^{2}b_{i}b_{j}\right) +2\phi \left( 4b_{i}b_{j}{-}\gamma ^{2}\right) \left( 4b_{i}b_{j}{+}\gamma ^{2}\right) +\left( 4b_{i}b_{j}{-}\gamma ^{2}\right) ^{2}>0.$

If $\gamma <0,$ the indirect effect is negative. For the case $\gamma >0$, observe that the sign of the indirect effect is the same as the sign of the numerator. Assume it is negative, then

$$\begin{aligned} \frac{A_{1}A_{3}}{8b_{j}b_{i}\phi ^{3}\gamma ^{3}}<2b_{j}\frac{\left( a_{i}-c_{i}\right) }{\left( a_{j}-c_{j}\right) } \end{aligned}$$

(20)

Conditions (20) and (15) are in contradiction since $\frac{A_{1}A_{3}}{8b_{j}b_{i}\phi ^{3}\gamma ^{3}}> \frac{\gamma \left( 4b_{i}b_{j}-\gamma ^{2}\left( 1-\phi \right) \right) }{ \left( 4b_{i}b_{j}\left( 1+\phi \right) -\gamma ^{2}\right) }$. Therefore, the indirect effect of a $\phi$ increase on total equilibrium output of a firm is positive if $\gamma >0$.$\square$

Proof of Proposition 8

This is based on the same idea as above, namely one can check (after some computations) that the demand effect outweighs the indirect effect when both firms are active in the market in equilibrium (computational details available upon request).$\square$

Proof of Proposition 9

Let $\gamma >0$. From Lemma 3 the sum of the demand effects is positive and from Proposition 7, the indirect effects are positive as well. If $\gamma <0$  we conclude by Proposition 8 that the sum of both firms’ outputs increases when $\phi$ increases.$\square$

Proof of Proposition 10

The derivative of equilibrium profit with respect to $\phi$ is given by

$$\begin{aligned} \frac{\partial }{\partial \phi }\varPi ^{i}(p_{1}^{*},p_{2}^{*})= & {} (p_{i}^{*}-c_{i})\left( D^{i}-\frac{1}{2}d_{i}+\phi D_{j}^{i}p_{j}^{*\prime }(\phi )\right) \\= & {} (p_{i}^{*}-c_{i})\frac{ A_{5}\left( a_{i}-c_{i}\right) -2b_{i}\gamma \left( a_{j}-c_{j}\right) A_{6} }{8A_{7}^{2}\left( 4b_{j}b_{i}-\gamma ^{2}\right) b_{i}}, \end{aligned}$$

where

$$\begin{aligned} A_{5}= & {} \left( 4b_{i}b_{j}+\gamma ^{2}\right) \left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{4}+7\phi ^{4}\left( 4b_{i}b_{j}+\gamma ^{2}\right) \left( \gamma ^{4}+16b_{i}^{2}b_{j}^{2}\right) \\&\times \left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}-4\gamma ^{2}b_{i}b_{j}\right) \\&+\,4\phi \left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}-4\gamma ^{2}b_{i}b_{j}\right) \left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{3} \\&+\,4\phi ^{3}\left( 4b_{i}b_{j}-\gamma ^{2}\right) \left( \gamma ^{4}-2\gamma ^{2}b_{i}b_{j}+16b_{i}^{2}b_{j}^{2}\right) \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2} \\&+\,2\phi ^{2}\left( 4b_{i}b_{j}+\gamma ^{2}\right) \left( 3\gamma ^{4}+2\gamma ^{2}b_{i}b_{j}+48b_{i}^{2}b_{j}^{2}\right) \left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{2}>0,\\ A_{6}= & {} \left( 4b_{i}b_{j}\left( 1+\phi \right) -\gamma ^{2}\left( 1-\phi \right) \right) 2\left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{3}+\phi ^{3}\left( 4b_{i}b_{j}+\gamma ^{2}\right) \\&\times \left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}-4\gamma ^{2}b_{i}b_{j}\right) \\&+\,5\phi \left( 4b_{i}b_{j}+\gamma ^{2}\right) \left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{2}\\&+\,4\phi ^{2}\left( 2b_{i}b_{j}+\gamma ^{2}\right) \left( 8b_{i}b_{j}+\gamma ^{2}\right) \left( 4b_{i}b_{j}-\gamma ^{2}\right) >0,\\ A_{7}= & {} -\left( \left( 4b_{i}b_{j}+\gamma ^{2}\right) ^{2}-4\gamma ^{2}b_{i}b_{j}\right) \phi ^{2}-2\left( 4b_{i}b_{j}-\gamma ^{2}\right) \left( \gamma ^{2}+4b_{i}b_{j}\right) \phi \\&-\left( 4b_{i}b_{j}-\gamma ^{2}\right) ^{2}<0. \end{aligned}$$

$\frac{\partial }{\partial \phi }\varPi ^{i}(p_{1}^{*},p_{2}^{*})$ is positive if $\gamma <0$, or when the goods are complements. If $\gamma >0,$ $D_{j}^{i}>0$ and $p_{j}^{*\prime }(\phi )<0$, hence a negative demand effect is enough for $\frac{\partial }{\partial \phi }\varPi ^{i}(p_{i}^{*},p_{j}^{*})<0.$ $\square$

Proof of Proposition 11

We have $\frac{d}{d\phi }W(p_{1}^{*}(\phi ),p_{2}^{*}(\phi ),\phi )=T_{1}+T_{2},$ where

$$\begin{aligned} T_{1}= & {} (p_{i}^{*}-c_{i})\left[ \phi (D_{i}^{i}p_{i}^{*\prime }+D_{j}^{i}p_{j}^{*\prime })+\frac{1-\phi }{2}d_{i}^{\prime }p_{i}^{*\prime }\right] \\&+(p_{j}^{*}-c_{j})\left[ \phi (D_{i}^{j}p_{i}^{*\prime }+D_{j}^{j}p_{j}^{*\prime })+\frac{1-\phi }{2}d_{j}^{\prime }p_{j}^{*\prime }\right] \end{aligned}$$

and

$$\begin{aligned} T_{2}=U(D^{i},D^{j})-c_{i}D^{i}-c_{j}D^{j}-\frac{1}{2} [U(d_{i},0)-c_{i}d_{i}]-\frac{1}{2}[U(0,d_{j})-c_{j}d_{j}]. \end{aligned}$$

$T_{1}$ consists of the indirect effects of $\phi$ on total equilibrium outputs of the firms, weighted by their margins. Hence, when $\gamma >0,$ it is positive and when $\gamma <0,$ it is negative.

$T_{2}$ is always positive (all necessary computations can be provided upon request), and to show this we need to use the fact that (15) must hold for both firms.

Hence, if $\gamma>0\,,\ T_{1}+T_{2}>0$. If $\gamma <0$, $T_{1}+T_{2}>0$ as well, but showing this requires direct computation and condition (15) for both firms (computational details upon request).$\square$