Suppose $n$ bidders participate in a second price sealed-bid auction, in which one object is being sold. Each bidder $i$ values the object at $v_i$, and each $v_i$ is independently and uniformly distributed on $[0,1]$. The $v_i$'s are private information.

The rules of the auction is that the highest bidder, say $i$, wins the object and pays the a price equal to $\max_{j\ne i}v_j$. If one doesn't win the object, then payment is zero.

We know that it's optimal for each bidder to submit a bid equal to their true value, i.e. $b_i=v_i$ for all $i=1,\dots,n$.


For a bidder $i$ with value $v_i$, what is his expected payment?

What I've done...

I know that the seller's expected revenue is just the the second highest order statistic from the $n$ independent draws from the uniform distribution, and I have no problem in calculating that to be $(n-1)/(n+1)$.

From a bidder's perspective, expected payment should be $$ \mathbb E\left[\max_{j\ne i}v_j\middle\vert v_j\le v_i,\;j\ne i\right]\Pr(v_j\le v_i,\;j\ne i) $$

It is computing the conditional expectation that I'm having trouble with. In particular, I'm having difficulty deriving the conditional distribution for the second highest order statistic given that $v_i$ is the highest one.

Any help would be appreciated.


1 Answer 1


Hint: The probability of bidder $i$ winning is the probability that all other valuations are lower than his. What is that?. Given that he wins, the other valuations are now uniform on $[0,v_i)$. What is the expected maximum of $n-1$ of these?

  • $\begingroup$ Thanks! So conditional on $v_i$ being the highest, $\Pr(v_j\le x|v_j\le v_i)=F(x)=v_i^{-1}x$. The distribution of the highest conditional order statistic would be $F^{n-1}(x)=\left(v_i^{-1}x\right)^{n-1}$. The conditional expectation is $\int_0^{v_i} xdF^{n-1}(x)=(n-1)v_i/n$. $\endgroup$
    – Herr K.
    Commented Sep 30, 2013 at 3:41
  • $\begingroup$ @KevinC: what is $x?$ I would claim that the probability that a given bid is lower is $v_i$, so the chance that they all are is $v_i^{n-1}$ $\endgroup$ Commented Sep 30, 2013 at 3:50
  • $\begingroup$ Perhaps I wasn't careful enough with my notations? The answer to your first question in the hint is $\Pr(v_j\le v_i,\;j\ne i)=v_i^{n-1}$. My comment actually answers the second question in the hint. $F(x)$ is the distribution of the rest of the $v_j$'s conditional on $v_i\ge v_j\;\forall j$, which is, like you said, uniform on $[0,v_i]$ $\endgroup$
    – Herr K.
    Commented Sep 30, 2013 at 4:07

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