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I have been pondering about the following question:

Given a time-dependent function $f(t)$, is it possible to show that its autocorrelation function will generally follow a decaying exponential behavior for some time $\tau$, i.e. $\langle f(t)f(t+dt) \rangle_t \rightarrow e^{-dt/{\tau}}$?

EDIT: assuming $f(t)$ varies and has an autocorrelation function independent of initial time

I was thinking about the following. To first order:

$$\dot{f}(t) \approx \frac{f(t+dt) - f(t)}{dt}$$

so $$\frac{d}{dt}[f(t)f(t+dt)] - f(t)\dot{f(t+dt)} = \dot{f}(t) f(t+dt) \approx \frac{f(t+dt)^2 - f(t)f(t+dt)}{dt}$$

with $$f(t) \dot{f(t+dt)} \approx f(t)\frac{f(t+2dt) - f(t+dt)}{dt} = \frac{f(t)f(t+2dt) - f(t)f(t+dt)}{dt}$$

Assuming that the correlation with $t+2dt$ is approximately zero (first term in last equation), I obtain

$$\frac{d}{dt}[f(t)f(t+dt)] \approx \frac{f(t+dt)^2}{dt} - \frac{2}{dt}f(t)f(t+dt)$$

which kind of looks what I am trying to achieve. Are my approximations defensible? What about the $\frac{f(t+dt)^2}{dt}$ term? Or did I follow the wrong way?

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  • $\begingroup$ Might Mathematics be better suited for this question? $\endgroup$
    – Kyle Kanos
    Mar 16 '18 at 13:20
  • $\begingroup$ Can you be more precise about what is $f(t)$? If not, the answer to your question is trivially 'no'. For example, pick $f(t) = 1$ and get $<f(t) f(t+dt)> = 1$. $\endgroup$ Mar 16 '18 at 13:51
  • $\begingroup$ I would try it for a general time-dependent function, that is $f(t) \neq f(t+dt)$, and assuming it is a function for which the autocorrelation does not depend on the initial time $t$ (i think of correlated noise). I want to prove that for any such function, one can find a time $\tau$ (it can be small) for which the autocorrelation is exponentially decaying. $\endgroup$
    – M. Hennes
    Mar 16 '18 at 13:58
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No. Witness the Wiener-Khinchin theorem. Arbitrary power spectral densities are possible and observable with physical processes. Accordingly, their Fourier transforms (i.e. autocorrelation functions) can also be pretty much arbitrary.

An exponential autocorrelation function is the special case when the power spectral density function is a Lorentzian function of spatial frequency:

$$H(f) \propto \frac{1}{1+\frac{(f-f_0)^2}{f_B^2}}$$

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  • $\begingroup$ I agree that the Lorentzian PSD will give an exponential decaying ACF, for all dt. My question is rather: can i find a $\tau$ (however small it is), for which some time varying function will decay exponentially on this small time scale. $\endgroup$
    – M. Hennes
    Mar 16 '18 at 13:28
  • $\begingroup$ @M.Hennes Well, one can always match any differentiable-with-time decay to exponential decay over a short enough time scale - just match the initial gradients. $\endgroup$ Mar 16 '18 at 13:33
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    $\begingroup$ Yes this is what i am trying to do. What do you mean by match the gradients? Are my calculations close? $\endgroup$
    – M. Hennes
    Mar 16 '18 at 13:37
  • $\begingroup$ @M.Hennes Well, $R(0)=\sigma^2$ and so, for small $t$, $R(t) = \sigma^2- R^\prime(0)\,t$. Also $\sigma^2\,\exp(-t/\tau)=\sigma^2-\sigma^2\,t/\tau+\cdots$, so just choose $\tau$ s.t. $R^\prime(0) = \sigma^2/\tau$. But it won't work if $Rˆ\prime(0)=0$: then you are going to need a Gaussian approximation. $\endgroup$ Mar 16 '18 at 14:05
  • $\begingroup$ But here you approximate the $exp$ by its first order linear term. So you show that for an even smaller time, the ACF decays linearly in time, with constant slope $R'(0)$. My question relates to the full exp. $\endgroup$
    – M. Hennes
    Mar 16 '18 at 14:15

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