<p>The suggested pure strategies do not create an equilibrium, because if $X_2=X_1=2$, there is a tie. Whichever tie-breaking rule is employed, one of the players would want to deviate to $\beta_i=1+\epsilon$ to avoid the tie and win with certainty. If tie breaking is always in favor of the second player, player 1 deviates to $0+\epsilon$.</p>

<p>Discrete values in this setting do suggest a mixed strategy equilibrium (non-degenerate), because the probability of a tie in valuations is non-zero. This is why the pure strategy equilibrium from the continuous case is missing here.</p>

<p>The more general case of this problem is in <a href="http://dx.doi.org/10.1515/bejte-2012-0014" rel="nofollow" title="Doni &amp; Menicucci &#40;2013&#41;">Doni &amp; Menicucci (2013)</a>.</p>

CC BY-SA 3.0
2016-04-26T19:21:08.040Z
2016-04-26T21:02:51.790Z
2016-04-26T21:02:51.790Z
kitsune
<p>Any pure strategy Nash equilibrium is implicitly a mixed-strategies Nash equilibrium. Since the valuations vary, it's a good indicator we want to consider mixed-strategies. The fact that the problem tells us this is a stronger indicator, though I'm sure not the axiomatic justification you are seeking. :-)</p>

<p>Consider player $1$. We have player $1$'s expected profit: $\mathbb{E}[\Pi_{1}(b)] = (2-b) Pr[b \geq \beta_{2}(v_{2})]$, where $b$ is player $1$'s bid, $\beta_{2}$ is player $2$'s bidding strategy, and $v_{2}$ is player $2$'s valuation. We can assume $\beta_{2}(0) = 0$ (because if $\beta_{2}(0) &gt; 0$, player $2$ can improve upon this by decreasing his bid). Since we are only considering two potential valuations for player $2$, we can assume $\beta_{2}(v) = av$, for some constant $a \in \mathbb{R}_{++}$. (That is, given the two points $(0, 0)$ and $(2, \beta_{2}(2))$, we just draw a line between them). </p>

<p>Observe that $Pr[b \geq av] = Pr[v \leq \frac{b}{a}] = \frac{b}{2a}$, with the last inequality since we have a 50-50 chance on the valuation of player $2$. </p>

<p>Now for a Nash equilibrium, player $1$ seeks to maximize his expected value. This is given by the following optimization problem:</p>

<p>$$\max_{b} (2-b) \cdot (\frac{b}{2a})$$</p>

<p>This yields the first order conditions:</p>

<p>$\frac{1}{2a} \cdot (2 - 2b) = 0$, and we obtain that $b = 1$ is our only solution for player $1$. This answer should be reasonably intuitive.</p>

<p>Now player $2$ only wins if his valuation is $2$. So he players $\beta_{2}(2) = 1$ and $\beta_{2}(0) = 0$. </p>

CC BY-SA 3.0
2016-04-26T21:00:00.403Z
This part
$$
Pr[v \leq \frac{b}{a}] = \frac{b}{2a}
$$
seems incorrect. For any given $b$ the probability could take three values: 0, $\frac{1}{2}$, 1. It is not continuous in $a$.
Giskard
CC BY-SA 3.0
2015-09-09T21:23:00.840Z
2015-09-09T21:23:00.840Z
ml0105
<p>Exercise 4.5 in Vijay Krishna's <em>Auction Theory</em>:</p>

<blockquote>
  <p>Suppose that bidder 1 always has value $X_1=2$, while bidder 2's value, $X_2$ is equally likely to be $0$ or $2$. Find equilibrium bidding strategies in a first-price auction. (Note that since values are discrete, the equilibrium will be in mixed strategies.)</p>
</blockquote>

<ul>
<li><p>I don't understand why it is that since values are discrete, the equilibrium will be in mixed strategies. Is there some rule/principle to this?</p></li>
<li><p>How do I set up for the problem? Since equilibrium are in mixed strategies, I can suppose that player 1 bids $b_1,\ldots,b_k$ with probabilities $p_1,\ldots,p_k$ respectively. But that seems too general to make any progress.</p></li>
</ul>

