According to Wikipedia, the transitional probability of the Ornstein–Uhlenbeck process

$$ \mathrm dx_t = -\theta(\mu-x_t)\,\mathrm dt + \sigma\,\mathrm dW_t $$


$$ P(x, t\,|\,x_0, t_0) = \sqrt{-\frac{\theta}{\sigma^2(1-e^{-2\theta(t-t_0)})}}\exp\left(-\frac{\theta}{\sigma^2}\frac{(x-x_0e^{-\theta(t-t_0)})^2}{\sigma^2(1-e^{-2\theta(t-t_0)})}\right) $$


$$ x_t\,|\,x_0 = N\left(x_0e^{-\theta(t-t_0)}, -\frac{1}{\theta}\sigma^2(1-e^{-2\theta(t-t_0)})\right) $$

without any proof. It mentions that this is derived from the Fokker–Planck representation, but would there be a way to derive this directly? Perhaps from the formal solution?

$$ x_t - x_0 e^{-\theta(t-t_0)} = \mu(1-e^{-\theta(t-t_0)}) + \sigma\int_{t_0}^t e^{-\theta(t-s)}\,\mathrm dW_s $$

Any references to literature with details would also be appreciated.


1 Answer 1


There are a few different direct derivations. One can use the fact that Itô integral processes with non-random integrands are Gaussian. (Another solution is to compute the transition density from the formal solution by use of the Markov property).

First, note that for the OU process with drift (note the sign change to agree with the usual definition!) $$d X_t= \theta(\mu−X_t) dt + \sigma dW_t$$ one has the solution $$X_t = x_0 e^{-\theta (t - t_0)} + \mu (1 - e^{-\theta (t - t_0)}) + \sigma \int_{t_0}^t e^{-\theta(t-r)}\,dW_r$$

Now, since $$\int_{t_0}^t e^{-2\theta(t-r)}\,dr = \frac{1 - e^{-2\theta (t-t_0)}}{2\theta}$$ it follows the Itô integral process is $\mathcal N\left(0, \frac{1 - e^{-2\theta (t-t_0)}}{2\theta}\right)$. Hence by a scaling argument $$ X_t \mid X_0 \sim N\left(x_0 e^{-\theta (t - t_0)} + \mu (1 - e^{-\theta (t - t_0)}),\ \frac{\sigma^2}{2\theta}(1 - e^{-2\theta (t-t_0)})\right)$$ from which the transition density can be easily deduced (thus there is a typo in the proposed solution in the question).


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