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

Question

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:

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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?

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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.

Answer 1

For part 1, I think you pretty much got everything right. The only thing that is surprising is the period around 100ps where the scaling of the distance is greater than 1. This implies that there is some long term correlation between the movement of the two domains. Since I can't read the paper, I can't say why this is. The later period, where alpha is less than 1, is what you'd expect since the domains are attached so that at some point meaning the distance between them saturates and the scaling must approach 0.

Part 2 is really just a basic autocorrelation function. If the lag is large the two terms in the integral are random, and can be both positive and negative, and thus should average out roughly to zero. At short times they will be identical. One could replace the integrand in this case with a square, which is strictly positive, and thus results in a non-zero positive answer. If you plot it, it should start at a maximum, and slowly decay to zero with increasing lag. The graph essentially gives a visual idea of how long the given quantity is correlated.