TITLE: Confidence band for Brownian Motion with uniformly distributed hitting position
QUESTION [8 upvotes]: Let $(B_t)$ denote the standard Brownian motion on the interval $[0,1]$. For a given confidence level $\alpha \in (0,1)$ a confidence band on $[0,1]$ is any function $u$ with the property that 
$$
P(\omega; |B_t(\omega)| < u(t), \quad \forall t\in [0,1])=\alpha.
$$
In other words, the probability that a path of the Brownian motion stays within a confidence band is $\alpha$. Additionally the boundary hitting position for those paths leaving the band must be uniformly distributed on $[0,1]$. This condition can be stated using the stopping time 
$$\tau(\omega) = \inf [ t \in [0,1], |B_t(\omega)|=u(t) ].
$$
Then $\tau $ is the time of the first hitting, and one asks that $\tau$ is uniformly distributed on $[0,1]$ conditionally on the event that $\tau$ is finite.
I am interested in

References and links to literature or papers considering this or similar problems
Thoughts, ideas, discussion

The context of the problem is a rather boring one, so will not state it here. The problem itself seem to be non-trivial and interesting.

REPLY [2 votes]: If I understand you correctly, you are looking for a curve $u(t)$ with $t \in [0,1]$ so that the probability the absolute value of a standard Wiener process does not cross the curve is $\alpha$ and that the probability density of the first crossing is a constant $1-\alpha$.
The following simulation in R may help indicate the shape of $u_\alpha(t)$:
##simulated boundary for standard Wiener process 
##time for absolute value to cross boundary first time 
##uniformly distributed on [0,1] given crosses boundary
steps     <- 100                      #how many steps in (0,1] 
cases     <- 100000                   #how many processes to simulate 
alpha     <- 0.00                     #probability  does not cross boundary
normmat   <- matrix(rnorm(steps*cases), ncol=steps)
brown     <- normmat/sqrt(steps)      #for var=1 after all steps
for (i in 2:steps){brown[,i] <- brown[,i-1] + brown[,i]}      #cumulative sum
absbrown  <- abs(brown) 
boundary  <- rep(0,steps)
for (i in 1:steps){
     boundary[i] <- quantile(absbrown[,i], 
                     probs = (steps-i*(1-alpha))/(steps-(i-1)*(1-alpha)),
                     names = FALSE)
     absbrown    <- absbrown[!(absbrown[,i] > boundary[i]), ] #del crossed
    }
plot( c(0,(1:steps)/steps), c(0,boundary), type="l", xlab="t", 
      ylab="boundary", main=paste("simulated boundary for alpha =",alpha) )
abline(h=0)
abline(v=0)  

Here are an example with $\alpha =0$.  The actual curve will be smoother. 

Here is another.  If George Lowther is correct then this is simply the first half of the previous curve stretched upwards.  

Added for comment: Taking the left hand half of the first graph (black below) and taking a shrunken version of the second (red below, dividing $t$ by $2$ and the boundary by $\sqrt{2}$), there is a very good match, except for the $y$ axis which may be a rounding effect in the simulation. So George Lowther looks correct.