# Common Value Second Price Auction - Winner's Expected Payment

How would you find the winner's expected payment in a second price auction with common values?

For example, suppose we have the case where two players are random variables $x_1$ and $x_2$ and are uniformly distributed on $[0,1]$ and independent. Their valuations are $v_1 = v_2 = x_1 + x_2$ so they are effectively the same.

I know the BNE is now $b_i(x_i) = 2x_i$ instead of the truthful bidding of the regular second price auction but how can we calculate the winner's expected payment using this new information? Is it the same procedure with simply a new BNE?

• Is this winner's payment or seller's revenue?
– mlc
Commented Apr 2, 2017 at 18:16
• Winner's payment which I think is conditional on being the winner.
– guy
Commented Apr 2, 2017 at 18:33

Hint:

If you can show that the players each bid twice their own value then, since this is still a second price auction, the expected payment is the lower bid, which is twice the minimum of the two random variables (i.e. twice the expected payment of a regular second price auction)

• That makes sense. I'll try to work it out and post here when I get something.
– guy
Commented Apr 2, 2017 at 18:16

(This answer partly recycles material from my answer to Second Price Auction with Reserve Prices - Expected Payment of Winner. You also asked that question, and I am sorry you did not see how to apply it here, especially after the hint by @Henry.)

The equilibrium bidding function is $b_i (v_i) = 2v_i$. Then the seller's revenue $R$ (equal to the winner's payment) is a function of $v_1,v_2$: $$R = \left\{ \begin{array}{ll} v_2 & \mbox{if } v_2 < v_1\\ v_1 & \mbox{if } v_1 < v_2\\ \end{array}\right.$$ 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 $$E(R) = 2 \left[ \int_0^1 \int_0^{v_1} v_2 dv_2 dv_1 \right]= \frac{1}{3}$$ For the expected payment conditional on being the winner divide this by the probability that $i$'s bid $b_i$ beats $v_j$ (the bid from the opponent), which is $1/2$, to find $$E \left(\mbox{ payment of } i \mid \mbox{ i is the winner} \right) = \frac{2}{3}$$