Creating a winner’s curse via jump bids

Abstract

We show that jump bids can be used by a bidder to create a winner’s curse and preserve an informational advantage that would otherwise disappear in the course of an open ascending auction. The effect of the winner’s curse is to create allocative distortions and reduce the seller’s expected revenue. Two novel features of equilibrium jump bids are derived. First, the jump bid may fail to hide completely the value of the common value component. Second, a bidder with a higher type might jump bid less frequently than a bidder with a lower type.

Appendix

1.1 Proof of Proposition 2

Consider the following strategies:

• Bidder 1. Never calls a price, stays active up to s (whether no price is called or a price lower than s is called), and leaves the auction if a price higher than s is called.

• Bidder 2. Always calls price $p=1$ and stays active up to $s+t_2$ afterwards. If a bidder calls a price higher than $p=1$, stays active up to $s+t_2$.

• Bidder 3. Never calls a price. If no price is called, leaves the auction at a price equal to $q+t_3$, q being the price at which Bidder 1 leaves the auction if it is in the interval [0, 1]. If Bidder 1 does not leave the auction at a price lower than 1, Bidder 3 stays active up to $1+t_3$.

  If a price p is called in the first stage:

  • (a) If Bidder 2 calls a price p strictly lower than 1, Bidder 3 stays active after the jump bid. Then, if Bidder 1 stays active after the jump bid, Bidder 3 behaves as in the case without jump bid. If Bidder 1 does not stay active after the jump bid, Bidder 3 stays active up to $t_3+p$. If Bidder 2 calls a price p strictly higher than 1, Bidder 3 stays active after the jump bid up to $1+t_3$. If Bidder 2 calls a price $p=1$, Bidder 3 stays active after the jump bid. Then, if Bidder 3 has type $t_3=t_l$, he stays active up to $t_l$; else, he stays active up to $t_h+1$.

  • (b) If Bidder 1 calls a price $p<1$, Bidder 3 stays active and then leaves the auction at a price equal to $q+t_3$, q being defined as before. If Bidder 1 does not leave the auction at a price lower than 1, Bidder 3 stays active up to $1+t_3$. If Bidder 1 calls a price $p \ge 1$, Bidder 3 stays active up to $1+t_3$.

We also suggest a belief function for Bidder 3 which is coherent with these strategies.

Bidder 3’s belief conditional on observing that Bidder 2 calls price 1 and Bidder 1 does not stay active after must be the same as its prior since Bidder 2 always calls price 1 and Bidder 1 never stays active after such a jump bid at the equilibrium. With such a belief and assuming that Bidder 2 stays active up to his valuation for the good, Bidder 3 cannot obtain more than what he obtains following the proposed strategy (if $t_3=t_l$, he cannot derive any profit and leaving at price $t_l$ is a best response and if $t_3=t_h$, leaving at price $t_h+1$ is also a best response).

Now, since at the equilibrium, Bidder 2 never calls a price different from 1, we can propose many beliefs following a jump bid to a price different from 1: they will not be in contradiction with the actual distribution of s conditional on the jump bid. We assume that Bidder 3 believes that if Bidder 2 calls a price $p\ne 1$ and Bidder 1 does not stay active after the jump, $s=\min (p,1)$. This is no incoherent with Bidder 1 and 2’s strategy and Bidder 3’s strategy is coherent with this belief.

If Bidder 3 observes that Bidder 1 leaves the auction at a price p, he believes that $s=\min (p,1)$.

If bidders choose these strategies, their behaviors coincide with what we describe in Proposition 2. Now we need to show that these strategies constitute an equilibrium.

Bidder 1: Staying active beyond (or calling a price higher than) his valuation is weakly dominated. Further, considering Bidder 2 and Bidder 3’s strategies, Bidder 1 cannot make a profitable deviation with a jump bid lower than his valuation.

Bidder 2: Whether a price is called or not, in the second part of the auction, staying active up to his valuation for the good is a weakly dominant strategy. Therefore, in order to find a profitable deviation, we need to focus on the jump bidding part of the strategy, assuming that after any possible jump bid, he will stay active up to his valuation for the good.

If Bidder 2 does not call a price or calls a price lower than s, Bidder 3 discovers the value of s by observing the price at which Bidder 1 leaves the auction and Bidder 3 stays active up to $s+t_3$. We consider separately the different possible values of $t_2$.

$t_2=t_l$. If Bidder 2 calls price $p=1$, he obtains an expected payoff equal to s / 2. If he does not call a price or calls a price strictly lower than s, he obtains 0. If he calls a price $p\in [s,1)$, he cannot obtain more than what he obtains when he calls price $p=1$. If he calls a price strictly higher than 1, he obtains 0. Hence, there is no profitable deviation.

