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I have just recently begun studying Brownian motion and stochastic calculus at the level of an undergraduate or beginning graduate student of applied mathematics. (Textbooks I've looked at are by Mikosch, Gardiner, Kloeden and Platen.)

Back when I first started thinking about stochastic dynamics, I was a cognitive science student who was simulating neural decision making circuits. We had a differential equation to model our system, and my advisor suggested that I try "adding in some noise."

Intuitively, I would have written a discretized equation to model the evolution of some state variable $X$ like this: $$\Delta X = [ a(t,X)+ R ] \Delta t $$ where a is a coefficient that depends upon the state variable $x$ and time $t$, and where $R$ denotes a random variable (probably taken to be normal and centered at zero with variance $\sigma^2$)

In other words, at some level, I would have expected an "intuitive stochastic differential equation" to be expressed in the general form $$ \Delta X = [a(t,X,R)] \Delta t $$ where a is now any function of a random variable.

However, crucially for the theory of stochastic differential equations and the models I see (at least the ones that come up in modeling neural decisions -- e.g. Ornstein-Uhlenbeck), the discretized equation has the stochastic part scaled with $\sqrt{\Delta t}$, rather than just $\Delta t$. For instance, Brownian motion is discretized as $$\Delta X = R \sqrt{ \Delta t}$$ where R denotes a standard normal.

Finally my question: Why is it so desirable, for modeling stochastic dynamics generally speaking, to have the stochastic part of a stochastic difference equation scale with the square root of the change in time rather than simply linearly with the change in time?

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You are aware that $R\sqrt{\Delta t}$ follows a normal distribution with variance $\Delta t$, right?. So the "variation" of this term is of order $\Delta t$. –  Stefan Hansen Jan 23 '13 at 8:07
    
Yes, but why not have the variation be of order ($\Delta t)^2$? Somehow that seems more natural to me (on a total, wildly naive level), because it would preserve a common order for the differential (or time increment) across both the deterministic and stochastic parts of the expression. –  Mike Wojnowicz Jan 23 '13 at 8:09
    
I actually don't know - I'm sure someone else can answer your question. +1 for a good question. –  Stefan Hansen Jan 23 '13 at 8:15

1 Answer 1

up vote 13 down vote accepted

The fundamental reason is that variances are additive, while standard deviations are not. If you didn't make the variance proportional to the time step, your discretizations would not behave consistently when you changed the time step.

To be explicit, let's take $X_{i+1}=X_i+R_{i+1}\sqrt{\Delta t}$, where the $R_i$ are independent and standard normal. After two steps, we have $$X_2=X_0+(R_1+R_2)\sqrt{\Delta t}.$$ But $R_1+R_2$ is normal with variance $2$, so we can write it as $\sqrt2 R_{12}$, where $R_{12}$ is again standard normal. Then we get $$X_2=X_0+\sqrt2 R_{12}\sqrt{\Delta t}=X_0+R_{12}\sqrt{2\Delta t},$$ so it's exactly the same as if we took a single time step of twice the length. This doesn't happen if you take the stochastic component to be $R_i\Delta t$ instead (try it).

What's really going on is that you're adding a constant variance, say $\sigma^2$, to the stochastic part at each time step. So after $n$ time steps you will have accumulated a variance of $n\sigma^2$. If you were to take a single time step of length $n\Delta t$ instead, you'd want your stochastic part to have the same variance $n\sigma^2$, and that corresponds to scaling by $\sqrt n$, not by $n$.

In other words: $\text{variance}\propto\text{time step}$; $\text{scale}\propto\sqrt{\text{variance}}$; so $\text{scale}\propto\sqrt{\text{time step}}$.

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I really like this answer. That makes a lot of sense to me. Thank you for clarifying something I had been wondering about for a long while. –  Mike Wojnowicz Jan 23 '13 at 8:40
    
@Mike: You're right, but it's gone now... I rewrote my answer to be more concise. I think it's better, but let me know if you think it's not. –  Rahul Jan 23 '13 at 8:48
    
As a beginner, I actually really liked the concrete example a LOT. I saw the abstraction in it. It captured your point. But your new answer adds in additional nice commentary, so I definitely wouldn't want you to scrap it. Maybe you could synthesize them into a single answer? Anyways thank you again. –  Mike Wojnowicz Jan 23 '13 at 9:04
    
Happy to oblige! Thanks a lot for your comments. I always have a hard time telling what the right balance of explicitness vs. abstraction is. –  Rahul Jan 23 '13 at 9:18
    
No problem & likewise. Now there is both so everyone will be happy. –  Mike Wojnowicz Jan 23 '13 at 9:29

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