Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Question:
Let $U_{1},U_{2},...$ be independent RVs, each uniformly distributed over (0,1]. These random variables represent successive bids on an asset that you're trying to sell, and that you must sell by time t = 1. As a strategy you adopt a secret number $\theta$, and you will accept the first offer that is greater than $\theta$. For example, you accept the second offer if $U_{1} \le \theta$ while $U_{2} > \theta$. Suppose that the offers arrive according to a unit rate Poisson process ($ \lambda = 1$).
(1) What is the probability that you sell the asset by time t =1?
(2) What is the value for $\theta$ that maximizes your expected return? (You get nothing if you don't sell the asset by t = 1).
(3) To improve your return you adopt a new strategy, which is to accept an offer at time t if it exceeds $ \theta(t) = (1-t)/(3-t) $. What are your new chances of selling the asset, and what is your new expected return?

Just for the first part (1) I'm still quite unsure what to do. What seems somewhat reasonable to me is to calculate $ Pr[X_{T}(t) > \theta | X(t) > 0] $ where $ X_{T}(t) = max(U_{1},...,U_{X(t)}) $. Because the max value would have to be greater than $\theta$ conditioned on there being more than zero bids sent in. I'm not sure how to calculate this probability either though. Hints and clarifications would be very much appreciated thanks.

share|improve this question
add comment

1 Answer

So an offer is accepted with probability $1-\theta$, so you imagine you define a new 'acceptable offer' process, such that it jumps with probability $\theta$ if the 'offer arrival' process jump. the probability the asset is sold by time 1 is simply the probability your 'acceptable offer' process jumped before time 1. so what is this, by the property of Poisson processes? (Or waht is the 'acceptable offer' process?)

For the second part, sure you are maximising $\theta p$ where $p$ is the probability you got in the first part, so differentiation?

For part 3, I am not qutie sure, and it is getting late in England and I am getting up in 6 hours to go to France, sorry can't be bothered to think about it right now

share|improve this answer
Thanks for the suggestion, but I'm still not sure how to formalize this idea. –  rhl Mar 29 '13 at 2:54
do you get 1 and 2 now? –  Lost1 Mar 29 '13 at 19:26
Unfortunately no. Maybe I could have an indicator of sorts defined as: $X(t)=X_{0}(t) + X_{1}(t)$ // $X_{1}(t) = \sum_{k=1}^{\infty}U_{k}I_{k} $ // where $ I_{k} = \begin{cases} 1, & \text{if $U_{k} > \theta$} \\ 0, & \text{if $U_{k} \le \theta$} \\ \end{cases} $ –  rhl Mar 29 '13 at 23:32
Yes $X_0(t)$ is a poisson process with parameter $\theta$ This is the thinning property of Poisson process. Look this up. Needless to say $X_1(t)$ is a poisson process with parameter 1-$\theta$ –  Lost1 Mar 30 '13 at 8:09
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.