So with all the fourier and laplace transforms I'm also confused why we need them to handle the processes. Is it the case that somehow, we only have the fourier transform at hand in practice, and need to conclude from this properties of the correlation function?
@Thorgott nowhere, thank god
Because afterwards we go all out fourier and define properties that ensure that a function from the upper half pflane to itself (+ the real line) is actually a the fourier transform of a correlation function, kinda similarly to what we did with Bochner-Khintchin
It's basically all about finding formulas like $i \hat{\phi}(z)=\frac{1}{\pi} \int_{\Bbb R} \frac{\operatorname{Re} \hat{\phi}(\omega+i 0)}{\omega-z} d \omega, \quad \Im z>0$
(It's amazing that automatic latex readers recognize my profs handwriting)
Here, this has got to do with the "Riesz-Herglotz representation of a Nevanlinna function"
Then we go more specific and consider processes that aren't just stationary, but also symmetric under time reversal, i.e. $\{-X_t\}$ has the same law as $\{X_t\}$
We say that such a process is in equilibrium
And have again formulas like $i \Phi(z)=\int_{R} \frac{z}{\omega^{2}-z^{2}} d F(\omega)$. Here $F$ is some distribution function so this is just a stieltjes integral. And again I wonder what this tells me
Apparently, Fourier inversion is easier in equilibrium, so that we get the correlation function
Our next concern then is finding out, from the fourier transform, so-called memory kernels
They satisfy $\frac{d}{d t} \phi(t)+\gamma \phi(t)+\int_{0}^{t} D\left(t-t^{\prime}\right) \phi\left(t^{\prime}\right) d t^{\prime}=0, \quad \phi(0)=1$
I should just be able to differentiate this and get a second order ODE right? Is this some spring type stuff?
Anyways it is apparent that from the fourier transform, and by approximating it right at short and high frequencies, we can say a lot about the process. I just don't see the natural setting
For example, we get a "harmonic oscillator with retarded friction" describing the differential equation with the memory kernel, in one example. What's that?
Afterwards it gets a little more hands-on again and we work with the "velocity auto-correlation function", look at integro-differential equations comparable to the one for memory kernels, only this time they're called "friction kernels", and the ODE is approx the same, but with $\gamma = 0$
the kernels we obtain again via the fourier transform, which makes a bit more sense now to me, since this directly comes from our knowledge of the process, and still purposefully uses the fourier transform
Does laplace-transforming just generally make handling the processes easier? What are the properties that we forget when we fourier-transform, which make it easier to handle the information? That might be a bit too general a question, but maybe there's a good answer
One of the thing's we're able to show then is that glassy dynamics are non ergodic, since the friction kernel does not go to zero. Pic incoming
Here $Z$ is the velocity auto-correlation function $Z(t) = \frac{d^2}{dt^2} \Bbb E[(R_t - R_0)^2]$ for some particle trajectory $R_t$
And the friction kernel is the $\zeta$ that shows up in the demoninator of the fourier-transform
I understand what cross-correlation does given a kernel and an input image, but the formula confuses me a little. Given here in Goodfellow's Deep Learning (page 329), I can't quite understand what $m$ and $n$ are. Are they the dimensions of the kernel along the height and width dimensions?
$$S(i,...
