TITLE: The Bruss-Yor conjecture about an iterated integral
QUESTION [17 upvotes]: 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).

REPLY [9 votes]: Here is a sketch of a full proof that can be (at least, in principle) verified by a human.
Claim 1. $\gamma_n=(\frac{n-1}{n})^{n-1}$ is a decreasing sequence.
This is well known and easy to prove.
Claim 2. The integral in question is just $\frac n{n+1}$ times the probability $P_n$ that $Y_k\le \frac k{k+1}$ for all $k=1,\dots,n-1$ where $Y_k$ is the $k$-th value in the list obtained by sampling $n-1$ independent random variables uniformly distributed on $[0,1]$ and putting them in the increasing order.
This has already be mentioned by the OP.
Claim 3. The joint distribution of $Y_{n-1},\dots,Y_1$ is the same as of $X_1^{\frac 1{n-1}},X_1^{\frac 1{n-1}}X_2^{\frac 1{n-2}},\dots,X_1^{\frac 1{n-1}}X_2^{\frac 1{n-2}}\cdots X_{n-1}$ where $X_j$ are i.i.d. random variables uniformly distributed on $[0,1]$.
Just condition on the maximum and use induction. From now on, we will just assume that $Y_k=X_1^{\frac 1{n-1}}\dots X_{n-k}^{\frac 1{k}}$.
Claim 4. For all $k\ge 2$, we have
$$
(\frac{k-1}{k})^{\frac{k-1}k}(\frac{k}{k+1})^{\frac 1{k(k+1)}}\le \frac{k}{k+1}
$$
Rewrite it as $(\frac{k^2-1}{k^2})^{k^2-1}\le (\frac{k}{k+1})^{k}$ and use Claim 1.
Claim 5. Let $2\le k\le n-2$. If $X_1^{\frac 1{n-2}}\dots X_{n-k}^{\frac 1{k-1}}\le \frac{k-1}{k}$ and $X_1^{\frac 1{n-2}}\dots X_{n-k-1}^{\frac 1{k}}\le \frac{k}{k+1}$, then
$X_1^{\frac 1{n-1}}\dots X_{n-k}^{\frac 1{k}}\le \frac{k}{k+1}$.
Raise the first inequality to the power $\frac{k-1}{k}$, the second to the power $\frac 1{k(k+1}$, multiply, use the fact that $X^t$ is decreasing in $t$ for $0<X<1$, and apply Claim 4.
Claim 6. 
$$
P_n\ge \frac{\gamma_n}{\gamma_{n-1}}P_{n-1}-(1-\frac 2n)^{n-1}2^{-(n-1)}
$$
Claim 5 allows one to conclude that replacing a sample $X_1,X_2,\dots$ satisfying the desired set of inequalities for $n-1$ instead of $n$ by $X_1'=\frac{\gamma_n}{\gamma_{n-1}}X_1, X_2'=X_2, X_3'=X_3,\dots$, we get a sample satisfying all the desired inequalities for $n$ except, perhaps, the very last one. Thus, the probability $Q_n$ that all inequalities except the last one are satisfied is at least $\frac{\gamma_n}{\gamma_{n-1}}P_{n-1}$ (the factor comes from the bound for the density of $X_1'$ relative to the density of $X_1$). What we need to subtract is at most the probability that the minimum of $n-1$ independent random variables uniformly distributed on $[0,1]$ is at least $1/2$ while their maximum is at most $\frac{n-1}n$. This is the last term in the inequality.
Claim 7. We have
$$
\frac{n}{n+1}P_n\ge \frac{n}{n+1}\frac{n}{n-1}\frac{\gamma_n}{\gamma_{n-1}}\frac{n-1}n P_{n-1}-\frac{n}{n+1}(1-\frac 2n)^{n-1}2^{-(n-1)}
$$  
This is just Claim 6 rewritten.
Assuming that $\frac{n-1}nP_{n-1}\ge\frac 5{16}$, we conclude that we will be able to run the induction if 
$$
[\frac{n}{n+1}(1+\frac{1}{n(n-2)})^{n-2}-1]*\frac{5}{16}\ge \frac{n}{n+1}(1-\frac 2n)^{n-1}2^{-(n-1)}
$$
Using the first three terms in the Newton's formula on the left, we see that to pass from $n-1$ to $n$, it is enough to have
$$
\frac{5}{16}\frac{n-3}{2n^2(n-2)}\ge (1-\frac 2n)^{n-1}2^{-(n-1)}
$$
Claim 8. This inequality holds for $n\ge 7$.
For $n=7$, it reduces to $8\cdot 7^4>5^6$, i.e., $(2-\frac 1{25})^2>4-\frac 78$, which is fairly obvious by Bernoulli. It remains to note that when $n$ goes up by $1$, the left hand side is multiplied by at least $(\frac 78)^2$\ge $\frac 34$ while the right hand side is multiplied by $1/2$ because of the power of $2$ and at most $(1-\frac{n-1}{n(n-2})^{-1}\le \frac{35}{29}$, which is the Bernoulli bound for the factor by which $(1-\frac 2n)^{n-1}$ differs from the decreasing sequence $(\frac{n-1}{n})^2(n-1)$. Now, $\frac 12\cdot\frac {35}{29}\le \frac{18}{29}\le \frac 34$, so we can run the induction.
The most difficult (=time and effort consuming) part of this proof is 
Claim 9. $P_1$ through $P_6$ are fine for the base of induction. 
I managed to compute $P_4$ by hand without an arithmetic error, but not $P_5$. That's why I would still prefer to see an argument that doesn't require a direct computation of any value beyond $P_3$.