Competition in online markets with auctions and posted prices

Abstract

The paper studies an online consumer-to-consumer market with limited supply, where sellers may list their items by posted prices or auctions. I show that when there is competition among sellers, they use only posted prices in the equilibrium. This result contrasts with the findings for a monopolistic seller listing objects by auctions and posted prices on markets with infinite supply, where using both mechanisms is the equilibrium. The model helps to explain the trends documented in Einav et al. (J Polit Econ 126(1):178–215, 2018).

Appendices

Appendix A: Proof of Lemma 1

Without loss of generality, consider a buyer with valuation v. If x is the first order statistic among his opponents, and y is the second order statistic, then there are several possibilities for their realizations:¹²

(a) $x,y<c$. In this situation, x and y cannot do better but to bid their valuations. Both of the opponents drop out when the auction price reaches their valuations, and the buyer in question wins the auction paying the second highest bid: $\int _0^{c}\int _0^{x} (v-x)2{\mathrm{d}}y{\mathrm{d}}x$.

(b) (i) $c<x<p$ and $y<c$. In this scenario, when the auction price reaches y, this buyer drops out. At c, the buyer with valuation v observes that one of his opponents has dropped out from the auction and that the good outside is still available, so he no longer exits the auction. It is easy to check that his best response is to bid $\text {min}[v_i,p]$ instead. Because $x<p$, the first order statistic stays in the auction until its price reaches the value of x. Hence, the buyer in question pays the second highest bid: $\int _{c}^{p}\int _0^{c} (v-x)2{\mathrm{d}}y{\mathrm{d}}x$.

(ii) $c<x<p$ and $c<y<x$. In this case, when the auction price reaches the exit price of the buyer in question, he leaves the auction and buys the good at the posted price: $\int _{c}^{p}\int _{c}^{x} (v-p)2{\mathrm{d}}y{\mathrm{d}}x$.

(c) (i) $p<x<z^{-1}(c)$ and $y<c$. This situation is similar to the one described in b(i): the auction continues until the auction price reaches y: second order statistic drops out from the auction, and the remaining bidders play according to $\text {min}[v_i,p]$. Bidding continues until the auction price matches the outside posted price. At p, one bidder leaves the auction and buys the good at the posted price, while the other buyer wins the auction and pays the same amount: $\int _{p}^{z^{-1}(c)}\int _0^{c} (v-p)2{\mathrm{d}}y{\mathrm{d}}x$.

(ii) $p<x<z^{-1}(c)$ and $c<y<p$. In this case, when the auction price reaches c, the buyer in question leaves the auction and buys the good at the posted price: $\int _{p}^{z^{-1}(c)}\int _{c}^{p} (v-p)2{\mathrm{d}}y{\mathrm{d}}x$.

(iii) $p<x<z^{-1}(c)$ and $p<y<x$. Nothing changes from c(ii)—buyer in question is still the first to leave the auction and buy the good at the posted price: $\int _{p}^{z^{-1}(c)}\int _{p}^{x} (v-p)2{\mathrm{d}}y{\mathrm{d}}x$.¹³

(d) (i) $z^{-1}(c)<x<v$ and $y<z(x)$. In this case, y drops out before the exit price of the first order statistic, and the remaining buyers (v and x) bid according to $\text {min}[v_i,p]$. The auction continues until its price matches the outside posted price. At p, one bidder leaves the auction and buys the good at the posted price, while the other buyer wins the auction and pays the same amount: $\int _{z^{-1}(c)}^v\int _{0}^{z(x)} (v-p)2{\mathrm{d}}y{\mathrm{d}}x$.

(ii) $z^{-1}(c)<x<v$ and $z(x)<y<x$. In this scenario, the first order statistic leaves the auction and buys the good at the posted price. The remaining buyers bid according to $\text {min}[v_i,p]$. When the auction price reaches y, the buyer in question wins the auction and receives a payoff of $v-y$. Since the good outside of the auction is no longer available it is readily seen that the same outcome will also be observed when $c<y<x$, hence: $\int _{z^{-1}(c)}^v\int _{z(x)}^{x} (v-y)2{\mathrm{d}}y{\mathrm{d}}x$.

