Confidence band for Brownian Motion with uniformly distributed hitting position

Question

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.

Answer 1

Thoughts, ideas and discussions (not answers):

Consider the random variable $X(\omega)= sup_{t \in [0,1]} |B_t(\omega)|$. Let $c^*$ be the upper $1-\alpha$ quantile of this distribution; i.e, $P(\{\omega \in \Omega | X(\omega)>c^*\})=1-\alpha$.

The first $u(t)$ that came to my mind was the following: $u(t)=c^*$ for all $t \in [0,1]$; so that the confidence band is a constant. This works, right? Also, this is not conservative as

$\{\omega | |B_t(\omega)|\leq c^* \quad \forall \quad t \in [0,1]\} \subseteq \{\omega | X(\omega) \leq c^* \}$.

but the other inclusion also holds.

Seems like an interesting problem. Presumably there are different confidence regions that you can build. Any ideas on what are the properties that you would like to have? I'd be interested in the context of this inference problem!

Answer 2

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.