Limits of price competition: cost asymmetry and imperfect information

Abstract

In a class of asymmetric-cost duopoly price competition games, price-elastic individual demand reveals a threshold of informed buyers below which equilibrium is in monopoly prices. Even if the threshold is met, unless all buyers are informed, monopoly prices are listed and are more likely between less alike sellers. An increase in the efficient seller’s cost weakly reduces its rival’s price (in a first order stochastic dominated sense) without disturbing its profit. Lastly, increasing the number of sellers increases competitive pricing incentives.

Appendices

Appendix 1: Proof of the Theorem

Proof

(a) Let $f \le \hat{f}$; therefore, $\underline{p}_2\in [p_1^m, p_2^m]$. Seller 2 undercuts its rival only if $p_1>\underline{p}_2$ (by definition); its best response to all lower prices is $p_2^m$. Seller 1’s best response to $p_2 \ge \underline{p}_2$ is $p_1^m$. The unique equilibrium therefore is $(p_1^*=p_1^m, p_2^*=p_2^m)$. $\square$

Proof

(b) Let $f \in (\hat{f}, 1)$.

Step 1: There can be no pure strategy equilibrium.⁸

Proof

$p_1 \ngtr \underline{p}_2$ in pure strategy equilibrium, by definition 1. But, $\forall p_1 \le \underline{p}_2$, seller 2’s best response is $p_2^m$, to which seller 1’s best response is $p_1^m$ which exceeds $\underline{p}_2$ because $f \in (\hat{f}, 1) \Leftrightarrow \underline{p}_2 \in (c_2, p_1^m)$. $\square$

Let $G_1(p)$ and $G_2(p)$ be the respective (cumulative) probability distribution functions in mixed strategy equilibrium, with supports $S_1$ and $S_2$; and let $P_i(p)$ denote any positive probability on price p by seller i, if p has a point mass of probability.

Step 2: $\underline{p}_2$ is the lower bound of $S_i, \; i=\{1,2\}$; and $\forall p \in S_1 \cap S_2$, the probability of a tie is zero.

Proof

If there is no common lower bound ($p^l$), then one seller can profitably increase its lowest price. If $p^l>\underline{p}_2$, then seller 2 can profitably undercut because $\underline{p}_2>c_2$ from $f<1$; and $p^l<\underline{p}_2$ is (by definition 1) dominated by $p_2^m$ for seller 2. Therefore, $p^l=\underline{p}_2$. Moreover, $\underline{p}_1<\underline{p}_2$ from (2), and therefore from any tie with positive probability in $S_1 \cap S_2$ seller 1 is better off deviating probabilistically to a marginally lower price. $\square$

For all prices in $S_i$, seller i’s expected profit is as follows⁹:

$$\begin{aligned} \pi _i(p; G_j(p))= & {} [1-G_j(p)]\frac{(1+f)}{2} \pi _i^m(p)+[G_j(p)-P_j(p)] \frac{(1-f)}{2} \pi _i^m(p) + P_j(p)\frac{1}{2} \pi _i^m(p) \nonumber \\&\implies \pi _i(p; G_j(p)) = \pi _i^m(p)\left[ \frac{1}{2} + f\left( \frac{1}{2} - G_j(p) + \frac{P_j(p)}{2}\right) \right] , \; \forall \; p \in S_i . \end{aligned}$$

(7)

Step 3: $p_2^m \in S_2$, with $G_2(p_2^m)=1$; $G_1(p_2^m)=1$; and $G_1(\underline{p}_2)=P_1(\underline{p}_2)=P_1(p_2^m)=0$.

