SMU H3 Notes Game TheoryGamesSMU H3

War of Attrition / Second-Price All-Pay Auction

Game theory analysis: War of Attrition / Second-Price All-Pay Auction.


Setup

Definition:

War of Attrition / Second-Price All-Pay Auction

  • Players: Two players: player 11 and player 22.
  • Strategies: Each player chooses a stopping time or bid xi[0,)x_i\in[0,\infty); In equilibrium, each player uses a continuous mixed strategy.

Rules

Derivation (Verification by Expected Payoff FOC)

E(π1)=0x(v1y)f2(y)dyx(1F2(x))E(\pi_1) =\int_0^x (v_1-y)f_2(y)\,dy-x(1-F_2(x)) E(π1)=v1F2(x)0xyf2(y)dyx(1F2(x))E(\pi_1) =v_1F_2(x)-\int_0^x yf_2(y)\,dy-x(1-F_2(x)) E(π1)x=v1f2(x)xf2(x)[(1F2(x))xf2(x)]=0\frac{\partial E(\pi_1)}{\partial x} =v_1f_2(x)-xf_2(x)-\left[(1-F_2(x))-xf_2(x)\right]=0 v1f2(x)=1F2(x)v_1f_2(x)=1-F_2(x) F2(x)1F2(x)=1v1\frac{F_2'(x)}{1-F_2(x)}=\frac{1}{v_1}

Derivation (Nash Equilibrium)

α1=1v1\alpha_1=\frac{1}{v_1} xln(1F2(x))=F2(x)1F2(x),\frac{\partial}{\partial x}\ln(1-F_2(x)) =-\frac{F_2'(x)}{1-F_2(x)},

the differential equation becomes

xln(1F2(x))=α1\frac{\partial}{\partial x}\ln(1-F_2(x))=-\alpha_1 ln(1F2(x))ln(1F2(a))=α1(xa)\ln(1-F_2(x))-\ln(1-F_2(a))=-\alpha_1(x-a) F2(x)=1eα1(xa)F_2(x)=1-e^{-\alpha_1(x-a)} F2(x)=1ex/v1F_2(x)=1-e^{-x/v_1} F1(x)=1ex/v2F_1(x)=1-e^{-x/v_2}

Nash Equilibrium

Result:

The mixed-strategy equilibrium is

F1(x)=1ex/v2,F2(x)=1ex/v1,x[0,)F_1(x)=1-e^{-x/v_2}, \qquad F_2(x)=1-e^{-x/v_1}, \qquad x\in[0,\infty)

Social Optimum

Insights

Insight:

  • Each player’s distribution is chosen to make the opponent indifferent.
  • If v1>v2v_1>v_2, then F2(x)<F1(x)F_2(x)<F_1(x) for x>0x>0.
  • Player 11 bids less on average:
0xf1(x)dx<0xf2(x)dx\int_0^\infty x f_1(x)\,dx < \int_0^\infty x f_2(x)\,dx
  • Equivalently, expected bids are v2v_2 for player 11 and v1v_1 for player 22.
  • Player 11 wins with probability
P(P1 wins)=0f1(x)F2(x)dx=v2v1+v2P(\text{P1 wins}) =\int_0^\infty f_1(x)F_2(x)\,dx =\frac{v_2}{v_1+v_2}
  • If v1>v2v_1>v_2, then P(P1 wins)<12P(\text{P1 wins})<\frac{1}{2}.
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