A dynamic model of bidding patterns in sponsored search auctions

Abstract

We develop an infinite horizon alternative-move model of the unique second-price sponsored search auction. We use this model to explain two distinguishable bidding patterns observed in our bidding data: bidding war cycle and stable bid. With examples, we show that only a small portion of the value generated may be extracted by search engines, if advertisers are engaged in bidding war cycles. Finally, we show the impact of auction design on advertiser bids and search engine revenue.

Appendices

Appendix A

Proof of Proposition 2

Following Maskin and Tirole [12] and Eckert [4] the proof proceeds as follows. First, we define the parameters $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ and $\bar{b}$ and present the conditions that they must satisfy; then we show that the mixing probabilities μ( · ), θ, ρ are well defined given $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ and $\bar{b}.$ And then, we show that no profitable one-period deviation exist for X. In other words, if X chooses a different bid than the one specified in the equilibrium strategy for one period and then follow the equilibrium strategy, this deviation must not be more profitable than its payoff when she follows the equilibrium strategy without any deviations. Since we consider stationary Markov strategies, Maskin and Tirole [11] Proposition 1 implies that this set of conditions is sufficient for proving the symmetric strategy profile in (6) forms an MPE.

1.1 Expected continuation payoffs

We define three value functions to help with the exposition of the proof. Recall that there are two phases of the BWC equilibrium. In the bidding war phase, advertisers outbid each other to get the first position. In the ensuing war of attrition phase, each advertiser outbid the opponent with a probability less than one hoping the opponent would relent first. When the opponent relents, the advertiser enjoys the first position with the lowest payment (l + k) for two periods.

Let V ^relent($\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$) be X’s expected discounted total future payoff after X relents to l from state $\left( { \cdot ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right),$

$$ \begin{aligned} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = & \Uppi_{2} + \delta V^{X,Y} \left( {l,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)\quad {\text{and}} \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = & {\frac{{\Uppi^{X} \left( {l,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi^{X} \left( {l,l + k} \right) + \delta^{2} \Uppi^{X} \left( {l + 2k,l + k} \right) + \cdots + \delta^{2t + 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)}}{{1 - \delta^{2t + 2} }}}, \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = & {\frac{{\Uppi_{2} + \delta \Uppi_{2} + \delta^{2} \Uppi_{1} \left( {l + 2k} \right) + \cdots + \delta^{2t + 1} \Uppi_{2} }}{{1 - \delta^{2t + 2} }}}. \\ \end{aligned} $$

(A1)

for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} = l + \left( {2t + 1} \right)k$ with t a positive integer.⁹ $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ represents X’s expected continuation payoff from relenting. Next, we define the continuation payoff when an advertiser fails at forcing the opponent to relent. That is, $V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ represents X’s expected continuation payoff when the opponent Y outbids X’s bid b ^X, and then X relents.

$$ \begin{gathered} V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi^{X} \left( {b^{X} ,b^{X} + k} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi_{2} + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ \end{gathered} $$

(A2)

Similarly, we now define the continuation payoff when an advertiser is successful at forcing the opponent to relent. Accordingly, $V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ is the X’s expected continuation payoff when Y relents against X’s bid b ^X.

$$ \begin{gathered} V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi^{X} \left( {b^{X} ,l} \right) + \delta \Uppi^{X} \left( {l + k,l} \right) + \delta^{2} V^{X,Y} \left( {l + k,l} \right) \hfill \\ V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi_{1} \left( {l + k} \right) + \delta \Uppi_{1} \left( {l + k} \right) + \delta^{2} V^{X,Y} \left( {l + k,l} \right) \hfill \\ \end{gathered} $$

(A3)

for b ^X > l. Note that $V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$ The expected continuation payoff functions $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right),\,V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right),\,V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ depend on $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ because a different parameter value represents a different BWC. For example, $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + 2k} \right)$ represents the continuation value from relenting in a cycle where the bidding war phase continues until the bid $(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + 2k) \, + k.$

1.2 Definition of $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$

$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$is the bid that marks the end of the bidding war phase. $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ is defined by the following conditions.¹⁰

• 1. X weakly prefers {outbidding $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k$ and then failing} to relenting. Formally,

$$ \begin{gathered} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} + \delta^{2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} }}{{1 - \delta^{2} }}} \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

• 2. X prefers relenting to {outbidding $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ and then failing}. Formally,

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta \Uppi_{2} + \delta^{2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta \Uppi_{2} }}{{1 - \delta^{2} }}}. \hfill \\ \end{gathered} $$

Putting both of the conditions together produces the equilibrium condition (A4).

