I would like to know what bidder i's expected payoff looks like in a second-price auction with $N=\{1,2,...,n\}$ bidders, where each bidder $i\in N$ has independent and uniform distributed valuations $v_i\sim U(0,1)$, and the auctioneer sets a reserve price, $r$.
I assume that each bidder already knows $r$ like she knows her own true valuation, but only knows the distribution of the other bidders' valuations, and she assumes that all other bidders bid their true valuations $b_{-i}^*=v_{-i}$

Hence, bidder i's payoff function is: $u_i(b_i,b_{-i},r) = \left\{ \begin{array}{lr} v_i-\max\{b_{-i}\} & \text{if } b_i>\max\{b_{-i}\}>r \\ v_i-r & \text{if } b_i>r>\max\{b_{-i}\}\\ 0 & \text{if } b_i<r \vee b_i<\max\{b_{-i}\} \end{array} \right.$

I am unsure exactly what the expected payoff looks like, but I have assumed that it would be something like: $ \mathbb{E}[u_i] = \mathbb{P}(b_i>\max\{b_{-i}\}>r)\cdot \mathbb{E}\left[ v_i-\max\{b_{-i}\} | b_i>\max\{b_{-i}\}>r \right] \\ +\mathbb{P}(b_i>r>\max\{b_{-i}\})\cdot \mathbb{E}\left[ v_i-r | b_i>r>\max\{b_{-i}\} \right]$
but I don't know where to go from here..

Any help is highly appreciated!

  • 1
    $\begingroup$ Yes, so the reserve price r is set by the auctioneer to maximize his expected revenue. Thus, each bidder just sees the reserve price as if it was an additional bidder's bid. Then one could argue that it is still optimal for the bidder to bid their true values, which is not necessarily equal to the reserve price. But now there is a probability that the second-highest bidder bids below the reserve price, in which case the highest bidder doesn't pay the second-highest bid, but pays the reserve price instead. If I understand your question correctly? $\endgroup$ Commented Nov 2, 2021 at 8:33
  • $\begingroup$ Yes, I meant to ask why you do not assume that $i$ bids her valuation. I mixed up the notations, sorry. $\endgroup$
    – Giskard
    Commented Nov 2, 2021 at 8:36

1 Answer 1


Let $n \geq 2$. Observe that $$ \mathbb{E}[u_i] = \mathbb{E}[u_i|v_i > r]P(v_i > r) + \mathbb{E}[u_i|v_i \leq r]P(v_i \leq r) = \mathbb{E}[u_i|v_i > r]P(v_i > r)$$ since $\mathbb{E}[u_i|v_i \leq r] = 0$. Moreover, $P(v_i > r) = 1 - r$ if the values are standard uniform. So to compute the expected payoff $\mathbb{E}[u_i]$, all that remains is to compute $\mathbb{E}[u_i|v_i > r]$.

Let $p$ denote the highest bid submitted by your opponents. We can then decompose $$\mathbb{E}[u_i|v_i > r] = \mathbb{E}[u_i|v_i > r, p \leq r]P(p \leq r) + \mathbb{E}[u_i|v_i > r, p > r]P(p > r)$$ Starting with the first term, $$ \mathbb{E}[u_i|v_i > r, p \leq r] = \mathbb{E}[v_i - r|v_i > r] = \frac{1+r}{2} - r = \frac{1-r}{2}$$ More obviously, $P(p \leq r) = P(v_j \leq r)^{n-1} = r^{n-1}$ and so $P(p > r) = 1 - r^{n-1}$. Using symmetry of bidders and results on order statistics, one can see that

\begin{equation} \begin{split} \mathbb{E}[u_i|v_i > r, p > r] &&= \mathbb{E}[u_i|\text{win}, v_i > r, p > r] P(\text{win}) \\ &&= \left(r + (1-r)\left(\frac{n}{n+1}\right) - r - (1-r)\left(\frac{n-1}{n+1}\right) \right)\frac{1}{n} \\ &&= \frac{1-r}{n(n+1)} \end{split} \end{equation} Putting this all together, we have $$ \mathbb{E}[u_i|v_i > r] = \frac{1-r}{2} \times r^{n-1} + \frac{1-r}{n(n+1)} \times (1 - r^{n-1}) $$ and so (multiplying by $1 - r$) $$ \mathbb{E}[u_i] = \frac{(1-r)^2 r^{n-1}}{2} + \frac{(1-r)^2 (1 - r^{n-1})}{n(n+1)} $$ Observe that, if $r = 1$, then $\mathbb{E}[u_i] = 0$ as one would expect. Moreover, if $r = 0$ (no reservation price), then one obtains the (hopefully familiar?) expression $$ \mathbb{E}[u_i] = \frac{1}{n(n+1)} $$

  • $\begingroup$ I'm still not entirely sure if I get how $\mathbb{E}[u_i|v_i>r,p>r]$ becomes $(r+(1-r)\frac{n}{n+1}-r-(1-r)\frac{n-1}{n+1})\frac{1}{n}$, so I can't really figure out if there is a mistake in that part. But I would think that $\mathbb{E}[u_i|v_i>r,p<r]=v_i-r$ since the bidder knows her own valuation and the reserve price, and doesn't have to take the expectation of $v_i$. $\endgroup$ Commented Nov 2, 2021 at 13:44
  • $\begingroup$ Sure, if you want to also condition on $v_i$. But I thought you wanted the unconditional expected utility? (In which case you need to average over all possible $v_i$ as I have done.) PS I suspect that the error may indeed be with the first bit you mention, maybe you can have a go at that? $\endgroup$
    – afreelunch
    Commented Nov 2, 2021 at 14:50
  • $\begingroup$ OK mistake found! I somehow divided by $1-r$ at the end... $\endgroup$
    – afreelunch
    Commented Nov 2, 2021 at 15:18
  • $\begingroup$ Ahh, of course! you calculate the ex-ante expected utility, while I was originally thinking of the interim expected utility. Though on second thoughts, I do need both the interim and ex-ante expected utility. And the interim shouldn't be any problem for me to derive when I have the expression for the ex-ante expected utility. But would you mind spelling out how you got $E[u_i|v_i>r,p>r]$? And great that you found the mistake :D $\endgroup$ Commented Nov 2, 2021 at 15:21
  • 1
    $\begingroup$ It's a bit complicated actually.... The $1/n$ term is because of the symmetry: each player is equally likely to win. The other term comes from thinking about the expected value of the maximum and highest highest of $n - 1$ uniform RVs on the interval $[r, 1]$. $\endgroup$
    – afreelunch
    Commented Nov 2, 2021 at 15:23

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