Joint probability distribution of the Ornstein-Uhlenbeck process

The best known stationary Markov process is the Ornstein-Uhlenbeck process fully characterized via: $$p_1 (x_1) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x_1^{2}}, \quad p_{1|1}(x_2|x_1) = \frac{1}{\sqrt{2\pi (1 - e^{-2\tau})}} exp \left[ -\frac{(x_2 - x_1 e^{-\tau})^2} {2(1-e^{-2\tau})} \right],$$ $$p_1$$ is the probability to obtain the value $$x_1$$ at time $$t_1$$, while $$p_{1|1}$$ is the transtion probability, i.e., the conditioned probability of obataining $$x_2$$ at $$t_2$$ given $$x_1$$ at $$t_1$$. The transition probability depends only on the time difference $$\tau = t_2 - t_1 > 0$$. This process was originally derived to determine the velocity of a Brownian particle and is essentially the only process which is Gaussian, Markovian and stationary.

How to calculate the joint probability distribution $$p(x_1,x_2)$$ of the Ornstein-Uhlenbeck process?

-> How to calculate the autocorrelation function of the Ornstein-Uhlenbeck process via $$\langle \langle x_1x_2\rangle\rangle = \int dx_1 \int dx_2 x_1 x_2 p(x_1x_2)$$

Joint probability distribution $$p(x_1,x_2)$$ of the Ornstein-Uhlenbeck process. Using the definition, $$p(x_1, x_2) = p(x_2|x_1) p(x_1) \\ = \frac{1}{\sqrt{2\pi (1 - e^{-2\tau})}} exp \left[ -\frac{(x_2 - x_1 e^{-\tau})^2} {2(1-e^{-2\tau})} \right] . \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x_1^{2}} \\ = \frac{1}{2\pi \sqrt{(1 - e^{-2\tau})}} exp \left[ -\frac{1}{2} \left( \frac{(x_2 - x_1 e^{-\tau})^2} {2(1-e^{-2\tau})} - x_1^2 \right) \right]$$

How to proceed further or this is the final solution??

• The definition of a conditional probability is: $$p(x_1, x_2) = p(x_1|x_2)p(x_2)=p(x_2|x_1)p(x_1)$$ Commented Dec 7, 2020 at 11:11
• @Vadim Thanks for the response. So I need to just plug in the $p_1(x_1)$ values? Commented Dec 7, 2020 at 11:17
• Yes, this is it. Commented Dec 7, 2020 at 11:28
• @Vadim That seems like an answer Commented Dec 7, 2020 at 13:44
• Yes, this is it. Commented Dec 8, 2020 at 14:06

Actually, I have the same doubt. And I got the same Integration form. I will tell you the procedure that I followed. You can first integrate the $$x_2$$ part. Which will give you $$x_1exp(-(t_1-t_2))$$. Next, the integrand will be $$x_1^2exp(-(t_1-t_2)) dx_1$$ integrated over -infinity to infinity. Basically, leaves a term proportional to $$exp(-(t_1-t_2))$$. Since integrand is not taken over time. Hopefully, it clarifies your doubt.