1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Master equation for Ornstein-Uhlenbeck process

  1. May 31, 2012 #1
    Hi everybody...

    I've been working through N.G. Van Kampen's "Stochastic Processes in Physics and Chemistry" and have run into something that has got me sort of stumped. He defines the Ornstein-Uhlenbeck process (a stationary, Gaussian, Markovian random process) in terms of the transition probability between two values [itex]y_{1}[/itex] and [itex]y_{2}[/itex] separated by a time t:

    [itex]T_{t}(y_{2}|y_{1}) = (2π(1-e^{-2t})^{1/2}\exp(\frac{(y_{2}-y_{1}e^{-t})^{2}}{2(1-e^{-2t})})[/itex]

    and the probability distribution

    [itex]P_{1}(y_{1}) = (2π)^{-1/2}e^{-y_{1}^{2}/2}[/itex]

    (This is pg. 83 if you have the book.)

    A little bit later, he defines the master equation by expanding T[itex]_{t}(y_{2}|y_{1})[/itex] in powers of t and defining the coefficient of the linear term as [itex]W(y_{2}|y_{1})[/itex], the transition probability per unit time (pg. 96 if you have the book).

    The thing that's got me stuck here is that, as t→0, T[itex]_{t}(y_{2}|y_{1})[/itex]→δ[itex](y_{2}-y_{1})[/itex], since T[itex]_{t}(y_{2}|y_{1})[/itex] is a Gaussian. I can't seem to come up with a reasonable way to expand this in terms of small t, which makes it difficult to define the master equation.

    So... does anyone know how the Ornstein-Uhlenbeck process is formulated as a master equation? Any ideas would be greatly appreciated.

  2. jcsd
  3. Jun 1, 2012 #2


    User Avatar
    Science Advisor

    For small t, 1-e-2t ≈ 2t. This gives a Gaussian with variance ≈ 2t, which -> delta function as t -> 0.
  4. Jun 1, 2012 #3
    Exactly. But that still doesn't tell me how to get a small-t expansion for [itex]y_{1} ≠ y_{2}[/itex], which is what I need for a meaningful transition rate.
  5. Jun 2, 2012 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The point is that [itex]y[/itex] is a random variable. I don't know that book, but what he finally should get at, I guess, is that the Fokker-Planck equation for the time evolution of the distribution function is equivalent to a Langevin equation, which is an ordinary stochastic differential equation with Gaussian-distributed (white) noise.

    For a simple derivation (however for the relativistic case) see


    p. 41ff.
  6. Jun 2, 2012 #5


    User Avatar
    Science Advisor

    e-t≈1-t, so y1-y2e-t≈y1-y2-ty2. The net result for the integrand as t -> 0 is δ( y1-y2)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook