Random walk --> Brownian motion

The concept of Brownian motion is usually preceded by first explaining a random walk, where someone starts at 0 on a number line, and at each point in time takes one step forward (with probability 1/2) or one step backward (probability 1/2). Then, they say something like, imagine instead of moving 1 step for every 1 unit increase in time, you move sqrt(h) steps in every h period of time, and take the limit as h goes to zero.

This, the very beginning, is where I begin to get confused. From t=0 to t=2 we would travel two steps. From t=2 to t=4 we would travel two steps. Yet, in h = 4 periods, we were supposed to travel sqrt(4) = 2 steps, not 4!

How is this resolved? Do things just get different from you move down into infinitesimal changes in distance and time?

You call a cab and quit stumbling back and forth.

I’m reaching way back into the memory banks for this one, so it may be way off the mark. There is an effect associated with infinitesimal step size. You’re essentially modeling a continuous process with discrete steps, so you have to take the limit as x (step size) goes to zero. You also have to remember that Brownian motion by definition is random, so at any particular time, it’s a probability that the particle will be at any point rather than a deterministic value. I think that’s right, but I can’t guarantee the memory is working correctly anymore :smiley:

The actual derivation is a little bit more complicated than that. This page has an outline, although the coding is a little off. Wikipedia discusses it a bit, although without the details.

Is this for MFE/3F?

Inspired by, although the question is really just for my understanding.

Also I screwed up my example: 2 is supposed to sqrt(2), but still, sqrt(2) + sqrt(2) <> 2

Each individual step remains the same size. The total distance traveled (the amount of wear and tear on the soles of your shoes, say) goes up linearly with time. The square root isn’t for how far you travel; it’s for how far you get. How far you get is less than linear, because a lot of time, you’re just erasing previous progress. So, for instance, after you take 10,000 steps, you’re probably only about 100 steps away from where you started, because most of your steps just cancel each other out.

The simple answer is you don’t take sqrt(h) steps in h periods. You take h steps, it’s the size of the step that is sqrt(h) rather than one. Originally the variance per step was 0.51^2 +0.5(-1)^2 = 1. Since there was one step per unit time the variance per unit time is 1. Now the variance per step is h. There are 1/h steps in one unit of time and the steps are independent so the variance is the sum of 1/h h’s or 1. So the variance is still one per unit time.

Right, it makes sense that the displacement would shrink. I guess what I’m really saying is, when you integrate displacement, you get distance traveled. What happens when you integrate a standard Brownian motion? I seem to recall you get something like 1/2 * [W^2(t) - t] where W(t) is the Brownian motion. Not quite sure how to interpret that though.