$t_2=t_h$. If Bidder 2 calls price $p=1$, he obtains an expected payoff equal to $(t_h-t_l+s)/2$. If he does not call a price or calls a price strictly lower than s, he obtains $(t_h-t_l)/2$. If he calls a price $p\in [s,1)$, he cannot obtain more than what he obtains when he calls price $p=1$. If he calls a price strictly higher than 1, he obtains $\max (0,t_h+s-t_l-1)/2$. Hence, there is no profitable deviation.

Bidder 3: We first consider deviations that do not involve calling a price and consider separately the different possible values of $t_3$.

$t_3=t_l$. If Bidder 2 calls price $p=1$ (or any price greater or equal than s), he never leaves the auction for a price lower than $s+t_l$. Therefore, conditional on winning the auction, Bidder 3 can only make a negative profit. Leaving the auction at price $t_l$, Bidder 3 avoids winning and picks a strategy that is not dominated. If no price is called (or a price lower than s is called), Bidder 3 discovers the value of s by observing at which price Bidder 1 leaves the auction. Then, staying active up to $q+t_l$ is a weakly dominant strategy.

$t_3=t_h$. If Bidder 2 calls price $p=1$ (or any price greater or equal than s), he never leaves the auction at a price strictly higher than $s+t_h$, which means that Bidder 3 always wins and never makes a loss when winning. Thus, the proposed equilibrium strategy is not dominated. If no price is called (or a price lower than s is called), Bidder 3 discovers the value of s by observing at which price Bidder 1 leaves the auction. Then, staying active up to $q+t_h$ is a weakly dominant strategy.

Now, let us consider deviations that include jump bids.

Suppose that Bidder 3 calls a price lower than 1. This jump bid does not qualify as the highest jump bid in the first stage, so it yields the same outcome as not calling a price at all. Suppose that Bidder 3 calls a price $p\in [1,t_l]$. After the jump bid, Bidder 2 stays active up to $v_2$. Bidder 3’s information is the same as in the case when Bidder 2 is the bidder placing the highest bid in the first stage. Thus, calling a price $p\in [1,t_l]$ cannot be part of a profitable deviation. We can show with the same type of arguments that calling a price $p>t_l$ cannot be part of a profitable deviation either (Bidder 3 does not obtain more information when $t_2+s\ge p$ and if $t_2+s\ge p$, the jump bid makes him lose money). $\square$

1.2 Proof of Proposition 3

Consider the following strategies:

• Bidder 1. Never calls a price, stays active up to s and leaves the auction if a price higher than s is called.

• Bidder 2. If $t_2=t_l,t_m$ and if $t_2=t_h$ and $s>t_h-t_m$, calls price $p=1$ and stays active up to $s+t_2$ afterwards. If $t_2=t_h$ and $s< \underline{s}$, does not call a price and stays active up to $t_h+s$. If a bidder calls a price higher than 1, stays active up to $s+t_2$.

• Bidder 3. Never calls a price. If no price is called, Bidder 3 leaves the auction at a price equal to $q+t_3$, q being the price at which Bidder 1 leaves the auction if it is in the interval [0, 1]. If Bidder 1 does not leave the auction at price lower than 1, Bidder 3 stays active up to $1+t_3$.

  If a price p is called in the first stage:

  • (a) If Bidder 2 calls a price p strictly lower than 1, Bidder 3 stays active after the jump bid. Then, if Bidder 1 stays active after the jump bid, Bidder 3 behaves as in the case without jump bid. If Bidder 1 does not stay active after the jump bid, Bidder 3 stays active up to $t_3+p$. If Bidder 2 calls a price p strictly higher than 1, Bidder 3 stays active after the jump bid up to $1+t_3$. If Bidder 2 calls a price $p=1$, Bidder 3 stays active after the jump bid. Then, if Bidder 3’s type is $t_l$, he stays active up to $t_l$; if $t_3=t_m$, he stays active up to $2t_h-t_m$; and if $t_3=t_h$, he stays active up to $t_h+1$.

  • (b) If Bidder 1 calls a price $p<1$, Bidder 3 stays active and then leaves the auction at a price equal to $q+t_3$, q being defined as before. If Bidder 1 does not leave the auction at a price lower than 1, Bidder 3 stays active up to $1+t_3$. If Bidder 1 calls a price $p \ge 1$, Bidder 3 stays active up to $1+t_3$.

If bidders choose these strategies, their behaviors coincide with what we describe in Proposition 3. Now we need to show that these strategies constitute an equilibrium.

Bidder 1: Analogous argument as for the proof of Proposition 2.

