I am having trouble understanding a second price auction with a reserve price, i.e. a second price auction where each player’s valuation is uniformly distributed on $[0, 1]$, and the two valuations are independent random variables. Here we have a fixed reserve price $r$ > 0 that is common knowledge with the buyers: if the two bids are below the threshold $r$, then there’s no winner.

This thread about second price auctions explained fairly well the logic in finding the expected revenue for the auctioneer, however, I don't understand how we can find the expected payment for the winner.

To me it seems like we have 3 possible cases for the winner:

  1. Both players bid under $r$ and there is no winner, therefore, expected payment is zero.
  2. One player bids below $r$ and the other exceeds $r$, making the expected payment $r$.
  3. Both players bid above the reserve, making the game reduce to a regular second price auction, so the expected payment by the winner is the loser's bid.

Does my logic make sense? How then do I find the winner's payment on expectation? I having trouble setting these problems up so any advice on that is welcome as well.

  • $\begingroup$ In case 3., the winner pays the loser's bid. So far, your reasoning is correct. $\endgroup$
    – mlc
    Commented Apr 1, 2017 at 17:12
  • $\begingroup$ How then do I find the winner's payment on expectation? $\endgroup$
    – guy
    Commented Apr 1, 2017 at 18:07
  • $\begingroup$ Your description of (2) suggests the reserve is in fact a third bid, made by the existing owner $\endgroup$
    – Henry
    Commented Apr 4, 2017 at 7:49

1 Answer 1


The only equilibrium in weakly dominant strategies is truthful bidding, so the bidding function is $b_i(v_i) = v_i$ for both players. The seller's revenue $R$ (equal to the winner's payment) is a function of $v_1,v_2,r$: $$R = \left\{ \begin{array}{ll} 0 & \mbox{if } r > \max\{v_1, v_2\}\\ r & \mbox{if } v_2 < r < v_1\\ r & \mbox{if } v_1 < r < v_2\\ v_2 & \mbox{if } r < v_2 < v_1\\ v_1 & \mbox{if } r < v_1 < v_2\\ \end{array}\right.$$ where I have ignored equalities because they occur with zero probability and thus do not affect the expected value. This function is symmetric around the bisector, so it suffices to compute the expected value for $v_1 > v_2$ and double it.

The expected revenue is thus $$E(R) = 2 \left[ \int_r^1 \int_0^r r dv_2 dv_1 + \int_r^1 \int_r^{v_1} v_2 dv_2 dv_1 \right]= \frac{1+3r^2-4r^3}{3}$$

It has occurred to me that you might be interested in the expected payment of a given player (when he is a winner). The unconditional expectation is of course just half of the expected revenue, or $E(R)/2$. To find his expected payment conditional on being the winner divide this by the probability that his bid beats both $r$ and $v_j$ (the bid from the opponent), which is $(1-r^2)/2$, to find $$E \left(\mbox{ payment of } i \mid \mbox{ $i$ is the winner} \right) = \frac{E(R)}{2(1-r^2)}$$

  • $\begingroup$ Thanks for the clear explanation. For some reason it wasn't clear in my head that the expected revenue is equal to the expected payment. $\endgroup$
    – guy
    Commented Apr 1, 2017 at 19:18
  • $\begingroup$ I have added the computation for an alternative interpretation of "winner's payment". In auction theory, this latter is usually ignored, but it is easily deduced. $\endgroup$
    – mlc
    Commented Apr 1, 2017 at 19:23
  • $\begingroup$ Thank you, I knew there was something that I was missing and that is exactly it. Do you know any good references that explain auction theory like you? $\endgroup$
    – guy
    Commented Apr 1, 2017 at 19:25
  • $\begingroup$ Try Chapter 9 (Auction Theory) in the third edition of the Advanced Economic Theory textbook by Jehle and Reny. (Beware: this is a textbook for advanced graduate students or first-year doctoral students.) $\endgroup$
    – mlc
    Commented Apr 1, 2017 at 20:08
  • $\begingroup$ If $r \approx 1$ then the conditional expected payment at the end should be close to $1$, but seems to be close to $0$. I think your $(1+r)/2$ might need revisiting $\endgroup$
    – Henry
    Commented Apr 4, 2017 at 7:54

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