$$ {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta \Uppi_{2} }}{{1 - \delta^{2} }}} < V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \le {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} }}{{1 - \delta^{2} }}} .$$

(A4)

1.3 Definition of $\bar{b}$

$\bar{b}$is the bid that marks the end of the war of attrition phase. $\bar{b}$ is defined by the following conditions.¹¹

• 1. X weakly prefers {outbidding $\bar{b}$ and then succeeding} to relenting. Formally, $\Uppi^{X} \left( {\bar{b} + k,\bar{b}} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$

• 2. X prefers relenting to {outbidding $\bar{b} + k$ and then succeeding}. Formally, $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi^{X} \left( {\bar{b} + 2k,\bar{b} + k} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$

Putting these conditions together produces the following condition that defines $\bar{b}$

$$ \begin{gathered} \Uppi^{X} \left( {\bar{b} + k,\bar{b}} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi^{X} \left( {\bar{b} + 2k,\bar{b} + k} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right), \hfill \\ \Uppi_{1} \left( {\bar{b} + k} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi_{1} \left( {\bar{b} + 2k} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

(A5)

1.4 War of attrition

1.4.1 Definition of μ(b)

In the war of attrition phase, in which $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{Y} \le \bar{b},$ X and Y outbid each other with some positive probability hoping the opponent will relent. For X to use a mixed strategy, X and Y must be indifferent between continuing the bidding war and relenting throughout the war of attrition phase. Thus, for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b \le \bar{b},$ μ(b) satisfies:

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi^{Y} \left( {b - k,b} \right) + \mu \left( b \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( b \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right), \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi_{1} \left( b \right) + \mu \left( b \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( b \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

(A6)

(A6) indicates that X bids b + k at state (b−k, b) with probability μ(b) to make Y indifferent between continuing the war of attrition and relenting at state (b−k, b−2 k). Next, we show that the probability of continuing the war of attrition, μ(b) for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b \le \bar{b},$ is well-defined, i.e. 0 ≤ μ(b) ≤ 1, due to the conditions (A4) and (A5).

When $\mu \left( b \right) = 1$ for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b \le \bar{b},$ we must have $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge \Uppi_{1} \left( b \right) + \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$ This holds because in deriving (A4), we use the condition $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$ This condition and $\Uppi_{1} (b) < \Uppi_{1} (\hat{b})$ for $b > \hat{b}$ imply that $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > \Uppi_{1} \left( b \right) + \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$

When μ(b) = 0 for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b \le \bar{b},$ we must have $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \le \Uppi_{1} \left( b \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$ This follows from the left-hand side of (A5), $\Uppi_{1} \left( {\bar{b} + k} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$, and $\Uppi_{1} (b) < \Uppi_{1} (\hat{b})$ for $b > \hat{b}.$

In order to show that μ(b) decreases linearly with b, consider μ(b + k), from (A6) it satisfies

$$ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi_{1} \left( {b + k} \right) + \mu \left( {b + k} \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( {b + k} \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) $$

Subtracting this from (A6), we obtain

$$ \begin{gathered} \Uppi_{1} \left( b \right) - \Uppi_{1} \left( {b + k} \right) + \left\{ {\mu \left( b \right) - \mu \left( {b + k} \right)} \right\}\delta \left\{ {V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right\} = 0, \hfill \\ \mu \left( {b + k} \right) - \mu \left( b \right) = {\frac{{\Uppi_{1} \left( {b + k} \right) - \Uppi_{1} \left( b \right)}}{{\delta \left\{ {V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right\}}}} = {\frac{{ - \alpha_{1} k}}{{\delta \left\{ {V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right\}}}}. \hfill \\ \end{gathered} $$

(A7)

Because $V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) > V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ and $V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ and $V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ are independent of b, μ(b) decreases linearly with b.

1.4.2 Definition of ρ

In the event that Y’s bid reaches $\bar{b} + k$, X has no incentive to outbid Y. However, X cannot relent with probability one, since Y would always bid $\bar{b} + k$ and always occupy the first position. In fact, relenting with probability one can never be part of a bidding war cycle equilibrium. Therefore, X uses a mixed strategy between bidding $\bar{b}$ and relenting to l; X bids $\bar{b}$ at state $\left( {\bar{b},\bar{b} + k} \right)$with probability ρ and l with probability 1 − ρ to make Y indifferent between choosing $\bar{b} + k$ and l (relenting) at states $\left( {\bar{b},\bar{b} - k} \right)$ and $\left( {\bar{b},b^{Y} } \right)$ for $b^{Y} \ge \bar{b} + k.$ Hence, ρ satisfies:

