# Exponential autocorrelation function by approximation of derivative

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?

• Might Mathematics be better suited for this question? Mar 16 '18 at 13:20
• 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 $<f(t) f(t+dt)> = 1$. Mar 16 '18 at 13:51
• 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. Mar 16 '18 at 13:58

$$H(f) \propto \frac{1}{1+\frac{(f-f_0)^2}{f_B^2}}$$
• 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. Mar 16 '18 at 13:28
• @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. Mar 16 '18 at 14:05
• 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. Mar 16 '18 at 14:15