Efficiency levels in sequential auctions with dynamic arrivals

Abstract

In an environment with dynamic arrivals of players who wish to purchase only one of multiple identical objects for which they have a private value, we analyze a sequential auction mechanism with an activity rule. If the players play undominated strategies then we are able to bound the efficiency loss compared to an optimal mechanism that maximizes the total welfare. We have no assumptions on the underlying distribution from which the players’ arrival times and valuations for the object are drawn. Moreover we have no assumption of a common prior on this distribution.

Appendices

Appendices

Appendix 1: The need for an activity rule via an example

To exemplify the need to add an activity rule to the auction, consider a setting of two items and three bidders that arrive in time $1$. In the second auction, it is rather immediate that the strategy of “remaining until price reaches value” weakly dominates all other strategies¹⁹. Throughout the paper we have focused on two strategies:

• 1. In both auctions, remain until price reaches value, i.e. for $t=1,2$

  $$\begin{aligned} b_{i}^{\text {value}}((r_{i},v_{i}),h(t,p,k))=D,\quad \hbox {iff}\ \ p\ge v_{i}. \end{aligned}$$

• 2. In the first auction, remain until exactly one other bidder remains, and in the second auction, remain until your value. Formally,

  $$\begin{aligned} b_{i}^{EPD}((r_{i},v_{i}),h_{1}(p,k))=D,\quad \hbox {iff}\ \ \{|I_{1}(p,k)|=1\quad \hbox {or}\quad p\ge v_{i}\}, \end{aligned}$$

  and

  $$\begin{aligned} b_{i}^{EPD}((r_{i},v_{i}),h_{1}(p_{1}^{*},s_{p_{1}^{*}}^{1}),h_{2}(p,k))=D,\quad \text { iff }\ p\ge v_{i} \end{aligned}$$

  (where $EPD$ stands for Earliest Possible Dropping).

We note that these two strategies do not weakly dominate all other strategies, in the sequential auction without the activity rule. Suppose three bidders $1,2,3$ arrive at time $1$, with $v_{1}>v_{2}>v_{3}$, and let us consider the strategy $b_{1}^{0}$ for bidder $1$, in which she drops at price $0$ in the first auction, and remains until her value in the second auction²⁰. None of the above two strategies dominates $b_{1}^{0}$ , due to the following reasoning.

Consider first the strategy $b_{1}^\mathrm{value}$. This strategy performs strictly worse than $b_{1}^{0}$ in case both $2$ and $3$ choose to do the same (remain until their value in both auctions, i.e. play $b_{2}^\mathrm{value},b_{3}^\mathrm{value}$), due to the following. By playing $b_{1}^\mathrm{value}$, bidder $1$ will win the first auction and will pay $v_{2}$. By playing $b_{1}^{0}$, bidder $1$ will lose the first auction and will win the second auction for a lower price of $v_{3}$.

Consider next the strategy $b_{1}^\mathrm{EPD}$. This strategy performs strictly worse than $b_{1}^{0}$ when bidder $3$ plays the strategy $b_{3}^\mathrm{value}$ and $2$ uses the following strategy (conditional drop): In the first auction, if bidder $1$ drops at price $0$ then bidder $2$ continues until her value, and if bidder $1$ does not drop at price $0$ then bidder $2$ drops at price $\epsilon$ (for some small fixed $\epsilon$). In the second auction, bidder $2$ remains until her value. In this case, if bidder $1$ follows $b_{1}^{0}$ and drops at $0$ then bidder $2$ will win the first auction, bidder $1$ will win the second auction, and will pay $v_{3}$. If bidder $1$ follows $b_{1}^\mathrm{EPD}$ then bidder $2$ drops, and bidder $1$ drops immediately after that (since now only bidder $3$ remains besides $1$). Thus, bidder $3$ wins the first auction, bidder $1$ again wins the second auction, but this time pays $v_{2}$ which is larger than $v_{3}$.