(e) (i) $v<x<1$ and $y<z(x)$. In this case, the second order statistic drops out from the auction before the exit price of the first order statistic, and the remaining buyers bid according to $\text {min}[v_i,p]$, resulting in both buyers getting the objects at p: $\int _{v}^1\int _{0}^{z(x)} (v-p)2{\mathrm{d}}y{\mathrm{d}}x.$

(ii) $v<x<1$ and $z(x)<y<v$. In this scenario, the first order statistic leaves the auction and buys the good at the posted price. The remaining buyers bid according to $\text {min}[v_i,p]$. When the auction price reaches y, the buyer in question wins the auction and receives a payoff of $v-y$. Since the good outside of the auction is no longer available, it is readily seen that the same outcome will also be observed when $c<y<v$: $\int _{v}^1\int _{z(x)}^{v} (v-y)2{\mathrm{d}}y{\mathrm{d}}x$.

(iii) When $x,y>v$ , the buyer in question gets a zero payoff.

Taking the first order condition from the sum of the preceding integrals with respect to the cutoff price level (c) and substituting $z(c)=v$ and $c=z(v)$ (which must hold in the equilibrium) produces the following differential equation:

$$\begin{aligned} (p-z(v))^{2}+\frac{(v-z(v))(v+z(v)-2p)}{z^{\prime }(v)}=0. \end{aligned}$$

(11)

Rewrite Eq. (11) as follows (to simplify notation use z instead of z(v), also note that $(p-z)^2=(z-p)^2$):

$$\begin{aligned} (z-p)^2z'+(v-z)(v+z-2p)=0. \end{aligned}$$

(12)

Substituting $Z=z-p$ and $V=v-p$ gives:

$$\begin{aligned}&Z^2Z'+(V-Z)(V+Z)=0 \nonumber \\&\frac{{\mathrm{d}}Z}{{\mathrm{d}}V}=Z'=\frac{Z^2-V^2}{Z^2}. \end{aligned}$$

(13)

Replace $Z=VF(V)$ to get:

$$\begin{aligned} V\frac{{\mathrm{d}}F}{{\mathrm{d}}V}+F= & {} \frac{F^2-1}{F^2} \ \rightarrow \ V\frac{{\mathrm{d}}F}{{\mathrm{d}}V}=\frac{F^2-1-F^3}{F^2} \nonumber \\ \frac{{\mathrm{d}}V}{V}= & {} \frac{F^2}{-F^3+F^2-1}{\mathrm{d}}F \ \rightarrow \ \ln (V)=\int \frac{F^2}{-F^3+F^2-1}{\mathrm{d}}F. \end{aligned}$$

(14)

The denominator $-F^3+F^2-1$ can be decomposed using three roots—two complex and one real:

$$\begin{aligned} \int \frac{F^2}{-(F-r_1)(F-r_2)(F-r_3)}{\mathrm{d}}F= & {} -\frac{r_1^2\ln \left| {F-r_1}\right| }{(r_1-r_2)(r_1-r_3)} \nonumber \\&-\frac{r_2^2\ln \left| {F-r_2}\right| }{(r_2-r_1)(r_2-r_3)}-\frac{r_3^2\ln \left| {F-r_3}\right| }{(r_3-r_1)(r_3-r_2)}+c. \end{aligned}$$

(15)

$$\begin{aligned} V= & {} C\ e^{\Bigl (-\frac{r_1^2\ln \left| {F-r_1}\right| }{(r_1-r_2)(r_1-r_3)} -\frac{r_2^2\ln \left| {F-r_2}\right| }{(r_2-r_1)(r_2-r_3)}-\frac{r_3^2\ln \left| {F-r_3}\right| }{(r_3-r_1)(r_3-r_2)}\Bigr )}\nonumber \\ V= & {} C (\left| {F-r_1}\right| )^{\frac{-r_1^2}{(r_1-r_2)(r_1-r_3)}}\nonumber \\&\times (\left| {F-r_2}\right| )^{\frac{-r_2^2}{(r_2-r_1)(r_2-r_3)}}(\left| {F-r_3}\right| )^{\frac{-r_3^2}{(r_3-r_1)(r_3-r_2)}}. \end{aligned}$$

(16)

In the solution above $F\ne r_1$, $F\ne r_2$ and $F\ne r_3$ since it was implicitly assumed that these values are not equal to zero when dividing by the corresponding polynomial. From the parametric expression it is seen that when F approaches either of the roots, the value of the function becomes infinitely large, which may not be a solution to the bidding function. Hence, it is necessary to examine whether the roots of the polynomial are the solution to Eq. (13). In other words, check if $Z=kV$ solves the following equation:

$$\begin{aligned} k=\frac{(kV)^2-V^2}{(kV)^2} \Rightarrow k^3=k^2-1. \end{aligned}$$

The last expression is exactly the equation with the roots found previously. Considering only the real root the solution becomes $k=-0.7549$ or $Z=-0.7549V$.