Bidder 2: We stress only the parts that differ from the proof of Proposition 2.

$t_2=t_l$. Same argument as in the proof of 2 except that now the expected payoff in equilibrium is $s/3+\max (0,t_l+s-2t_h+t_m)/3$.

$t_2=t_m$. If Bidder 2 calls price $p=1$, he obtains an expected payoff equal to $(t_m-t_l+s)/3+\max (0,2t_m+s-2t_h)/3$ and $(t_m-t_l)/3$ if he does not call a price or call a price strictly lower than s. If he calls a price $p\in [s,1)$, he cannot obtain more than what he obtains when he calls price 1 and if he calls a price strictly higher than 1, he derives $\max (0,t_m+s-t_l-1)/3$. Hence, there is no profitable deviation when $t_2=t_m$.

$t_2=t_h$. If Bidder 2 calls price $p=1$, he obtains an expected payoff equal to $(t_h-t_l+s)/3+\max (0,t_m+s-t_h)/3$ and $(t_h-t_l)/3+(t_h-t_m)/3$ if he does not call a price or call a price strictly lower than s. If he calls a price $p\in [s,1)$, he cannot obtain more than what he obtains when he calls price 1 and if he calls a price strictly higher than 1, he derives $\max (0,t_h+s-t_l-1)/3+\max (0,t_h+s-t_m-1)/3$. Therefore, Bidder 2’s best choice are either calling price 1 or not calling a price. The first alternative gives him $(t_h-t_l+s)/3+\max (0,t_m+s-t_h)/3$ and the second one $(t_h-t_l)/3+(t_h-t_m)/3$. When $s \ge \underline{s}$, the first alternative gives him a higher payoff and calling price $p=1$ is a better response. When $s<\underline{s}$, the second alternative gives him a higher payoff and not calling any price is a better response. Hence, there is no profitable deviation when $t_2=t_h$.

Bidder 3: We first consider deviations that do not involve calling a price and consider separately the different possible values of $t_3$.

$t_3=t_l$. Same argument as in the proof of Proposition 2.

$t_3=t_m$. If Bidder 2 calls price $p=1$ and leaves the auction at a price below $t_h+\underline{s}$, the probability that $t_2=t_h$ is zero. Therefore, $v_2 \le v_3$ and since Bidder 2 leaves the auction at a price equal to $v_2$, staying active up to $t_h+\underline{s}$ is not costly and it may be profitable. Therefore, Bidder 3 cannot profitably deviate leaving the auction at a price lower than $t_h+\underline{s}$. Now, suppose that Bidder 3 considers leaving the auction at a price strictly higher than $t_h+\underline{s}$. Since the expected value of $v_3$ conditional on Bidder 2’s leaving the auction at a price p strictly higher than $t_h+\underline{s}$ is strictly lower than p (since $t_h-t_m>t_m-t_l$), such a deviation cannot be profitable either. If no price is called, Bidder 3 discovers the value of s by observing at which price Bidder 1 leaves the auction. Then, staying active up to $q+t_m$ is a weakly dominant strategy.

$t_3=t_h$. Same argument as in the proof of Proposition 2.

Now, let us consider deviations that include jump bids.

Since Bidder 2 always leaves the auction at a price equal to $s+t_2$, Bidder 3 cannot make any profitable deviation even if it includes a jump bid when $t_3=t_l$ (he cannot derive any profit) and when $t_3=t_h$ (he cannot win the auction at a price strictly lower than $t_2+s$) so that we only need to consider $t_3=t_m$.

Suppose that Bidder 3 calls a price lower than 1. This jump bid could only affect the auction when $t_2=t_h$ and $s< \tilde{s}$. However, in that case, Bidder 2 stays active up to $t_h+s$ after the jump bid and Bidder 3 cannot obtain any strictly positive profit.

Suppose that Bidder 3 calls a price $p\in (1,t_l]$. After the jump bid, Bidder 2 stays active up to $v_2$. Bidder 3’s information is the same as in the case when Bidder 2 calls price $p=1$ except that he can no longer learn the event “$t_2=t_h$ and $s< \tilde{s}$”. Therefore, Bidder 3 is better off leaving the auction at a price equal $t_h$ rather than staying active up to $t_h+\tilde{s}$. Hence, calling price p lowers his expected payoff by $\min (t_l+1-t_h,\tilde{s})(t_m-t_l)$. Calling a price $p\in (1,t_l]$ cannot be part of a profitable deviation.

The same type of arguments applies for a jump bid $p>t_l$ so that it cannot be part of a profitable deviation either. $\square$

1.3 Proof of Proposition 4

Consider the following strategies:

• Bidder 1. Never calls a price, stays active up to s and leaves the auction if a price higher than s is called.

