# Expected revenue obtained by the Vickery auction with reserve price $1/2$

I would like to prove that the expected revenue of the Vickery auction with reserve price $1/2$ is $5/12$ when there are one item and two bidders

the distribution of valuations are uniformly between 0 to 1.

I computed the revenue with out the reverse price above to be $1/3$. How should I act with the reverse price? The probability for a sell is where the value of the losing bidder is $p(v)>1-(0.5+x)$, so I thought that it should be $\int_{0}^{1}(0.5-x)^2 dx$ but the answer is $1/12$ while it should be $5/12$.

• What are you assuming about the distribution of valuations? Jun 24, 2014 at 20:53
• @SergioParreiras I edited the question. Thanks! Jun 24, 2014 at 21:49

We have two bidders who play a second-price auction (or Vickrey auction if you prefer). Let the reserve price be $r$. The bidder knows his own valuation but sees the valuation of the rival as uncertain and distributed uniformly in the unit interval.
Conjecture: Bidders with valuations below $r$ bid zero and bidders with valuation above $r$ bid their valuation. The expected utility of bidder 1 of bidding below $r$ is zero and the utility of bidding $b>r$ when his valuation is $v_1$ and bidder 2 follows this strategy is: $$U_1(b|v_1)=\int_{0}^{r} (v_1-r) dy + \int_{r}^{b} (v_1-y) dy\, ,$$ clearly, the best response of player 1 is to choose $b=v_1$ if and only if $v_1>r$. A symmetric reasoning establishes the same for player 2, hence the strategy described in the conjecture is a symmetric, Bayesian-Nash equilibrium (there are others but they are in weakly dominated strategies).
The revenue is then: \begin{align*}R&=\underbrace{\int_0^r\int_0^r 0\, dx \,dy}_{\text{no one bids}}\quad+\underbrace{2\int_0^r\int_r^1 r\,dx\,dy}_{\text{one of the bidders stays out, winner pays }r}+\underbrace{\int_r^1\int_r^1 \min(x,y)\,dx\,dy}_{\text{both active}}\\&=0+2r(r-0)(1-r)+\int_r^1\int_r^y x \,dx\,dy+\int_r^1\int_y^1y\,dx\,dy=\\&=2r^2(1-r)+\int_r^1\left.\dfrac{x^2}{2}\right|_{x=r}^y\,dy+\int_r^1 y(1-y)dy=\\&=2r^2(1-r)+\int_r^1\dfrac{y^2-r^2}{2}\,dy+\int_r^1 y(1-y)dy=\\&=2r^2(1-r)-\frac{(1-r)r^2}{2}+\int_r^1 \left(y-\frac{y^2}{2}\right)\,dy\\&= \frac 32 r^2(1-r)+\left.\left(\frac {y^2}2-\frac {y^3}6\right)\right|_{y=r}^1=\frac 32 r^2(1-r)+\left(\frac 12-\frac 16\right)-\left(\frac {r^2}2-\frac {r^3}6\right)=\\ &=\frac 13 +r^2-\frac 43 r^3\end{align*} $$r=\frac 12\Longrightarrow R=\frac 5{12}\quad\blacksquare$$
Remark: $R$ is increasing at $r=0$ (as in Myerson (1983) 's optimal auction, the optimal reserve price is positive)
• Does this mean that the BNE for a second price auction with a reserve price is still $b_i = v_i$ ? i.e. the best response for each player is still to bid their true valuation (assuming it's greater than $r$)