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi^{Y} \left( {\bar{b},\bar{b} + k} \right) + \rho \delta \left( {\Uppi^{Y} \left( {\bar{b},\bar{b} + k} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) + \left( {1 - \rho } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right), \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi_{1} \left( {\bar{b} + k} \right) + \rho \delta \left( {\Uppi_{1} \left( {\bar{b} + k} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) + \left( {1 - \rho } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

(A8)

For ρ to be well-defined, we must have $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \le \Uppi_{1} \left( {\bar{b} + k} \right) + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right),$ when ρ = 0. This directly follows from the left-hand side of (A5). Likewise, when ρ = 1, we must have

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge \Uppi_{1} \left( {\bar{b} + k} \right) + \delta \Uppi_{1} \left( {\bar{b} + k} \right) + \delta^{2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right), \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge {\frac{{\Uppi_{1} \left( {\bar{b} + k} \right) + \delta \Uppi_{1} \left( {\bar{b} + k} \right).}}{{1 - \delta^{2} }}} \hfill \\ \end{gathered} $$

Since $\bar{b} + k \ge \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k$ and from the left-hand side of (A4), this condition is guaranteed to hold, if

$$ \Uppi_{1} \left( {\bar{b} + k} \right) \le \Uppi_{2} $$

Because $\Uppi_{1} \left( {b^{*} + k} \right) < \Uppi_{2} ,$ a sufficient condition for this inequality is

$$ \begin{gathered} \bar{b} + k \ge b^{*} + k \hfill \\ \bar{b} \ge b^{*} \hfill \\ \end{gathered} $$

(A9)

1.4.3 Definition of θ

For an advertiser to mix between $\bar{b}$ and l, she must be indifferent between these two bids. As a result, we need another mixed strategy for establishing this indifference. X bids $\bar{b} + k$ at states $\left( {b^{X} ,\bar{b}} \right)$ for $b^{X} \ge \bar{b} + k$ with probability θ and l with probability 1−θ to make Y indifferent between choosing $\bar{b}$ and l at state $\left( {\bar{b} + k,\bar{b}} \right).$ Hence, θ satisfies:

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi^{Y} \left( {\bar{b} + k,\bar{b}} \right) + \theta \delta \left( {\Uppi^{Y} \left( {\bar{b} + k,\bar{b}} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) + \left( {1 - \theta } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right), \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \Uppi_{2} + \theta \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \theta } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

(A10)

For θ to be well-defined, we must have $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \le \Uppi_{2} + \delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right),$ when θ = 0. This is guaranteed to hold, if (A9) holds, due to the left-hand side of (A5).

Likewise, when θ = 1, we must have

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge \Uppi_{2} + \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right), \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge {\frac{{\Uppi_{2} + \delta \Uppi_{2} }}{{1 - \delta^{2} }}}. \hfill \\ \end{gathered} $$

From the left-hand side of (A4), this is guaranteed to hold, if

$$ \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) \ge \Uppi_{2} $$

Because $\Uppi_{1} \left( {b^{*} } \right) \ge \Uppi_{2} ,$ a sufficient condition for this inequality is

$$ \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{*} . $$

(A11)

Putting (A9) and (A11) together produces the sufficient condition in Proposition 2

$$ \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{*} \le \bar{b}. $$

We have shown that given $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{*} \le \bar{b},$ all the parameters in Proposition 2 are well defined.

1.5 One-period deviation conditions

In this section, we prove that there is no one-period profitable deviation and thus the symmetric strategy profile of Proposition 2 is indeed a Markov perfect equilibrium (Maskin and Tirole [11]). We consider each possible deviation and show that this deviation is not profitable as a response to any $b^{Y}$. The static symmetric reaction function (6) consists of two types of bids: the bids on the equilibrium path and the bids that return the system to the equilibrium path from off-equilibrium states. In Lemma A1, we show that all the bids that return the system to the equilibrium path from off-equilibrium states are equivalent to relenting.

Lemma A1

The expected continuation payoffs from the following bids at the corresponding states are equal to the expected continuation payoff from relenting to l, $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$

$$ R^{X} \left( {b^{X} ,b^{Y} } \right) = \left\{ {\begin{array}{*{20}c} {b^{Y} + k\quad \quad \quad {\text{for }}b^{Y} = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,} \hfill \\ {b^{Y} + k\quad \quad \quad {\text{for }}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{Y} < \bar{b}{\text{ and }}b^{X} \ne b^{Y} - k,} \hfill \\ {\bar{b} + k\quad \quad \quad \;{\text{for }}b^{Y} = \bar{b}{\text{ and }}b^{X} \le \bar{b} - 2k ,} \hfill \\ {\bar{b}\;\;\,{\text{ for }}b^{Y} \ge \bar{b} + k{\text{ and }}b^{X} \ne \bar{b}.} \hfill \\ \end{array} } \right. $$

Proof of Lemma A1

We prove the claim for each case one by one. The expected continuation payoff from $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k$ for $b^{Y} = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ is

$$ \begin{gathered} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \mu \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \hfill \\ \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \mu \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

from (A6).

