Equivalence and revenue comparison among identical-item auctions

Abstract

We compare outcomes of the common identical-item auction formats in an independent private value environment. In single-unit demand cases, discriminatory auctions, uniform-price auctions, sequential auctions with disclosing only the winner’s identity, and sequential auctions with disclosing both the winner’s identity and winning price, are revenue equivalent. The seller’s expected revenue is the lowest in sequential auctions with disclosing all bids due to bidders’ faking behaviors. In multi-unit demand cases, discriminatory auctions generate the highest revenue, followed by sequential auctions with three different information disclosure rules. The seller’s revenue varies from the highest to the lowest in uniform-price auctions owing to the existence of multiple equilibria. Interestingly, we find that with more information disclosed, a bidder’s expected payoff may increase, decrease, or even stay unchanged, which is different from the conventional wisdom “the more information, the better”.

Appendix

Proof of lemma 1of lemma 1.

(1) $\alpha = \phi$ means that bidders are symmetric and their equilibrium distributions of bids are identical, that is, $F_{A} \left( x \right) = F_{{\Omega }} \left( x \right),x \in \left[ {\underline {b} ,\overline{b}} \right]$, where $\left[ {\underline {b} ,\overline{b}} \right]$ is the strategy space. Besides, the belief of bidders regarding their opponent as being of a low-value type is $1 - \alpha$. Therefore, if a high-value bidder (Bidder A or any high-value bidder in ${\Omega }$) bids $x \in \left[ {\underline {b} ,\overline{b}} \right]$, then her expected payoff is: $\pi \left( x \right) = \left( {1 - \alpha + \alpha F_{A} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right)$. From Maskin and Riley (1985, pp. 151), the infimum of bids made by a high-value bidder is $v_{L}$ and a high-value bidder’s expected payoff is $\left( {1 - \alpha } \right)^{n - 1}$. Moreover, from the property of the equilibrium in mixed strategies, that is, all bids in the strategy space produce the same expected payoff, we get $\left( {1 - \alpha + \alpha F_{A} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) = \left( {1 - \alpha } \right)^{n - 1}$. It then follows that $F_{A} \left( x \right) = \frac{{1 - {\upalpha }}}{{\upalpha }}\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }} - 1} \right),x \in [v_{L} ,v_{H} - \left( {1 - {\upalpha })^{n - 1} } \right].$

(2) Suppose $1 = \phi > \alpha$. The high-value Bidder A’s expected payoff from bidding $x \in \left[ {b_{{\text{A}}} ,b^{{\text{A}}} } \right]$ is $\pi_{A} \left( x \right) = \left( {1 - \alpha + \alpha F_{\Omega } \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right)$. We then consider any high-value bidder in $\Omega$, and her expected payoff from bidding $x \in \left[ {b_{{\Omega }} ,b^{{\Omega }} } \right]$ is $\pi_{{\Omega }} \left( x \right) = F_{A} \left( x \right)\left( {1 - \alpha + \alpha F_{{\Omega }} \left( x \right)} \right)^{n - 2} \left( {v_{H} - x} \right)$. First, we prove that $b_{{\text{A}}} = b_{{\Omega }}$. Suppose $b_{{\text{A}}} < b_{{\Omega }}$, then $\forall x \in \left( {b_{{\text{A}}} ,b_{{\Omega }} } \right), \pi_{A} \left( {b_{{\text{A}}} } \right) = \left( {1 - \alpha } \right)^{n - 1} \left( {v_{H} - b_{{\text{A}}} } \right) > \pi_{A} \left( x \right) = \left( {1 - \alpha } \right)^{n - 1} \left( {v_{H} - x} \right)$. This contradicts the property of the equilibrium in mixed strategies. Otherwise, suppose $b_{{\text{A}}} > b_{{\Omega }}$, then $\pi_{{\Omega }} \left( {b_{{\Omega }} } \right) = 0$. However,$\forall x \in \left( {b_{{\text{A}}} ,v_{H} } \right),\pi_{{\Omega }} \left( x \right) = F_{A} \left( x \right)\left( {1 - \alpha + \alpha F_{{\Omega }} \left( x \right)} \right)^{n - 2} \left( {v_{H} - x} \right) > 0,$ which also leads to a contradiction. Accordingly, $b_{{\text{A}}} = b_{{\Omega }} \ge v_{L}$. Next, we show that $b_{{\text{A}}} = v_{L}^{ + }$. Since $\pi_{A} \left( {v_{L} } \right) = \frac{1}{n}(1 - \alpha )^{n - 1} < \pi_{A} \left( {v_{L}^{ + } } \right) = (1 - \alpha )^{n - 1}$, thus $b_{{\text{A}}} > v_{L}$. Also notice that $\pi_{A} \left( {b_{{\text{A}}} } \right) = (1 - \alpha )^{n - 1} \left( {v_{H} - b_{{\text{A}}} } \right)$ is decreasing with $b_{{\text{A}}}$. That is, the high-value Bidder A never bids $v_{L}$, and she also has incentive to lower the infimum of her strategy space. Accordingly, $b_{{\text{A}}} = v_{L}^{ + }$. Finally, since $\pi_{A} \left( {v_{L}^{ + } } \right) = \left( {1 - \alpha } \right)^{n - 1}$, and combined with the property of the equilibrium in mixed strategies, we obtain that $F_{{\Omega }} \left( x \right) = \frac{1 - \alpha }{\alpha }\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }} - 1} \right),x \in [v_{L}^{ + } ,v_{H} - \left( {1 - \alpha )^{n - 1} } \right].$ Moreover, Narasimhan (1988) proved that $F_{A} \left( x \right)$ and ${ }F_{{\Omega }} \left( x \right)$ have the same supremum in their strategy spaces. So, we get $\pi_{{\Omega }} \left( x \right) = { }\pi_{{\Omega }} \left( {v_{H} - (1 - \alpha )^{n - 1} } \right) = (1 - \alpha )^{n - 1}$ and $F_{A} \left( x \right) = \frac{1 - \alpha }{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }},x \in [v_{L}^{ + } ,v_{H} - \left( {1 - \alpha )^{n - 1} } \right]$.□

Proof of Proposition 1 We use the backward induction technique by first considering the second auction. The belief of bidders regarding their opponents as being of a low-value type is ${ }1 - q = \frac{1 - \theta }{{1 - \theta + \theta F_{1}^{{\text{I}}} \left( {b_{w} } \right)}}$. It then follows from Lemma 1 that a high-value bidder in $\Delta$ bids randomly according to $F_{2}^{{\text{I}}} \left( x \right) = \frac{1 - q}{q}\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 2}}} }} - 1} \right), x \in \left[ {v_{L} ,v_{H} - \left( {1 - q} \right)^{n - 2} } \right]$, and her expected payoff is $\left( {1 - q} \right)^{n - 2}$. Next, we turn to the first auction and denote $s_{1}^{{\text{I}}}$ as the supremum of bids made by high-value bidders. A high-value bidder’s total expected payoff across the two auctions from bidding $x \in \left[ {v_{L} ,s_{1}^{{\text{I}}} } \right]$ in the first auction is:

$$ \pi^{I} \left( x \right) = \left( {1 - \theta + \theta F_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {1 - \left( {1 - \theta + \theta F_{1}^{I} \left( x \right)} \right)^{n - 1} } \right)E\left[ {\left( {1 - q} \right)^{n - 2} |x \le b_{w} } \right] = \left( {1 - \theta + \theta F_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - F_{1}^{I} \left( x \right)} \right). $$

