What does it mean to have delta-correlated process physically? I am reading about Langevin dynamics, and I see the following equation:
$$\frac{dx}{dt} = -\frac{1}{\xi} \frac{\partial U}{\partial x} + g(t)$$
Then, they claim that the average $$\langle g(t) \rangle = 0$$
Which makes sense. The distribution has a mean of zero, over time, so
$$\int_0^{\infty} g(t) dt = 0$$.
Then they claim that the variance is
$$\langle g(t)g(t') \rangle = 2D\delta (t-t')$$
I don't understand what this means.
Isn't variance for a distribution $f$ defined as
$$\int _{-\infty}^{\infty} f(x)(x-\mu)^2 dx?$$
What is with the other variable, $t'$ popping in?
What is the physical behind a delta-correlated function, and how do I mathematically write it down?
 A: $t'$ is popping in because it appears on the left-hand side of the equation. Your calculation is not so much a variance as a covariance between two different random variables, $g(t), g(t')$ This is saying that $g(t)$ has mean zero, which you say makes sense and that there is no correlation between $g()$ at two different times $t, t'$, which is why the right-hand side vanishes for $t \ne t'$. However, the fact that it's a Dirac delta function rather than a true zero means that $g(t)$ is not always zero, that it fluctuates about zero in a way that's entirely random.
I find it easier sometimes to think about such things in terms of discrete time intervals rather than in the continuum. Suppose that we have some set of discrete times $t_i$ and that for any time $t_i$ $g(t_i)$ could equal $g_0$ or $-g_0$ with equal probability. Further, suppose that the values of $g$ at the different times are completely uncorrelated. Then, we would have:
$$\langle g(t_i) \rangle = 0,$$
$$\langle g(t_i) g(t_i') = g_0^2 \delta_{t_i t_i'},$$
where the final symbol, the Kronecker delta is one if the two times are equal and zero otherwise. Hopefully this case feels a bit more intuitive. When we take our time intervals smaller and smaller, it turns into the continuous-time case that you introduced.
A: Consider a Gaussian white-noise Langevin  equation
$$
\dot x= \eta(t), \quad \langle \eta(t)\rangle=0, \quad \langle \eta(t)\eta(t')\rangle = \frac{2}{\beta} \delta(t-t').
$$
We   have
$$
x(t) = \int_0^t \eta(\tau)d\tau
$$
and deduce
$$
\langle{x^2(t)}\rangle =  \langle  \int_0^t  \eta(\tau) d\tau  \int_0^t \eta(\tau')d\tau' \rangle  \nonumber\\ 
= \int_0^t\int_0^t  \langle { \eta(\tau) \eta(\tau')\rangle } d\tau d\tau'\nonumber\\
= \frac 2{\beta} \int_0^t\int_0^t \delta(\tau-\tau')d\tau d\tau'\nonumber\\
= \frac 2{\beta} t.\nonumber
$$
The position $x(t)$ is a sum of Gaussian random variables, so its  probability distribution must also be a Gaussian with the above expression as its  variance. In  other words
$$
P[x(t)=x|x(0)=0] = \sqrt{\frac{\beta}{4\pi t}} \exp\left\{-\frac{\beta x^2}{4t}\right\}.
$$