Proof

From Step 2, $\underline{p}_2 \in S_1$; and because $f<1$, seller 2’s best response to $p_1 = \underline{p}_2$ is $p_2^m$. Also from Step 2, $\underline{p}_2 \in S_2$. But from Definition 1, for seller 2 the maximum profit from $\underline{p}_2$ equals the minimum profit from $p_2^m$; i.e. $\frac{(1+f)}{2} \pi _2^m(\underline{p}_2) = \frac{(1-f)}{2} \pi _2^m(p_2^m)>0$, where its positive sign follows from $f<1$. Therefore, $p_2^m \in S_2$,¹⁰ and the expected profits for seller 2 from $\underline{p}_2$ and $p_2^m$ should be equal. Using (7) and the above then

$$\begin{aligned} f\left( -G_1(\underline{p}_2)+\frac{P_1(\underline{p}_2)}{2}\right) =0, \end{aligned}$$

(8)

and

$$\begin{aligned} \frac{1}{2}-G_1(p_2^m)+\frac{P_1(p_2^m)}{2} = -\frac{1}{2}. \end{aligned}$$

(9)

From Step 2, we have $G_1(\underline{p}_2)=P_1(\underline{p}_2)$; but then (8) implies $G_1(\underline{p}_2)=P_1(\underline{p}_2)=0$. Moreover (9) implies $1 = G_1(p_2^m) - \frac{P_1(p_2^m)}{2}$. But $G_1(p_2^m) \le 1$ and $P_1(p_2^m) \ge 0$ as probability measures; therefore it must be that $P_1(p_2^m)=0$, and $G_1(p_2^m)=1$. But this in turn implies that prices $\{p:p>p_2^m\}$ are dominated by $p_2^m$ for seller 2; therefore $G_2(p_2^m)=1$. $\square$

Step 4: $S_1=[\underline{p}_2, \bar{p}] = S_1 \cap S_2$.

Proof

$\underline{p}_2 \in S_1 \cap S_2$ (from Step 2), and $P_1(\underline{p}_2)=0$ (from Step 3) together give $\exists \;p \in (\underline{p}_2,p_2^m) \; s.t. \; [\underline{p}_2, p] \subseteq S_1$. Call the largest such price $\bar{p}$.

Suppose $\exists \; p \in [\underline{p}_2, \bar{p}]$, but $p \notin S_2$ which implies $P_2(p)=0$.

If $p<p_1^m$, take the smallest price in $S_2$ greater than p, which exists because $G_2(p)<1, \; \forall p<p_2^m$ from Step 3 and because $p_1^m<p_2^m$; call this $p'$. Then there exists $p''$ such that $p_1^m>p''$; $p'>p'' > p$; $p''\notin S_2$; and $G_2(p)=G_2(p'')$. If, however, $p>p_1^m$, then take the largest price in $S_2$ that is smaller than p and call that $p'$; then there exists $p''$, such that $p_1^m<p''$; $p'<p'' < p$; $p''\notin S_2$; and $G_2(p)=G_2(p'')$. In either case, $p''$ dominates p for seller 1, from (7), contradicting $p \in S_1$. Thus, if $p \ne p_1^m$ and $p \in [\underline{p}_2, \bar{p}]$, then $p \in S_2$.

And if $p_1^m \in [\underline{p}_2, \bar{p}]$ but $p_1^m \notin S_2$, then take the largest price in $S_2$ that is smaller than $p_1^m$ and call it $p'$. Because $\underline{p}_2<p_1^m$ from $f>\hat{f}$, continuity of $[\underline{p}_2, \bar{p}]$ implies that $\exists p''$ such that $p'<p'' < p_1^m$ and $p'' \in [\underline{p}_2, \bar{p}]$ but $p'' \notin S_2$. This contradicts: if $p \ne p_1^m$ and $p \in [\underline{p}_2, \bar{p}]$, then $p \in S_2$. Therefore, $[\underline{p}_2, \bar{p}] \subseteq S_1 \cap S_2$.

Lastly, to establish $[\underline{p}_2, \bar{p}]=S_1$. Suppose not; then $\exists \; p>\bar{p}$ such that $p\in S_1$ but $(\bar{p},p) \notin S_1$. There are two possible cases for $\bar{p}$, and in each a contradiction is established as follows, implying $[\underline{p}_2, \bar{p}]=S_1$.