(6)

The derivation of Formula (6) is as follows. The conditional distribution of the winning bidding is:${\text{ F}}\left( {b_{w} {|}x \le {\text{b}}_{w} } \right) = \frac{{\left( {1 - \theta + \theta F_{1}^{{\text{I}}} \left( {b_{w} } \right)} \right)^{n - 1} - \left( {1 - \theta + \theta F_{1}^{{\text{I}}} \left( x \right)} \right)^{n - 1} }}{{1 - \left( {1 - \theta + \theta F_{1}^{{\text{I}}} \left( x \right)} \right)^{n - 1} }}$, and thus

$$ \begin{aligned} E\left[ {\left( {1 - q} \right)^{n - 2} |x \le b_{w} } \right] = & \mathop \smallint \limits_{x}^{{s_{1} }} \left( {1 - q} \right)^{n - 2} f\left( {b_{w} |x \le b_{w} } \right)db_{w} \\ = & \frac{{\mathop \smallint \nolimits_{x}^{{s_{1} }} \left( {1 - q} \right)^{n - 2} \left( {n - 1} \right)\left( {1 - \theta + \theta F_{1}^{I} \left( {b_{w} } \right)} \right)^{n - 2} \theta f_{1}^{I} \left( {b_{w} } \right)db_{w} }}{{1 - \left( {1 - \theta + \theta F_{1}^{I} \left( x \right)} \right)^{n - 1} }} \\ = & \frac{{\left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - F_{1}^{I} \left( x \right)} \right)}}{{1 - \left( {1 - \theta + \theta F_{1}^{I} \left( x \right)} \right)^{n - 1} }}. \\ \end{aligned} $$

From the property of separating equilibria, that is, $F_{1}^{{\text{I}}} \left( {v_{L} } \right) = 0$, and combined with the fact that the infimum of bids made by a high-value bidder in the first auction is $v_{L}$, we obtain that $\pi^{{\text{I}}} = \pi^{{\text{I}}} \left( {v_{L} } \right) = \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right).$ Because all bids in the strategy space produce the same expected payoff, $F_{1}^{{\text{I}}} \left( x \right)$ should satisfy the following implicit equation:

$$ \left( {1 - \theta + \theta F_{1}^{{\text{I}}} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - F_{1}^{{\text{I}}} \left( x \right)} \right) = \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right),\;\;\forall x \in \left[ {v_{L} ,s_{1}^{{\text{I}}} } \right]. $$

(7)

Moreover, from the facts of $\pi^{{\text{I}}} \left( {s_{1}^{{\text{I}}} } \right) = \pi^{{\text{I}}}$ and $F_{1}^{{\text{I}}} \left( {s_{1}^{{\text{I}}} } \right) = 1$, we get $s_{1}^{{\text{I}}} = v_{H} - \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right)$ via substituting $s_{1}^{{\text{I}}}$ into Formula (7).

At last, we compute the seller’s expected revenue, which is just the sum of the ex-ante expected payments from bidders. Further, the sum of bidders’ ex-ante expected payoffs is the expected allocation valuations of the two items minus the sum of bidders’ ex-ante expected payments. Therefore, the seller’s expected revenue is:${ }\Pi^{{\text{I}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + C_{n}^{1} \theta \left( {1 - \theta } \right)^{n - 1} \left( {v_{H} + v_{L} } \right) + \mathop \sum \limits_{j = 2}^{j = n} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} 2v_{H} - n\theta \pi^{{\text{I}}} = 2\left( {v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \frac{{n\left( {n - 1} \right)}}{2}\theta^{2} \left( {1 - \theta } \right)^{n - 2} } \right)$. □

Proof of Proposition 2 In the second auction, given that the high-value bidder ${ }i(i \in \Delta , i \ne k)$ bids $b_{i}^{2} = \beta \left( {F_{1}^{{\text{I}}} \left( {b_{i}^{1} } \right),n - 2} \right) = v_{H} - \frac{{\left( {1 - \theta } \right)^{n - 2} }}{{\left( {1 - \theta + \theta F_{1}^{{\text{I}}} \left( {b_{i}^{1} } \right)} \right)^{n - 2} }}$ or bids randomly over $\left[ {v_{L} ,v_{H} - \left( {1 - q} \right)^{n - 2} } \right]$ according to $F_{2}^{{\text{I}}} \left( x \right)$, we analyse the optimal bid of the high-value bidder $k \in \Delta$. Since $P_{r} \left\{ {b_{i}^{2} \le x{|}b_{i}^{1} \le b_{w} } \right\} = \frac{{\frac{1 - \theta }{\theta }\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 2}}} }} - 1} \right)}}{{F_{1}^{{\text{I}}} \left( {b_{w} } \right)}} = \frac{1 - q}{q}\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 2}}} }} - 1} \right) = F_{2}^{{\text{I}}} \left( x \right)$, it means that bidder k believes her high-value opponents are mixing their bids according to $F_{2}^{{\text{I}}} \left( x \right)$. Then bidder k’s expected payoff from bidding $x \in \left[ {v_{L} ,v_{H} - \left( {1 - q} \right)^{n - 2} } \right]$ is $\left( {1 - q + qF_{2}^{{\text{I}}} \left( x \right)} \right)^{n - 2} \left( {v_{H} - x} \right) = \left( {1 - q} \right)^{n - 2}$. So, bidder k is indifferent between all bids in $\left[ {v_{L} ,v_{H} - \left( {1 - q} \right)^{n - 2} } \right]$ and, in particular, it is optimal for her to bid $b_{k}^{2} = \beta \left( {F_{1}^{{\text{I}}} \left( {b_{k}^{1} } \right),n - 2} \right)$ or to bid randomly according to $F_{2}^{{\text{I}}} \left( x \right)$. Therefore, on the one hand, the behavioral strategy profile $\beta \left( {F_{1}^{{\text{I}}} \left( {b_{k}^{1} } \right),n - 2} \right)$, for every $k \in \Delta$ and every $b_{w}$, constitutes an equilibrium of the second of sequential auctions with the incomplete information policy. On the other hand, the equilibrium in behavioral strategies and the equilibrium in mixed strategies satisfy the rectangular property. □

Proof of Proposition 3 With the hidden information policy, only the winner’s identity is disclosed at the end of the first auction, the posterior belief of ${ }1 - q = \frac{1 - \theta }{{1 - \theta + \theta F_{1}^{{\text{H}}} \left( {b_{w} } \right)}}$ is not commonly known. First, let us consider the second auction. Given that the high-value bidder ${ }i\left( {i \in \Delta , i \ne k} \right)$ bids ${ }b_{i}^{2} = \beta \left( {F_{1}^{{\text{H}}} \left( {b_{i}^{1} } \right),n - 2} \right)$, we analyse the optimal decision of the high-value bidder $k \in \Delta$. Because $P_{r} \left\{ {b_{i}^{2} \le x} \right\} = \frac{1 - \theta }{\theta }\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 2}}} }} - 1} \right)$ and $b_{i}^{1} \le b_{w}$ is equivalent to $b_{i}^{2} \le v_{H} - \left( {1 - q} \right)^{n - 2}$, so $P_{r} \left\{ {b_{i}^{2} \le x|b_{i}^{1} < b_{w} } \right\} = \frac{1 - q}{q}\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 2}}} }} - 1} \right)$. It then follows from the proof of Proposition 2 that bidder $k$ is indifferent between all bids in [$v_{L} , v_{H} - \left( {1 - q} \right)^{n - 2}$] and, in particular, it is optimal for her to bid $b_{k}^{2} = \beta \left( {F_{1}^{{\text{H}}} \left( {b_{k}^{1} } \right),n - 2} \right)$ (even if she does not know what $1 - q$ is).

