Basic doubt regarding Markov Processes

Question

Take the Langevin equation for the position of a particle in Brownian motion.

$$ m\frac{d^2x}{dt^2} = -\gamma\frac{dx}{dt} + \eta(t) $$

My professor wrote this as the following in the class: $$ \lim_{\Delta t\to 0} \frac{x(t+\Delta t) + x(t-\Delta t)-2x(t)}{(\Delta t)^2} = \lim_{\Delta t\to 0} \frac{-\gamma}{m} \frac{x(t+\Delta t)-x(t)}{\Delta t} + \frac{\eta(t)}{m} $$ where $\gamma$ stands for the frictional co-efficient, m is the mass of the particle, and $\eta(t)$ stands for the force due to collisions which is stochastic in nature. The above can be written as:
$$ x(t+\Delta t) = f (t, t-\Delta t) $$ for some $\text{f(t)}.$ Now, my professor claims that the above process where X(t) is the stochastic variable is non-markovian in nature since to evaluate the position at the instant x(t+$\Delta$t), one needs information about not only the previous instant (x(t)) but also about the one before that (x(t-$\Delta$t)). This means that the process still has some memory of its past.

Now I'm not sure I'm convinced. What is a suitable numeric time in the past until when the process is allowed have memory and all instants before that there must be no memory for the process to be termed memoryless and hence markovian. I don't think there's anything like that but then what marks the distinction between the process that is memoryless and one that isn't? Here, $\Delta$t is arbitrarily small and hence to get a position at a time t+$\Delta$t, one can obtain it by making $\Delta$t arbitrarily small in which case it would look like one can evaluate the position by only knowing position at arbitrarily close times before the current position, making the process seem memoryless.

Or is it that we need two instants in the past to calculate the position at x(t+ $\Delta$t) it and hence the process has a memory as opposed to only one instant in the past? I'm not sure if I'm thinking in the right direction. Please help me out.

Answer 1

The transition probability of a Markovian process obeys Smolukhovsky equation: $$P(x,t|x',t') = \int dy P(x,t|y, \tau)P(y, \tau|x',t'),$$ where $t> \tau > t'$.

Now back to your problem.

• First of all, the argument given by your professor is somewhat hand-waving: there is no equivalency between the second and the third equations, since the former is valid in the limit $\Delta t \rightarrow 0$, whereas the latter is a finite difference equation.
• Secondly, the usual way to go about this problem is to reduce it to the system of two first-order equations by introducing $v(t)=\frac{dx(t)}{dt}$ and then solving for $P(x,v,t|x',v',t')$ either solving the Langevin equation directly and then averaging over the noise or by writing the corresponding Fokker-Planck equation. (which however already presupposes that the process is Markovian).
• If one tried to bypass the reduction to the first-order system of equations and tried to solve for $x(t)$ directly and average over the noise, the non-Markovian nature would surface in the fact that one has to integration constants (initial position and initial velocity: $x', v'$) rather than a single one ($x'$)

To summarize: your professor is right, his/her argument is not rigorous, and the non-Markovian is not really a problem here.