First consider $\bar{p} \ge p_1^m$. Because $p>\bar{p}$ implies $G_2(p) \ge G_2(\bar{p}) \ge G_2(p_1^m)$; from (7), $p\in S_1$ only if $P_2(p)>0$, because otherwise seller 1 could profitably replace p with $p_1^m$. But if $P_2(p)>0$, then from (7) seller 1 profits by deviating from p to a marginally lower price, call it $\hat{p}$, because $G_2(\hat{p}) \le G_2(p) - P_2(p)$ and $\pi _1^m(\hat{p}) > \pi _1^m(p)$, contradicting $(\bar{p},p) \notin S_1$.

Consider the other possibility: $\bar{p}<p_1^m$. Because $(\bar{p},p) \notin S_1$, for any two prices, $p_a,p_b \in (\bar{p},p)$, we have $P_1(p_a)=P_1(p_b)=0$, and $G_1(p_a)=G_1(p_b)=G_1(\bar{p})$. Therefore if $p_a<p_b$, then from (7), $\pi _2(p_b; G_1(p_b))>\pi _2(p_a;G_1(p_a))$. From any price in $(\bar{p},p)$ then, seller 2 has a profitable price hike; i.e. $(\bar{p},p) \notin S_2$. But then it must be that $P_2(\bar{p})>0$, because otherwise from $\bar{p}<p_1^m$ and (7) seller 1 prefers a price marginally higher than $\bar{p}$ to $\bar{p}$ itself, contradicting $(\bar{p},p) \notin S_1$. But $P_2(\bar{p})>0$ implies $P_1(\bar{p})=0$, from Step 2; and $P_1(\bar{p})=0$ and $G_1(\bar{p})=G_1(p_a), \; \forall p_a \in (\bar{p},p)$ together imply a profitable deviation for seller 2 from $\bar{p}$ to a marginally higher price in $(\bar{p}, p)$; i.e. $\bar{p} \notin S_2$, contradicting $[\underline{p}_2, \bar{p}] \subseteq S_1 \cap S_2$ proven above. $\square$

Step 5: $\bar{p}=p_1^m$; and therefore, $(p_1^m, p_2^m) \notin S_2$, and $P_2(p_2^m)>0$.

Proof

Because $S_1=[\underline{p}_2,\bar{p}]$ from Step 4, $(\bar{p},p_2^m) \notin S_2$ because these prices are dominated by $p_2^m$ for seller 2. Also using Step 2, let seller i be the only seller listing $\bar{p}$ with positive probability in equilibrium; in case neither lists it with positive probability, let seller i be a randomly chosen seller. Therefore, if $\bar{p}<p_1^m$, then $G_j(\bar{p})=G_j(p_1^m)$ and $P_j(\bar{p})=P_j(p_1^m)=0$, which implies using (7) that seller i has a profitable deviation from $\bar{p}$ to $p_1^m$.

If, however, $\bar{p}>p_1^m$, then because $G_2(\bar{p}) - P_2(\bar{p}) > G_2(p_1^m)$ and $\pi _1^m(\bar{p}) < \pi _1^m(p_1^m)$, we have $\pi _1^m(p_1^m) [\frac{1}{2} + f(\frac{1}{2} - G_2(p_1^m) + \frac{P_2(p_1^m)}{2})] > \pi _1^m(\bar{p}) [\frac{1}{2} + f(\frac{1}{2} - G_2(\bar{p}) + \frac{P_2(\bar{p})}{2})]$.¹¹ Using (7), therefore $\pi _1(\bar{p}; G_2(\bar{p})) < \pi _1(p_1^m; G_2(p_1^m))$; which in turn implies $\bar{p} \notin S_i$, a contradiction to Step 4.

Therefore, $\bar{p}=p_1^m$. It is straightforward then that $(p_1^m, p_2^m) \notin S_2$ because these prices are dominated by $p_2^m$; and because from Step 3, we have $p_2^m \in S_2$, it must be that $P_2(p_2^m)>0$. $\square$

All prices in $S_2$ must give the same expected profit; therefore from Steps 3, 4, and 5, we have

$$\begin{aligned} \pi _2(p;G_1(p))=\frac{(1-f)}{2} \pi _2^m(p_2^m), \quad \forall p\in S_2. \end{aligned}$$

(10)

Step 6: $P_i(p)=0, \; \forall p\in [\underline{p}_2, p_1^m), \quad \forall i=1,2$.