Substituting back $Z=z-p$ and $V=v-p$ produces $z-p=-\frac{3}{4}(v-p)$. Solving for z provides the final equation for the exit function:

$$\begin{aligned} z(v)=\frac{7}{4}p-\frac{3}{4}v. \end{aligned}$$

(17)

The solution is defined only for $v>p$. Function z(v) gives each valuation $v\in (p,1]$ a corresponding exit price c. Note that the function is decreasing, so buyers with higher valuations exit the auction first. Hence, following this strategy, a buyer has a zero probability of obtaining the object outside if there is a buyer with a higher valuation. However, if several buyers exit the auction before it starts, then a buyer has strictly positive probability of obtaining the object outside. This results in a well-documented discontinuity of the cutoff function at the point of an indifferent buyer with valuation ${\hat{v}}$ (e.g., Hidvegi et al. 2006; Reynolds and Wooders 2009). Moreover, conditional on the posted price, an indifferent buyer may either exist or not. Let this price be ${\tilde{p}}$. If $p>{\tilde{p}}$ then an indifferent buyer does not exist. If $p<{\tilde{p}}$ then an indifferent buyer exists, and his valuation is equal to ${\hat{v}}$. $\square$

Appendix B: Proof of Lemma 2

Here, the construction of the expected profit is done for the case when an indifferent buyer exists (i.e., for $p\le \frac{3}{5}$). It is straightforward to adjust it to the case when an indifferent buyer does not exist. For exposition purposes I derive cumulative profit, which can then be decomposed into the profits gained from the auction and the posted price. Recall that $x_1$ is the first order statistic among buyers’ valuations, $x_2$—second order statistic, and $x_3$—third order statistic. There are several possibilities of realization of their valuations.

(a) $x_1,x_2,x_3<p$. We know that the profit of the seller using a standard second-price auction is equal to the second highest valuation. Since bidders cannot afford the good at the posted price, the profit of the other seller using the posted price is zero: $\int _0^p\int _0^{x_1}\int _0^{x_2} (x_2+0)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(b) $p<x_1<\frac{5}{3}p$ and $x_3<x_2<c$. In this case, both second and third order statistics drop out before the auction price reaches the exit price of the first order statistic, and $x_1$ wins the auction paying the second highest valuation. The good outside remains unsold: $\int _p^{\frac{5}{3}p}\int _0^c\int _0^{x_2} (x_2+0)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(c) (i) $p<x_1<\frac{5}{3}p$, $c<x_2<p$ and $x_3<c$. In this case, the third order statistic drops out before the auction price reaches the exit price of the first order statistic. Remaining buyers continue bidding in the auction until the second order statistic drops out. His valuation is the revenue of the seller using the auction. The seller with the good outside receives nothing: $\int _p^{\frac{5}{3}p}\int _c^p\int _0^c (x_2+0)6dx_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(ii) $p<x_1<\frac{5}{3}p$, $c<x_2<p$ and $c<x_3<x_2$. In this case, the first order statistic exits the auction and buys the good at the posted price. The auction is won by the second order statistic who pays the price equal to the valuation of the third order statistic: $\int _p^{\frac{5}{3}p}\int _c^p\int _c^{x_2} (x_3+p)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(d) (i) $p<x_1<\frac{5}{3}p$, $p<x_2<x_1$ and $x_3<c$. In this case, the third order statistic drops out before the auction price reaches the exit price of the first order statistic, and the other buyers continue to bid in the auction according to $\text {min}[v_i,p]$. The auction price rises until it matches the posted price outside. One of the remaining bidders wins the auction, the other one buys the good at the posted price; they both pay the price p: $\int _p^{\frac{5}{3}p}\int _p^{x_1}\int _0^c (p+p)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(ii) $p<x_1<\frac{5}{3}p$, $p<x_2<x_1$ and $c<x_3<p$. In this case, the first order statistic exits the auction at the exit price and buys the good outside. The auction continues with the other buyers bidding up until the valuation of the third order statistic: $\int _p^{\frac{5}{3}p}\int _p^{x_1}\int _c^p (x_3+p)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(iii) $p<x_1<\frac{5}{3}p$, $p<x_2<x_1$ and $p<x_3<x_2$. This case is identical to the previous case: $\int _p^{\frac{5}{3}p}\int _p^{x_1}\int _p^{x_2} (x_3+p)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