<game-theory><auctions>
First-price auction with discrete values
Alexi
# First-price auction with discrete values

(Asked by: username_0 on 2015-08-31)

Exercise 4.5 in Vijay Krishna's *Auction Theory*:

> Suppose that bidder 1 always has value $X\_1=2$, while bidder 2's value, $X\_2$ is equally likely to be $0$ or $2$. Find equilibrium bidding strategies in a first-price auction. (Note that since values are discrete, the equilibrium will be in mixed strategies.)

* I don't understand why it is that since values are discrete, the equilibrium will be in mixed strategies. Is there some rule/principle to this?
* How do I set up for the problem? Since equilibrium are in mixed strategies, I can suppose that player 1 bids $b\_1,\ldots,b\_k$ with probabilities $p\_1,\ldots,p\_k$ respectively. But that seems too general to make any progress.

**Tags:** game-theory, auctions

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## 2 answers

### Answer by username_1 on 2016-04-26. Score: 3

The suggested pure strategies do not create an equilibrium, because if $X\_2=X\_1=2$, there is a tie. Whichever tie-breaking rule is employed, one of the players would want to deviate to $\beta\_i=1+\epsilon$ to avoid the tie and win with certainty. If tie breaking is always in favor of the second player, player 1 deviates to $0+\epsilon$.

Discrete values in this setting do suggest a mixed strategy equilibrium (non-degenerate), because the probability of a tie in valuations is non-zero. This is why the pure strategy equilibrium from the continuous case is missing here.

The more general case of this problem is in [Doni & Menicucci (2013)](http://dx.doi.org/10.1515/bejte-2012-0014 "Doni & Menicucci (2013)").


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### Answer by username_2 on 2015-09-09. Score: 0

Any pure strategy Nash equilibrium is implicitly a mixed-strategies Nash equilibrium. Since the valuations vary, it's a good indicator we want to consider mixed-strategies. The fact that the problem tells us this is a stronger indicator, though I'm sure not the axiomatic justification you are seeking. :-)

Consider player $1$. We have player $1$'s expected profit: $\mathbb{E}[\Pi\_{1}(b)] = (2-b) Pr[b \geq \beta\_{2}(v\_{2})]$, where $b$ is player $1$'s bid, $\beta\_{2}$ is player $2$'s bidding strategy, and $v\_{2}$ is player $2$'s valuation. We can assume $\beta\_{2}(0) = 0$ (because if $\beta\_{2}(0) > 0$, player $2$ can improve upon this by decreasing his bid). Since we are only considering two potential valuations for player $2$, we can assume $\beta\_{2}(v) = av$, for some constant $a \in \mathbb{R}\_{++}$. (That is, given the two points $(0, 0)$ and $(2, \beta\_{2}(2))$, we just draw a line between them).

Observe that $Pr[b \geq av] = Pr[v \leq \frac{b}{a}] = \frac{b}{2a}$, with the last inequality since we have a 50-50 chance on the valuation of player $2$.

Now for a Nash equilibrium, player $1$ seeks to maximize his expected value. This is given by the following optimization problem:

$$\max\_{b} (2-b) \cdot (\frac{b}{2a})$$

This yields the first order conditions:

$\frac{1}{2a} \cdot (2 - 2b) = 0$, and we obtain that $b = 1$ is our only solution for player $1$. This answer should be reasonably intuitive.

Now player $2$ only wins if his valuation is $2$. So he players $\beta\_{2}(2) = 1$ and $\beta\_{2}(0) = 0$.

> **Comments:**
> **username_3**: This part $$ Pr[v \leq \frac{b}{a}] = \frac{b}{2a} $$ seems incorrect. For any given $b$ the probability could take three values: 0, $\frac{1}{2}$, 1. It is not continuous in $a$.  

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