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?
Best Answer
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 $P(A)$.