# Expected value of $2$nd highest draw from uniform dist out of n draws

Jane wants to auction off an item, but does not know where to go to find bidders. David offers to find bidders for her, but will charge her $$\10$$ per bidder he gets to show up. Each bidder will uniformly value the item between $$[500, 1000)$$. The highest bidder will win the item and pay the second-highest bidder's price (Vickrey auction). How many bidders should Jane pay David to find?

I need to maximize the difference between $$E[\text{second largest bid out of n bidders}]$$ and $$10n$$. I'm not sure how to find $$E[\text{2nd largest big}]$$

• A guess: around 10 bidders.
– mjw
Sep 4, 2019 at 21:34
• You can try to solve this numerically using a Monte Carlo simulation.
– mjw
Sep 4, 2019 at 21:35
• This topic is discussed in depth in Casella and Berger Section 5.4. I've put the relevant text in an imgur post imgur.com/a/UFm12aK .
– Mark
Sep 4, 2019 at 21:47

This topic is discussed in depth in Casella and Berger Section 5.4. I've put the relevant text in an imgur post https://i.stack.imgur.com/UFfYC.jpg .

The second largest number is the n-1st order statistic, $$X_{(n-1)}$$. Theorem 5.4.4 says that $$E[X_{(n-1)}]=(n-1)/(n+1)$$ for a uniform(0,1), but I believe that if you follow the proof you will find that $$E[X_{(n-1)}]=500+ ((n-1)/(n+1))(1000-500)$$ for a unif(500,1000) (I'm assuming you are talking about a continuous uniform distribution, you may mean discrete since you specified which endpoints were included, which would change the answer a bit, but since you just need integer precision in n, the final answer may be the same).

You are looking to maximize $$E[X_{(n-1)}]-10n$$. So you can differentiate in $$n$$ and find the local maxima and check the 2 nearest integer values of $$n$$ to see which is larger.

Instead of differentiating, you know you want to find where $$[(n)/(n+2)-(n-1)/(n+1)]*500=10$$ (i.e. adding one person costs 10 dollars but also increases the expected value by of the second highest bid by 10). That happens at n=8.5

For 8 bidders, $$E[X_{(7)}]= 500+\frac{7}{9}500=888.89$$

For 9 bidders, $$E[X_{(8)}]= 500+\frac{8}{10}500=900$$

For 10 bidders, $$E[X_{(9)}]= 500+\frac{9}{11}500=909.09$$

So, the 10th bidder only provided \$9.09 worth of value. Thus you should have 9 bidders and you would expect to make \$890.

• I see that makes sense. I'm doing this question for quant trading interviews and i only have around 90 seconds to answer it. This would require I have this n-1 order statistic memorized right? Theres no other way to do it in that time frame? Differentiating $$500 + ((n-1)/(n+1))(1000-500) - 10n$$ Gets me $$- 10 (n^2 +2n -99)/(n+1)^2$$ Sep 4, 2019 at 22:02
• Can you explain where you get the function $$[(n)/(n+2) - (n-1)/(n+1)] * 500 = 10$$ Where does the $$(n)/(n+2)$$ come froM? Sep 4, 2019 at 22:09
• I think that the method I proposed is definitely uses stronger tools than necessary. Since it's uniform you can "know" that the EV for the highest bidder will be 500+ $n/(n+1)*500$ and the second highest bidder is $(n-1)/(n+1)*500$. Then you can do the part where I said instead of differentiating..., basically finding the cutoff when the next guy provides less than \$10 worth of value.
– Mark
Sep 4, 2019 at 22:11