<blockquote>
  <p>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 <span class="math-container">$[0,1]$</span>. Suppose the auction has a buy-it-now price of <span class="math-container">$B \ge 1/2$</span>; either bidder can end the auction by paying <span class="math-container">$B$</span>. Assume that there's an equilibrium when the bidding reaches <span class="math-container">$p(x)$</span> that a bidder with value <span class="math-container">$x\geq B$</span> will pay <span class="math-container">$B$</span> (i.e. <span class="math-container">$p(x)$</span> is the cutoff when a bidder with value <span class="math-container">$x$</span> pays the buy it now price). Also assume also that <span class="math-container">$p'(x)&lt;0$</span>. i) What price, conditional on winning, does a buyer with valuation <span class="math-container">$x$</span> expect to pay (as a function of <span class="math-container">$x$</span> and <span class="math-container">$p(x)$</span>)? </p>
</blockquote>

<p>It seems reasonable to assume that bidders will not bid more than <span class="math-container">$p(x)$</span>. Hence, <span class="math-container">$b_1,b_2\in[0,p(x)]$</span>. Then, it makes sense to consider two cases: where <span class="math-container">$x&lt;p(x)$</span>. Here the expected payment is simply the expected payment of the lower valuation bidder, namely, <span class="math-container">$\frac{p(x)}{2}$</span>. </p>

<p>Secondly, we should consider the case where <span class="math-container">$x&gt;p(x)$</span>: the expected payment is here, the probability of facing a below-<span class="math-container">$p(x)$</span>-valuation opponent, times their bid, plus the probability of facing an above-<span class="math-container">$p(x)$</span>-valuation opponent times the probability of winning (ties are broken randomly) times <span class="math-container">$p(x)$</span></p>

<p><span class="math-container">$$p(x)*\frac{p(x)}{2}+(1-p(x)*\frac{1}{2}*p(x))=\frac{1}{2}p(x)$$</span></p>

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

<p>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 <span class="math-container">$p(x)$</span>.</p>

CC BY-SA 3.0
2018-04-05T00:06:48.867Z
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
<game-theory><auctions>
Auction with "buy-it-now" option
user526463
# Auction with "buy-it-now" option

(Asked by: username_0 on 2018-04-04)

> 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 $p(x)$.

**Tags:** game-theory, auctions

> **Comments:**
> **username_1**: 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.  

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## 0 answers