Next, we turn to the first auction. Since $b_{k}^{2}$ is monotonically increasing with $b_{k}^{1}$, the top two bidders in the first auction win the two items. Besides, the bidder with the first-highest bid $b_{F}^{1}$ pays her bid, while the bidder with the second-highest bid $b_{s}^{1}$ pays $\beta \left( {F_{1}^{{\text{H}}} \left( {b_{s}^{1} } \right),n - 2} \right)$. So, a high-value bidder’s total expected payoff across the two auctions from bidding $x \in \left[ {v_{L} ,s_{1}^{{\text{H}}} } \right]$, where $s_{1}^{{\text{H}}}$ is the supremum of bids, is $\pi^{{\text{H}}} \left( x \right) = \left( {1 - \theta + \theta F_{1}^{{\text{H}}} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - F_{1}^{{\text{H}}} \left( x \right)} \right)$. It then follows from the proof of Proposition 1 that $F_{1}^{{\text{H}}} \left( x \right)$ should satisfy equation (7) and $s_{1}^{{\text{H}}} = s_{1}^{{\text{I}}}$. The total expected payoff for a high-value bidder and the expected revenue for the seller can be similarly obtained. □

Proof of Lemma 2 Suppose that a separating equilibrium exists in the first auction, that is, a high-value bidder bids randomly according to $F_{1}^{{\text{C}}} \left( x \right)$, where $F_{1}^{{\text{C}}} \left( {v_{L} } \right) = 0$. Thus, a high-value bidder’s total expected payoff is $\pi^{{\text{C}}} = \left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2}$. However, let us consider that one high-value bidder deviates from the separating equilibrium and bids $v_{L}$ in the first auction. After the first auction, if all losing bids are equal to $v_{L}$, then she bids $v_{L}^{ + }$; otherwise, bids $v_{L} + \frac{1}{2}$. In this case, the high-value bidder’s expected payoff is $\left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} + C_{n - 1}^{2} \theta^{2} \left( {1 - \theta } \right)^{n - 3} \frac{1}{2} > \pi^{{\text{C}}}$. The first two terms are the payoffs when at most one high-value opponent exists. When two high-value opponents exist, one wins the first item and drops out, while the other believes that all losers are of low-value types and subsequently bids $v_{L}^{ + }$. So, the deviating bidder bids $v_{L} + \frac{1}{2}$ to win the second item and her payoff is the third term. Therefore, there does not exist any separating equilibrium in the first auction. Suppose that a pooling equilibrium exists in the first auction, that is, all high-value bidders bid $v_{L}$. In this case, a high-value bidder’s expected payoff is $\pi^{{\text{C}}} = \frac{1}{n} + \left( {1 - \frac{1}{n}} \right)\left( {1 - \theta } \right)^{n - 2}$. If one high-value bidder deviates and bids $v_{L}^{ + }$, then she will win the first item for sure and her expected payoff is $1 > \pi^{{\text{C}}}$. Therefore, there does not exist any pooling equilibrium in the first auction. □

Proof of Proposition 4 First, let us consider the second auction. We suppose the semi-pooling equilibrium in the first auction is that a high-value bidder bids $v_{L}$ with probability $p$, where $0 < p < 1$. After the first auction, all bids are disclosed. One of the following three cases will arise: (i) More than two losing bids are greater than $v_{L}$. In this case, the second auction corresponds to a Bertrand game since it is common knowledge that at least two bidders are of high-value types. So, a high-value bidder in $\Delta$ will bid $v_{H}$. (ii) Only one losing bid is greater than $v_{L}$. Asymmetric information is created in this case. The bidder with that losing bid $( > v_{L} )$ believes that each of her opponents being a low-value type with probability $1 - \delta = \frac{1 - \theta }{{1 - \theta + \theta p}}$. Other high-value bidders who bid $v_{L}$ in the first auction confirm the bidder with the losing bid ($> v_{L}$) is a high-value type, but believe the other opponents being a low-value type with probability $1 - \delta$. The equilibrium immediately follows from Lemma 1, where $\alpha = \delta$ and $\phi = 1$. (iii) All losing bids are equal to $v_{L}$. Each bidder’s posterior belief of an opponent being a low-value type is $1 - \delta$, which is common knowledge. The equilibrium also follows from Lemma 1, where $\alpha = \phi = \delta$.

Next, we turn to the first auction, a high-value bidder’s total expected payoff across the two auctions from bidding $v_{L}$ in the first auction is:

$$ \begin{aligned} \pi^{C} \left( {v_{L} } \right) = & \left( {1 - \theta + \theta p} \right)^{n - 1} \left( {\frac{1}{n} + \left( {1 - \frac{1}{n}} \right)\left( {1 - \delta } \right)^{n - 2} } \right) + \left( {1 - \delta } \right)^{n - 2} \left( {\mathop \sum \limits_{j = 1}^{j = n - 1} C_{n - 1}^{j} \left( {1 - \theta } \right)^{n - 1 - j} \theta^{j} j\left( {1 - p} \right)p^{j - 1} + \mathop \sum \limits_{j = 2}^{j = n - 1} C_{n - 1}^{j} \left( {1 - \theta } \right)^{n - 1 - j} \theta^{j} \frac{{j\left( {j - 1} \right)}}{2}\left( {1 - p} \right)^{2} p^{j - 2} } \right) \\ = & \left( {1 - \theta + \theta p} \right)^{n - 1} \frac{1}{n} + \frac{n - 1}{n}\left( {1 - \theta } \right)^{n - 2} \left( {1 - \theta + \theta p} \right) + \left( {n - 1} \right)\left( {1 - p} \right)\theta \left( {1 - \theta } \right)^{n - 2} + C_{n - 1}^{2} \frac{{\left( {1 - p} \right)^{2} \left( {1 - \theta } \right)^{n - 2} \theta^{2} }}{1 - \theta + \theta p}. \\ \end{aligned} $$

(8)

