In general, for a given physical quantity $f(t)$ that is a smooth function of time $t$, we can define a self or auto-correlation by $$C=\langle f(t)f(t')\rangle.$$ Often it is said that correlations of type $C$ are most useful when expressed in terms of the convolution of $f$ with a suitable function. Is there a simple way of showcasing how auto-correlation functions and convolutions are related?


As $\langle f(t)f(t')\rangle$ only depends on the difference $s = t' - t$ (if we integrate over $\Bbb R^n$), we can write

$$\begin{split}C(s) & = \langle f(t)f(t + s)\rangle \\ & \equiv \int f(t)f(t + s) dt \\ & = \int f(t-s)f(t)dt \\ & = \int \tilde f(s - t)f(t)dt \\ & \equiv (\tilde f\ast f)(s)\end{split}$$

where $\tilde f(t) \equiv f(-t)$.

  • $\begingroup$ It should be taken as the definition. Depending on the domain you may want to multiply by a constant factor. If the domain is a torus (finite volume) or finite, we could multiply the integrand by the actual probability density, which is constant for our uniform domain, to get an actual expectation, but on $\Bbb R^n$ we cannot normalize and we have to work with relative probabilities, a bit like wave functions of free particles, which cannot be normalized. $\endgroup$ – doetoe Jun 19 '17 at 9:45
  • $\begingroup$ thanks a lot for getting back to me, fair points. Can we also interpret $\langle \rangle$ as a time average? $\endgroup$ – user929304 Jun 19 '17 at 10:56
  • $\begingroup$ If you interpret $t$ as time, then this is essentially the time average of the product of the values of $f$ at time $t$ and at time $t + s$, where $s$ is a fixed time shift. In your original notation you essentially do the same for any combination of $t$ and $t'$, and then you see that it only depends on $t - t'$, the relative displacement in time. $\endgroup$ – doetoe Jun 19 '17 at 12:46

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