Proof

From (7), $[\frac{1}{2}+f(\frac{1}{2}-G_j(p)+\frac{P_j(p)}{2})]$ should be a continuous function of p throughout the support $[\underline{p}_2, p_1^m]$, because $\pi _i^m(p), \forall i$ is continuous and each price in $S_i$ gives equal expected profit. Therefore, any atoms in the support must lie on the end points of $[\underline{p}_2, p_1^m]$. From Step 3, we already know that $P_1(\underline{p}_2)=0$. If $P_2(\underline{p}_2)>0$, then seller 1 can profitably marginally undercut $\underline{p}_2$, contradicting Step 2. $\square$

Step 7: $S_1=[\underline{p}_2, p_1^m]$, with $G_1(p)= {\left\{ \begin{array}{ll} \frac{1+f}{2f}-\frac{(1-f)\pi _2^m(p_2^m)}{2f\pi _2^m(p)}, \forall p \in [\underline{p}_2, p_1^m) \\ 1, \; if \; p=p_1^m \end{array}\right. }$, and $P_1(p_1^m)=\frac{(1-f)}{f} [\frac{\pi _2^m(p_2^m)-\pi _2^m(p_1^m)}{\pi _2^m(p_1^m)}]$; and $S_2=[\underline{p}_2,p_1^m] \cup \{p_2^m\}$, with $G_2(p)= {\left\{ \begin{array}{ll} \frac{(1+f)}{2f}[1- \frac{\pi _1^m(\underline{p}_2)}{\pi _1^m(p)}], \forall p \in [\underline{p}_2, p_1^m] \\ 1, \; if \; p=p_2^m \end{array}\right. }$, and $P_2(p_2^m)=[\frac{(1+f)\pi _1^m(\underline{p}_2)-(1-f)\pi _1^m(p_1^m)}{2f\pi _1^m(p_1^m)}]$.

Proof

Steps 2, 3, 4, and 5 define $S_1$ and $S_2$. Using (7) to expand the left hand side (LHS) of (10), we have:

$$\begin{aligned} \pi _2^m(p)\left[ \frac{1}{2} + f\left( \frac{1}{2} - G_1(p) + \frac{P_1(p)}{2}\right) \right] =\frac{(1-f)}{2} \pi _2^m(p_2^m); \end{aligned}$$

(11)

and substituting $p=p_1^m$ and $G_1(p_1^m)=1$ (from Steps 4 and 5), gives $P_1(p_1^m)=\frac{(1-f)}{f} [\frac{\pi _2^m(p_2^m)-\pi _2^m(p_1^m)}{\pi _2^m(p_1^m)}]>0$.

But then Step 2 implies $P_2(p_1^m)=0$, in equilibrium. Using Step 6 then, $P_2(p)=0, \forall p\in S_1$. Therefore, substituting $p=\underline{p}_2$ in (7), we have for seller 1 for all prices in $S_1$:

$$\begin{aligned} \pi _1^m(\underline{p}_2)\frac{(1+f)}{2} = \pi _1^m(p)\left[ \frac{1}{2} + f\left( \frac{1}{2} - G_2(p) \right) \right] . \end{aligned}$$

(12)

Therefore, $G_2(p)=\frac{(1+f)}{2f}[1- \frac{\pi _1^m(\underline{p}_2)}{\pi _1^m(p)}], \forall p \in [\underline{p}_2, p_1^m]$. Substituting $p=p_1^m$ gives:

$$\begin{aligned} P_2(p_2^m)= 1-G_2(p_1^m) = \frac{(1+f)\pi _1^m(\underline{p}_2)-(1-f)\pi _1^m(p_1^m)}{2f\pi _1^m(p_1^m)} >0. \end{aligned}$$