Denote $s_{1}^{{\text{c}}}$ as the supremum of bids made by high-value bidders. Then the high-value bidder’s expected payoff from bidding $x$, where $v_{L} < x \le s_{1}^{{\text{c}}}$, is:

$$ \pi^{C} \left( x \right) = \left( {1 - \theta + \theta F_{1}^{C} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {n - 1} \right)\theta \left( {1 - F_{1}^{C} \left( x \right)} \right)\left( {1 - \theta + \theta p} \right)^{n - 2} \left( {1 - \delta } \right)^{n - 2} . $$

(9)

From the property of the equilibrium in mixed strategies, we have $\pi^{{\text{C}}} \left( {v_{L} } \right) = \pi^{{\text{C}}} \left( x \right),\forall x \in \left( {v_{L} ,s_{1}^{{\text{c}}} } \right]$, and in particular, $\pi^{{\text{C}}} \left( {v_{L} } \right) = \pi^{{\text{C}}} \left( {v_{L}^{ + } } \right)$. Therefore, from Formula (8) and (9), and combined with $F_{1}^{{\text{C}}} \left( {v_{L}^{ + } } \right) = p$, we have,

$$ \left( {1 - \theta + \theta p} \right)^{n} = \left( {1 - \theta } \right)^{n - 2} \left( {\left( {1 - \theta + \theta p} \right)^{2} + \frac{{n\left( {n - 2} \right)}}{2}\left( {1 - p} \right)^{2} \theta^{2} } \right). $$

(10)

Also, $F_{1}^{{\text{C}}} \left( x \right)$ should satisfy the following equation:

$$ \left( {1 - \theta + \theta F_{1}^{C} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - F_{1}^{C} \left( x \right)} \right) = \left( {1 - \theta + \theta p} \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - p} \right),\;\;\forall x \in \left[ {v_{L} ,s_{1}^{c} } \right]. $$

(11)

Similar to the proof of Proposition 1, we get $s_{1}^{{\text{c}}} = v_{H} - \left( {1 - \theta + \theta p} \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - p} \right)$. Accordingly, a high-value bidder in the first auction bids $v_{L}$ with probability $p$, and with probability $1 - p$, bids randomly over $\left( {v_{L} ,s_{1}^{{\text{c}}} } \right]$ according to the distribution function $F_{1}^{{\text{C}}} \left( x \right)$. $p$ and $F_{1}^{{\text{C}}} \left( x \right)$ are solutions to Equation (10) and (11).

At last, we compute the seller’s expected revenue of $\Pi^{{\text{C}}}$. Clearly, $\pi^{{\text{C}}} = \left( {1 - \theta + \theta p} \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {1 - p} \right).$ Similar to the computation of $\Pi^{{\text{I}}}$, we get that $\Pi^{{\text{C}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + C_{n}^{1} \theta \left( {1 - \theta } \right)^{n - 1} \left( {v_{H} + v_{L} } \right) + \mathop \sum \limits_{j = 2}^{j = n - 1} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} \left( {p^{j} \frac{n - j}{n}\left( {v_{L} + v_{H} } \right) + \left( {1 - p^{j} \frac{n - j}{n}} \right)2v_{H} } \right) + C_{n}^{n} \theta^{n} 2v_{H} - n\theta \pi^{{\text{C}}} .$ Through simplifying, we have $\Pi^{{\text{C}}} = 2v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {p\left( {1 + \left( {n - 1} \right)\theta } \right) - n\theta } \right) - \left( {1 - \theta + \theta p} \right)^{n - 1} \left( {1 + \left( {n - 1} \right)\theta } \right).$□

Proof of Proposition 5 First, let us consider the second auction. The winner W of the first auction believes that each of her $n - 1$ opponents is of a low-value type with probability $1 - \omega = \frac{1 - \theta }{{1 - \theta + \theta G_{1}^{{\text{I}}} \left( {b_{w} } \right)}}$. Also, a high-value bidder in $\Delta$ confirms that W is a high-value type and believes that any other opponent is of a low-value type with probability ${ }1 - \omega$. Accordingly, the equilibrium immediately follows from Lemma 1, where $1 = \phi > \alpha = \omega$.

Next, we turn to the first auction and denote $t_{1}^{{\text{I}}}$ as the supremum of bids. If a high-value bidder wins the first auction with bid $x$, where $x \in \left[ {v_{L} ,t_{1}^{{\text{I}}} } \right]$, then her posterior belief can be denoted as ${ }1 - \omega \left( x \right) = \frac{1 - \theta }{{1 - \theta + \theta G_{1}^{{\text{I}}} \left( x \right)}}$, and her total expected payoff across the two auctions from bidding $x \in \left[ {v_{L} ,t_{1}^{{\text{I}}} } \right]$, in the first auction is:

$$ \begin{aligned} \tau^{I} \left( x \right) = & \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {1 - \omega \left( x \right)} \right)^{n - 1} + \mathop \smallint \limits_{x}^{{\overline{s}}} \left( {1 - \omega \left( y \right)} \right)^{n - 1} \mathop \sum \limits_{k = 1}^{n - 1} c_{n - 1}^{k} \theta^{k} (1 - \theta )^{n - 1 - k} kg_{1}^{I} \left( y \right)\left( {G_{1}^{I} \left( y \right)} \right)^{k - 1} dy \\ = & \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {1 - \theta } \right)^{n - 1} + \mathop \smallint \limits_{x}^{{\overline{s}}} \left( {1 - \omega \left( y \right)} \right)^{n - 1} \left( {n - 1} \right)\theta g_{1}^{I} \left( y \right)\left( {1 - \theta + \theta G_{1}^{I} \left( y \right)} \right)^{n - 2} dy \\ = & \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {1 - \theta } \right)^{n - 1} \left( {1 - \left( {n - 1} \right)\log \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)} \right) \\ \end{aligned} $$

Similar to the proof of Proposition 1, we obtain that $\tau^{{\text{I}}} \left( x \right) = \tau^{{\text{I}}} \left( {v_{L} } \right) = \left( {1 - \theta } \right)^{n - 1} \left( {2 - \left( {n - 1} \right)\log \left( {1 - \theta } \right)} \right),\forall x \in \left[ {v_{L} ,t_{1}^{{\text{I}}} } \right]$. Thus, $G_{1}^{{\text{I}}} \left( x \right)$ should satisfy the following implicit equation:

$$ \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)^{n - 1} \left( {v_{H} - x} \right) + \left( {1 - \theta } \right)^{n - 1} \left( {1 - \left( {n - 1} \right)\log \left( {1 - \theta + \theta G_{1}^{I} \left( x \right)} \right)} \right) = \left( {1 - \theta } \right)^{n - 1} \left( {2 - \left( {n - 1} \right)\log \left( {1 - \theta } \right)} \right),\;\;\;\forall x \in \left[ {v_{L} ,t_{1}^{I} } \right]. $$

(12)

Substitute $t_{1}^{{\text{I}}}$ into Formula (12), we get $t_{1}^{{\text{I}}} = v_{H} - \left( {1 - \theta } \right)^{n - 1} \left( {2 - \left( {n - 1} \right){\text{log}}\left( {1 - \theta } \right)} \right)$. At last, the seller’s expected revenue is $\Gamma^{{\text{I}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + \mathop \sum \limits_{j = 1}^{j = n} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} 2v_{H} - n\theta \tau^{{\text{I}}} = 2v_{H} - 2\left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} \left( {2 - \left( {n - 1} \right)l{\text{og}}\left( {1 - \theta } \right)} \right)$. □