More generally, $b_1^0$ is not weakly dominated by any other strategy (and is thus an undominated strategy of player 1). To see this, suppose towards a contradiction that $b_1^0$ is weakly dominated by some strategy $\bar{b}_i$. Note that $\bar{b}_i$ cannot drop at zero in the first auction, because this is an unconditional drop making $\bar{b}_i$ identical to $b_1^0$. Now, suppose player 1 plays $\bar{b}_i$, player 2 plays conditional drop, and player 3 bids up to her value in both auctions. Then, player 2 drops at $\epsilon$ regardless of the values $v_1, v_2, v_3$ (as long as all values are larger than $\epsilon$). Consider the range of values where $v_1 > \max (v_2,v_3)$. If in the first auction player 1 drops before player 3, the resulting utility of player 1 is $v_1 - v_2$. For the range of values $v_2 > v_3$, playing $b_1^0$ would result in a utility $v_1 - v_3 > v_1 - v_2$, contradicting the assumption that $\bar{b}_i$ weakly dominates $b_1^0$. Thus, in the first auction player 3 must drop before player 1. I.e., player 1 must win the first auction with a resulting utility of $v_1 - v_3$. For the range of values $v_2 < v_3$, playing $b_1^0$ would result in a utility $v_1 - v_2 < v_1 - v_3$, contradicting once again the assumption that $\bar{b} _i$ weakly dominates $b_1^0$.

Appendix 2: Ex-post mechanisms cannot do better

In a “detail-free” /“robust” setting, the literature commonly uses the solution concepts of dominant-strategies (for direct mechanisms) and ex-post equilibria (for indirect mechanisms). A natural question is whether such mechanisms can obtain higher worst-case efficiency than our sequential auction with an activity rule. In this appendix we give a negative answer to this question, and show that every dominant-strategy mechanism for our setting can obtain, in the worst-case, at most half of the optimal welfare.  Lavi and Nisan (2005) and subsequently Hajiaghayi et al. (2005) and Cole et al. (2008) prove similar results for slightly different settings. In particular they all rely on the fact that players have a departure time to prove the impossibility. The argument that we devise here does not rely on this assumption and is therefore suitable for our sequential auction setting.

By the direct-revelation principle, we can focus on direct mechanisms in which truthful reporting of the type is a dominant-strategy. We term these “truthful mechanisms”. We additionally assume ex-post Individual Rationality, i.e. that a winner is never required to pay more than her declared value. We show the impossibility for the very restrictive setting of two items and three players, where it is common knowledge that players $1$ and $2$ arrive for the first auction and player $3$ arrives for the second auction. Clearly, this only strengthens the impossibility, since if one can freely determine the number of items and players and their arrival times then one can replicate this limited setting.²¹

Definition 13

(A direct-mechanism for a limited dynamic setting) A direct mechanism is a set of four functions: $w_{1}(v_{1},v_{2})$ determines the winner (either $1$ or $2$) of the first item, and she pays a price $p_{1}(v_{1},v_{2})$, where $p_{1}(v_{1},v_{2}) \le v_{w_{1}(v_{1},v_{2})}$. $w_{2}(v_{1},v_{2}, v_{3})$ determines the winner of the second item (either $1$, $2$, or $3$, but not $w_{1}(v_{1},v_{2})$), and she pays a price $p_{2}(v_{1},v_{2}, v_{3})$, where $p_{2}(v_{1},v_{2},v_{3}) \le v_{w_{2}(v_{1},v_{2},v_{3})}$. Such a mechanism is truthful if it is a dominant-strategy of every player to report her true type.

Theorem 14

Every truthful mechanism for the limited dynamic setting obtains in the worst-case at most half of the optimal social welfare.

Proof

Fix any $\frac{1}{2}\ge \epsilon >0$. Suppose by contradiction that their exists a truthful mechanism $M=(w_{1}(\cdot ,\cdot ),p_{1}(\cdot ,\cdot ),w_{2}(\cdot ,\cdot ,\cdot ),p_{2}(\cdot ,\cdot ,\cdot ))$ that always obtains at least $\frac{1}{2}+\epsilon$ of the optimal social welfare. We reach a contradiction via a series of three claims. $\square$

Claim 15

If $v_{2} > \frac{v_{1}}{2\epsilon }$ then $w_{1}(v_{1},v_{2})=2,$ i.e. player $2$ must be the winner of the first auction.

