Jane Street Interview Question: was asked this question today... | Glassdoor.co.uk

Interview Question

Trader Interview London, England

was asked this question today...you want buy a car at an

  auction whos price is uniformly distributed between 0-1000 if you bid more than the car you win it at the price you bid.if you bid less that the cars proce you dont win it but dont lose anything. you can sell the car for 1.5 time the value of which you bought it. what should you bid on the car to maximumise your profit?
Answer

Interview Answer

12 Answers

0

i told the interviewer that the expected value is negitve so you should bid 0 to maximum your profit.(0 is greater that any negitive number)

but the answer which i was guided towards wasE(X)= X-(X/2)*1.5
letting X = your bid

so if you bid 500 E(x) = 500 - 250(1.5) = 125

pretty weird question if you ask me. comming up with a formula for expected value when i was orgionaly asked what would be my bid to maxiumise my profit.

any1 have an thoughts? maybe this was obvious?

Interview Candidate on 27 Apr 2011
0

If you bid X then your expected gain is (X/2)*1.5 So (X/2)*1.5 - X should be your profit... I.E. negative expected value.

Could it be the interview just wanted to see how you can mathematically back up your original statement?

M on 27 Apr 2011
0

yeah good point! that is most likely what he was testing. i suppose i just assumed it was obviously neg EV.

Anonymous on 28 Apr 2011
2

im sorry but can u guys explain why is it X/2 *1.5 and not x*1.5

Also, where have you guys taken into acct that the gain is only when your bid is successful. If its not successful u gain ntn.
Hence in my opinion it should be
E(X) = (X/1000) * (1.5*X - X)

h on 25 May 2011
0

its x/2 because the expected value of all the other possible outcome when you win is x/2.the expected profit is negative.

say you bid 500 and win then the expected value is 0(1/501) + 1(1/501).....500(1/501) that where the x/2 come from

canidate on 25 May 2011
4

Shouldn't it be bid 1000? then you will win the car for sure, and you can make a 1000*50%=$500 profit. (theoretically, assuming you can sell the car for 1.5 times the purchase price no matter what the purchase price is, you should bid as large as possible, then your profit is purchase price*50%)

e on 23 Jun 2011
0

Looks like an options contract.

Anonymous on 11 Aug 2011
3

I agree with "e". This doesn't seem like it should be more complicated than it is.
if you can sell the car for 1.5 times the value of which you bid for, then no matter what the bid is, your profit will be 50%. You want to maximize profit, which implies the fact that you "want to win the bid" so that you can then sell it for more. Bidding less wins you nothing, and so you might as well bid the largest amount, i.e. $1000, to ensure you win, so you can then spin it off for a 50% profit. Granted, this explanation is only valid if it's a "sure" thing that no matter what you pay for the car, you can 100% sell it for 1.5 times the price you paid. Thoughts?

Stephen on 10 Sep 2011
1

You dont sell it for 1.5 times you paid you sell it for a 1.5 times it's true value (a number uniformly distributed between 0 and your bid (which expected value 0.5 times your bid).

Anonymous on 4 Nov 2011
5

assuming the question is wrongly worded, and we can sell it for 1.5 * its actual value.

bid x dollars, between 0 and 1000:

expected profit = probability you get the car * (1.5 * expected value of car - bid)

p = (x/1000) * (1.5 * 500 - x)

maximise w.r.t x to get p = 140.6 at x = 375

b on 7 Nov 2012
1

a=actual price
b=bidding price
------------------
If profit = 1.5b - a
answer is easy, bid 1000
-------------------
If question was wrongly worded &
profit = 1.5a - b

to win, b>a
to make profit, b<1.5a
so to win and make profit: a

Me on 5 May 2014
1

b is close to the right answer, however, the expected value of the car isn't locked @ 500, it is dependent on x (the amount bid). If one were to bid 1, the expected value of the car isn't going to be 500.

Consider a fixed x, [0,1000], the expected value of the car will be x / 2. In essence, we are shrinking the bounds of a uniform distribution.

Thus, profit = (x/1000) * ( (x/2) * (3/2) - x)
profit = -x^2 / 4000

Thus, the correct bid is 0.

A quick sanity check would be to bid 1. Half the time, the car is worth 0, the other half it is worth 3/2. Thus, if we would win the car we would pay 1 dollar and expect to get 3/4 of value back. This is negative EV. It can be seen that as the bid increases, the EV becomes further negative.

c on 30 Oct 2019

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