Proof of Proposition 6 Suppose that the high-value bidder $k \in {\Delta }$ bids ${ }b_{k}^{2} = \beta \left( {G_{1}^{{\text{I}}} \left( {b_{k}^{1} } \right),n - 1} \right)$ or bids randomly according to $G_{2L}^{{\text{I}}} \left( x \right)$. Since $P_{r} \left\{ {b_{k}^{2} \le x{|}b_{k}^{1} \le b_{w} } \right\} = G_{2L}^{{\text{I}}} \left( x \right)$, so winner W believes that bidder $k$ is mixing bids according to $G_{2L}^{{\text{I}}} \left( x \right)$. Therefore, W’s expected payoff from bidding $x \in [v_{L}^{ + } ,v_{H} - \left( {1 - \omega )^{n - 1} } \right]$ is $\tau_{2W}^{{\text{I}}} \left( x \right) = \left( {v_{H} - x} \right)(1 - \omega + \omega G_{2L}^{{\text{I}}} \left( x \right))^{n - 1} = (1 - \omega )^{n - 1}$. Thus, W is indifferent between all bids in $[v_{L}^{ + } ,v_{H} - \left( {1 - \omega )^{n - 1} } \right]$ and in particular, it is optimal for her to bid randomly according to $G_{2W}^{{\text{I}}} \left( x \right)$.

Given that W bids randomly according to $G_{2W}^{{\text{I}}} \left( x \right)$, and the high-value bidder $i \left( {i \in \Delta , i \ne k} \right)$ either bids $b_{i}^{2} = \beta \left( {G_{1}^{{\text{I}}} \left( {b_{i}^{1} } \right),n - 1} \right)$ or bids randomly according to $G_{2L}^{{\text{I}}} \left( x \right)$, let us consider the optimal decision of the high-value bidder $k \in \Delta .$ Notice that bidder k believes each high-value bidder i in $\Delta$ is mixing bids according to $G_{2L}^{{\text{I}}} \left( x \right)$. So, bidder k’s expected payoff from bidding $x \in [v_{L}^{ + } ,v_{H} - \left( {1 - \omega )^{n - 1} } \right]$ is: $\tau_{2L}^{{\text{I}}} \left( x \right) = (v_{H} - x){ }G_{2W}^{{\text{I}}} \left( x \right)\left( {1 - \omega + \omega G_{2L}^{{\text{I}}} \left( x \right)} \right)^{n - 2} = (1 - \omega )^{n - 1}$. Thus, bidder k is indifferent between all bids in $\left[ {v_{L}^{ + } ,v_{H} - (1 - \omega )^{n - 1} } \right]$ and, in particular, it is optimal to bid $b_{k}^{2} = \beta \left( {G_{1}^{{\text{I}}} \left( {b_{k}^{1} } \right),n - 1} \right)$ or to bid randomly according to $G_{2L}^{{\text{I}}} \left( x \right)$. Therefore, the winner W bids randomly according to $G_{2W}^{{\text{I}}} \left( x \right)$ and any high-value bidder $k \in \Delta$ bids $\beta \left( {F_{1}^{{\text{I}}} \left( {b_{k}^{1} } \right),n - 2} \right)$, for every $b_{w}$, constitutes an equilibrium of the second auction. In addition, the equilibrium in behavioral strategies and the equilibrium in mixed strategies satisfy the rectangular property. □

Proof of Proposition 7 In the second auction and with the hidden information policy, winner W believes that each of her $n - 1$ opponents is of a low-value type with probability $1 - \omega = \frac{1 - \theta }{{1 - \theta + \theta G_{1}^{{\text{H}}} \left( {b_{w} } \right)}}$. A high-value bidder in ${\Delta }$ confirms that winner W is a high-value type but she is unsure of the other opponents’ types. First, suppose that every high-value bidder $k \in {\Delta }$ bids $b_{k}^{2} = \beta \left( {G_{1}^{{\text{H}}} \left( {b_{k}^{1} } \right),n - 1} \right)$. It then follows from the proof of Proposition 6 that it is optimal for winner W to bid randomly according to ${ }G_{2W}^{{\text{I}}} \left( x \right) = \frac{1 - \omega }{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }},x \in [v_{L}^{ + } ,v_{H} - \left( {1 - \omega )^{n - 1} } \right]$.

Next, we suppose that winner W bids randomly according to ${ }G_{2W}^{{\text{I}}} \left( x \right)$ and the high-value bidder $i(i \in \Delta , i \ne k)$ bids $b_{i}^{2} = \beta \left( {G_{1}^{{\text{H}}} \left( {b_{i}^{1} } \right),n - 1} \right)$. Then we consider the optimal decision of the high-value bidder k. Notice that bidder k believes that the high-value bidder i in $\Delta$ is mixing bids according to ${ }G_{2L}^{{\text{I}}} \left( x \right) = \frac{1 - \omega }{\omega }\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }} - 1} \right)$. It then follows from the proof of Proposition 6 that bidder $k$ is indifferent between all bids in [$v_{L}^{ + } , v_{H} - (1 - \omega )^{n - 1}$] and, in particular, it is optimal to bid $b_{k}^{2} = \beta \left( {G_{1}^{{\text{H}}} \left( {b_{k}^{1} } \right),n - 1} \right)$(even if she does not know what $1 - \omega$ is). Also, it immediately follows from the proof of Proposition 5 that a high-value bidder’s equilibrium distribution of bids in the first auction should satisfy equation (12). The total expected payoff for a high-value bidder and the expected revenue for the seller can be similarly obtained. □

Proof of Proposition 8 See Kannan (2012) and we supplement the computation of the seller’s expected revenue. If $\theta \le \frac{1}{n + 1}$, then the allocation is efficient and the seller’s expected revenue is: $\Gamma^{{\text{C}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + \mathop \sum \limits_{j = 1}^{j = n} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} 2v_{H} - n\theta \tau^{{\text{C}}} = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} ).$ If $\theta > \frac{1}{n + 1}$, then the allocation of the first item is inefficient and the seller’s expected revenue is: $\Gamma^{{\text{C}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + \mathop \sum \limits_{j = 1}^{j = n - 1} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} \left( {\gamma^{j} \frac{n - j}{n}\left( {v_{L} + v_{H} } \right) + \left( {1 - \gamma^{j} \frac{n - j}{n}} \right)2v_{H} } \right) + C_{n}^{n} \theta^{n} 2v_{H} - n\theta \tau^{{\text{C}}} = 2v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \left( {1 - \theta + \theta \gamma } \right)^{n - 1} \left( {1 + \left( {n - 1} \right)\theta } \right)$. □

