Is the sequence $$w_n=n! \int_0^{1/2} \int_{x_1}^{2/3} \cdots\int_{x_{n-2}}^{\frac{n-1}{n}} \int_{\frac{n}{n+1}}^1 dx_n dx_{n-1} \cdots dx_1$$ increasing for $n\ge 3$?

This is a conjecture of F. Thomas Bruss and Marc Yor in a recent paper Stochastic Processes with Proportional Increments and The Last-Arrival Problem. It is presented here with the permission of them.

Background

In the mentioned paper Bruss and Yor introduce stochastic processes with proportional increments to deal with the so called last arrival problem (l.a.p.) where an unknown number of independent random random variables $X_1,\ldots,X_N$ (which are uniformly distributed on the interval $[0,1]$) is observed and "our objective is to stop online with exactly one stop on the very last of these points, i.e. at their largest order statistics $X_{\langle N;N\rangle}$" (cited from Bruss and Yor). Bruss and Yor give an interpretation of the problem (it is not so obvious what it means that $N$ is unknown) and then provide the optimal strategy. The numbers $w_n$ are then the win probability conditioned on $N=n$. Bruss and Yor state that $w_n \to 1/e$.

Some relations

The inner integral clearly is $1/(n+1)$ and the remaining integral is thus up to the factor $n/(n+1)$ the probability that the order statistic $X_{\langle 1;n-1\rangle},\ldots,X_{\langle n-1;n-1\rangle}$ of $n-1$ independent on $[0,1]$ uniformly distributed random variables has values in $[0,1/2] \times [0,2/3] \times \cdots \times [0,\frac{n-1}{n}]$. It would be thus natural if a similar problem had appeared in some statistical context.

One can write down the latter probability using multinomial sums (which looks very ugly). Nevertheless this suggests that some clever combinatorial arguments might help.

Finally the order statistics are closely related to Poisson processes (however, as the authors are experts in stochastic processes they have most probably checked that area).