The next region is tricky, because it is the region affected by the employed rationing scheme. Recall that buyers with $v>{\hat{v}}$ attempt to buy the good at the posted price at the beginning of the auction. Since every buyer is equally likely to obtain the good, the profits of the seller using the auction will change conditional on who of the buyers succeeds:

(f) (i) $\frac{5}{3}p<x_1<1$, $0<x_2<\frac{5}{3}p$ and $0<x_3<x_2$. In this case, the first order statistic does not participate in the auction and buys the good at the posted price right away. The auction is carried out between the other two buyers who bid their valuations. The auction ends at the valuation of the third order statistic: $\int _{\frac{5}{3}p}^1\int _0^{\frac{5}{3}p}\int _0^{x_2} (x_3+p)6dx_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(ii) $\frac{5}{3}p<x_1<1$, $\frac{5}{3}p<x_2<x_1$ and $0<x_3<\frac{5}{3}p$. In this case, both first and second order statistics attempt to buy the object outside at the beginning of the auction, and one of them succeeds. The other one returns to the auction and bids against the third order statistic eventually winning the auction and paying the price equivalent to the value of the latter: $\int _{\frac{5}{3}p}^1\int _{\frac{5}{3}p}^{x_1}\int _0^{{\frac{5}{3}p}} (x_3+p)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

(iii) $\frac{5}{3}p<x_1<1$, $\frac{5}{3}p<x_2<x_1$ and $\frac{5}{3}p<x_3<x_2$. When all three buyers’ values are above the valuation of the indifferent buyer, they all attempt to buy the object outside at the beginning of the auction. Since every buyer has equal probability of obtaining the object, in 1/3 of the cases, the buyer with the lowest value (third order statistic) gets the object. The remaining buyers return to the auction and bid against each other. The auction ends at the valuation of the second order statistic. It is readily seen that in 2/3 the of cases, either $x_1$ or $x_2$ gets the object. Hence, the auction ends at the valuation of the third order statistic: $\int _{\frac{5}{3}p}^1\int _{\frac{5}{3}p}^{x_1}\int _{\frac{5}{3}p}^{x_2} (\frac{1}{3}x_2+\frac{2}{3}x_3+p)6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

When an indifferent buyer does not exist, the last three cases disappear, and the domain of integration for ${\mathrm{d}}x_1$ with $x_1>p$ becomes “$\int _p^1$” instead of “$\int _p^{\frac{5}{3}p}$”.

Hence, decomposing and summarizing, the cumulative payoff for the case when an indifferent buyer exists produces the following profit functions:

$$\begin{aligned} \pi _A|p\le \frac{3}{5}&= 6\Bigg [\int _0^p\int _0^{x_1}\int _0^{x_2} x_2{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1+\int _p^{\frac{5}{3}p}\int _0^{\frac{7}{4}p-\frac{3}{4}x_1}\int _0^{x_2} x_2{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\ \quad + \int _p^{\frac{5}{3}p}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p\int _0^{\frac{7}{4}p-\frac{3}{4}x_1} x_2{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^{\frac{5}{3}p}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p\int _{\frac{7}{4}p-\frac{3}{4}x_1}^{x_2} x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\ \quad + \int _p^{\frac{5}{3}p}\int _{p}^{x_1}\int _0^{\frac{7}{4}p-\frac{3}{4}x_1} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^{\frac{5}{3}p}\int _{p}^{x_1}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\ \quad + \int _p^{\frac{5}{3}p}\int _{p}^{x_1}\int _p^{x_2} x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _{\frac{5}{3}p}^1\int _{0}^{\frac{5}{3}p}\int _0^{x_2} x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\ \quad + \int _{\frac{5}{3}p}^1\int _{\frac{5}{3}p}^{x_1}\int _0^{\frac{5}{3}p} x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1+ \int _{\frac{5}{3}p}^1\int _{\frac{5}{3}p}^{x_1}\int _{\frac{5}{3}p}^{x_3} \Bigg (\frac{1}{3}x_2+\frac{2}{3}x_3\Bigg ) {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1\Bigg ] \\ = \frac{1}{3}-\frac{5p}{9}+\frac{25p^2}{18}-\frac{125p^3}{81}+\frac{8251p^4}{3888}. \\ \pi _P|p\le \frac{3}{5}&= 6\Bigg [\int _p^{\frac{5}{3}p}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p\int _{\frac{7}{4}p-\frac{3}{4}x_1}^{x_2} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\ \quad + \int _p^{\frac{5}{3}p}\int _{p}^{x_1}\int _0^{\frac{7}{4}p-\frac{3}{4}x_1} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^{\frac{5}{3}p}\int _{p}^{x_1}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p p{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\ \quad + \int _p^{\frac{5}{3}p}\int _{p}^{x_1}\int _p^{x_2} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1+ \int _{\frac{5}{3}p}^1\int _{0}^{x_1}\int _0^{x_2} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1\Bigg ] \\ = 6\Bigg [\frac{p^4}{36}+\frac{4p^4}{27}+\frac{2p^4}{27}+\frac{4p^4}{81}+\Bigg (\frac{p}{6}-\frac{125p^4}{162}\Bigg )\Bigg ]=p-\frac{17p^4}{6}. \end{aligned}$$

When an indifferent buyer does not exist, the same procedure results in:

$$\begin{aligned} \pi _A|p>\frac{3}{5}&= 6\Bigg [\int _0^p\int _0^{x_1}\int _0^{x_2} x_2 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^1\int _0^{\frac{7}{4}p-\frac{3}{4}x_1}\int _0^{x_2} x_2 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\&\quad + \int _p^1\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p\int _0^{\frac{7}{4}p-\frac{3}{4}x_1} x_2 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^1\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p\int _{\frac{7}{4}p-\frac{3}{4}x_1}^{x_2} x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\&\quad + \int _p^1\int _p^{x_1}\int _0^{\frac{7}{4}p-\frac{3}{4}x_1} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^1\int _p^{x_1}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\&\quad + \int _p^1\int _{p}^{x_1}\int _p^{x_2} x_3 {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1\Bigg ]\\&= -\frac{71}{256}+\frac{135p}{64}-\frac{213p^2}{128}+\frac{7p^3}{64}+\frac{57p^4}{256}. \\ \pi _P|p>\frac{3}{5}&= 6\Bigg [ \int _p^1\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p\int _{\frac{7}{4}p-\frac{3}{4}x_1}^{x_2} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\&\quad + \int _p^1\int _p^{x_1}\int _0^{\frac{7}{4}p-\frac{3}{4}x_1} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 + \int _p^1\int _p^{x_1}\int _{\frac{7}{4}p-\frac{3}{4}x_1}^p p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1 \\&\quad + \int _p^1\int _{p}^{x_1}\int _p^{x_2} p {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1\Bigg ]=\frac{25p}{16}-\frac{27p^2}{16}-\frac{21p^3}{16}+\frac{23p^4}{16}. \end{aligned}$$

Appendix C: Proof of Proposition 3

Recall that $x_1$ is the first order statistic among buyers’ valuations from the seller’s perspective, $x_2$ is the second order statistic, and $x_3$ is the third order statistic. First, consider a case when sellers use the same prices. Then, in expectation, they will each get half of the cumulative profit, which is equal to p when at least one buyer’s valuation is above the posted prices and 2p when at least two buyers’ values are above the posted prices:

$$\begin{aligned} \pi _{1}(p)= & {} \pi _2(p)=\pi (p)=\frac{1}{2}\Big [\int _{p}^{1}\int _0^p p f^{(n)}_{1:2}(x_1,x_2){\mathrm{d}}x_2{\mathrm{d}}x_1\nonumber \\&+\int _{p}^{1}\int _{p}^{x_1}2p f^{(n)}_{1:2}(x_1,x_2){\mathrm{d}}x_2{\mathrm{d}}x_1\Big ]= \frac{1}{2}(p^4-3p^3+2p) \end{aligned}$$

(18)

where $f^{(n)}_{1:2}(x_1,x_2)=6x_2^2$ is the joint density of two highest order statistics.

