12 Noted that the distribution for fixed lambda as N -> inf has a mean somewhere between L/2 and L, not necessarily at L/2

Please imagine that one populates a finite line of unit length, or circle with unit length contour (to avoid edge-effects), with $N$ one-dimensional 'rods' such that their LHS-ends, at positions $(p_1, ..., p_k, ..., p_N) \in P$, are placed in accordance with a uniform random distribution over [0, 1]. Here, the rod lengths, $(l_1, ..., l_k, ..., l_N) \in L$, are exponentially distributed according to some rate parameter $\lambda$ - i.e. the random variable $l_k$ has distribution $l_k$ ~ Exp($\lambda$), giving a probability density function for rod length of $\lambda e^{\lambda l}$. One would similarly expect an exponential distribution for the distances between adjacent points in the set $P$.

We have the following two rules for handling overlaps between rods:

(1) - If the 'contour' of one rod (say, 'Rod A') completely covers another (say, 'Rod B'), i.e. where (Rod A-LHS) < (Rod B-LHS) and (Rod A-RHS) > (Rod B-RHS), we remove 'Rod B' from from the line and no longer consider it.

(2) - If there is only a partial overlap in the contours of two rods, 'Rod A' and 'Rod B', the length of this overlap is split evenly and each half is added to the contours of 'Rod A' and 'Rod B', respectively.

Starting from our initial exponential distribution of rod lengths, $(l_1, ..., l_k, ..., l_N)$, after this overlap-splitting process what is the new probability distribution for the length of some rod, $l_k$?

A few observations:

As $\lambda \rightarrow \infty$, the number of rods left on the line (after overlaps are handled) should increase, and the mean rod length should decrease.

As $N \rightarrow \infty$, the number of overlap-processed rods left on the line should increase, and the mean rod length should decrease. Intuitively I would expect that the number of rods remaining on the line after overlap processing will increase ever more slowly with $N$ after some threshold/'saturation' value is reached (presumably where the line is completely covered with rods).

As $\lambda \rightarrow -\infty$, there should be fewer rods left remaining on the line after overlap processing, and the mean rod length should increase. At some sufficiently large value of $\lambda$, we should be left with only a single rod on the line which has the left-most/smallest LHS-side. If we also have that $N \rightarrow \infty$, the mean length of the rod should approach the unit length of the line.

As $N \rightarrow 0$, there should be fewer rods, and an increasing mean rod length.

Inspired by Joseph O'Rourke's answer, and some simulation results of mine, if one fixes $\lambda$ and lets $N \rightarrow \infty$, one appears to converge to a rod length distribution centered around a mean value of somewhere between $\frac{L}{2}$, \frac{L}{2}$and$L$, where$L$is the original mean length of the rods before overlap processing. However, this distribution appears to be Gaussian, not uniform. Do we actually converge to a Gaussian distribution? How does the distribution and its variance change with increasing$N$? 11 added 9 characters in body Please imagine that one populates a finite line of unit length, or circle with unit length contour (to avoid edge-effects), with$N$one-dimensional 'rods' such that their LHS-ends, at positions$(p_1, ..., p_k, ..., p_N) \in P$, are placed in accordance with a uniform random distribution over [0, 1]. Here, the rod lengths,$(l_1, ..., l_k, ..., l_N) \in L$, are exponentially distributed according to some rate parameter$\lambda$- i.e. the random variable$l_k$has distribution$l_k$~ Exp($\lambda$), giving a probability density function for rod length of$\lambda e^{\lambda l}$. One would similarly expect an exponential distribution for the distances between adjacent points in the set$P$. We have the following two rules for handling overlaps between rods: (1) - If the 'contour' of one rod (say, 'Rod A') completely covers another (say, 'Rod B'), i.e. where (Rod A-LHS) < (Rod B-LHS) and (Rod A-RHS) > (Rod B-RHS), we remove 'Rod B' from from the line and no longer consider it. (2) - If there is only a partial overlap in the contours of two rods, 'Rod A' and 'Rod B', the length of this overlap is split evenly and each half is added to the contours of 'Rod A' and 'Rod B', respectively. Starting from our initial exponential distribution of rod lengths,$(l_1, ..., l_k, ..., l_N)$, after this overlap-splitting process what is the new probability distribution for the length of some rod,$l_k$? A few observations: As$\lambda \rightarrow \infty$, the number of rods left on the line (after overlaps are handled) should increase, and the mean rod length should decrease. As$N \rightarrow \infty$, the number of overlap-processed rods left on the line (after overlaps are handled) should increase, and the mean rod length should decrease. Intuitively I would expect that the number of rods remaining on the line after overlap processing will increase ever more slowly with increasing$N$after some threshold/'saturation' value is reached (presumably where the line is completely covered with rodsrods). As$\lambda \rightarrow -\infty$, there should be fewer rods left remaining on the line after overlap processing, and the mean rod length should increase. At some sufficiently large value of$\lambda$, we should be left with only a single rod on the line which has the left-most LHS-endleft-most/smallest LHS-side. If we also have that$N \rightarrow \infty$, the mean length of the rod should approach the unit length of the line. As$N \rightarrow 0$, there should be few fewer rods, and a larger an increasing mean for the rod lengthslength. Inspired by Joseph O'Rourke's answer, and some simulation results of mine, if one fixes$\lambda$and lets$N \rightarrow \infty$, one appears to converge to a rod length distribution centered around a mean value of$\frac{L}{2}$, where$L$is the original mean length of the rods before overlap processing. However, this distribution appears to be Gaussian, not uniform. Do we actually converge to a Gaussian distribution? On average, how large How does the distribution and its variance change with increasing$N$have to be for this to occur?N$?

10 Added some notes about my intuition for what happens at limits of lambda & N, added note about simulation results for fixed lambda and increasing N

Why is this interesting? Simulations seem to indicate that one quickly converges towards a more uniform distribution

A few observations:

As $\lambda \rightarrow \infty$, the number of rods for appropriate left on the line (after overlaps are handled) should increase, and the mean rod length should decrease.

As $\lambda$, something I accidentally discovered while simulating an unrelated systemN \rightarrow \infty$, the number of rods left on the line (after overlaps are handled) should increase, and the mean rod length should decrease. I'd like to know a bit Intuitively I would expect that the number of rods remaining on the line after overlap processing will increase ever more about slowly with increasing$N$after some threshold/'saturation' value is reached where the mechanism... though unfortunately this line is only of recreational interest to mecompletely covered with rods. Update As$\lambda \rightarrow -I would also \infty$, there should be interested in a random sequential absorption version of this systemfewer rods left remaining on the line after overlap processing, where and the mean rod length should increase. At some sufficiently large value of$\lambda$, we should be left with only a single rod on the RHS ends line which has the left-most LHS-end. If we also have that$N \rightarrow \infty$, the mean length of each sequentially deposited the rod are allowed to overlap with should approach the LHS ends unit length of any previously deposited the line. As$N \rightarrow 0$, there should be few rods, and a larger mean for the rod lengths. Inspired by Joseph O'Rourke's answer, and some simulation results of mine, if one fixes$\lambda$and lets$N \rightarrow \infty$, one appears to converge to a rod length distribution centered around a mean value of$\frac{L}{2}$, where$L$is the original mean length of the rods before overlap processing. However, this distribution appears to be Gaussian, not uniform. Do we actually converge to a Gaussian distribution? On average, how large does$N\$ have to be for this to occur?