From Steps 4, 5 and 6, there are no more point masses. And from (11) we have $G_1(p)=\frac{1+f}{2f}-\frac{(1-f)\pi _2^m(p_2^m)}{2f\pi _2^m(p)}, \forall p \in [\underline{p}_2, p_1^m)$. Expected profits are therefore, $\pi _1=\frac{(1+f)}{2} \pi _1^m(\underline{p}_2)$, and $\pi _2=\frac{(1-f)}{2} \pi _2^m(p_2^m)=\frac{(1+f)}{2} \pi _2^m(\underline{p}_2)$, from (7), (12), and (10). $\square$

Proof

(c) Let $f=1$. This is equivalent to $\underline{p}_2=c_2$. Therefore, seller 2’s best response for any $p_1>c_2$ is to undercut. This reduces the game to an asymmetric Bertrand game that has mixed strategy equilibria as found by Blume (2003), with seller 1 selling to the entire market alone at $p_1^*=c_2$, and seller 2 using a mixed strategy on the support $[c_2, c_2+\gamma ]$ with $P_2(c_2)=0$. $\square$

Appendix 2: Proofs for Section 4

Lemma 1.

Proof

(i) $\frac{\partial P_1(p_1^m)}{\partial c_1} = \frac{(1-f)}{f} [\frac{- \pi _2^m(p_2^m) \pi _2^{m'}(p_1^m) \frac{dp_1^m}{dc_1}}{(\pi _2^m(p_1^m))^2}]<0$. And using (5) and simplifying, $\frac{\partial P_2(p_2^m)}{\partial c_1} = \frac{-(1+f)q(\underline{p}_2) q(p_1^m) (p_1^m - \underline{p}_2)}{2f \{\pi _1^m(p_1^m)\}^2} <0$.

(ii) Using (5), and simplifying, $\frac{\partial P_1(p_1^m)}{\partial c_2} = \frac{(1-f) q(p_1^m) q(p_2^m) [p_2^m-p_1^m]}{f[\pi _2^m(p_1^m)]^2} >0$. And $\frac{\partial P_2(p_2^m)}{\partial c_2}=\frac{(1+f)}{2f \pi _1^m(p_1^m)} \frac{\partial \underline{p}_2}{\partial c_2} \pi _1^{m'}(\underline{p}_2) >0$, using (2) and the fact that equilibria of type (b) result when $f\in (\hat{f},1)$ which is equivalent to $\underline{p}_2\in (c_2,p_1^m)$, implying $\pi _1^{m'}(\underline{p}_2)>0$. $\square$

Lemma 2.

Proof

From the theorem, we have $G_1(p)= \frac{1+f}{2f}-\frac{(1-f)\pi _2^m(p_2^m)}{2f\pi _2^m(p)}$, and $G_2(p)= \frac{(1+f)}{2f}[1- \frac{\pi _1^m(\underline{p}_2)}{\pi _1^m(p)}]$. Therefore, using (5), $\frac{\partial G_1(p|c_2)}{\partial c_2}= -\frac{(1-f)}{2f} \frac{q(p)q(p_2^m)(p_2^m-p)}{(\pi _2^m (p))^2}< 0, \; \forall p \in (\underline{p}_2 (c_2^h), p_1^m)$. Secondly, (2) gives $\underline{p}_2 (c_2^h)>\underline{p}_2 (c_2^l)$, and therefore $G_1(p|c_2^h) =0 \le G_1(p|c_2^l), \forall p\in [\underline{p}_2 (c_2^l),\underline{p}_2 (c_2^h)]$. Thirdly, $G_1(p_1^m|c_2^h) = 1 = G_1(p_1^m|c_2^l)$. Together, this proves, in equilibrium, $p_1 |c_2^h \; \textit{FOSD} \; p_1 |c_2^l$.