Proof

Suppose by contradiction that there exists an instance $(v_{1},v_{2},v_{3})$ such that $v_{2}>\frac{v_{1}}{2\epsilon }$ and $w_{1}(v_{1},v_{2})=1$. Consider another instance $(\tilde{v}_{1},\tilde{v}_{2},\tilde{v}_{3})$, where $\tilde{v}_{1}=v_{1},\tilde{v}_{2}=v_{2}$, and $\tilde{v}_{3}=v_{2}$. The optimal social welfare in this instance is $2v_{2}$. We have $w_{1}( \tilde{v}_{1},\tilde{v}_{2})=w_{1}(v_{1},v_{2})=1$, and therefore the social welfare that the mechanism obtains is $v_{1}+v_{2}$. But

$$\begin{aligned} \frac{v_{1}+v_{2}}{2v_{2}}<\frac{1}{2}+\epsilon \end{aligned}$$

which contradicts the fact that the mechanism always obtains at least $\frac{ 1}{2}+\epsilon$ of the optimal social welfare. $\square$

Claim 16

In the instance $(v_{1}=1, v_{2} > \frac{1-2\epsilon }{ 1+2\epsilon }, v_{3}=0),$ the winners must be players $1$ and $2$.

Proof

Any other set of winners has welfare strictly less than a fraction of $\frac{ 1}{2}+\epsilon$ of the optimal social welfare of this instance. $\square$

Claim 17

If $v_{1}=1, v_{2} > \frac{1-2\epsilon }{1+2\epsilon },$ and $w_{1}(v_{1},v_{2})=2,$ then $p_{1}(v_{1},v_{2}) \le \frac{1-2\epsilon }{ 1+2\epsilon }$ (note that $\frac{1-2\epsilon }{1+2\epsilon } < 1$ ).

Proof

Suppose a contradicting instance $(v_{1},v_{2},v_{3})$ where $p_{1}(v_{1},v_{2}) > \frac{1-2\epsilon }{1+2\epsilon } + \delta$ for some $\delta > 0$. Note that player $2$ wins item $1$ and pays the same price in the instance $(v_{1},v_{2},0)$ (call this “instance $2$”). Consider a third instance $(\tilde{v}_{1},\tilde{v}_{2},\tilde{v}_{3})$, where $\tilde{v }_{1} = 1, \tilde{v}_{2} = \frac{1-2\epsilon }{1+2\epsilon } + \delta$, and $\tilde{v}_{3} = 0$. By claim 16 player $2$ must be a winner in the third instance, and by individual rationality she pays at most $\frac{ 1-2\epsilon }{1+2\epsilon } + \delta$. Therefore, in instance $2$, player $2$ has a false announcement ($\frac{1-2\epsilon }{1+2\epsilon } + \delta$ instead of $v_{2}$) that strictly increases her utility, a contradiction to truthfulness. $\square$

We can now reach a contradiction and conclude the proof of the theorem. Consider the instance $(1,1,5)$. Suppose without loss of generality that $w_{1}(1,1)=1$. To obtain at least half of the optimal social welfare we must have $w_{2}(1,1,5)=3$. Thus player $2$ loses and has zero utility. However if player $2$ will declare some $\tilde{v}_{2}>\frac{1}{2\epsilon }$ instead of her true type $v_{2}=1$ then by claim 15 she will win the first item and by claim 17 she will pay a price of at most $\frac{1-2\epsilon }{1+2\epsilon }<1$. Thus player $2$ is able to strictly increase her utility by some false declaration, contradicting truthfulness.

Appendix 3: Proof of Proposition 10

We need to show that, for any $n,r$, and $0\le p<1$,

$$\begin{aligned} \frac{E_{F_{p}}[\tilde{A}_{r,n}]}{E_{F_{p}}[OPT_{r,n}]}=\frac{ 2-p^{r}-p^{n}-r(1-p)p^{r-1}}{2-p^{r}-p^{n}-r(1-p)p^{n-1}} \ge \frac{\sqrt{2}}{ 2}\simeq 0.70711. \end{aligned}$$

(4)

We first differentiate this expression with respect to $n$, to show that it decreases as $n$ increases.

$$\begin{aligned}&\frac{d}{dn}\left( \frac{\left( 2-p^{r}-p^{n}-r\left( 1-p\right) p^{r-1}\right) }{\left( 2-p^{r}-p^{n}-r\left( 1-p\right) p^{n-1}\right) } \right) \\&\quad =rp^{n-2}\left( \ln p\right) \left( 1-p\right) \frac{ \left( 2p-rp^{r}+p^{r+1}\left( r-2\right) \right) }{\left( -rp^{n-1}-p^{r}+\left( r-1\right) p^{n}+2\right) ^{2}}. \end{aligned}$$

