# Ornstein-Uhlenbeck process and Markov property

There isn't a similar question in the forum, so here it goes.

Firstly, let the O-U velocity process be defined as $$dV_t = -\beta V_t dt + \sigma dB_t$$ with $V_0 = v$, and $B = (B_t), t \geq 0$ a Brownian motion. Additionally, let the O-U position process be defined as $$dX_t = V_tdt$$ with $X_0 = x_0$.

In the literature, it is widely taken for granted that $V_t$ is a Markov process, but $X_t$ is not.

How can these two things be proved?

And a bonus question: if we add the velocity process to the position process, thus obtaining an O-U position-velocity process $(X,V) = ((X_t,V_t))_{t \geq 0}$, why is this Markov?

Thanks very much

• Try to prove this when $V=B$. The idea is the same hence if you have trouble finding the proof that X is not Markov but (X,V) is Markov in this simpler setting, this is the case you should be looking at.
– Did
May 9 '14 at 11:12

Hint: You can usually write the unique strong solution of the first SDE as : $V_t = e^{\beta t}V_0 + \sigma \int_0^t e^{\beta s}dB_s =e^{\beta t}v + \sigma \int_0^t e^{\beta s}dB_s$ Then what do you think about the second term in the RHS? The question about the non-Markovianity of $X$ follows from the previous calculus
• if anything, $V_t = e^{-\beta t}v + \sigma e^{-\beta t}\int_0^t \! e^{\beta s} \, \mathrm{d}B_s$, so I'm afraid your expression for $V_t$ is not correct. Moreover, you answer is not very rigorous so it unfortunately does not really suit me.
• Adam, i'm afraid I supposed your question was well-posed. Actually, you wrote the SDE $dV_t =\beta V_tdt + \sigma dB_t$ . My answer is correct regarding this equation, but I agree that OU-Process is usually defined as the solution of $dV_t =-\beta V_tdt + \sigma dB_t$, in which case your answer is right...
• @Adam so if $- \beta$ is $<0$, we have that the process is non markovian, otherwise it is ? I am asking because there is no answer to that question Jan 19 '20 at 15:48