Also, $\frac{\partial G_2(p|c_2)}{\partial c_2} = -\frac{(1+f)}{2f \pi _1^m(p)} \frac{\partial \underline{p}_2}{\partial c_2} \pi _1^{m'}(\underline{p}_2) <0, \; \forall p \in (\underline{p}_2 (c_2^h), p_1^m]$. Also, as above, $\underline{p}_2 (c_2^h)>\underline{p}_2 (c_2^l)$, and therefore $G_2(p|c_2^h) =0 \le G_2(p|c_2^l), \forall p\in [\underline{p}_2 (c_2^l),\underline{p}_2 (c_2^h)]$. And lastly, because $p_2^m(c_2^h)>p_2^m(c_2^l),$ therefore $G_2(p_2^m(c_2^l)|c_2^h) < 1 =G_2(p_2^m(c_2^l)|c_2^l)$, while $G_2(p_2^m(c_2^h)|c_2^h)= G_2(p_2^m(c_2^h)|c_2^l) = 1$. Together this proves, in equilibrium, $p_2 |c_2^h \; \textit{FOSD} \; p_2 |c_2^l$. $\square$

Lemma 3.

Proof

Notice that neither $\underline{p}_2$ nor $G_1(p |c_1), \forall p\in [\underline{p}_2, p_1^m)$ are functions of $c_1$; however, $\frac{\partial p_1^m}{\partial c_1} > 0$. Therefore, $G_1(p |c_1^l)= G_1(p |c_1^h), \forall p\in [\underline{p}_2, p_1^m(c_1^l))$, but $\forall p \in [p_1^m(c_1^l), p_1^m(c_1^h)), \; G_1(p |c_1^l) =1 > G_1(p |c_1^h)$; and $G_1(p_1^m(c_1^h) |c_1^l)= G_1(p_1^m(c_1^h) |c_1^h) = 1$. Therefore, in equilibrium, $p_1 |c_1^h \; \textit{FOSD} \; p_1 |c_1^l$.

And using (5), we have

$$\begin{aligned} \frac{\partial G_2(p |c_1)}{\partial c_1} = \frac{-(1+f)}{2f} \frac{q(p)q(\underline{p}_2) \{\underline{p}_2-p\}}{\{\pi _1^m(p)\}^2} ={\left\{ \begin{array}{ll} 0, \textit{if} \; p=\underline{p}_2, \\ >0, \; \forall p\in (\underline{p}_2, p_1^m(c_1)] \end{array}\right. }. \end{aligned}$$

(13)

That is, $G_2(p |c_1^h) \ge G_2(p |c_1^l), \; \forall p\in [\underline{p}_2, p_1^m(c_1^l))$. Also because $p_1^m(c_1^l)< p_1^m(c_1^h)< p_2^m$, Lemma 1 implies that $G_2(p_1^m(c_1^h)|c_1^l) = 1-P_2(p_2^m|c_1^l) < G_2(p_1^m(c_1^h)|c_1^h) = 1-P_2(p_2^m|c_1^h)$.

Lastly, $\forall p \in [p_1^m(c_1^l), p_1^m(c_1^h)), \; G_2(p |c_1^l) = G_2(p_1^m(c_1^l) |c_1^l) = \frac{(1+f)}{2f} [1-\frac{q(\underline{p}_2)(\underline{p}_2 - c_1^l)}{q(p_1^m(c_1^l))(p_1^m(c_1^l) -c_1^l)}]$, and $G_2(p |c_1^h) = \frac{(1+f)}{2f}[1- \frac{q(\underline{p}_2)(\underline{p}_2-c_1^h)}{q(p)(p-c_1^h)}]$. The condition $[1-\frac{q(\underline{p}_2)(\underline{p}_2 - c_1^l)}{q(p_1^m(c_1^l))(p_1^m(c_1^l) -c_1^l)}] \le [1- \frac{q(\underline{p}_2)(\underline{p}_2-c_1^h)}{q(p)(p-c_1^h)}],$ can be simplified to $q(p)(p-c_1^h) \ge \frac{(\underline{p}_2-c_1^h) q(p_1^m(c_1^l))(p_1^m(c_1^l) -c_1^l)}{(\underline{p}_2 - c_1^l)}.$ The LHS increases in p within the given range of prices whereas the RHS doesn’t change; therefore, this condition would be satisfied iff it is satisfied at the lower limit, i.e. at $p_1^m(c_1^l)$. Therefore, $G_2(p |c_1^l) \le G_2(p |c_1^h), \; \forall p \in [p_1^m(c_1^l), p_1^m(c_1^h))$ iff

$$\begin{aligned} q(p_1^m(c_1^l))(p_1^m(c_1^l)-c_1^h) \ge \frac{(\underline{p}_2-c_1^h) q(p_1^m(c_1^l))(p_1^m(c_1^l) -c_1^l)}{(\underline{p}_2 - c_1^l)}. \end{aligned}$$

