I've been struggling with proving a conjecture concerning order statistics of Brownian motions for a while. The conjecture I'm looking to prove is the following: (I have run Monte Carlo simulations that confirm the conjecture.)


Let $(q\_1(t), …, q\_n(t))$ be a vector of independent Brownian motions with $q\_i(0) = 0$ for all $i$. Let $T > 0$. Each $q\_i$ represents an object in a game, and each of $J < n$ players in the game can pick exactly one object $i$ at any time $t \in [0, T]$. Whenever a player picks an object, he picks the one that currently has the highest value amongst those that have not been picked before. A player's payoff is the value of the object at the time he picks it. Let $x\_j(t\_j; t\_1, …, t\_{j-1})$ denote the expected payoff (expectation taken at time 0) from picking at time $t\_j$, given that players $1, …, j-1$ will pick at times $t\_1, …, t\_{j-1}$. I want to show that $x\_j(t\_j; t\_1, …, t\_{j-1})$ is decreasing in $t\_1, …, t\_{j-1}$. 


I will need this for an economic theory paper I'm writing, and would be very grateful for any help!


Some things that I already thought through: 


* I have already shown that $x\_j(t\_j; t\_1, …, t\_{j-1})$ is increasing in $t$ and jointly continuous in all arguments.
* The case $j=2$ is quite tractable, and for the case $j = 3$ I have the intuition on how to prove it. That intuition, however, doesn't extend to the general case.



Order statistics of Brownian motions
The strategy of the other $n-1$ players can be viewed as a single adversary trying to minimize the payoff of the $n$th player. Picking the highest unpicked object is optimal for that adversary by coupling. 


If $t\_{i-1} \le t\_i' \lt t\_i$ then picking at time $t\_i'$ instead of $t\_i$ does an imperfect job of choosing the unpicked object which will be highest at time $t\_i$. You can view picking at $t\_i'$ as sometimes playing optimally at time $t\_i$, and sometimes choosing a lower object at time $t\_i$.


If you want to consider decreasing $t\_i'$ past $t\_{i-1}$ then you can view this as decreasing $t\_i'$ to $t\_{i-1}$, then replacing $t\_{i-1}$ with a smaller $t\_{i-1}'$. 


