Consider the buy-it-now price often included in online auctions. Suppose that 2 bidders in an ascending oral (English) auction bid for an object. Each has values i.i.d uniformly on $$[0,1]$$. Suppose the auction has a buy-it-now price of $$B \ge 1/2$$; either bidder can end the auction by paying $$B$$. Assume that there's an equilibrium when the bidding reaches $$p(x)$$ that a bidder with value $$x\geq B$$ will pay $$B$$ (i.e. $$p(x)$$ is the cutoff when a bidder with value $$x$$ pays the buy it now price). Also assume also that $$p'(x)<0$$. i) What price, conditional on winning, does a buyer with valuation $$x$$ expect to pay (as a function of $$x$$ and $$p(x)$$)?

It seems reasonable to assume that bidders will not bid more than $$p(x)$$. Hence, $$b_1,b_2\in[0,p(x)]$$. Then, it makes sense to consider two cases: where $$x. Here the expected payment is simply the expected payment of the lower valuation bidder, namely, $$\frac{p(x)}{2}$$.

Secondly, we should consider the case where $$x>p(x)$$: the expected payment is here, the probability of facing a below-$$p(x)$$-valuation opponent, times their bid, plus the probability of facing an above-$$p(x)$$-valuation opponent times the probability of winning (ties are broken randomly) times $$p(x)$$

$$p(x)*\frac{p(x)}{2}+(1-p(x)*\frac{1}{2}*p(x))=\frac{1}{2}p(x)$$

ii) Assuming risk neutrality, using the fact that the expected payment with valuation $$x$$ is the same without the buy-it-now option (namely, $$1/2x$$), what is $$p(x)?$$ (Note: you should get a quadratic equation; one root will be $$p(x)=x$$, but this is not the solution, given $$p'(x)<0$$)

I'm not sure how to solve this part, and I'm doubting my answer to the first part given that I don't know how it enables me to derive $$p(x)$$.

• It makes that $p(x)$ should be decreasing in $x$: the higher a bidder's valuation, the earlier she will accept to pay the buy-now price. For higher valuation bidders, there is more room for the price to increase beyond the buy-now price. – pafnuti Apr 5 '18 at 0:06

Here is how I would approach the first question:

• We are conditioning on the fact that you have won. Given the symmetry, this implies that your valuation $$x$$ is greater than your opponent's valuation $$y$$.

• Given that you have won, you either pay the 'auction price' or $$B$$. If you don't pay $$B$$, you pay somewhere between $$0$$ and $$p(x)$$. The expected auction price is the expected value of the second highest draw, so actually $$p(x)/3$$ - not $$p(x)/2$$ as you seem to be saying (intuitively, the expected valuation of the lowest bidder must be below the unconditional expected valuation).

• The probability that you pay the auction price is the probability that your opponent's valuation $$y$$ is below your cutoff $$p(x)$$. This is because your opponent will drop out of the bidding as soon as the price exceeds her valuation. $$Pr(y < p(x)) = p(x)$$ since $$y$$ is uniform on $$[0, 1]$$.

• Therefore, assuming you win, the probability that you pay $$B$$ is $$1 - p(x)$$.

• Putting these claims together, I conclude that your expected payment conditional on winning equals: $$(p(x)/3)p(x) + B(1 - p(x))$$.

• To check the plausibility of this, set $$p(x) = 1$$ so effectively the buy it now option disappears, and yields an expected payment of $$1/3$$. This is the standard result for $$2$$ buyers and a uniform distribution.