Simplifying and reducing, we get $\underline{p}_2 \le p_1^m(c_1^l)$. But this is automatically satisfied when equilibrium is of type (b); in other words $f > \hat{f}(c_1^l)$ gives $\underline{p}_2 < p_1^m(c_1^l)$. Therefore in equilibrium, $p_2 |c_1^l \; \textit{FOSD} \; p_2 |c_1^h$. $\square$

Proposition 1.

Proof

A higher $c_2$ raises $p_2^m$ from (1), raises $\underline{p}_2$ from (2), and raises $\hat{f}$ from (6). Let $c_2$ increase from $c_2^l$ to $c_2^h$; from (6) and the theorem, one of the following captures the shifting equilibrium due such a change.

• (i) $f \le \hat{f}(c_2^l)<\hat{f}(c_2^h)<1$: equilibrium prices change from $p_1^m; p_2^m (c_2^l)$ to $p_1^m; p_2^m (c_2^h)$, and because $p_2^m (c_2^h)>p_2^m (c_2^l)$, market price at $c_2^h$ first order stochastically dominates that at $c_2^l$ making buyers worse off. Expected profit changes only for the inefficient seller, from $\frac{(1-f)}{2} \pi _2^m(p_2^m(c_2^l))$ to $\frac{(1-f)}{2} \pi _2^m(p_2^m(c_2^h))$, which is a decrease from (5). Aggregating therefore, market welfare decreases.

• (ii) $\hat{f}(c_2^l)<f \le \hat{f}(c_2^h)<1$: seller 1’s equilibrium price increases from a mixed strategy with support $S_1=[\underline{p}_2 (c_2^l),p_1^m]$ to a pure strategy at price $p_1^m$, thus improving its expected profit from $\frac{(1+f)}{2} \pi _1^m(\underline{p}_2 (c_2^l))$ to $\frac{(1+f)}{2} \pi _1^m(p_1^m)$. Seller 2’s price increases from a mixed strategy with support $S_2=[\underline{p}_2 (c_2^l),p_1^m] \bigcup \{p_2^m (c_2^l)\}$ to a pure strategy at price $p_2^m (c_2^h)$, i.e. its expected profit changes from $\frac{(1-f)}{2} \pi _2^m(p_2^m(c_2^l))$ to $\frac{(1-f)}{2} \pi _2^m(p_2^m(c_2^h))$, which is a decrease from (5). Market price at $c_2^h$ first order stochastically dominates that at $c_2^l$ because both sellers’ prices increase, making buyers worse off.

• (iii) $\hat{f}(c_2^l)<\hat{f}(c_2^h)<f<1$: both sellers continue to use mixed strategy pricing, and from lemma 2, equilibrium market price at $c_2^h$ first order stochastically dominates that at $c_2^l$ making buyers worse off. Expected profit for seller 1 improves from $\frac{(1+f)}{2} \pi _1^m(\underline{p}_2 (c_2^l))$ to $\frac{(1+f)}{2} \pi _1^m(\underline{p}_2 (c_2^h))$; and expected profit for seller 2 decreases from $\frac{(1-f)}{2} \pi _2^m(p_2^m(c_2^l))$ to $\frac{(1-f)}{2} \pi _2^m(p_2^m(c_2^h))$ from (5).

• (iv) $\hat{f}(c_2^l)<\hat{f}(c_2^h)<f=1$: the only price at which sales occur ($p_1 = c_2$) increases from $c_2^l$ to $c_2^h$, leading to a first order stochastically dominating market price, lowering buyer surplus and raising seller 1’s profit; while seller 2 remains indifferent earning zero profit.