Now, assume that sellers set different prices $p_l<p_h$. A seller who sells his item at an ex-post low price will always be able to sell it if at least the first order statistic of the buyers’ valuations is above this price:

$$\begin{aligned} \pi _{p_l}(p_l)=\int _{p_l}^1 p_l f(x_1) {\mathrm{d}}x_1=p_l-p_l^4 \end{aligned}$$

(19)

where $f(x_1)=3x_1^2$ is the marginal density of the first order statistic.

The seller with an ex-post high posted price will be able to sell the item if one buyer’s valuation is above it and this buyer fails to buy at a lower price or if at least two buyers’ valuations are above this price. This may happen in several ways:

• (a) $p_h<x_1<1$, $p_l<x_2<p_h$ and $0<x_3<p_l$. In this case, buyers $x_1$ and $x_2$ first attempt to buy the object at the lower price. With 1/2 probability the first order statistic fails to buy it at $p_l$ and instead buys at $p_h$: $\int _{p_h}^{1}\int _{p_l}^{p_h}\int _{0}^{p_l}\frac{1}{2}p_h 6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

• (b) $p_h<x_1<1$, $p_l<x_2<p_h$ and $p_l<x_3<x_2$. In this case, all buyers first attempt to buy the object at the lower price. With probability 2/3 the first order statistic fails to buy it at $p_l$ and instead buys at $p_h$: $\int _{p_h}^{1}\int _{p_l}^{p_h}\int _{p_l}^{x_2}\frac{2}{3}p_h 6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

• (c) $x_1,x_2>p_h$. In this case, the seller with an ex-post high posted price always sells the item in possession: $\int _{p_h}^{1}\int _{p_h}^{x_1}\int _{0}^{x_2}p_h 6{\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1$.

Hence, the profit of the seller selling at an ex-post high posted price is:¹⁴

$$\begin{aligned} \pi _{p_h}(p_l,p_h)= & {} 6\Bigg [\int _{p_h}^{1}\int _{p_l}^{p_h}\int _{0}^{p_l}\frac{1}{2}p_h {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1+\int _{p_h}^{1}\int _{p_l}^{p_h}\int _{p_l}^{x_2}\frac{2}{3}p_h {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1\nonumber \\&+\int _{p_h}^{1}\int _{p_h}^{x_1}\int _{0}^{x_2}p_h {\mathrm{d}}x_3{\mathrm{d}}x_2{\mathrm{d}}x_1\Bigg ]=(p_h-1)p_h(p_l-1)(1+p_l+p_h). \end{aligned}$$

(20)

Observe that $\pi _{p_l}(p_l)$ is a singe-peaked function, because it is a product of increasing and decreasing continuous functions $p_l$ and $1-p_l^3$. Hence, $\frac{{\mathrm{d}}\pi _{p_l}(p_l)}{{\mathrm{d}}p_{l}}> 0|p_l< p_l^*$ and $\frac{{\mathrm{d}}\pi _{p_l}(p_l)}{{\mathrm{d}}p_{l}}< 0|p_l> p_l^*$. Function $\pi _{p_l}(p_l)$ reaches its maximum when $\frac{d\pi _{p_l}(p_l)}{dp_{l}}=0$ or at $p_l^*=0.623$. Moreover, $\pi _{p_h}(p_l,p_h)$ is also a single-peaked function by the same argument (the product of two increasing and two decreasing continuous functions). However, the maximum is conditional on the ex-post low price. Furthermore, $\frac{\partial \pi _{p_h}(p_l,p_h)}{\partial p_l}=(p_h-1)p_h(p_h+2p_l)\le 0\ \forall p_l$ meaning that the seller with an ex-post high price always prefers ex-post low price to be as low as possible.

