# Sub-/Super Diffusive Processes, Time-averaged MSD and Autocorrelation. Help with interpretation?

This is a two part question, cross-posted from the Math-SE becuase I thought you people might be able to help a little more with intuition:

Part 1:

In a (totally fascinating) paper studying the distance between two domains of a protein in an Molecular Dynamics simulation, the time-averaged mean square displacement of the distance between the two protein domains, $R(t)$, is given by:

$$\overline{\delta^{2}(\Delta;t)} = \frac{1}{t-\Delta} \int_{0}^{t-\Delta} [R(t'+ \Delta)-R(t')]^{2} \ dt'$$

where $\Delta$ is the "lag time" and $t$ is the total observation time of the simulation.

I'm familiar with the normal average value of a function $f(x)$ between $a$ and $b$:

$$\overline{f(x)}_{(a,b)} = \frac{1}{b-a}\int_{a}^{b} f(x) \ dx,$$

and I understand that $[R(t'+\Delta)-R(t')]^2$ is the (squared) measure of the difference in the distance between the two domains after a certain "lag time" $\Delta$, and that $\overline{\delta^{2}(\Delta;t)}$ is measuring the average of the this measure over a time $(t-\Delta)$.

For example, in a simulation with total observation time $t=100$ ps, $\overline{\delta^{2}(\Delta;t)} \propto \Delta^{1.5}$, with $\Delta \in (10^{-1}\text{ps},10^1\text{ps})$.

At the lower end, the integral is averaging the distance between the domains after a lag time of $10^{-1}$ps over ~$100$ ps. At the upper end, the integral is averaging the distance between the domains after a lag time of $10$ps, over $90$ps. What information does the relationship between $\Delta$ and $\overline{\delta^2(\Delta;t)}$ convey? The fact that the time-averaged mean square displacement goes up as the lag time goes up means... what exactly? And what would it mean if this proportionality was $\overline{\delta^{2}(\Delta;t)} \propto \Delta^{\alpha}, \text{with} \alpha<1$

To sum up a bit, what I don't understand is this:

• What information does this integral convey?

• What's the purpose of changing the time over which the square displacement is averaged in the formula?

• I get that diffusive processes have MSDs that are linear in time (i.e. $\text{MSD} \propto t$), but this process being sub-/super-diffusive means what exactly? It's parts move apart slower/faster than would be expected by just random diffusion? One of the main take-aways from the paper seems to be that the domain-separation distance is subdiffusive, which they say leads to some sort of "ageing" behavior. I must admit I don't really get why this would be surprising... The two domains are connected to each other, wouldn't it make sense that they don't move apart from one another at the same rate as simple brownian motion (diffusion) would carry them apart from one another?

Part 2:

The same paper defines the normalized auto-correlation function of the distance between the domains as: $C(\Delta;t) = C'(\Delta;t)/C'(0;t)$, where:

$$C'(\Delta;t)=\frac{1}{t-\Delta}\int_{0}^{t-\Delta}\delta R(t')\delta R(t' + \Delta) \ dt'$$

where $\delta R(t)=R(t)-\langle R \rangle$; in other words, how far the distance between the domains is from the average inter-domain distance.

Here, I have to admit more ignorance than in Part 1. I understand that the auto-correlation function is supposed to be some measure of the similarity of a function to itself at different times, but I don't understand how this function achieves that measure. I wish I had a more pointed question to ask, but I'm hoping that someone can help anyway. I understand if it's too broad.