We concentrate on the term

$$\begin{aligned} G\left( p,r\right) =2p-rp^{r}+p^{r+1}(r-2) \end{aligned}$$

and claim that it is nonnegative for every $2\le r\le n$ and $p\in [0,1]$. For $p=0$ we have $G(0,r)=0$ and for $p=1$ we have $G(1,r)=0$. Moreover

$$\begin{aligned} \frac{d^{2}}{dp^{2}}G\left( p,r\right) =rp^{r-2}\left( \left( r+1\right) \left( r-2\right) p-r\left( r-1\right) \right) \end{aligned}$$

and since

$$\begin{aligned} \frac{r\left( r-1\right) }{\left( r+1\right) \left( r-2\right) }>1 \end{aligned}$$

for every $r\ge 2$ we know that $\frac{d^{2}}{dp^{2}}G\left( p,r\right) \le 0$ and $G\left( p,r\right)$ is concave in $p$. We thus conclude that $G\left( p,r\right) \ge 0$ for every $2\le r\le n$ and $p\in \left[ 0,1 \right]$ and consequently that $\frac{d}{dn}\left( \frac{\left( 2-p^{r}-p^{n}-r\left( 1-p\right) p^{r-1}\right) }{\left( 2-p^{r}-p^{n}-r\left( 1-p\right) p^{n-1}\right) }\right) \le 0$. We take $n$ to infinity and get that

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( \frac{\left( 2-p^{r}-p^{n}-r\left( 1-p\right) p^{r-1}\right) }{\left( 2-p^{r}-p^{n}-r\left( 1-p\right) p^{n-1}\right) }\right) =1-\frac{r\left( 1-p\right) p^{r-1}}{\left( 2-p^{r}\right) } \end{aligned}$$

We wish to find the minimum of

$$\begin{aligned} 1-\frac{r\left( 1-p\right) p^{r-1}}{\left( 2-p^{r}\right) } \end{aligned}$$

which will give us a lower bound for (4), for every $n$, since we obtained that (4) decreases towards the limit as $n$ increases.

Equivalently, we look for the maximum of

$$\begin{aligned} H\left( p,r\right) =\frac{r\left( 1-p\right) p^{r-1}}{\left( 2-p^{r}\right) } . \end{aligned}$$

Now

$$\begin{aligned} \frac{d}{dr}H\left( p,r\right)&= -\frac{p^{r-1}\left( 1-p\right) \left( -2r\ln p+p^{r}-2\right) }{\left( 2-p^{r}\right) ^{2}}\\ \frac{d}{dp}H\left( p,r\right)&= rp^{r-2}\frac{\left( 2r\left( 1-p\right) +p^{r}-2\right) }{\left( 2-p^{r}\right) ^{2}}. \end{aligned}$$

Therefore if there exists a global maximum at $0<p<1$ and $2<r$ then we must have

$$\begin{aligned} -2r\ln p=2-p^{r} \end{aligned}$$

and

$$\begin{aligned} 2r\left( 1-p\right) =2-p^{r} \end{aligned}$$

but this is not possible since for every $0<p<1$ we have $-\ln p>1-p$. We thus conclude that the maximum is achieved on the boundary. Now for $p=0$, we have $H\left( 0,r\right) =0$ and for $p=1$, we have $H\left( 1,r\right) =0$; therefore we conclude that the maximum is achieved on the boundary where $r=2$. We find $p$ that solves

$$\begin{aligned} \max _{p\in \left( 0,1\right) }H\left( p,2\right) =\max _{p}\frac{2\left( 1-p\right) p}{\left( 2-p^{2}\right) } \end{aligned}$$

and the solution is

$$\begin{aligned} p^{*}=2-\sqrt{2}. \end{aligned}$$

Finally, for $r=2,p^{*}=2-\sqrt{2}$ and $n\rightarrow \infty$ we have

$$\begin{aligned} \frac{E_{F_{p}}[\tilde{A}_{r,n}]}{E_{F_{p}}[OPT_{r,n}]}= \frac{\sqrt{2}}{2} \simeq 0.707\,11. \end{aligned}$$