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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.

edited Dec 19 at 2:15

Scott Morrison♦
2,879●8●28

asked Dec 18 at 19:46

Mark4483
89●1

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You don't seem to have defined W. – Qiaochu Yuan Dec 18 at 20:39

$W_t$ is the Wiener process. – Steve Huntsman Dec 18 at 20:43

Steve Huntsman · Answer 1 · 2009-12-18 20:39:09Z UTC

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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 $s < t$.

answered Dec 18 at 20:39

Steve Huntsman
3,202●4●22

	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 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 at 20:44

Units in Ornstein-Uhlenbeck(OU) process

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