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