# Finding ways to bid for items, which has a normal distributed price

This is an interview questions, I am not quite sure how to solve. The question is stated as this:

Suppose that you want to buy a specific type of car, but you don't know anything about cars (you can't find the underlying quality or the real value). But dealers know about cars and they know exactly how much a car is worth (real value).

Now you are given a list of prices that different dealers bought at, but the list doesn't show which dealer bought which car. The price is normal distributed. Assume there are infinite number of dealers and the price dealers bought at is considered as the true value of the car.

If you can only bid once for each dealer, and that dealer either accept or decline your offer. Dealer will only accept your offer if and only if your offer is above their buying price.

The question is, how to bid or what price to bid, so that you minimize the loss (the price you offer minus the real value). Assume that you only need to buy one car so you can bid as many times as you want until your offer get accepted.

(Interviewer mentioned mean and standard deviation, but I still couldn't figure out.)

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Why wouldn't bidding $x-1$ always be better than bidding $x$? Don't you have to introduce some penalty for not buying anything? –  Douglas Zare Aug 26 '12 at 0:35
I modify the question a bit to clarify your question. –  Negative Zero Aug 26 '12 at 5:06
It still looks to me like bidding $x-1$ is always better than bidding $x$. So what if it takes you $10^{50}$ attempts if you save a dollar when your bid is accepted? You need a more complete model or else you aren't going to have any optimal strategy. –  Douglas Zare Aug 26 '12 at 8:29
@Negative Zero :There is something wrong... you say "...you only need to buy one car so you can bid as many times as you want until your offer get accepted" If this is true, bidding something close to 0 will do. As the real value given by any dealer follows a Normal distribution and that you can deal as many times as you want, there will always be a delaer for which the car value is 0 (or close enough to it) I tend to agree with Douglas –  JDav Aug 26 '12 at 18:02
I doubt this is a faithful reproduction of the question. It seems likely that real question is closer to one in which the bidder can only bid once with a single dealer. –  cardinal Aug 26 '12 at 19:28

This is my guess...

The loss function: The price you offer minus the real value $P-R$ , ($R\sim N(\mu,\sigma^2)$) may be related to a risk function. In econometrics and statistics this is rather standard: http://en.wikipedia.org/wiki/Risk_function

Something I'm assuming is that you are interested in minimizing the (expected) deviation with respect to the real value i.e. minimizing the losses yes but also the gains. This implies minimizing the positive expected difference:

$R(P,V)= E(P-R)^2$ (The mean squared error http://en.wikipedia.org/wiki/Mean_squared_error)

It can be shown that this is equivalent to

$R(P,V)= \sigma^2 + (P-\mu)^2$

where we don't have control over the first term (the variance) so to minimize $R(P,V)$ P should be as close as posible to $\mu$, this implies bidding P=mean(prices in the list).

But still I'm not sure as he states that the real value is normally distributed !Moreover, we can consider R to be fixed and P to be our estimator... I guess the answer would be the same for a different reason

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I am having a hard time seeing how one might justify a symmetric loss function here. In particular, why should I penalize myself for underpaying assuming I get my offer accepted? (Implicit in this comment is a strong hint on how I think one might model this situation.) –  cardinal Aug 27 '12 at 8:29
Please read "justify" as "motivate" above. You've been very nicely explicit about your assumptions. I'd be interested in seeing a model of this situation where your assumed loss appears "naturally". –  cardinal Aug 27 '12 at 8:44
@cardinal : There are several interpretations of the question. As it is written I agree with you, I also wrote a comment where I guess that the best bid should be something close to 0. But also guess that it can't be so easy or simple... there is something missing in the question... I'm not proud of my answer, it's as ambiguous as the questions itself... ;) –  JDav Aug 29 '12 at 0:07