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My problem.

Consider the following auction for a single object. There are $n \geq 2$ bidders. They submit their bids simultaneously. The object is allocated to the player who submits the largest bid. If the winner's bid is $b$ he pays the amount $\alpha b$ where $\alpha$ is a positive number. The losers do not pay anything. Ties are broken randomly, with equal probabilities among all the players who submit the largest bid. The bidders' valuations for the object are private information. In particular, each player $i$ knows his own valuation $v_{i}$ which is distributed uniformly over the unit interval. The valuations are distributed independently across the players.

Construct the symmetric BNE of the game. (Assume that the bidding strategy $b :[0,1]$ $\rightarrow$ $\mathbb{R}$ is increasing).

Solution

Let $b :[0,1]$ $\rightarrow$ $\mathbb{R}$ R denote the equilibrium bidding strategy. Then for every $v \in [0, 1]$, we must have:

$v=arg$ $max_{w}((v-\alpha b(w))w^{n-1}$

We compute the first order conditions at $v$ and obtain:

$-\alpha b^{'}(v)v^{n-1}+(n-1)(v-\alpha b(v))v^{n-2}=0$

Which we can simplify as:

$-\alpha b^{'}(v)v+(n-1)(v-\alpha b(v))=0$

The solution to this differential equation is linear: $b (v) = Av$ where $A$ satisfies

$-\alpha Av+(n-1)(v-\alpha Av)=0$

Thus, the equilibrium bidding strategy is

$b(v)=\frac{n-1}{n \alpha}v$

Is there ant alternative way to calculate the BNE for the problem presented above

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There is another way to compute the symmetric BNE in increasing strategy.

Let $U(v)$ denote the expected utility of a player in equilibrium when his type is $v$: Given that the bidding strategy is increasing, a player with type $0$ will get the good with probability zero.

Thus he/she must bid zero and $U(0) = 0$. For any other $v > 0$, the probability that the player gets the good is $Q (v) = v^{n-1}$ (this is the probability that all the other players have a type lower than $v$) From the classes on mechanism design, we know that:

$U(v) = U(0) + \int_{0}^{v} Q(x) \,dx = \int_{0}^{v} x^{n-1} \,dx = \frac{v^{n}}{n}$

On the other hand, we can write $U(v)$ as

$U(v)= (v-\alpha b(v)))v^{n-1}$

Therefore

$(v-\alpha b(v))v^{n-1} = \frac{v^{n}}{n}$

and

$b(v) = \frac{n-1}{n \alpha} v$

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  • $\begingroup$ "Thus he/she must bid zero" Why though? If everyone with valuation $0$ would bid $1/10$ and follow an increasing strategy, bidding $1/10$ would still get you the item with probability zero. It seems like this line is not necessary to declare $U(0) = 0$. The player wins with probability $0$, thus they get an expected utility of $0$. $\endgroup$
    – Giskard
    Jul 23 at 17:05

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