I am having a problem on which I have 4 players bidding in second price auction with reserved price (r). I need to find the expected value of 2nd bigger number when there are 3 players or 4 bidding over r, bid>r. For example, if 3 players bids over r, I need to find the expected value of the second bigger bid. So far I've found that when 2 players bids over r the expected value is equal to $a + (b-a)/3$ when we have uniform distribution in [a,b]. So in my example in [r,1] will be equal to $ r + (1-r)/3 $ I need the expected value of 2nd bigger value of n values. Haven't thought a way to do so. Found this but didn't help me Expected value of 2nd-smallest out of 3 random variables. Thanks in advance and sorry if my question is amateur.
Game Theory – Expected Value of 2nd-Bigger Number in Uniform Distribution
auction-theorygame theoryorder-statistics
Related Solutions
First, we want to see which strategies the bidders will play: From a real bidders perspective, you can look at the reserve price as if it were the bid of a third bidder, regarding both allocation and payment rule. Therefore, as we know that the sealed bid second price auction is truthful, we can expect the bidders to bid their true value.
The probabilities of the bidders value profiles ($v_i, v_2$) are:
You can count the profiles if it helps you to understand the following case distinction:
- Assume we set $r > 2$, then the expected revenue is 0 (as item is always destroyed).
- Assume we set $2 \geq r > 1$, then the expected revenue is $$ E_{rev} = \frac{1}{9} * 2 + \frac{4}{9} * r + \frac{4}{9} * 0 $$ which is highest at r = 2 and takes the value $1.11$.
- Assume we set $1 \geq r \geq 0$ , then the expected revenue is $$ E_{rev} = \frac{1}{9} * 2 + \frac{3}{9} * 1 + \frac{4}{9} * r + \frac{1}{9} * 0$$ which is highest at r = 1 and takes the value $1$.
Therefore, the expected revenue is maximized with $r = 2$.
Note that I assumed your "above r" is actually "greater or equal than r", otherwise, the case, where the highest bid equals the reserve price, would not be defined.
Best, miweiss
The only equilibrium in weakly dominant strategies is truthful bidding, so the bidding function is $b_i(v_i) = v_i$ for both players. The seller's revenue $R$ (equal to the winner's payment) is a function of $v_1,v_2,r$: $$R = \left\{ \begin{array}{ll} 0 & \mbox{if } r > \max\{v_1, v_2\}\\ r & \mbox{if } v_2 < r < v_1\\ r & \mbox{if } v_1 < r < v_2\\ v_2 & \mbox{if } r < v_2 < v_1\\ v_1 & \mbox{if } r < v_1 < v_2\\ \end{array}\right.$$ 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 $$E(R) = 2 \left[ \int_r^1 \int_0^r r dv_2 dv_1 + \int_r^1 \int_r^{v_1} v_2 dv_2 dv_1 \right]= \frac{1+3r^2-4r^3}{3}$$
It has occurred to me that you might be interested in the expected payment of a given player (when he is a winner). The unconditional expectation is of course just half of the expected revenue, or $E(R)/2$. To find his expected payment conditional on being the winner divide this by the probability that his bid beats both $r$ and $v_j$ (the bid from the opponent), which is $(1-r^2)/2$, to find $$E \left(\mbox{ payment of } i \mid \mbox{ $i$ is the winner} \right) = \frac{E(R)}{2(1-r^2)}$$
Best Answer
I think the expected value of the second highest value in $n$ uniform random variables on $[a,b]$ is $a + (b-a)\frac{n-1}{n+1}$.
You can first consider $n$ uniform random variables on $[0,b-a]$, then the pdf of the second highest value is $$f(x) = \frac{n!}{(n-2)!}(\frac{x}{b-a})^{n-2}(1-\frac{x}{b-a}),$$ which has mean value $(b-a)(\frac{n-1}{n+1})$.
Maybe you can see https://i.sstatic.net/UFfYC.jpg and Expected value of $2$nd highest draw from uniform dist out of n draws for detailed discussion.