EDIT: I have tried to rephrase the problem, title, and context to my solution
I am wondering about expanding a problem I have to the continuous domain. The problem is defined as such:
Problem
Given $N$ disks confined to a 2 dimensional domain $D$. Each disk has a center of mass, and will randomly walk each step. Physically, this can be modelled using a random variable $R$, which defined the step magnitude, and $\Theta$, the step angle. The center of mass is defined as it's position in space. This is necessary in context in the field of particle tracking, or object localization.
Note that the 2D domain is treated as a projection, such as that of a telescope, or a camera. In this case, depth doesn't really matter as the volume of the confinement is small, but overlap is free to occur.
One issue when facing such disks is counting them, or knowing how many total disks exists in the region. Being able to count these disks is typically a task of particle tracking. In confinement, the exploration area, or set of possible locations that the disk can uptake is physically limited. This problem is asking about how this limit is related with measurable values.
Thus, given $N$ disks, randomly walking (diffusing) in time steps $\Delta t$, with magnitude and angle $R, \Theta$, what is the probability that $k$ of these disks are overlapping? Define an overlap event as two disks $epsilon$ distance from each other. This scales as $k$ overlaps if all disks have a separation of at least $epsilon$
In other words, how often, given a set of measured centre of masses (or particle detection), will there be a detection where two of the disks are overlapping.
Attempt
I have tried to do this in course-grained model for a simple set of points on a lattice. I am having trouble seeing how to make these problems bodies, but, here is the solution I calculated for the very discrete case.
The probability of a total overlap is simply the number of ways to orient $N-l$ particles, divided by the total positions of the $N$ particles:
The logic behind this is that the area can be treated as a set of sites where particles can occupy space. Thus, by segmenting $D$, the problem is restated in terms of combinatorics.
$$\mathbb{P(overlap)} = \frac{\sum_{k=1}^{N-1} \binom{D}{k}}{\sum_{l=1}^{N} \binom{D}{l}}$$
This is then expanded using the binomial summation formula:
$$\frac{\sum_{k=1}^{N-1} \binom{D}{k}}{\sum_{l=1}^{N} \binom{D}{l}} = \sum_{k=1}^{N-1} \frac{1}{k!(D-k)!}\sum_{l=1}^{N} l!(D-l)! = \left[\sum_{k=2}^{N}\frac{1}{(k-1)!(D-k+1)!} \right]\left[(D-1)!+\sum_{l=2}^{N}l!(D-l)!\right] $$
This can sort of be transformed to sit right in a semi-analytical form*: $$\frac{\sum_{k=1}^{N-1} \binom{D}{k}}{\sum_{l=1}^{N} \binom{D}{l}} = \frac{\left[\sum_{k=0}^{N} \binom{D}{k}\right] - \binom{D}{0} - \binom{D}{N}}{\left[\sum_{l=0}^{N} \binom{D}{l}\right ]- \binom{D}{0}} \rightarrow \text{lim}_{N\rightarrow D} = \frac{2^D-2}{2^D-1}$$
*: I think I may have made a mistake somewhere in these sums, as my simulations are not agreeing with this.
This formula gives pretty useful information in what I am using it for, however, if possible, a better model is always sought after. Thus, if someone can point me into the right direction for how to improve or continue this problem, I would be very grateful.
attempt 2
One simple way to get this estimate even better is to simulate this, however, although aware of this, this problem has the problem of classifying what an "overlap" is. This makes things non-trivial from the get-go.
In this sense, how can these "classification" problems be attempted? this is not the first time I have encountered such a problem and am not sure how to continue.
Thanks,