Proof of Proposition 9 (1) In discriminatory auctions when bidders have single-unit demand, the bidders with the top two bids win and respectively pay their own winning bids. Denote $s_{D}$ as the supremum of bids. Thus, a high-value bidder’s expected payoff from bidding $b_{1} = x$, where $x \in \left[ {v_{L} ,s_{D} } \right]$, is:

$$ \pi^{{\text{D}}} \left( x \right) = \left( {\left( {1 - \theta + \theta F^{{\text{D}}} \left( x \right)} \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - F^{{\text{D}}} \left( x \right)} \right)\left( {1 - \theta + \theta F^{{\text{D}}} \left( x \right)} \right)^{n - 2} } \right)\left( {v_{H} - x} \right) $$

From the property of the equilibrium in mixed strategies and combined with $F^{{\text{D}}} \left( {v_{L} } \right) = 0$, we have $\pi^{{\text{D}}} \left( x \right) = \pi^{{\text{D}}} \left( {v_{L} } \right) = \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right),\forall x \in \left[ {v_{L} ,s_{D} } \right]$. Accordingly, $F^{{\text{D}}} \left( x \right)$ should satisfy the following equation:

$$ \left( {\left( {1 - \theta + \theta F^{D} \left( x \right)} \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - F^{D} \left( x \right)} \right)\left( {1 - \theta + \theta F^{D} \left( x \right)} \right)^{n - 2} } \right)\left( {v_{H} - x} \right) = \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right),\;\;\forall x \in \left[ {v_{L} ,s_{D} } \right]. $$

Similar to the proof of Proposition 1, we get $s_{D} = v_{H} - \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right)$. Moreover, the seller’s expected revenue is $\Pi^{{\text{D}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + n\theta \left( {1 - \theta } \right)^{n - 1} \left( {v_{H} + v_{L} } \right) + \mathop \sum \limits_{j = 2}^{j = n} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} 2v_{H} - n\theta \pi^{{\text{D}}} = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \frac{{n\left( {n - 1} \right)}}{2}\theta^{2} \left( {1 - \theta } \right)^{n - 2} )$.

(2) In uniform-price auctions when bidders have single-unit demand, it is weakly dominant for a high-value bidder to bid $b_{1} = v_{H}$. Thus, a high-value bidder’s expected payoff $\pi^{{\text{U}}} = \left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} = \left( {1 - \theta } \right)^{n - 2} \left( {1 + \left( {n - 2} \right)\theta } \right).$ Since the allocation is efficient in both discriminatory auctions and uniform-price auctions and $\pi^{{\text{U}}} = \pi^{{\text{D}}}$, $\Pi^{{\text{U}}} = \Pi^{{\text{D}}}$ holds. □

Proof of Proposition 10 (1) Let us consider discriminatory auctions when bidders have multi-unit demand. In this case, given that other high-value bidders bid $b_{1}$ randomly according to $G^{{\text{D}}} \left( x \right) = \frac{1 - \theta }{\theta }\left( {\frac{1}{{(v_{H} - x)^{{\frac{1}{n - 1}}} }} - 1} \right),x \in \left[ {v_{L} ,v_{H} - \left( {1 - \theta } \right)^{n - 1} } \right]$ and bid $b_{2} = b_{1}$, we then consider the optimal decision of a specific high-value bidder. Her expected payoff from bidding $\left( {b_{1} ,b_{2} } \right)$, where $v_{L} \le b_{2} \le b_{1} \le v_{H} - \left( {1 - \theta } \right)^{n - 1}$, is: $\tau^{{\text{D}}} \left( {b_{1} ,b_{2} } \right) = \left( {1 - \theta + \theta G^{{\text{D}}} \left( {b_{2} } \right)} \right)^{n - 1} \left( {2v_{H} - b_{1} - b_{2} } \right) + \left( {\left( {1 - \theta + \theta G^{{\text{D}}} \left( {b_{1} } \right)} \right)^{n - 1} - \left( {1 - \theta + \theta G^{{\text{D}}} \left( {b_{2} } \right)} \right)^{n - 1} } \right)\left( {v_{H} - b_{1} } \right) = 2\left( {1 - \theta } \right)^{n - 1}$. Thus, the high-value bidder is indifferent between all bids over $\{ \left( {b_{1} ,b_{2} } \right)|v_{L} \le b_{2} \le b_{1} \le v_{H} - \left( {1 - \theta } \right)^{n - 1} \}$, and in particular, it is optimal to bid $b_{1}$ according to $G^{{\text{D}}} \left( x \right)$ and to bid $b_{2} = b_{1}$. Accordingly, a high-value bidder’s expected payoff is $\tau^{{\text{D}}} = 2\left( {1 - \theta } \right)^{n - 1}$. The seller’s expected revenue is $\Gamma^{{\text{D}}} = \left( {1 - \theta } \right)^{n} 2v_{L} + + \mathop \sum \limits_{j = 1}^{j = n} C_{n}^{j} \theta^{j} \left( {1 - \theta } \right)^{n - j} 2v_{H} - n\theta \tau^{{\text{D}}} = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} )$.

(2) In uniform-price auctions when bidders have multi-unit demand, it is weakly dominant for a high-value bidder to bid $b_{1} = v_{H}$. Next, we show that for every $v^{*} \in \left[ {v_{L}^{ + } ,v_{H} } \right]$, a high-value bidder bids $\left( {b_{1} ,b_{2} } \right) = \left( {v_{H} ,v^{*} } \right)$ is an equilibrium. Suppose that other high-value opponents bid $\left( {b_{1} ,b_{2} } \right) = \left( {v_{H} ,v^{*} } \right)$, then a specific high-value bidder’s expected payoff from bidding $\left( {v_{H} ,b_{2} } \right)$, where $v_{L} < b_{2} \le v_{H}$, is: $\tau^{{\text{U}}} \left( {b_{2} } \right) = 2\left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {\left( {v_{H} - v^{*} } \right)1_{{\left\{ {b_{2} \le v^{*} } \right\}}} + \left( {v_{H} - b_{2} } \right)1_{{\left\{ {b_{2} > v^{*} } \right\}}} } \right) \le \tau^{U} \left( {v^{*} } \right) = 2\left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {v_{H} - v^{*} } \right)$. Thus, it is optimal for her to bid $\left( {v_{H} ,v^{*} } \right)$. Accordingly, for every $v^{*} \in \left[ {v_{L}^{ + } ,v_{H} } \right]$, a high-value bidder bids $\left( {b_{1} ,b_{2} } \right) = \left( {v_{H} ,v^{*} } \right)$ is an equilibrium. In the equilibrium of $\left( {v_{H} ,v^{*} } \right)$, a high-value bidder’s expected payoff is $\tau^{{\text{U}}} \left( {v^{*} } \right) = 2\left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {v_{H} - v^{*} } \right)$, and the seller’s expected revenue is $\Gamma^{{\text{U}}} \left( {v^{*} } \right) = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \frac{{n\left( {n - 1} \right)}}{2}\theta^{2} \left( {1 - \theta } \right)^{n - 2} \left( {v_{H} - v^{*} } \right))$. □