The expected continuation payoff from $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k$ for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{Y} < \bar{b}{\text{ and }}b^{X} \ne b^{Y} - k$ is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + k,b^{Y} } \right) + \mu \left( {b^{Y} + k} \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( {b^{Y} + k} \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \hfill \\ \Uppi_{1} \left( {b^{Y} + k} \right) + \mu \left( {b^{Y} + k} \right)\delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \mu \left( {b^{Y} + k} \right)} \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

from (A6).

The expected continuation payoff from $R^{X} \left( {b^{X} ,b^{Y} } \right) = \bar{b} + k$ for $b^{Y} = \bar{b}{\text{ and }}b^{X} \le \bar{b} - 2k$ is

$$ \begin{gathered} \Uppi^{X} \left( {\bar{b} + k,\bar{b}} \right) + \rho \delta \left( {\Uppi^{X} \left( {\bar{b} + k,\bar{b}} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) + \left( {1 - \rho } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \hfill \\ \Uppi_{1} \left( {\bar{b} + k} \right) + \rho \delta \left( {\Uppi_{1} \left( {\bar{b} + k} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) + \left( {1 - \rho } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

from (A8).

The expected continuation payoff from $R^{X} \left( {b^{X} ,b^{Y} } \right) = \bar{b}$ for $b^{Y} \ge \bar{b} + k{\text{ and }}b^{X} \ne \bar{b}$ is

$$ \begin{gathered} \Uppi^{X} \left( {\bar{b},\bar{b} + k} \right) + \theta \delta \left( {\Uppi^{X} \left( {\bar{b},\bar{b} + k} \right) + \delta V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) + \left( {1 - \theta } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \hfill \\ \Uppi_{2} + \theta \delta V^{fail} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \left( {1 - \theta } \right)\delta V^{success} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

from (A10). Q.E.D.

We proceed by showing that when an advertiser relents, she relents to l not to a higher bid. We use $\tilde{R}^{X}$ to represent deviation bids.

Lemma A2

$R^{X} \left( {b^{X} ,b^{Y} } \right) = l$ for $b^{Y} \ge \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k$.

Proof of Lemma A2

There are four possible deviations.

Deviation A2.1

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b$ where $l < b \le \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k.$We, first, show that $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l + k$ is not a profitable deviation. Then, we show that $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l + k$ is a better response than $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l + 2k.$ Hence, the result follows by induction on k. The difference between the expected continuation payoff from relenting, $R^{X} \left( {b^{X} ,b^{Y} } \right) = l,$ and $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l + k$ is

$$ \begin{gathered} \Uppi^{X} \left( {l,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + {{\updelta}}\Uppi^{X} \left( {l,l + k} \right) + \cdots + \delta^{2t + 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \hfill \\ \left( {\Uppi^{X} \left( {l + k,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {l + k,l + 2k} \right) + \cdots + \delta^{2t} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t + 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta^{2t + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) = \hfill \\ \delta^{2} \left( {\Uppi_{1} \left( {l + 2k} \right) - \Uppi_{1} \left( {l + 3k} \right)} \right) + \cdots + \delta^{2t} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) - \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) > 0. \hfill \\ \end{gathered} $$

The difference between the expected continuation payoff from $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l + k$ and $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l + 2k$ is

$$ \begin{gathered} \Uppi^{X} \left( {l + k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + {{\updelta}}\Uppi^{X} \left( {l + k,l + 2k} \right) + \cdots + \delta^{2t} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t + 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta^{2t + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \hfill \\ \left( {\Uppi^{X} \left( {l + 2k,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {l + 2k,l + 3k} \right) + \cdots + \delta^{2t - 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) = \hfill \\ \delta^{2} \left( {\Uppi_{1} \left( {l + 3k} \right) - \Uppi_{1} \left( {l + 4k} \right)} \right) + \cdots + \hfill \\ \delta^{2t - 2} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) - \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right)} \right) + \delta^{2t} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + {{\updelta}}\Uppi_{2} - \left( {1 - \delta^{2} } \right)V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) > 0. \hfill \\ \end{gathered} $$

The inequality follows from the right-hand side of (A4).

