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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 $E[\text{2nd largest big}]$

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  • $\begingroup$ A guess: around 10 bidders. $\endgroup$
    – mjw
    Sep 4, 2019 at 21:34
  • $\begingroup$ You can try to solve this numerically using a Monte Carlo simulation. $\endgroup$
    – mjw
    Sep 4, 2019 at 21:35
  • $\begingroup$ 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 . $\endgroup$
    – Mark
    Sep 4, 2019 at 21:47

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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.

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  • $\begingroup$ 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 $$ - 10 (n^2 +2n -99)/(n+1)^2 $$ $\endgroup$
    – rkk1995
    Sep 4, 2019 at 22:02
  • $\begingroup$ 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? $\endgroup$
    – rkk1995
    Sep 4, 2019 at 22:09
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Sep 4, 2019 at 22:11

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