i am trying to follow the following tutorial. I keep seeing $\delta$ over functions such as $\delta F(x)=F(x)-\langle F(x)\rangle_t$ (Eq 14.4) in this and in other tutorials and questions here. What does the $\delta$ represent here?

Also, is it correct to say that the autocorrelation of such a function $\delta F(x,t)$ Is given by $\langle \delta F(x,t),\delta F(x,t+\tau)\rangle$? If so, how do I develop this? Is that the covariance somehow?

Googled it and could not find anything.


The $\delta$ is just a common notation for indicating that the quantity after the $\delta$ is a deviation from an average; it doesn't really mean or do anything other than indicate that.

If the original quantity under consideration (say $F(x)$) is a random variable, then $\delta F(x)$ is also a random variable, which is usually the case—unfortunately the document you cite appears to use this notation to describe deterministic quantities in the form of differences between different averages.

In any case, $\langle \delta F(x,t),\delta F(x,t+\tau)\rangle$ is the auto-covariance of $F(x)$ whereas $\langle F(x,t),F(x,t+\tau)\rangle$ is the auto-correlation of $F(x)$.

  • $\begingroup$ Can you have a look in this paper and comment on equation 4? Here, it is the autocorrelation? $\endgroup$ – havakok Jan 1 '19 at 9:10
  • 1
    $\begingroup$ It is the autocorrelation of the flourescence signal, yes, but not of the point spread function (PSF). $\endgroup$ – aghostinthefigures Jan 1 '19 at 9:17
  • $\begingroup$ So, as autocorrelation, shouldn't it be $\langle F(r,t+\tau)F(r,t)\rangle $? why is it $\langle \delta F(r,t+\tau)\delta F(r,t)\rangle $? $\endgroup$ – havakok Jan 1 '19 at 9:50
  • 1
    $\begingroup$ The auto-covariance of $F(x)$ is the auto-correlation of $\delta F(x)$. $\endgroup$ – GiorgioP Jan 1 '19 at 10:24
  • $\begingroup$ Quick correction: it is the /autocovariance/ of the flourescence signal that is represented by G2, not the autocorrelation (don't know how I missed those deltas!). $\endgroup$ – aghostinthefigures Jan 1 '19 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.