Consider an Ornstein-Uhlenbeck process

$$ dX_t = \kappa( \theta -X_t) d t + \sigma d W_t, \quad X_0 = x $$ and $W$ a Brownian Motion. Consider also its time integral $$ Y(t) = \int_0^t X(s) d s. $$ One can show/knows that they are both Gaussian and $X$ is also Markov. I've read the claim that $Y$ is not Markov, but $(X,Y)$ is. Is there an easy proof of this? - I would mostly be interested in a probabilistic one.


1 Answer 1


It takes only some little algebra to get the result.

Using the fact that $X_t= X_s.e^{-\kappa(t-s)}+\theta.(1-e^{-\kappa(t-s)})+\int_s^t \sigma e^{-\kappa(t-u)}.dW_u$ equation (1),

$Y_t$ can be rewritten for any $s<t$ as :

$$Y_t=Y_s +\int_s^t X_udu$$

Now using equation (1) :

$$Y_t=Y_s + X_s.\int_s^t e^{-\kappa(u-s)}du + \theta.\int_s^t(1-e^{-\kappa(u-s)}) du + \sigma.\int_s^t\int_s^u e^{-\kappa(u-v)}dW_v.du $$ So now everything becomes clear once we use stochastic Fubini in the last integral (equation (2)):

$$Y_t=Y_s + X_s.g(s,t) + f(s,t) + \sigma.\int_s^t[\int_u^t e^{-\kappa(u-v)}du].dW_v $$

where $f$ and $g$ are suitable deterministic functions. As $X$ is known to be Markovian (classical result form solution of SDE equations gives this result see for example Protter), we have that the couple (X_t,Y_t) is Markovian using equation (2).

Best regards.

  • $\begingroup$ Any idea where this will be written up? Also looking for full mean and covar calculations. I see a few papers covering parts of this but it must be fairly common. $\endgroup$
    – mathtick
    Mar 10 at 14:19

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