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.

  • $\begingroup$ What, exactly, is your question? $\endgroup$ – Mike Spivey Oct 30 '10 at 5:49
  • $\begingroup$ The question is, how does $u$ look like? Does it exists for all values of alpha? If yes, is it unique? $\endgroup$ – Julian Wergieluk Oct 30 '10 at 17:33
  • $\begingroup$ @wergrew: I edited your question. Please check the result suits you. $\endgroup$ – Did Mar 20 '11 at 18:05
  • 2
    $\begingroup$ My guess is that $u$ exists, is unique, and is continuous. For $\alpha=1$ it should satisfy $u(t)\sim\sqrt{2t\log\log(1/t)}$ as $t\to0$ (because of the law of the iterated logarithm) and $u(1)=0$. Then the solution for arbitrary $\alpha$ should follow from rescaling. $u_\alpha(t)=u_1(\alpha t)/\sqrt{\alpha}$. $\endgroup$ – George Lowther Mar 20 '11 at 21:53
  • $\begingroup$ My comment above is taking $\alpha$ to be the probability that te barrier is hit (rather than the probability that it is not hit). $\endgroup$ – George Lowther Mar 20 '11 at 22:01

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) )

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.


  • $\begingroup$ You might wish to show the superposition of the second graph and of the streched half of the first? $\endgroup$ – Did Jan 16 '12 at 7:25
  • $\begingroup$ @Didier: The added match looks good enough for me $\endgroup$ – Henry Jan 16 '12 at 8:54
  • $\begingroup$ Quite conclusive match indeed. $\endgroup$ – Did Jan 16 '12 at 17:00

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!

  • $\begingroup$ Note that the second requirement was that, the hitting time of the boundary is uniformly distributed on $[0,1]$. Can you prove this for $u(t)=c^*$? $\endgroup$ – Julian Wergieluk Nov 30 '11 at 20:57

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