Proof of Proposition 11 In the case of single-unit demand and from Propositions 1 and 3, the equilibrium distribution of bids in the first of sequential auctions satisfies $F_{1}^{{\text{I}}} \left( x \right) = F_{1}^{{\text{H}}} \left( x \right)$. When the equilibrium bid in the second of sequential auctions with the hidden information policy is given by $b_{k}^{2} = \beta \left( {F_{1}^{{\text{H}}} \left( {b_{k}^{1} } \right),n - 2} \right), k \in \Delta$, its distribution conditional on $b_{k}^{1} < b_{w}$ is $F_{2}^{{\text{I}}} \left( x \right)$. This result follows from the proof of Proposition 3. Therefore, the equilibrium distribution of bids in the second of sequential auctions under the two information policies is also identical. Accordingly, the weak strategic equivalence holds. In the case of multi-unit demand and from Propositions 5 and 7, we have $G_{1}^{{\text{I}}} \left( x \right) = G_{1}^{{\text{H}}} \left( x \right)$, $P_{r} \left\{ {\beta \left( {G_{1}^{{\text{H}}} \left( {b_{k}^{1} } \right),n - 1} \right) \le x{|}b_{k}^{1} < b_{w} } \right\} = G_{2L}^{{\text{I}}} \left( x \right)$, and $G_{2W}^{{\text{I}}} \left( x \right) = G_{2W}^{{\text{H}}} \left( x \right)$. Therefore, the weak strategic equivalence also holds. □

Proof of Proposition 12 (1) In single-unit demand cases, we obtain that $F_{1}^{{\text{I}}} \left( x \right) = F_{1}^{{\text{H}}} \left( x \right),x \in \left[ {v_{L} , s_{1}^{{\text{I}}} } \right]$ satisfy Equation (7) and $F_{1}^{{\text{C}}} \left( x \right),x \in \left[ {v_{L} , s_{1}^{{\text{C}}} } \right]$ satisfies Equation (9). Notice that $F_{1}^{{\text{C}}} \left( {v_{L} } \right) = p > 0 = F_{1}^{{\text{I}}} \left( {v_{L} } \right)$, and if there exists a $x_{n}^{\theta } > v_{L}$ satisfies $F_{1}^{{\text{I}}} \left( {x_{n}^{\theta } } \right) > F_{1}^{{\text{C}}} \left( {x_{n}^{\theta } } \right)$, it then follows that there exists a $y_{n}^{\theta } \in \left( {v_{L} ,x_{n}^{\theta } } \right)$ such that $F_{1}^{{\text{I}}} \left( {y_{n}^{\theta } } \right) = F_{1}^{{\text{C}}} \left( {y_{n}^{\theta } } \right)$. By substituting $y_{n}^{\theta }$ into Equations (7) and (9), we get $p = 0$, which is a contradiction. Therefore, $F_{1}^{{\text{C}}} \left( x \right) \ge F_{1}^{{\text{I}}} \left( x \right), \forall x \ge v_{L}$.

(2) In multi-unit demand cases, we know that $G_{1}^{{\text{I}}} \left( x \right) = G_{1}^{{\text{H}}} \left( x \right), x \in \left[ {v_{L} , t_{1}^{{\text{I}}} } \right]$ satisfy Equation (12). When $\theta \in \left( {0,\frac{1}{n + 1}} \right]$, it follows from Proposition 8 that $G_{1}^{{\text{C}}} \left( x \right) = \frac{1 - \theta }{\theta }\left( {\frac{1}{{\left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }} - 1} \right), x \in \left[ {v_{L} ,v_{H} - \left( {1 - \theta } \right)^{n - 1} } \right]$. Combined with Equation (12), we have that $\forall x \in \left[ {v_{L} , t_{1}^{{\text{I}}} } \right]$, $\left( {\left( {1 - \theta + \theta G_{1}^{{\text{I}}} \left( x \right)} \right)^{n - 1} - \left( {1 - \theta + \theta G_{1}^{{\text{C}}} \left( x \right)} \right)^{n - 1} } \right)\left( {v_{H} - x} \right) = \left( {n - 1} \right)\left( {1 - \theta } \right)^{n - 1} \left( {\log \left( {1 - \theta + \theta G_{1}^{{\text{I}}} \left( x \right)} \right) - \log \left( {1 - \theta } \right)} \right) \ge 0$. Thus $G_{1}^{{\text{I}}} \left( x \right) \ge G_{1}^{{\text{C}}} \left( x \right)$. Besides, $\forall x \in \left[ {t_{1}^{{\text{I}}} ,v_{H} - \left( {1 - \theta } \right)^{n - 1} } \right]$, $G_{1}^{{\text{I}}} \left( x \right) = 1 \ge G_{1}^{{\text{C}}} \left( x \right)$. Therefore, $\forall x \ge v_{L} , G_{1}^{{\text{I}}} \left( x \right) \ge G_{1}^{{\text{C}}} \left( x \right)$ holds.

When $\theta \in \left( {\frac{1}{n + 1},1} \right),$ we have $G_{1}^{{\text{C}}} \left( x \right) = \frac{1 - \theta + \theta \gamma }{{\theta \left( {v_{H} - x} \right)^{{\frac{1}{n - 1}}} }} - \frac{1 - \theta }{\theta }, x \in \left[ {v_{L} ,v_{H} - \left( {1 - \theta + \theta \gamma } \right)^{n - 1} } \right]$ and $\gamma$ satisfies $\left( {1 - \theta + \theta \gamma } \right)^{n} = n\theta \left( {1 - \theta } \right)^{n - 1} \left( {1 - \gamma } \right)$. Notice that $G_{1}^{{\text{C}}} \left( {v_{L} } \right) = \gamma > { }G_{1}^{{\text{I}}} \left( {v_{L} } \right) = 0$ and $G_{1}^{{\text{I}}} \left( {t_{1}^{{\text{I}}} } \right) = 1 > G_{1}^{{\text{C}}} \left( {t_{1}^{{\text{I}}} } \right)$. Thus, there exists a $x_{n}^{\theta } \in \left( {v_{L} ,t_{1}^{{\text{I}}} } \right)$ such that $G_{1}^{{\text{C}}} \left( {x_{n}^{\theta } } \right) = G_{1}^{{\text{I}}} \left( {x_{n}^{\theta } } \right)$. From Equation (12), we know that $\forall x \in \left[ {v_{L} , t_{1}^{{\text{I}}} } \right]$, $\left( {\left( {1 - \theta + \theta G_{1}^{{\text{I}}} \left( x \right)} \right)^{n - 1} - \left( {1 - \theta + \theta G_{1}^{{\text{C}}} \left( x \right)} \right)^{n - 1} } \right)\left( {v_{H} - x} \right) + \left( {1 - \theta } \right)^{n - 1} \left( {1 - \left( {n - 1} \right)\log \left( {1 - \theta + \theta G_{1}^{{\text{I}}} \left( x \right)} \right)} \right) = \left( {1 - \theta } \right)^{n - 1} \left( {2 - \left( {n - 1} \right)\log \left( {1 - \theta } \right)} \right) - \left( {1 - \theta + \theta \gamma } \right)^{n - 1}$. Clearly, the right side is independent of x. Thus, if there is a $y_{n}^{\theta } \in \left( {v_{L} ,t_{1}^{{\text{I}}} } \right)$ that also satisfies $G_{1}^{{\text{C}}} \left( {y_{n}^{\theta } } \right) = G_{1}^{{\text{I}}} \left( {y_{n}^{\theta } } \right)$, then it holds that $G_{1}^{{\text{I}}} \left( {y_{n}^{\theta } } \right) = G_{1}^{{\text{I}}} \left( {x_{n}^{\theta } } \right)$ and $y_{n}^{\theta } = x_{n}^{\theta }$. Accordingly, there exists a unique $x_{n}^{\theta } \in \left( {v_{L} ,t_{1}^{{\text{I}}} } \right)$ such that $G_{1}^{{\text{C}}} \left( x \right) \ge { }G_{1}^{{\text{I}}} \left( x \right)$ for $x \in \left[ {v_{L} ,x_{n}^{\theta } } \right]$ and $G_{1}^{{\text{I}}} \left( x \right) \ge G_{1}^{{\text{C}}} \left( x \right)$ for $x > x_{n}^{\theta }$. □

