Learning, Mean Field Approximations, and Phase Transitions in Auction Models

Abstract

In this paper, we study an agent-based model for multi-round, pay as bid, sealed bid reverse auctions using techniques from partial differential equations and statistical mechanics tools. We assume that in each round a fixed fraction of bidders is awarded, and bidders learn from round to round using simple microscopic rules, adjusting myopically their bid according to their performance. Agent-based simulations show that bidders coordinate in the sense that they tend to bid the same value in the long-time limit. Moreover, this common value is the true cost or the ceiling price of the auction, depending on the level of competition. A discontinuous phase transition occurs when half of the bidders win. We establish the corresponding rate equations, and we obtain a system of ordinary differential equations describing the dynamics. Finally, we derive formally the kinetic equations modeling the dynamics, and we study the asymptotic behavior of solutions of the corresponding first-order, nonlinear partial differential equation satisfied by the distribution of agents.

Appendix

Proposition 8

When $f_0 = 1$ in [0, 1], the solution of (5.6) is given by

• if $p\le \frac{1}{4}$:

  $$\begin{aligned} f_t = {\left\{ \begin{array}{ll} 1_{[t,1-t]} + 2t\delta _{a(t)}, \\ \qquad a(t)=p &{}\quad 0\le t\le p \\ 1_{[a(t),1-t]} + 2\sqrt{pt}\delta _{a(t)}, \\ \qquad a(t)=2\sqrt{pt}-t, &{}\quad p\le t\le T:=4p. \end{array}\right. } \end{aligned}$$

• if $\frac{1}{4}< p<\frac{1}{2}$:

  $$\begin{aligned} f_t = {\left\{ \begin{array}{ll} 1_{[t,1-t]} + 2t\delta _{a(t)}, \\ \qquad a(t)=p &{}\quad 0\le t\le p \\ 1_{[a(t),1-t]} + (t+a(t))\delta _{a(t)}, \\ \qquad a(t)=2\sqrt{pt}-t, &{}\quad p\le t\le \frac{1}{4p}, \\ \delta _{a(t)}, \\ \qquad a(t)=(2p-1)(t-\frac{1}{4p}) + 1-\frac{1}{4p}, &{} \quad \frac{1}{4p}\le t\le T:=\frac{1}{2(1-2p)}. \end{array}\right. } \end{aligned}$$

• if $p=\frac{1}{2}$:

  $$\begin{aligned} f_t = {\left\{ \begin{array}{ll} 1_{[t,1-t]} + 2t\delta _{a(t)}, \\ \qquad a(t)=p &{}\quad 0\le t\le \frac{1}{2} \\ \delta _{a(t)}, \\ \qquad a(t)=\frac{1}{2} &{}\quad t\ge \frac{1}{2} \end{array}\right. } \end{aligned}$$

• if $\frac{1}{2}\le p\le \frac{3}{4}$:

  $$\begin{aligned} f_t = {\left\{ \begin{array}{ll} 1_{[t,1-t]} + 2t\delta _{a(t)}, \\ \qquad a(t)=1-p &{}\quad 0\le t\le 1-p \\ 1_{[t, a(t)]} + 2\sqrt{(1-p)t} \delta _{a(t)}, \\ \qquad a(t)=1-2\sqrt{(1-p)t}+t, &{}\quad 1-p\le t\le \frac{1}{4(1-p)}, \\ \delta _{a(t)}, \\ \qquad a(t)=-(1-2p)(t-\frac{1}{4(1-p)}) + \frac{1}{4(1-p)}, &{}\quad \frac{1}{4(1-p)}\le t\le T:=\frac{1}{2(2p-1)}. \end{array}\right. } \end{aligned}$$