Deviation A2.2

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b$ where $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k \le b \le b^{Y} - k.$The difference between the expected continuation payoff from relenting and this deviation is

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \left( {\Uppi^{X} \left( {b,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {b,b + k} \right) + \delta^{2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) = \hfill \\ \left( {1 - \delta^{2} } \right)V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \left( {\Uppi_{2} + {{\updelta}}\Uppi_{2} } \right) \ge 0. \hfill \\ \end{gathered} $$

The inequality follows from (A11).

Deviation A2.3

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b,$ where $b^{Y} + k < b \le \bar{b}.$The difference between the expected continuation payoff from relenting and this deviation is

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \left( {\Uppi^{X} \left( {b,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {b,b + k} \right) + \delta^{2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) = \hfill \\ \left( {1 - \delta^{2} } \right)V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \left( {\Uppi_{1} \left( {b^{Y} + k} \right) + {{\updelta}}\Uppi_{2} } \right) > 0. \hfill \\ \end{gathered} $$

This inequality follows from the left-hand side of (A4).

Deviation A2.4

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b$ where $b^{Y} + k < b$ and $b \ge \bar{b} + k.$The difference between the expected continuation payoff from relenting and this deviation is

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \left( {\Uppi^{X} \left( {b,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {b,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) = \hfill \\ \left( {1 - \delta^{2} } \right)V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - \left( {\Uppi_{1} \left( {b^{Y} + k} \right) + {{\updelta}}\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right)} \right) \ge 0. \hfill \\ \end{gathered} $$

This inequality follows from the left-hand side of (A4) and (A9). The result follows from Lemma A1, (A6), (A8), and (A10). Q.E.D.

We, now, show that there is no profitable one-period deviation from the bidding war phase of the equilibrium in Lemma A3.

Lemma A3

$R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k$ for $b^{Y} \le \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} .$

Proof of Lemma A3

There are four possible deviations. We need to consider two cases when $b^{Y} - l$ is an odd $(b^{Y} = l + (2r + 1)k)$ and even $(b^{Y} = l + 2rk)$ multiple of k. r is a positive integer in the rest of the proof for Deviations A3.1 and A3.2.

Deviation A3.1

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l.$ In Deviation A2.1 we show that if an advertiser drops her bid below the opponent’s bid, the best bid is l. Therefore, we only consider relenting as a deviation by bidding below Y’s bid.

Case A3.1.1

$$ b^{Y} = l + (2r + 1)k $$

For $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k$ (optimal reaction function), X’s expected continuation payoff is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + k,b^{Y} } \right) + \delta \, \Uppi^{X} \left( {b^{Y} + k,b^{Y} + 2k} \right) + \cdots \hfill \\ \, + \delta^{2t - 2r - 2} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) + \delta^{2t - 2r - 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t - 2r} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \hfill \\ \Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r - 1} \Uppi_{2} + \delta^{2t - 2r} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

The continuation payoff of $\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = l$ is $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right).$ Thus, the difference is

$$ \begin{gathered} \Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r - 1} \Uppi_{2} + \delta^{2t - 2r} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge 0 \hfill \\ {\frac{{\Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r - 1} \Uppi_{2} }}{{1 - \delta^{2t - 2r} }}} \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

This holds because

$$ \begin{gathered} {\frac{{\Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r - 1} \Uppi_{2} }}{{1 - \delta^{2t - 2r} }}} > {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t - 2r - 1} \Uppi_{2} }}{{1 - \delta^{2t - 2r} }}} \hfill \\ = {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} }}{{1 - \delta^{2} }}} \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

The last inequality follows from (A4).

Case A3.1.2

$$ b^{Y} = l + 2rk $$

For $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k,$ X’s expected continuation payoff is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + k,b^{Y} } \right) + \delta \, \Uppi^{X} \left( {b^{Y} + k,b^{Y} + 2k} \right) + \cdots \hfill \\ \, + \delta^{2t - 2r} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r + 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta^{2t - 2r + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = \hfill \\ \Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r + 1} \Uppi_{2} + \delta^{2t - 2r + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

The difference between this quantity and $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)$ is

$$ \begin{gathered} \Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r + 1} \Uppi_{2} + \delta^{2t - 2r + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \ge 0 \hfill \\ {\frac{{\Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r + 1} \Uppi_{2} }}{{\delta^{2t - 2r + 2} }}} \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

This holds, similarly from (A4)

$$ \begin{gathered} {\frac{{\Uppi_{1} \left( {b^{Y} + k} \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r + 1} \Uppi_{2} }}{{\delta^{2t - 2r + 2} }}} > {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} + \cdots + \delta^{2t - 2r} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t - 2r + 1} \Uppi_{2} }}{{\delta^{2t - 2r + 2} }}} \hfill \\ = {\frac{{\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta \Uppi_{2} }}{{1 - \delta^{2} }}} \ge V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

Deviation A3.2

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b$ where $b^{Y} + k < b < \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k.$This deviation stands for jump bidding – i.e. outbidding the rival more than k in the bidding war. We show that X chooses $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k$ rather than $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + 2k,$ then the result follows by induction.