$\square$

Proposition 2.

Proof

A higher $c_1$ raises $p_1^m$, but leaves $\underline{p}_2$ unchanged (from definition 1), and lowers $\hat{f}$ (from (6)); therefore, its impact on prices is not straightforward. Let $c_1$ increase from $c_1^l$ to $c_1^h$; from (6) and the theorem, one of the following captures the shift in equilibrium.

• (i) $f \le \hat{f}(c_1^h)<\hat{f}(c_1^l)<1$: equilibrium prices change from $p_1^m (c_1^l); p_2^m$ to $p_1^m (c_1^h); p_2^m$; i.e. $p_1|c_1^h$ first order stochastically dominates $p_1|c_1^l$, while $p_2$ doesn’t change, making buyers worse off. Expected profit changes only for the efficient seller, from $\frac{(1+f)}{2} \pi _1^m(p_1^m(c_1^l))$ to $\frac{(1+f)}{2} \pi _1^m(p_1^m(c_1^h))$, which is a decrease from (5). Aggregating therefore, market welfare decreases.

• (ii) $\hat{f}(c_1^h)<f \le \hat{f}(c_1^l)<1$: seller 1’s price changes from a pure strategy at $p_1^m (c_1^l)$ to a mixed strategy with support $S_1=[\underline{p}_2 ,p_1^m(c_1^h)]$; because $\hat{f}(c_1^h)<f \le \hat{f}(c_1^l)$ is equivalent to $p_1^m(c_1^l) \le \underline{p}_2 < p_1^m(c_1^h)$, this implies $p_1|c_1^h$ first order stochastically dominates $p_1|c_1^l$. Seller 2’s price decreases from a pure strategy at $p_2^m$ to a mixed strategy with support $S_2=[\underline{p}_2 ,p_1^m(c_1^h)] \bigcup \{p_2^m\}$, i.e. $p_2|c_1^l$ first order stochastically dominates $p_2|c_1^h$. Seller 2’s expected profit does not change despite the equilibrium moving to mixed strategies and remains at $\frac{(1-f)}{2} \pi _2^m(p_2^m)$, while that of seller 1 changes from $\frac{(1+f)}{2} \pi _1^m(p_1^m (c_1^l))$ to $\frac{(1+f)}{2} \pi _1^m(\underline{p}_2|c_1^h)$. Because $\hat{f}(c_1^h)<f$ is equivalent to $\underline{p}_2 < p_1^m(c_1^h)$, we have $\pi _1^m(\underline{p}_2|c_1^h) < \pi _1^m(p_1^m(c_1^h))$; but from (5) we have $\pi _1^m(p_1^m(c_1^h)) < \pi _1^m(p_1^m(c_1^l))$. Therefore $\pi _1^m(\underline{p}_2|c_1^h) < \pi _1^m(p_1^m(c_1^l))$, or that seller 1’s expected profit falls.

• (iii) $\hat{f}(c_1^h)<\hat{f}(c_1^l)<f<1$: both sellers continue to use mixed strategies in equilibrium. From lemma 3, $p_1|c_1^h$ first order stochastically dominates $p_1|c_1^l$, but $p_2|c_1^l$ first order stochastically dominates $p_2|c_1^h$. Expected profit for seller 2 does not change because neither $c_2$ nor $p_2^m$ changes, but expected profit for seller 1 falls despite no change in $\underline{p}_2$, from $\frac{(1+f)}{2} \pi _1^m(\underline{p}_2|c_1^l)$ to $\frac{(1+f)}{2} \pi _1^m(\underline{p}_2|c_1^h)$ because its cost has risen.

• (iv) $\hat{f}(c_1^h)<\hat{f}(c_1^l)<f=1$: the only price at which sales occur ($p_1 = c_2$) does not change and therefore neither does buyer surplus nor the profit of seller 2; however seller 1’s profit falls from $\pi _1^m(c_2|c_1^l)$ to $\pi _1^m(c_2|c_1^h)$. Aggregating therefore, market welfare decreases.

$\square$