• Bidder 2. If $s \in [0,1/2)$ calls price $p=1/2$ and stays active up to $s+t_2$ afterwards. If $s \in [1/2,1]$ calls price 1 and stays active up to $s+t_2$ afterwards. If a bidder calls a price higher than the price called by Bidder 2, stays active up to $s+t_2$.

• Bidder 3. Never calls a price. If no price is called, Bidder 3 leaves the auction at a price equal to $q+t_3$, q being the price at which Bidder 1 leaves the auction if it is in the interval [0, 1]. If Bidder 1 does not leave the auction at a price lower than 1, Bidder 3 stays active up to $1+t_3$.

  If a price p is called in the first stage:

  • (a) If Bidder 2 calls a price p strictly lower than 1 with $p \ne 1/2$, Bidder 3 stays active after the jump bid. Then, if Bidder 1 stays active after the jump bid, Bidder 3 behaves as in the case without jump bid. If Bidder 1 does not stay active after the jump bid, Bidder 3 stays active up to $t_3+p$. If Bidder 2 calls a price 1 / 2, Bidder 3 stays active after the jump bid. Then, if Bidder 1 stays active after the jump bid, Bidder 3 behaves as in the case without jump bid. If Bidder 1 does not stay active after the jump bid, Bidder 3 stays active up to $t_l+1/2$ and if $t_3=t_h$, he stays active up to $t_h+1$. If Bidder 2 calls a price $p=1$, Bidder 3 stays active after the jump bid. If Bidder 3 has type $t_3=t_l$, he stays active up to $t_l+1/2$; if $t_3=t_h$, he stays active up to $t_h+1$. If Bidder 2 calls a price $p=1$, Bidder 3 stays active after the jump bid. If Bidder 3 has type $t_3=t_l$, he stays active up to $t_l+1/2$; if $t_3=t_h$, he stays active up to $t_h+1$. If Bidder 2 calls a price p strictly higher than 1, Bidder 3 stays active after the jump up to $1+t_3$.

  • (b) If Bidder 1 calls a price $p<1$, Bidder 3 stays active and then leaves the auction at a price equal to $q+t_3$, q being defined as before. If Bidder 1 does not leave the auction at a price lower than 1, Bidder 3 stays active up to $1+t_3$. If Bidder 1 calls a price $p \ge 1$, Bidder 3 stays active up to $1+t_3$.

If bidders choose these strategies, their behaviors coincide with what we describe in Proposition 4. Now we need to show that these strategies constitute an equilibrium.

Bidder 1: Analogous arguments as in the proof of Proposition 2.

Bidder 2: We stress only the parts that differ from the proof of Proposition 2.

• (a) If $s \ge 1/2$

  $t_2=t_l$. If Bidder 2 calls price $p=1$, he obtains a payoff equal to $(s-1/2)/2$. If he does not call a price or call a price strictly lower than s, he obtains 0. If he calls a price $p\in [s,1)$, he does not obtain more than what he obtains when he calls price $p=1$. If he calls a price strictly higher than 1, he derives obtains 0. Hence, there is no profitable deviation.

  $t_2=t_h$. If Bidder 2 calls price $p=1$, he obtains a payoff equal to $(t_h-t_l+(s-1/2))/2$. If he does not call a price or call a price strictly lower than s, he obtains $(t_h-t_l)/2$. If he calls a price $p\in [s,1)$, he does not obtain more than what he obtains when he calls price $p=1$. If he calls a price strictly higher than 1, he derives $\max (0,t_h+s-t_l-1)/2$. Hence, there is no profitable deviation.

• (b) If $s < 1/2$

  $t_2=t_l$. If Bidder 2 calls price $p=1/2$, he obtains a payoff equal to s / 2. If he does not call a price or call a price strictly lower than s, he obtains 0. If he calls a price $p\in [s,1]$, he does not obtain more than what he obtains when he calls price $p=1/2$. If he calls a price strictly higher than 1, he obtains 0. Hence, there is no profitable deviation.

  $t_2=t_h$. If Bidder 2 calls price $p=1/2$, he obtains a payoff equal to $(t_h-t_l+s))/2$. If he does not call a price or call a price strictly lower than s, he obtains $(t_h-t_l)/2$. If he calls a price $p\in [s,1]$, he does not obtain more than what he obtains when he calls price $p=1/2$. if he calls a price strictly higher than 1, he derives $\max (0,t_h+s-t_l-1)/2$. Hence, there is no profitable deviation.

Bidder 3: $t_3=t_l$ and $t_3=t_h$. Same arguments as in the proof of Proposition 2. $\square$