Proof of Proposition 13 (1) In single-unit demand cases and from Propositions 1, 3, 4, and 8, we have $\Pi^{{\text{I}}} = \Pi^{{\text{H}}} = \Pi^{{\text{D}}} = \Pi^{{\text{U}}} = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \frac{{n\left( {n - 1} \right)}}{2}\theta^{2} \left( {1 - \theta } \right)^{n - 2} )$ and $\Pi^{{\text{c}}} = 2v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} + \left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {p\left( {1 + \left( {n - 1} \right)\theta } \right) - n\theta } \right) - \left( {1 - \theta + \theta p} \right)^{n - 1} \left( {1 + \left( {n - 1} \right)\theta } \right)$, where $0 < p < 1$. Since $\frac{{\partial \Pi^{{\text{c}}} }}{\partial p} = \left( {n - 1} \right)\theta \left( {1 + \left( {n - 1} \right)\theta } \right)\left( {\left( {1 - \theta } \right)^{n - 2} - \left( {1 - \theta + \theta p} \right)^{n - 2} } \right)$, thus $\frac{{\partial \Pi^{{\text{c}}} }}{\partial p} < 0, \forall p \in \left( {0,1} \right)$. Therefore, $\Pi^{{\text{C}}} < \mathop {\lim }\limits_{{p \to 0^{ + } }} \Pi^{{\text{C}}} = \Pi^{{\text{I}}}$.

(2) In multi-unit demand cases and from Propositions 5, 7, and 9, we have $\Gamma^{{\text{I}}} = \Gamma^{{\text{H}}} = 2v_{H} - 2\left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} \left( {2 - \left( {n - 1} \right)\log \left( {1 - \theta } \right)} \right), \Gamma_{max}^{{\text{U}}} = \Gamma^{{\text{D}}} = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} )$ and $\Gamma_{min}^{{\text{U}}} = 2(v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \frac{{n\left( {n - 1} \right)}}{2}\theta^{2} \left( {1 - \theta } \right)^{n - 2} )$. Besides, from Proposition 8, we get

$$ \Gamma^{{\text{C}}} = \left\{ {\begin{array}{*{20}l} {2\left( {v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} } \right), } \hfill & {{\text{if}}\;\theta \le \frac{1}{n + 1}} \hfill \\ {2v_{H} - \left( {1 - \theta } \right)^{n} - n\theta \left( {1 - \theta } \right)^{n - 1} - \left( {1 - \theta + \theta \gamma } \right)^{n - 1} \left( {1 + \left( {n - 1} \right)\theta } \right),} \hfill & {{\text{if}}\;\theta > \frac{1}{n + 1}} \hfill \\ \end{array} } \right., $$

(13)

where $\gamma \in \left( {0,1} \right)$ and $\gamma$ satisfies $\left( {1 - \theta + \theta \gamma } \right)^{n} = n\theta \left( {1 - \theta } \right)^{n - 1} \left( {1 - \gamma } \right)$.

We then compare the seller’s expected revenue. Notice that $\forall \theta \in \left( {0,1} \right), \Gamma^{{\text{I}}} < {\Gamma }^{{\text{D}}}$ and $\Gamma_{min}^{{\text{U}}} < \Gamma^{{\text{D}}} = \Gamma_{max}^{{\text{U}}}$ hold. Besides, $\Gamma^{{\text{C}}} = \Gamma^{{\text{D}}}$ for $\theta \in \left( {0,\frac{1}{n + 1}} \right]$ and $\Gamma^{{\text{C}}} \le \Gamma^{{\text{D}}}$ for $\theta \in \left( {\frac{1}{n + 1},1} \right)$. Then, we compare the undominant terms of $\Gamma^{{\text{I}}} , \Gamma_{min}^{{\text{U}}} ,$ and $\Gamma^{{\text{C}}}$. On the one hand, $\forall \theta \in \left( {0,1} \right), \Gamma^{{\text{I}}} - \Gamma_{min}^{{\text{U}}} = n\left( {n - 1} \right)\theta \left( {1 - \theta } \right)^{n - 2} \left( {\left( {1 - \theta } \right)\log \left( {1 - \theta } \right) + \theta } \right) > 0.$ On the other hand, $\Gamma^{{\text{C}}} - \Gamma_{min}^{{\text{U}}} = \left( {1 + \left( {n - 1} \right)\theta } \right)\left( {\left( {1 - \theta } \right)^{n - 1} - \left( {1 - \theta + \theta \gamma } \right)^{n - 1} } \right) + n\left( {n - 1} \right)\theta^{2} \left( {1 - \theta } \right)^{n - 2}$. For $\theta \in \left( {0,\frac{1}{n + 1}} \right]$, $\gamma = 0$ holds. Thus, $\Gamma^{{\text{C}}} - \Gamma_{min}^{{\text{U}}} > 0$. For $\theta \in \left( {\frac{1}{n + 1},1} \right)$, $\gamma$ satisfies equation (13). Denote $f\left( \theta \right) = \frac{{\Gamma^{{\text{C}}} - \Gamma_{min}^{{\text{U}}} }}{{\left( {1 - \theta } \right)^{n - 2} }} = \left( {1 + \left( {n - 1} \right)\theta } \right)\left( {1 - \theta } \right)\left( {1 - \left( {\frac{{n\theta \left( {1 - \gamma } \right)}}{1 - \theta }} \right)^{{\frac{n - 1}{n}}} } \right) + n\left( {n - 1} \right)\theta^{2} \ge \left( {1 + \left( {n - 1} \right)\theta } \right)\left( {1 - \theta } \right)\left( {1 - \left( {\frac{n\theta }{{1 - \theta }}} \right)^{{\frac{n - 1}{n}}} } \right) + n\left( {n - 1} \right)\theta^{2} > f\left( {\frac{1}{n + 1}} \right) = \frac{{n\left( {3n - 1} \right)}}{{\left( {n + 1} \right)^{2} }} > 0$. Thus,${ }\forall \theta \in \left( {0,1} \right)$, $\Gamma^{{\text{C}}} - \Gamma_{min}^{{\text{U}}} > 0$ holds. Moreover, the comparison of $\Gamma^{{\text{I}}}$ and $\Gamma^{{\text{C}}}$ follows from Kannan (2012). □