Next, consider the profit function of the seller with an ex-post high price p when another seller sets a price $p-\epsilon$:

$$\begin{aligned} \pi _{p_h}(p-\epsilon ,p)&=p(1-p)^2(1+2p)=p(p^2-2p+1)(1+2p) \\&=p^3(1+2p)+p(1-4p^2)=2p^4-3p^3+p. \end{aligned}$$

If another seller sets the same price p, then he gets $\pi (p)\ge \pi _{p_h}(p-\epsilon ,p)$, because $\frac{1}{2}(p^4-3p^3+2p)-(2p^4-3p^3+p)=\frac{3}{2}(p^3-p^4)\ge 0$. However, this seller can earn even more by setting $p-\epsilon$, since:

$$\begin{aligned} \pi _{p_l}(p-\epsilon )-\pi (p)=(p-p^4)-\left( \frac{1}{2}p^4-\frac{3}{2}p^3+p\right) =\frac{3}{2}(p^3-p^4) \ge 0 \quad \forall p. \end{aligned}$$

It follows that $\pi _{p_l}(p_l)\ge \pi (p_h)\ge \pi _{p_h}(p_l,p_h)\ \forall p_l=p_h-\epsilon \ \text {and}\ p_h\in (0,1]$. In other words, a seller is always better off by setting $p-\epsilon$ rather than matching the price.

Now, using backward induction, consider the best response of the second-arriving seller to an arbitrary price p set by the first-moving seller:

• (a) $p>p_l^*$. It is readily seen that the second-arriving seller sets $p_l^*$.

• (b) $p<p_l^*$. The second-arriving seller may either undercut the first seller (setting $p_l=p-\epsilon$) or price at $p_h^*$ depending on what is more profitable. Notice that because $\frac{\partial \pi _{p_h}(p_l,p_h)}{\partial p_l}<0$ the profit of the seller with an ex-post high price increases with the decrease in the ex-post low price. On the other hand, because $\frac{{\mathrm{d}}\pi _{p_l}(p_l)}{{\mathrm{d}}p_l}>0$ for $p_l<p_l^*$ the profit of the seller with an ex-post low price falls with the decrease in $p_l$. Thus, there must be such a price ${\bar{p}}_l\in (0,p_l^*)$, at which sellers’ profits coincide, and beyond which it will no longer be profitable to undercut. Hence, the following equality must hold at ${\bar{p}}_l$:

  $$\begin{aligned} \pi _{{\bar{p}}_l}({\bar{p}}_l)=\pi _{p_h}({\bar{p}}_l,p_h^*({\bar{p}}_l)). \end{aligned}$$

  (21)

To find $p_h^*({\bar{p}}_l)$ first take the derivative of the profit function from the ex-post high price with respect to this price and express it as a function of the ex-post low price:

$$\begin{aligned} \frac{\partial \pi _{p_h}({\bar{p}}_l,p_h)}{\partial p_h}&=({\bar{p}}_l-1)(3p_h^2-1+(2p_h-1){\bar{p}}_l)=0 \\&\implies p_h^*({\bar{p}}_l)=\frac{{\bar{p}}_l-{\bar{p}}_l^2-\sqrt{3-3{\bar{p}}_l-2{\bar{p}}_l^2+{\bar{p}}_l^3+{\bar{p}}_l^4}}{3({\bar{p}}_l-1)}. \end{aligned}$$

Plugging $p_h^*({\bar{p}}_l)$ into (21) and solving for ${\bar{p}}_l$ results in ${\bar{p}}_l=0.325$ with $\pi _{p_l}({\bar{p}}_l)=\pi _{p_h}(\bar{p_l},p_h^*)=0.314$. Using $\pi _{p_h}(\bar{p_l},p_h^*)=0.314$, it is straightforward to find the corresponding ex-post high price $p_h^*=0.565$.

It is easy to check that the first-arriving seller has no incentive to deviate from the equilibrium price ${\bar{p}}_l$. If he sets a price higher than ${\bar{p}}_l$, then the second-arriving seller undercuts and the former seller’s profit is $\pi _{p_h}(p_l,p_l)<\pi _{p_l}({\bar{p}}_l)$ for $p_l>{\bar{p}}_l$. If the first-arriving seller sets a price lower that ${\bar{p}}_l$, the second-moving seller sets $p_h^*$ and earns even higher profit, because $\frac{\partial \pi _{p_h}(p_l,p_h)}{\partial p_l}<0$. Hence, the first-arriving seller may increase his profit by pricing exactly at ${\bar{p}}_l$.