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
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