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 $x<p(x)$. 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 $p(x)$


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 $p(x)$.

  • $\begingroup$ 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. $\endgroup$
    – pafnuti
    Commented Apr 5, 2018 at 0:06

1 Answer 1


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.


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