# Expression of Dirac Delta Correlation

spatio-temporal white noise $$\xi(x,t)$$ is often expressed as

$$\langle\xi(x,t)\rangle=0,$$ $$\langle\xi(x_1,t_1)\xi(x_2,t_2)\rangle=\delta(t_2-t_1)\delta(x_2-x_1).$$

Now I understand that the first term means zero mean and the second term means Dirac delta correlation, i.e., zero correlation unless $$t_2=t_2$$ and $$x_2=x_1$$. First of all I think it is a strong assumption that there is zero correlation in space. However, this clearly depends on the problem.

My question is: What is the explanation of the second term? I understand that the RHS is always zero unless $$t_2=t_1$$ and $$x_1=x_2$$. But what about the LHS? Is this the mean of the product of two random variables? If yes, why should it be $$\neq0$$ if the mean of both variables is 0? Or does it mean something else?

• Mean, or expectated value, of two random variables multiplied is just their covariance (given the means of each here are zero). A normal variable has a mean of zero, but its variance certainly does not vanish. Dec 1, 2019 at 12:11
• I’m not sure that I’d say that white noise is “expressed as” but perhaps instead “has the properties of” would be better/more accurate Dec 1, 2019 at 12:15
• But the latter relationship is really saying that the process is uncorrelated with itself, no? That is, its autocorrelation is a delta function Dec 1, 2019 at 12:18
• When $t_1=t_2$ and $x_1=x_2$ you are calculating the average of something that is never negative, ie., the average of the square of the process $\xi(x,t)$, and that average cannot be zero unless the process itself is zero almost everywhere. Dec 1, 2019 at 12:58

Quite generally, a random noise is considered a generalized force (in Brownian motion: Collisions with the solvent molecules, e.g. water). The averaging you carry out is over all possible values of the force. A mean zero random noise is often a consequence of the assumption of an isotropic noise distribution -- any direction is equally likely, so $$\langle \xi(x,t) \rangle = \langle -\xi(x,t) \rangle$$ = 0. Obviously, this does not apply to the square of the force, $$\langle \xi^2(x,t) \rangle$$.
This is a matter of scales. Every process has its time and length scale. The assumption of $$\delta$$-correlation is sensible if the length and time scale associated with the fluctuation of your random force are negligible compared to the time and length scales associated with the motion of your particle (or random field, whatever your system is describing). Let's take two length scales, the microscopic length scale $$l_{\textrm{mic}}$$ and a mesoscopic length scale $$l_{\textrm{mac}}$$. Now, ignoring time dependence, the random noise correlation in your system may be of the form $$\langle\xi(x_1)\xi(x_2)\rangle = C e^{-|x_1 - x_2|/l_{\textrm{mic}}}$$. The constant may depend on $$l_{\textrm{mic}}$$. But for $$l_{\textrm{mic}} \ll l_{\textrm{mac}}$$, the exponential will drop off very fast on the length scale you are solving your Langevin equation (or what it may be) on. In principle, you will end up with integrals of the form $$\int_0^l \textrm{d}(|x_1 - x_2|)\langle\xi(x_1)\xi(x_2)\rangle = C\int_0^l \textrm{d}r e^{-r/l_{\textrm{mic}}} \ {\color{red}=}\ C\int_0^{{\color{red}\infty}} \textrm{d}r e^{-r/l_{\textrm{mic}}} = C l_{\textrm{mic}}.$$ The red equality is due to the fact that $$l \sim l_{\textrm{mac}}$$ will be a macroscopic length scale and we can extend the integral to infinity. Therefore, it is a reasonable approximation to just replace the correlation by a $$\delta$$-function with a proper coefficient.