# What is the Bruss-Yor concept of no information?

A few years ago, a question related to a paper of Thomas Bruss and Marc Yor on the so-called last arrival problem received some attention on this forum.

What I'd like to know now is:

What are the assumptions under which Theorem 5.1 of the Bruss-Yor paper is derived?

In the last arrival problem as I first thought about it, an unknown number $$N$$ of items arrive at independent times chosen from uniform distribution on the interval $$[0,1]$$, and a selector (who sees the arrival times in increasing order) tries to choose "online" the very last one.

The way I first thought about the problem (in the context of a variation of the secretary problem), there is no a priori random distribution of $$N$$, the number of items. Instead, the task is to find a policy that maximizes the probability of success under a worst case (aka adversarial) assumption. In other words we want a policy that succeeds with (at least) some fixed positive probability for every $$N$$.

For instance, one such policy that admits a rather straightforward analysis is to wait until for the first time half of the remaining time goes by without any new item arriving, and then to accept the next one. This policy succeeds for $$N$$ items with probability exactly $$\prod_{k=1}^N\left(1-\frac1{2^k}\right),$$ which is bounded from below by its limit of about $$0.288788$$.

In a paper from 2011, I showed (with a somewhat clumsy computation) that there is an optimal policy that wins for the selector with probability around 0.3529170002.

But the last arrival problem can also be considered in other settings. We might have just a point-process on the interval $$[0,1]$$. A type of point-process considered by Bruss and Yor is what they call a process of proportional increments, or p.i. This is a process where as soon as we have seen one item, the expected rate of future items will be (at all times) equal to the frequency of items that we have already seen.

Since the selector will not have to take any decision until they have seen the first item, the p.i. assumption seems natural in this context. When I first saw the Bruss-Yor paper I was excited to see that they seemed to have obtained an explicit optimal policy for the selector for a p.i. process. Their suggested policy (Theorem 5.1) is to accept the $$k$$:th item if it arrives after time $$1-\frac1{k+1}.$$ At least it appears from their Conclusion 3 of Section 2.2 that what they are considering, even in Section 5, are p.i.-processes.

But when I looked more closely, it seemed that the Bruss-Yor policy can't be optimal under the assumption of p.i.

Suppose for instance that we have seen exactly one item up to time $$1/2$$. Then the probability of no more item ought to be $$1/2$$: If we discretize by chopping up the unit interval in $$2m$$ slots, then after seeing one item in the first $$m$$ time-slots, the probability of no new item in the next slot is about $$m/(m+1)$$, and conditioning on that, the probability of no new item in the following slot is about $$(m+1)/(m+2)$$ and so on, in total $$\frac{m}{m+1}\cdot \frac{m+1}{m+2}\cdots \frac{2m-1}{2m} = \frac12.$$

So it seems that if the first item arrives at time $$1/2$$, the selector succeeds with probability $$1/2$$ if they accept it. But that also means that they can't possibly succeed with probability $$1/2$$ if they don't. If they reject, the probability of seeing another item is $$1/2$$, but there might be more than one and they still have the nontrivial task of picking the right one.

Therefore it seems that an optimal policy in the p.i. setting must accept the first item even if it arrives slightly before time $$1/2$$.

I haven't been able to follow the derivation of Theorem 5.1, but what I'd like to know is:

What is the model under which the authors claim that the $$(1-\frac1{k+1})$$-policy is optimal?

Looking back again at their Section 2.2, it seems in Conclusion 1 and the discussion around it that there is also an assumption that the process, conditioned on giving rise to exactly $$N$$ items, should behave as if the arrival times were independent and uniform on $$[0,1]$$.

I haven't been able to figure out if this is what the authors are assuming later on, but it seems to me that this would be incompatible with the p.i. assumption already for $$N=1$$: If the first item arrives at time $$T_1$$, then assuming p.i., the probability of no more item is equal to $$T_1$$. In order for $$T_1$$ to be uniform on $$[0,1]$$ after conditioning on $$N=1$$, it would therefore have to have density at $$T_1=t$$ proportional to $$1/t$$, which is impossible.