Case A3.2.1

$$ b^{Y} = l + (2r + 1)k $$

For $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k,$ X’s expected continuation payoff is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + k,b^{Y} } \right) + \delta \, \Uppi^{X} \left( {b^{Y} + k,b^{Y} + 2k} \right) + \cdots \hfill \\ \, + \delta^{2t - 2r - 2} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) + \delta^{2t - 2r - 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t - 2r} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

For $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + 2k,$ X’s expected continuation payoff is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + 2k,b^{Y} } \right) + \delta \, \Uppi^{X} \left( {b^{Y} + 2k,b^{Y} + 3k} \right) + \cdots \hfill \\ \, + \delta^{2t - 2r - 2} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r - 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta^{2t - 2r} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) \hfill \\ \end{gathered} $$

The difference is

$$ \begin{gathered} \delta^{2} \left( {\Uppi_{1} \left( {b^{Y} + 3k} \right) - \Uppi_{1} \left( {b^{Y} + 4k} \right)} \right) + \delta^{4} \left( {\Uppi_{1} \left( {b^{Y} + 5k} \right) - \Uppi_{1} \left( {b^{Y} + 6k} \right)} \right) + \cdots + \hfill \\ \delta^{2t - 2r - 2} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) - \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right) > 0 \hfill \\ \end{gathered} $$

This inequality clearly holds.

Case A3.2.2

$$ b^{Y} = l + 2rk $$

For $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + k,$ X’s expected continuation payoff is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + k,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {b^{Y} + k,b^{Y} + 2k} \right) + \cdots \hfill \\ \, + \delta^{2t - 2r} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t - 2r + 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} ,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k} \right) + \delta^{2t - 2r + 2} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

For $R^{X} \left( {b^{X} ,b^{Y} } \right) = b^{Y} + 2k$, X’s expected continuation payoff is

$$ \begin{gathered} \Uppi^{X} \left( {b^{Y} + 2k,b^{Y} } \right) + {{\updelta}}\Uppi^{X} \left( {b^{Y} + 2k,b^{Y} + 3k} \right) + \cdots \hfill \\ \, + \delta^{2t - 2r - 2} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) + \delta^{2t - 2r - 1} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + \delta^{2t - 2r} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). \hfill \\ \end{gathered} $$

The difference is

$$ \begin{gathered} \delta^{2} \left( {\Uppi_{1} \left( {b^{Y} + 3k} \right) - \Uppi_{1} \left( {b^{Y} + 4k} \right)} \right) + \hfill \\ \delta^{4} \left( {\Uppi_{1} \left( {b^{Y} + 5k} \right) - \Uppi_{1} \left( {b^{Y} + 6k} \right)} \right) + \cdots + \delta^{2t - 2r} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) + {{\updelta}}\Uppi_{2} - \left( {1 - \delta^{2} } \right)V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right)} \right){ > 0} .\hfill \\ \end{gathered} $$

This inequality follows from the right-hand side of (A4).By induction, X does not choose $b = b^{Y} + jk < \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k,$j > 1 since $\Uppi_{1}$ is a decreasing function.

Deviation A3.3

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b$ where $b^{Y} + k < b$ and $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k \le b \le \bar{b}.$ This deviation is equivalent to Deviation A2.3.

Deviation A3.4

$\tilde{R}^{X} \left( {b^{X} ,b^{Y} } \right) = b$ where $b^{Y} + k < b$ and $b \ge \bar{b} + k.$ This deviation is equivalent to Deviation A2.4. Q.E.D.

We have shown that there is no profitable one-period deviation from the equilibrium reaction function. This completes the proof of Proposition 2. Q.E.D.

