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Consider 2 individuals who are interested in one indivisible object. Each player $i$ has a valuation $v_i > 0$ for the object. Assume $v_1 \geq v_2$. In this scenario, each player submits a bid $b_i \geq 0$. The individual with the highest bid wins the auction and obtains the object at a price equal to the second highest bid, in case of a tie, the object goes to player 1. The payoff functions for each player are given by

$$u_1(b_1, b_2) = \left\{ \begin{array}{l l} v_1 - b_2 & \quad \text{if $b_1 \geq b_2$}\\ 0 & \quad \text{if $b_1 < b_2$} \end{array} \right. $$ and $$u_2(b_1, b_2) = \left\{ \begin{array}{l l} v_2 - b_1 & \quad \text{if $b_2 > b_1$}\\ 0 & \quad \text{if $b_2 \leq b_1$} \end{array} \right. $$

Am I correct if I say that in this case, the best reply functions are given by:

$$\beta_1(b_2) = \left\{ \begin{array}{l l} \{b_1 | b_1 \geq b_2\} & \quad \text{if $b_2 < v_1$}\\ \{b_1 | b_1 < b_2\} & \quad \text{if $b_2 > v_1$} \\ [0, \infty] & \quad \text{if $b_2 = v_1$} \end{array} \right. $$

and

$$\beta_2(b_1) = \left\{ \begin{array}{l l} \{b_2 | b_2 > b_1\} & \quad \text{if $b_1 < v_2$}\\ \{b_2 | b_2 \leq b_1\} & \quad \text{if $b_1 > v_2$} \\ [0, \infty] & \quad \text{if $b_1 = v_2$} \end{array} \right. $$

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2 Answers 2

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These look right to me. Why did you doubt them?

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The second price auction has a dominant strategy equilibrium in which each player bids their true valuation. Given the other player bids $b_{-i}$, consider the 3 possible cases for player $i$:

  1. $v_i > b_{-i}$. Player $i$ should bid any $b_i \in (b_{-i}, \infty)$ and get payoff $v_i - b_{-i}$. Otherwise he gets $0$.

  2. $v_i = b_{-i}$. Player $i$ is indifferent between any bids.

  3. $v_i < b_{-i}$. Player $i$ should bid any $b_i \in (-\infty, b_{-i})$ and get payoff $0$. Otherwise he gets negative payoff.

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  • $\begingroup$ This ignores player 1 wins in the case of a tie $\endgroup$ Dec 15, 2013 at 7:44
  • $\begingroup$ How does that change the strategy? In case 2, player $i$, $i = 1$ or $2$, is indifferent between winning and not winning the bid regardless of this tie-breaking rule. In case 1, one might nitpick and say $b_1 \in [b_2, \infty)$ but the payoff is the same. Player $1$ pays the auctioneer $b_2$ even if he ties the bid. $\endgroup$
    – Michael
    Dec 16, 2013 at 23:20
  • $\begingroup$ The question asked if his best responses were right, they were. The best responses posted here have small differences as you pointed. $\endgroup$ Dec 17, 2013 at 2:46

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