• if $\frac{3}{4}\le p\le 1$:

  $$\begin{aligned} f_t = {\left\{ \begin{array}{ll} 1_{[t,1-t]} + 2t\delta _{a(t)}, \\ \qquad a(t)=p &{}\quad 0\le t\le 1-p \\ 1_{[t, a(t)]} + 2\sqrt{(1-p)t}\delta _{a(t)}, \\ \qquad a(t)=1-2\sqrt{(1-p)t}+t, &{}\quad 1-p\le t\le T:=4(1-p). \end{array}\right. } \end{aligned}$$

The proof is a direct verification that $f_t$ solves (5.4) with $\sigma =0$. First notice that for any t, $F_t^{-1}(p)=a(t)$ so that

$$\begin{aligned} P_0[f_t](\mu ) = {\left\{ \begin{array}{ll} 1 &{}\quad {\text {if }} \mu <a(t), \\ 0 &{}\quad {\text {if }} \mu >a(t), \\ \frac{p-f_t([0,a(t))}{f_t(\{a(t)\})}&{} \quad {\text {if }} \mu =a(t). \end{array}\right. } \end{aligned}$$

For $0\le t\le p$, $\langle f_t,\phi \rangle =\int _t^{1-t}\phi + 2t \phi (p)$ so that

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t} \langle f_t,\phi \rangle =-\phi (1-t)-\phi (t)+2\phi (p). \end{aligned}$$

On the other hand, notice that $P_0[f_t](p)=\frac{1}{2}$ so that the right-hand side of (5.4) is

$$\begin{aligned} \langle f_t, (2P_0[f_t]-1)\phi ' \rangle = \int _t^p \phi ' - \int _o^{1-t}\phi ' = \frac{\textrm{d}}{\textrm{d}t}\langle f_t,\phi \rangle . \end{aligned}$$

This computation holds until there is no more mass on at least one side of $\delta _{a(t)}$. When $p\le \frac{1}{2}$ (resp. $p\ge \frac{1}{2}$), this holds when $1_{[t,p]}$ (resp. $1_{[p,1-t]}$) collapses to $\delta _p$ which occurs at time $t=p$ (resp. $t=1-p$).

Let us examine first the case $p\le \frac{1}{2}$. For $t\ge p$, we look for a solution of the form $f_t = 1_{[a(t),1-t]} + (t+a(t))\delta _{a(t)}$. Then $\langle f_t,\phi \rangle =\int _{a(t)}^{1-t} \phi + (t+a(t))\phi (a(t))$ so that

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}t}\langle f_t,\phi \rangle = \phi (a(t)) - \phi (1-t) + (t+a(t))\phi '(a(t))a'(t). \end{aligned}$$

On the other hand, with $P_0[f_t](a(t))=\frac{p}{t+a(t)}$, the right-hand side of (5.4) is

$$\begin{aligned} \langle f_t, (2P_0[f_t]-1)\phi ' \rangle = -\int _{a(t)}^{1-t}\phi ' + (t+a(t))\phi '(a(t)) \Big (\frac{2p}{t+a(t)}-1\Big ). \end{aligned}$$

Equating these two expressions, we see we must choose a satisfying $a'(t)=\frac{2p}{t+a(t)}-1$, $t>p$, with $a(p)=p$. Solving we obtain $a(t)=2\sqrt{pt}-t$. This computation holds until $1_{[a(t),1-t]}$ collapses to $\delta _{a(t)}$, i.e., $1-t=a(t)$ if and only if $t=\frac{1}{4p}$, or $a(t)=0$ if and only if $t=4p$, whatever occurs first. If $\frac{1}{4}<p<\frac{1}{2}$, then $1_{[a(t),1-t]}$ collapses to $\delta _{a(t)}$ before reaching 0. Then from $t\ge \frac{1}{4p}$, $f_t$ is a Dirac mass $\delta _{a(t)}$, and we can use Proposition 3, for $a'(t)=2p-1$. This holds until $a(t)=0$ which occurs at $t=\frac{1}{2p(1-2p)}$.

If $p\ge \frac{1}{2}$, the result follows from the case $p\le \frac{1}{2}$ and Proposition 5.