Lemma A4

Let $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} = l + \left( {2t + 1} \right)k,$ then $V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) \ge \, 0$ implies that

$$ \alpha_{1} k\left[ {2t + \delta^{4} \left( {2t - 4} \right) + \delta^{6} \left( {2t - 6} \right) + \cdots + \delta^{2t - 2} 2} \right] \le \alpha_{1} \left( {s_{1} - l} \right) - \alpha_{2} \left( {s_{2} - l} \right) $$

Proof of Lemma A4

Using (A1), for $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} = l + \left( {2t + 1} \right)k$ we can write:

$$ \begin{gathered} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) = {\frac{{\Uppi_{2} + \delta \Uppi_{2} + \delta^{2} \Uppi_{1} \left( {l + 2k} \right) + \cdots + \delta^{2t} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta^{2t + 1} \Uppi_{2} }}{{1 - \delta^{2t + 2} }}} \hfill \\ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) = {\frac{{\Uppi_{2} + \delta \Uppi_{2} + \delta^{2} \Uppi_{1} \left( {l + 2k} \right) + \cdots + \delta^{2t - 2} \Uppi^{X} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 3k} \right) + \delta^{2t - 1} \Uppi_{2} }}{{1 - \delta^{2t} }}} \hfill \\ \end{gathered} $$

Let $D = {\frac{1}{{1 - \delta^{2t + 2} }}}$ and $E = {\frac{1}{{1 - \delta^{2t} }}} > D,$ then we have

$$ \begin{aligned} V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) = & \left( {D - E} \right)\left( {\Uppi_{2} + \delta \Uppi_{2} + \delta^{2} \Uppi_{1} \left( {l + 2k} \right) + \cdots + \delta^{2t - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 3k} \right) + \delta^{2t - 1} \Uppi_{2} } \right) \\ \, & + D\delta^{2t} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta \Uppi_{2} } \right) \\ \end{aligned} $$

$V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) \ge 0$ implies

$$ \left( {\frac{E}{D} - 1} \right)\left( {\Uppi_{2} + \delta \Uppi_{2} + \delta^{2} \Uppi_{1} \left( {l + 2k} \right) + \cdots + \delta^{2t - 2} \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 3k} \right) + \delta^{2t - 1} \Uppi_{2} } \right) \le \delta^{2t} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) + \delta \Uppi_{2} } \right) $$

It is easy to show that

$$ \frac{E}{D} - 1 = {\frac{{\delta^{2t} (1 - \delta^{2} )}}{{1 - \delta^{2t} }}} = {\frac{{\delta^{2t} }}{{1 + \delta^{2} + \delta^{4} + \cdots + \delta^{2t - 2} }}} $$

From the last inequality and the last equation, we get

$$ \begin{gathered} \delta^{2} \left( {\Uppi_{1} \left( {l + 2k} \right) - \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right)} \right) + \delta^{4} \left( {\Uppi_{1} \left( {l + 4k} \right) - \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right)} \right) + \cdots + \hfill \\ \delta^{2t - 2} \left( {\Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 3k} \right) - \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right)} \right) \le \Uppi_{1} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - k} \right) - \Uppi_{2} . \hfill \\ \end{gathered} $$

Substituting in the one-period payoff function (2),

$$ \begin{gathered} \alpha_{1} k\left[ {\delta^{4} \left( {2t - 4} \right) + \delta^{6} \left( {2t - 6} \right) + \cdots + \delta^{2t - 2} 2} \right] \le \alpha_{1} \left( {s_{1} - (l + 2tk)} \right) - \alpha_{2} \left( {s_{2} - l} \right) \hfill \\ \alpha_{1} k\left[ {2t + \delta^{4} \left( {2t - 4} \right) + \delta^{6} \left( {2t - 6} \right) + \cdots + \delta^{2t - 2} 2} \right] \le \alpha_{1} \left( {s_{1} - l} \right) - \alpha_{2} \left( {s_{2} - l} \right). \hfill \\ \end{gathered} $$

(A12)

Q.E.D.

Proof of Corollary 1

First, we prove the result on $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} .$ Note that the minimum relenting bid $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ can also be characterized by the following inequalities:

$$ V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} - 2k} \right) \ge 0 > V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + 2k} \right) - V^{relent} \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} } \right). $$

(A13)

Let f (t) be the left-hand side of (A12) as a function of t and g (l) be the right-hand side of (A12) as a function of l. Note that f (t) is increasing in t and is g (l) decreasing in l, because α₁ > α₂.Let $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} = l^{L} + (2t^{L} + 1)k$ be the minimum relenting bid for l ^L, and $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}^{\prime} = l^{H} + (2t^{H} + 1)k$ be the minimum relenting bid for l ^H (l ^L < l ^H). Thus, we know that the following inequalities hold.

$$ f(t^{L} ) \le g(l^{L} );\,f(t^{L} + 1) > g(l^{L} )\,and\,(t^{H} ) \le g(l^{H} );f(t^{H} + 1) > g(l^{H} ) $$

from (A13). Because g(l) decreasing in l, $g(l^{H} ) < g(l^{L} ).$ Suppose t ^H ≥ t ^L +1, then it follows that

$$ f(t^{L} + 1) \le g(l^{H} ) < g(l^{L} ) \, \Rightarrow \, f(t^{L} + 1) < g(l^{L} ) $$

This contradicts with $f(t^{L} + 1) > g(l^{L} ).$ This implies that t ^H must be at most as large as t ^L, t ^H ≤ t ^L. Q.E.D.

