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I was asked a question during a job interview. I don't think I managed to solve it properly during the interview, so I would like someone to explain the answer. The question was as follow:

  • There is a good with value $S$ that is uniformly distributed on [0,1]
  • You are placing a bid $B$. If $B$ is larger than or equal to $S$, you receive the good
  • Immediately after your bid is placed it is determined whether you receive the good or not
  • Immediately after that the value of the good doubles to $2S$

How should you place the bid $B$ to maximise your expected return? (I.e. maximize $2S-B$)

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  • $\begingroup$ What expected value are you trying to maximize? If you just want to win the auction then you would bid $2$. $\endgroup$ – John Douma Mar 29 at 21:28
  • $\begingroup$ @JohnDouma I tried to clarify the question. You want to maximize the return of the auction. $\endgroup$ – highviolet Mar 29 at 21:33
  • $\begingroup$ Assume the expected ualue is for you "gain", i.e. $0$ if you did not receive the goods, and $2S-B$ is you get it. I got expectation $0$ no matter how you place the bid $B$. Probably I did not understand the problem correctly. $\endgroup$ – Yu Ding Mar 29 at 21:34
  • $\begingroup$ $\newcommand{\I}{\mathbb{I}}\newcommand{\E}{\mathbb{E}}$If you bid an amount $b$ (using lower case to avoid confusion with random variables), then your return (which is a random variable as it depends on the random $S$) is $(2S-b)\I(S\le b)$. Hence your expected profit is $$f(b):= \E[(2S-b)\I(S\le b)],$$ where $\I$ is an indicator function and $S\sim\mathsf{Unif}[0,1]$. Can you compute this expectation as a function of $b$ (use integration)? $\endgroup$ – Minus One-Twelfth Mar 29 at 21:43
  • $\begingroup$ @YuDing: I agree: your expected return is $0$, no matter what you bid. Perhaps this was the answer the interviewer wanted? $\endgroup$ – TonyK Mar 29 at 21:43
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Your profit is $2S-B$ when $S \lt B$ and $0$ otherwise. Integrating over $S$, we get $$\int_0^B (2S-B) dS=(S^2-BS)|_0^B=0$$ Bid whatever you like in the interval $[0,1]$ and your expected profit is the same, $0$.

This seems a surprising conclusion, so we should try some things to validate it. If you bid $0$ you will never get the object, so your profit is $0$. If you bid $1$ and the initial value is $0$, you lose $1$. If you bit $1$ and the initial value is $1$, you win $1$. It is linear in between, so your expectation at a bid of $1$ is $0$. The problem is scale invariant. If you bid $B$ you lose $B$ when the initial value is $0$ and win $B$ when you barely get the object, so your expected profit is $0$.

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Now that we all know the expected gain is $0$ no matter what you bid, here's a simpler explanation "after the fact"... :)

Conditioned on you winning with bid $B$, the initial value of the goods is $Uniform(0,B)$, so the final value is $Uniform(0,2B)$, with an expectation of $B$.

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