What is the expected payoff for a bidder in a second-price auction with N uniform distributed bidders, when the auctioneer sets a reserve price?

Question

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!

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 $$ \mathbb{E}[u_i] = \frac{1}{n(n+1)} $$