0
$\begingroup$

Consider a third-price sealed-bid auction auction with three bidder indexed $i=1,2,3$ and private, independent values $v_1 \sim U(0,1)$. Bidders simultaneously and independently make a bid $b_i$

The question then is:

Is it a BNE for each bidder to bid truthfully, i.e. $b_i=v_i$? Explain.

Based on this I formulated this answer:

No. Consider an example that shows that this is not a BNE. Lets assume $v_1 > v_2 > v_3$ and that all players bid their values: $(b_1,b_2,b_3) = (v_1,v_2,v_3)$. Then player 2 does not win the auction, so has payoff zero ($u_2 = 0$). If player 2 increases his bid to $b_2 = v_1 + \varepsilon$, where $\varepsilon > 0$, then player 2 wins the auction and gets payoff $u_2 = v_2 - v_3 > 0$, which is more then when he bids his value $v_2$. Hence, there is a profitable deviation for the player with the middle value, so the proposed equilibrium is not a BNE.

My doubt about this answer is that the bidder with the middle value does not this, since he does not know the values of the other bidders. For rejecting the proposed BNE, do I need to take into account the posterior beliefs of player 2 being the middle bidder (using the CDF of the uniform distribution) or is it enough that in one state of the world (when he is the middle bidder) bidding higher than his value, and higher than player 1's value, is a profitable deviation?

$\endgroup$
0
$\begingroup$

Since the bidders bid simultaneously, you don’t need posterior beliefs (at least in the sense in which I would use that term in this context, namely beliefs that the players form based on the actions of other players). You only need their prior beliefs about the private values, which are given by the uniform distributions.

Say I’m player $1$. If the other players bid their true values $v_2$ and $v_3$ and I bid $b_1$, then I win the auction with probability $b_1^2$ and the expected price in that case is $\frac{b_1}3$, so my expected profit is

$$ b_1^2\left(v_1-\frac{b_1}3\right)\;. $$

Setting the derivative with respect to $b_1$ to zero yields

$$ 2b_1v_1-b_1^2=0 $$

and thus $b_1=2v_1$. So the strategy profile $b_k=v_k$ is not in equilibrium.

Note that in a second-price auction with three bidders, the expected price would have been $\frac23b_1$, so the maximum expected profit would indeed have been at $b_1=v_1$ in that case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you!. Perhaps the term posterior beliefs was poorly chosen. What I meant was the beliefs of the player after learning his own value. But I see now that you do take this into account in calculating the probability of winning the auction $b_1^2$. In a second question, I calculated the optimal strategy by finding the expected utility by integrating over the probability distributions of the other players and then taking the FOC. This also gave $b_1 = 2v_1$. Does you solution rely in the assumption that the other players bid their true value or does it hold more generally? $\endgroup$ – AlakReis Feb 4 at 20:34
  • $\begingroup$ @avs: I start out with "If the other players bid their true values $v_2$ and $v_3$". I wouldn't call that an assumption -- your question was whether each player bidding their true value is a Bayesian Nash equilibrium, so by definition I had to check whether bidding my true value $v_1$ is the best response to the other two players bidding their true values $v_2$ and $v_3$, no? About integrating over the probability distributions: I did that, too, implicitly; it just so happens that the values are straightforward to see without actually performing the integrals; if you want I can spell them out. $\endgroup$ – joriki Feb 4 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.