Are stationarity, Markovianity and Gaussianity sufficient conditions to ensure that the random force on a Brownian particle is delta correlated?

In the Langevin model, if we make the assumption that the random force $$\eta(t)$$ acting on the Brownian particle is a stationary, Markovian, and gaussian process, does it automatically ensure that the autocorrelator $$\langle\eta(t_1)\eta(t_2)\rangle\propto \delta(t_1-t_2)$$?

The reason I am asking this is that deviation from delta-correlated autocorrelator is often regarded as a signature of non-Markovianity.

Since there is an explicit formula for the autocorrelator, I want to check under what assumptions, this becomes delta-correlated (and whether those assumptions have anything to do with Markovianity). To that end, we recall that for a stationary, Gaussian process the relevant probability density function to calculate $$\langle\eta(t_1)\eta(t_2)\rangle$$ is given by

$$p(\eta_2;t_2;\eta_1,t_1)=C\exp\left[-\sum\limits_{j=1}^{2}\sum\limits_{k=1}^{2}\alpha_{jk}(\eta_j-\bar{\eta})(\eta_k-\bar{\eta})\right]\tag{1}$$

where $$C$$ and $$\alpha_{jk}$$ depend only on $$(t_1-t_2)$$. Therefore, $$\langle \eta(t_1)\eta(t_2)\rangle=\int d\eta_1\int d\eta_2~~ \eta_1\eta_2p(\eta_2,t_2;\eta_1,t_1).\tag{2}$$

But using $$(2)$$ in $$(1)$$ does not seem to lead to a $$\delta(t_1-t_2)$$ dependence. I know that when the process is Markovian, we have, $$p(\eta_2,t_2;\eta_1,t_1)=p(\eta_2,t_2|\eta_1,t_1)p(\eta_1,t_1)$$ but I am not sure if that is going to help here. I have used the symbol '$$|$$' to denote conditional probability.

Delta-correlated process and Markovian process is not necessarily the same thing (although the terminology may vary - my discussion is based on the theory of random processes, but, e.g., in the theory of master equations Markovian is understood somewhat differently).

Markovian process
Markovian process is often said to be the process where the probability of an event depends only on the state of the process in the previous time moment. This is not literally the preceding time instant, but the moment where the process was last measured.

On a more technical level, let us consider a process $$x(t)$$ that was measured at times $$t_n, t_{n-1}, ... , t_2, t_1$$. The joint probability density that at these times the process took values $$x_n, x_{n-1}, ..., x_2, x_1$$ is $$p(x_n, t_n; ... ;x_2, t_2; x_1, t_1) = p(x_n, t_n| x_{n-1}, t_{n-1}; ... ;x_2, t_2; x_1, t_1)\times p(x_{n-1}, t_{n-1}| x_{n-2}, t_{n-2};... ;x_2, t_2; x_1, t_1)\times p(x_2, t_2| x_1, t_1)\times p(x_1, t_1)$$ Note that so far we have made no assumptions, but only used the definition of the conditional probability density. In particular, if the process is measured only at two time instants: $$p(x_2, t_2; x_1, t_1)=p(x_2, t_2| x_1, t_1)p(x_1, t_1)$$ - this is just a definition.

For a Markovian process the conditional probability densities in the equation above depend only on the latest time instant: $$p(x_n, t_n; ... ;x_2, t_2; x_1, t_1) = p(x_n, t_n| x_{n-1}, t_{n-1})\times p(x_{n-1}, t_{n-1}| x_{n-2}, t_{n-2})\times ...\times p(x_2, t_2| x_1, t_1)p(x_1, t_1)$$

Gaussian process
For a Gaussian process the probability density above is given by a multivariate Gaussian/normal distribution $$p(x_n, t_n; ... ;x_2, t_2; x_1, t_1) = \frac{1}{\sqrt{(2\pi)^n\det{\Sigma}}} e^{-\frac{1}{2}\sum_{i,j=1}^n\left(x_i-\bar{x(t_i)}\right)\left(\Sigma^{-1}\right)_{ij}\left(x_j-\bar{x(t_j)}\right)}$$ If the number of measurements goes to infinity, the distribution is replaced by a functional $$p\left[x(t)\right] \propto \exp\left[-\frac{1}{2}\int dt_1, dt_2 \left(x(t_2)-\bar{x(t_2)}\right)\left(\Sigma^{-1}\right)(t_2,t_1)\left(x(t_1)-\bar{x(t_1)}\right)\right]$$

White noise
White noise is a process with uniform spectrum, i.e., with a constant Fourier transform of its covariance function. In real space it means that $$\langle x(t_2)x(t_1)\rangle -\langle x(t_2)\rangle\langle x(t_1)\rangle =\sigma_x^2\delta(t_2-t_1)$$

Remark As I noted in the beginning, the term Markovian is sometimes used with somewhat different meaning. In statistical literature, the deviations of the correlation function from delta-function are usually referred to as colored noise (as opposed to white noise). In this case the process is no more diffusive and one cannot use usual Fokker-Planck equation - instead one has to resort to its non-local version, such as Kolmogorov-Feller equation for a jump process. This has parallel developments in terms of Langevin equation, which becomes non-local (an integral term appears).

Correlation function
Two-point correlation function is defined as $$\langle x(t_2)x(t_1)\rangle=\int dx_2\int dx_1 x_2 x_1p(x_2,t_2; x_1,t_1).$$ We thus see that the Markovian nature of the process, as defined above, does not impose any restructions on the correlation function: Markovian process may easily correspond to a colored noise.

