# Expected Revenue of 2nd Price Auction with Reserve Price

For a second price auction with two bidders, $n=2$, whose valuations are i.i.d on $U[0,1]$ and a reserve price $r$, I am trying to solve this problem using order statistics rather than the classic way shown here.

I calculated the first order statistic, $f_{X_1}(x)=2-2x$ representing the PDF for $X$, where here $X$ is the random variable representing the second highest bid. This also means the CDF of $X$ is $F_{X_1}(x)=2x-x^2$.

So, I thought the expected revenue would essentially be:$$E(R)=Pr[X\geq r]\times[\text{winner's payment if }X\geq r]+Pr[X\leq r]\times[\text{winner's payment if }X\leq r]+Pr[\text{both bids}\leq r]\times[\text{winner's payment if no one bids}] \tag{1}$$

Since the winner's payment is zero if neither bid hits the reserve price $r$, we can discard the 3rd term. I then rationalized that the winner's payment if the second highest bid is less than the reserve price, is simple $r$. Therefore the equation reduces to: $$E(R)=Pr[X\geq r]\times[\text{winner's payment if }X\geq r]+Pr[X\leq r]\times[r] \tag{2}$$

Using the CDF above, I can find $Pr[X\leq r]=F_{X_1}(r)=2r-r^2$. Similarly, $Pr[X\geq r]=1-F_{X_1}(r)=1-2r+r^2$. Our equation is now: $$E(R)=(1-2r+r^2)\times[\text{winner's payment if }X\geq r]+r(2r-r^2) \tag{3}$$

This part I was least confident what the winner's payment would be if the second highest random variable was greater than the reserve. I know the winner's payment is the second highest value and on expectation it is $\frac{1}{3}$ if no reserve price existed so I found the expected value of $X$ given $X \geq r$, i.e. $$\text{winner's payment if }X\geq r = E(X | X \geq r)=\frac{1-3r^2+2r^3}{3(1-2r+r^2)} \tag{4}$$ Putting all this together gave me: $$E(R)=\frac{1}{3}+r^2-\frac{1}{3}r^3$$ This is close to the correct answer of $E(R)=\frac{1}{3}+r^2-\frac{4}{3}r^3$ but not quite right. Where am I going wrong?

You need to be careful how you handle events in (1). The event $\{X \le r\}$ is the union of two events: A) there is one bid above $r$ and one bid below; B) both bids are below $r$.

You recognise B) correctly in the rest of your formula, but miscompute $P(X \le r)$ as $P (A \cup B)$ instead of $P(A)$.

• I still don't see it, the event ${X \leq r}$, that is the event that the second highest number is less than $r$ occurs when $A$ or $B$ occur as you have defined them. Therefore, $P(X \leq r)=P(A \cup B)=P(A) + P(B)$, we can disregard $P(B)$ since the payment when this occurs is zero, therefore we need to find $P(A)$. But how do we find the probability that one bid is below and one is above $r$ using order statistics?
– guy
Commented Apr 25, 2017 at 0:35
• @tbone: find $P(\min < r < \max)$.
– mlc
Commented Apr 25, 2017 at 5:16
• still not right because the $P(min<r<max)=P(X_{(1)}<r<X_{(2)})=P(X_{(1)}<r)P(r<X_{(2)})=P(X_{(1)}<r)[1-P(X_{(2)}<r]=(2r-r^2)(1-r^2)=2r-r^2-2r^3+r^4$ which is way off from the right answer.
– guy
Commented Apr 25, 2017 at 23:48
• You cannot assume independence between order statistics.
– mlc
Commented Apr 26, 2017 at 4:52