# Units in Ornstein-Uhlenbeck(OU) process

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.

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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

## 2 Answers

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).

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

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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