MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Take an OU process characterized by

X(0) = x

dX(t) = - a X(t) dt + b dW(t)

where a,b >0. The parameter a is usually interpreted a dissipative term, and b is a volatility term.

My question is this: What are the units of a and b? Is it true that a is (time) -1 , and b is unitless? Then how can one make sense of the variance which approaches (b 2 /(2 a)) as t goes to infinity?

Thanks for your help.

share|cite|improve this question
You don't seem to have defined W. – Qiaochu Yuan Dec 18 '09 at 20:39
$W_t$ is the Wiener process. – Steve Huntsman Dec 18 '09 at 20:43

Say $X$ is a displacement and is measured in meters. Then $a$ indeed has units $1/s$ and $b$ has units $m/\sqrt{s}$; $dt$ as usual has units $s$ and $dW$ has units $1/\sqrt{s}$.

This can be verified by looking at a physical model of the OU process, such as Hooke's law with damping and a noise term (see Wikipedia). Then $a = - k/\gamma$, $b^2 = 2 k_b T/\gamma$, where $k$ is Hooke's constant in kg/s^2, $\gamma$ the friction coefficient in kg/s, and $k_b T$ is in Joules (kg*m^2/s^2).

share|cite|improve this answer

In your setup the Wiener term carries units. Think of the fact that $W_t - W_s \sim \mathcal{N}(0,t-s)$ for $s < t$.

share|cite|improve this answer
And just to clarify: this means that $dW_t$ has units of (time)^{1/2}, so b has units of (time)^{-1/2}. Since as you believed a has units of (time)^{-1}, this means that the variance is dimensionless. – Steve Huntsman Dec 18 '09 at 20:42
Thinking of $dW_t$ as having units of (time)^{1/2} is a useful heuristic for checking if expressions in stochastic calculus are dimensionally consistent. – Michael Lugo Dec 18 '09 at 20:44
I always remember it as dW^2 = dt by Ito calculus, hence the 1/2. – Alex R. Jan 8 '11 at 3:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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