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edited Jul 17 2010 at 0:19 |
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 $N$?
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edited Jul 16 2010 at 23:17 |
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$?
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edited Jul 16 2010 at 23:10 |
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?
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edited Jul 16 2010 at 18:39 |
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, $l_k$?
Why is this interesting? Simulations seem to indicate that one quickly converges towards a more uniform distribution of rods for appropriate $\lambda$, something I accidentally discovered while simulating an unrelated system. I'd like to know a bit more about the mechanism... though unfortunately this is only of recreational interest to me.
Update - I would also be interested in a random sequential absorption version of this system, where only the RHS ends of each sequentially deposited rod are allowed to overlap with the LHS ends of any previously deposited rod.
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edited Jul 15 2010 at 20:44 |
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, $l_k$?
Why is this interesting? Simulations seem to indicate that one quickly converges towards a more uniform distribution of rods for appropriate $\lambda$ ~ $\frac{1}{N}$, \lambda$, something I accidentally discovered while simulating an unrelated system. I'd like to know a bit more about the mechanism... though unfortunately this is only of recreational interest to me.
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edited Jul 15 2010 at 20:28 |
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, $l_k$?
Why is this interesting? Simulations seem to indicate that one quickly converges towards a more uniform distribution of rods for appropriate $\lambda$, \lambda$ ~ $\frac{1}{N}$, something I accidentally discovered while simulating an unrelated system. I'd like to know a bit more about the mechanism... though unfortunately this is only of recreational interest to me.
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edited Jul 15 2010 at 14:28 |
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, $l_k$?
Why is this interesting? Simulations seem to indicate that one quickly converges towards a more uniform distribution of rods for appropriate $\lambda$.\lambda$, something I accidentally discovered while simulating an unrelated system. I'd like to know a bit more about the mechanism... though unfortunately this is only of recreational interest to me.
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edited Jul 15 2010 at 4:22 |
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, $l_k$?
Simulations seem to indicate that one converges towards a more uniform distribution .for appropriate $\lambda$.
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edited Jul 15 2010 at 0:41 |
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, $l_k$?
Simulations seem to indicate that one converges towards a more uniform distribution.
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edited Jul 15 2010 at 0:18 |
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 $\lambda_l$ \lambda$ - i.e. the random variable $l_k$ has distribution $l_k$ ~ Exp($\lambda_l$)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, $l_k$?
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edited Jul 15 2010 at 0:09 |
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 $\lambda_l$ - i.e. the random variable $l_k$ has distribution $l_k$ ~ Exp($\lambda_l$), giving a probability density function for rod length of $\lambda e^{\lambda l}$. We 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, $l_k$?
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asked Jul 15 2010 at 0:01 |
The consequence of overlap sharing for the length-distribution of rods randomly placed on a line
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 $\lambda_l$ - i.e. the random variable $l_k$ has distribution $l_k$ ~ Exp($\lambda_l$). We 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, $l_k$?
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