2
$\begingroup$

With a Brownian bridge, one can sample a Wiener process constrained to a specified initial value and a final value.

Can the same be done when the process is constrained also to have a specified maximum and minimum value over the time interval?

One could select a very small subset of enough samples (simulations) that happen to have the desired minimum and maximum values (and final value). But is there a way to explicitly construct such samples?

$\endgroup$

1 Answer 1

1
$\begingroup$

You can do this (within some small $\epsilon$) using the reflection principle.

Let's take the simpler case, where you only want to constrain the maximum. Adding the minimum is a fairly simple generalization.

Let's say you want your maximum to be 1.0 (and you have a standard BM starting at zero and running for 1 time unit).

First, construct sample paths until you have one whose total variation is greater than 1.0. Now we have two cases:

  1. The maximum on this sample path is greater than 1 at time $t$. In this case, at the point where the path exceeds 1, simply reflect the path after $t_1$ so that it turns back down. If the new path exceeds 1 at some later time $t_2>t_1$, reflect again.
  2. The maximum on the sample path does not exceed 1. Here, choose a subinterval on which the path has decreased, and flip all the signs in the subinterval. Repeat until the maximum exceeds 1 and revert to the case above.

This only comes within $\epsilon$ since your (finite) path sample times do not exactly identify the crossings of 1.

$\endgroup$
1
  • $\begingroup$ I thought about that, but the problem is that there would be a fraction of samples where the path hits the same "high" multiple times in the interval. The case where a sample has an extremum that occurs more than once in any time interval should essentially never, ever happen. With this construction, it happens all the time, with very high probability. Hence the resulting sample is not a true sample from a brownian motion. $\endgroup$
    – JoseOrtiz3
    Commented May 16 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.