Proof of Corollaries 2, 3 and 4

The same proof method used in Corollary 1’s proof easily applies to Corollaries 2, 3 and 4.

Appendix B

2.1 Calculation of the average first position bid on the BWC Equilibrium path

We define the following variables to help the exposition: any bid $b = l + nk$ and the equilibrium parameters $\bar{b} = l + \bar{n}k$ and $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} = l + \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{n} k.$ Moreover, we define the relenting bid, $b^{r} = l + n^{r} k,$ as the maximum value of bids that can be reached in a BWC. Note that this is a random variable.

We first characterize the distribution of the relenting bid and then calculate the payment per period for each realization of the relenting bid.

• 1. The discrete distribution of the relenting bid, $b^{r} ,$ is: $\Pr (b^{r} = b) = \left\{ {\begin{array}{*{20}c} \begin{gathered} 0\quad{\text{ for }}b < \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k{\text{ or }}b > \bar{b} + k, \hfill \\ (1 - \mu (b))\prod\limits_{i = 1}^{n - n - 1} {\mu \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + ik} \right)} {\text{ for }}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b \le \bar{b}, \hfill \\ \end{gathered} \\ {\prod\limits_{i = 1}^{{\bar{n} - n }} {\mu \left( {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + ik} \right)} {\text{ for }}b = \bar{b} + k} \\ \end{array} } \right.$

• 2. We define the function $S(b^{r} )$ as the first position bid function for each realization of $b^{r} .$

• a. If $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b^{r} \le \bar{b},$ then the average first position bid is $S(b) = {\frac{1}{{n^{r} + 1}}}\left( {l + k + \sum\limits_{i = 1}^{n} {l + ik} } \right){\text{ for }}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + k \le b \le \bar{b}$

• b. If $b^{r} = \bar{b} + k,$ then the average first position bid is:

• c.
  $$ \begin{gathered} (1 - \rho ){\frac{1}{{\bar{n} + 2}}}\left( {l + k + \sum\limits_{i = 1}^{{\bar{n} + 1}} {(l + ik)} } \right) + \rho (1 - \theta ){\frac{1}{{\bar{n} + 3}}}\left( {l + k + \sum\limits_{i = 1}^{{\bar{n} + 1}} {(l + ik)} + (\bar{b} + k)} \right) + \rho \theta (1 - \rho ){\frac{1}{{\bar{n} + 4}}}\left( {l + k + \sum\limits_{i = 1}^{{\bar{n} + 1}} {(l + ik)} + 2(\bar{b} + k)} \right) + \hfill \\ \rho^{2} \theta (1 - \theta ){\frac{1}{{\bar{n} + 5}}}\left( {l + k + \sum\limits_{i = 1}^{{\bar{n} + 1}} {(l + ik)} + 3(\bar{b} + k)} \right) + \cdots \hfill \\ \end{gathered} $$

Hence, the average first position bid can be written as:

$$ \begin{gathered} S(\bar{b} + k) = (1 - \rho ){\frac{1}{{\bar{n} + 2}}}\left( {l + k + \sum\limits_{i = 1}^{{\bar{n} + 1}} {(l + ik)} } \right) + \hfill \\ \, \sum\limits_{j = 1}^{\infty } {\rho^{{\left\lfloor {{\frac{j + 1}{2}}} \right\rfloor }} \theta^{{\left\lfloor {\frac{j}{2}} \right\rfloor }} (1 - \rho )^{{{\rm I}_{2} (j)}} (1 - \theta )^{{1 - {\rm I}_{2} (j)}} } {\frac{1}{{\bar{n} + 2 + j}}}\left( {l + k + \sum\limits_{i = 1}^{{\bar{n} + 1}} {(l + ik)} + j(\bar{b} + k)} \right) \hfill \\ \end{gathered} $$

where ${\rm I}_{2} (j) = \left\{ {\begin{array}{*{20}c} {1, \quad j{\text{ is an even number,}}} \\ {0, \quad j{\text{ is an odd number}} .} \\ \end{array} } \right.$

Since each individual term in the last summation expression tends to zero as j increase, we only calculate the sum of the first 10 terms in this expression. Finally, we can write the overall average first position bid as:

$$ S = \sum\limits_{i = 1}^{{\bar{n} - \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{n} + 1}} {S(\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + ik)} \Pr (b^{r} = \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b} + ik) $$