# Brownian motion integration/calculus

1. Nov 9, 2012

### operationsres

We all know that $\int_0^t dB(s) = B(t)$, where $B(t)$ is a standard Brownian Motion. However, is this identity true?

$\int_{t_1}^{t_2} dB(s) = B(t_2) - B(t_1)$

2. Nov 9, 2012

### Vargo

I only have a passing familiarity with Brownian motion. I assume that B(t) is a Lipschitz function. (Any real particle must have a Lipschitz path or else it is travelling faster than the speed of light).

Lipschitz functions are absolutely continuous, therefore the fundamental theorem of calculus holds as you have written it. So yes, it is true.

EDIT: Oops. I had doubts and looked it up on Wikipedia. it seems that B(t) is almost surely continuous everywhere but nowhere differentiable. Apparently I was way off base.

Last edited: Nov 9, 2012
3. Nov 9, 2012

### operationsres

4. Nov 10, 2012

### Ray Vickson

The short answer is YES, at least if t1 and t2 are non-random, or even if they are random but "unrelated" to {B(t)}.

The longer answer (or, rather, a series of questions) is: do you know what is meant by
$$\int_{t_1}^{t_2} dB(s)?$$ (Here, I am asking for an "intuitive" understanding of the concepts, not a 100% iron-clad presentation with all the proofs.) If you do not understand the meaning of the stochastic integral, then you have a lot of preliminary work to do, reading and absorbing the basics. If you do understand what it means, then just write it all out in detail and see what you get.

RGV

5. Nov 11, 2012

### operationsres

I don't really understand what's meant when we integrate with respect to a Wiener process. I was never taught this.I know all of the properties of $B(t)$ though.

My qualitative, uneducated guess as to what that integral is doing is adding up all the instantaneous changes in $B(t)$ over the interval $[t_1,t_2]$, and the summation of these infinitesimal changes is equal to $B(t_2)-B(t_1) \overset{d}{=} B(t_2-t_1) \sim \mathscr{N}(0,t_2-t_1)$. I can't really visualize this because of the fractal like property of $B(t)$. I guess I would have to really break down how $B(t)$ is constructed to understand.

This is less confusing than something like $\displaystyle \ \ \int_0^t r(s) dB(s)$, where r(s) is stochastic OR deterministic. What the heck does that integral mean ... intuitively?