• I am aware of the Markov process definition that you have written down. But if you look at this paper Quantum Langevin equation by G. W. Ford, J. T. Lewis, and R. F. O’Connell , it seems to say that in the quantum case, deviation of the autocorrelator from $\delta(t_1-t_2)$ is a sign of departure from Markovianity.
– SRS
Jun 14, 2021 at 7:51
• I have expanded the answer. As I noted in the beginning, the definition of Markovian is sometimes misused - I noticed it when dealing with people working in quantum optics. Jun 14, 2021 at 7:55
• @SRS it seems that there is linguistic ambiguity regarding the Markovian process as I described, and Markov approximation which implies no memory of previous events - essentially a delta-correlated process. Deviations from either can be referred to as non-Markovian. Jun 14, 2021 at 8:12
• @RogerVadim Exactly my point, I was just trying to clarify why the deviation from white noise is used to characterise non-Markovian nature in Langevin equation. My argument is that this is because if the driving noise in the Langevin equation is NOT white, then it necessarily leads to a non-Markovian dynamics in the driven process. I will probably try to summarise my comments in the form of an answer with equations for better clarity. Oct 28, 2021 at 9:22
• @user35952 As pointed in the early comments in this thread, the term markovian is used differently in some fields. If noise is not white, one cannot use an equation local in time, but it doesn't make a process non-markovian in statistical sense. Oct 28, 2021 at 9:29

In the Langevin model, if we make the assumption that the random force η(t) acting on the Brownian particle is a stationary, Markovian, and gaussian process, does it automatically ensure that the autocorrelator $$⟨η(t_1)η(t_2)⟩∝δ(t_1−t_2)$$?

No. The most general form of Langevin equation is given with coloured noise (which is NOT delta correlated). I will show how the Langevin equation pans out in the case of white noise (i.e. delta-correlated).

As I show below, the random force (or noise) acting on the particle need not be delta-correlated to be Markovian. For example, a stationary Gaussian distributed coloured noise with an exponential autocorrelation, $$⟨η(t_1)η(t_2)⟩∝\exp(\frac{t_1 - t_2}{\tau})$$, is Markovian.

The reason I am asking this is that deviation from delta-correlated autocorrelator is often regarded as a signature of non-Markovianity.

I think the signature implies that if the driving noise, $$f(t)$$ is coloured (NOT delta-correlated), then the driven process, $$x(t)$$, will essentially become non-Markovian.

I will try to illustrate this behaviour with an example of the Langevin equation.

In general, the Langevin-like Stochastic differential equations (SDEs) is given by (also called by generalised Langevin equation),

$$\frac{dx}{dt} = -\int_{0}^{t}\mu(t-t')x(t')dt' + f(t)$$

where I will refer to $$x(t)$$ as the driven process and $$f(t)$$ as the driving noise. Here, $$\mu(t)$$ is a memory kernel. Although the noise and memory kernel in this equation look independent, they have to be related to each other by a fluctuation-dissipation relationship given by,

$$\left = \Gamma\mu(t-t')$$ where $$\Gamma$$ is a constant to match the appropriate dimensional scaling.

Case of white noise

In this scenario, the driving noise $$f(t)$$ is delta correlated (let's say with a variance $$\Gamma$$), therefore the autocorrelation is given by,

$$\left = \Gamma\delta(t-t')$$

Using this in the first and second equations, the Generalised Langevin equation simply reduces the usual Langevin equation

$$\frac{dx}{dt} = -\Gamma x(t) + f(t)$$

The solution for this equation can be shown to be essentially Markovian, i.e. the driven process $$x(t)$$ is Markovian in nature if the driving noise $$f(t)$$ is White.

Case of coloured noise

A coloured noise is essentially a stochastic process with some finite correlation time (in the case of delta-correlated process, this is implicitly zero). Let's consider a stationary coloured noise with an arbitrary correlation function, such that,

$$\left = \Gamma \lambda(t-t';\tau)$$ where $$\tau$$ is the characteristic timescale associated with the decay of the coloured noise autocorrelation. Therefore in the case of coloured noise, the process $$f(t)$$ can be both Markovian or non-Markovian. A typical example of Markovian coloured noise can be defined through an exponential autocorrelation, $$\lambda(t) = \exp(-t/\tau)$$

But irrespective of the Markovianity of $$f(t)$$, if $$f(t)$$ is coloured, one can directly verify from the first two equations, that the driven process $$x(t)$$ becomes non-Markovian.

In summary

• If the driving noise, $$f(t)$$, is delta correlated, then the driven process, $$x(t)$$ is essentially Markovian.
• If the driving noise, $$f(t)$$, is NOT delta correlated (i.e. coloured noise), then the driven process $$x(t)$$ is non-Markovian.

I think it is in this context, signature of non-Markovianity is regarded through a deviation from delta-correlated autocorrelator.

To me, the white noise is an idealization that is defined by three properties: $$<\eta(t)>=0$$, $$<\eta(t)\eta(t')>=\delta(t-t')$$ and $$\eta(t)$$ is Gaussian. It is Markovian because of these properties. Physically, the white noise is a random sequence of negative and positive pulses and every pulse is independent from the previous one.

All of this is consistent with your equation (1) since you get a Dirac delta from a normal distribution making the standard deviation going to zero while keeping the area under the curve constant.