Timeline for An "inchworm-like" random walk on an integer interval
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 25, 2013 at 15:44 | |||||
| May 29, 2013 at 13:10 | answer | added | R W | timeline score: 7 | |
| May 29, 2013 at 6:03 | comment | added | Did | What are some motivations for this (quite exciting) model? | |
| May 29, 2013 at 3:06 | comment | added | Douglas Zare | The case $k=2$ should be solvable (we can even assume the stones alternate), but it misses some of the complexity. Consider the configuration with $d=2$, $k=3$ where the stones are at $0,1,3$. The first and last stones may only move left or stay at the same location. The middle stone may only move right. This means that the center of mass will move left on average. If $k=2$ the center of mass is a martingale. | |
| May 29, 2013 at 1:19 | answer | added | j.c. | timeline score: 2 | |
| May 29, 2013 at 0:13 | comment | added | j.c. | Have you figured out what happens for the case $k=2$? It seems simple enough to be tractable... | |
| May 28, 2013 at 23:17 | answer | added | Joseph O'Rourke | timeline score: 4 | |
| May 28, 2013 at 10:57 | comment | added | AmberWave | @Douglas Zare To clarify, the above comment, as stated in the problem description, is only meant to describe the state where all stones are within distance $d = 1$ of their nearest-neighbors. | |
| May 28, 2013 at 8:29 | history | edited | AmberWave | CC BY-SA 3.0 |
added 65 characters in body
|
| May 28, 2013 at 4:35 | comment | added | AmberWave | @Douglas Zare Regarding the "per step probability of $\frac{1}{k}$", I'm allowing for a "do nothing" step where you pick an internal stone and cannot move it. Thus, the probability you both select a stone that can be moved, and that you also select that it should move from its initial position (with probability $\frac{1}{2}$ if $d=1$), should be $\frac{1}{k}$ (i.e. pick either stone at the interval, then chose to move it away from its original position). | |
| May 28, 2013 at 4:28 | comment | added | AmberWave | @Douglas Zare Ah, sorry, I made a mistake and specified $d$ in two different ways. In the first paragraph, I state that $d$ is the maximum allowed distance to a nearest-neighbor, and in the second paragraph I defined it as a displacement from the original position. I meant the former - that the stone can be placed at any site which (a) is at most a distance $d$ from any of the stone's nearest-neighbors, and (b) does not perturb the ordering of the stones along the interval. I have corrected the question to clarify this. | |
| May 28, 2013 at 4:26 | history | edited | AmberWave | CC BY-SA 3.0 |
added 76 characters in body
|
| May 28, 2013 at 4:10 | comment | added | Douglas Zare | Yes, the distance between $-3$ and $-1$ is $2$. I pointed out that moving the stone from $-2$ to $-3$ is not prohibited with $d=1$ in the original operation, of selecting an unoccupied site at most a distance $d$ from the stone's original position $-2$. If you meant to rule this out for $d=1$, you needed to add some condition. I also don't know what you mean by "some per step probability of $\frac{1}{k}$. When $k=5$, what is supposed to have probability $\frac{1}{5}$? Suppose there are two legal destinations for a stone. Don't you move to each with probability $\frac{1}{2}$ instead? | |
| May 28, 2013 at 2:29 | history | edited | AmberWave | CC BY-SA 3.0 |
added 175 characters in body; added 114 characters in body; deleted 1 characters in body
|
| May 28, 2013 at 2:28 | comment | added | AmberWave | @Douglas Zare Just a minor point, but to clarify my method of counting: picking up a stone at -2 and placing it at -3 would make the distance to the closest stone Abs[(-3) - (-1)] = 2. You can do as you suggest if $d = 2$. I like your suggestion that the stones should be kept in order and I will amend the problem description to include this. | |
| May 28, 2013 at 1:33 | comment | added | Douglas Zare | If $d=1$ and you start at $(-2,-1,0,1,2)$, why can't you pick up the stone at $-2$ and put it down at $-3$? Are you requiring the stones to stay within $d$ of at least one other stone at all times, or did you mean the new location has to be strictly less than $d$ of the original location? If the former, do you want to keep the stones in order, requiring each stone to be within $d$ of both the previous and next stone? That would resemble a worm to me. | |
| May 28, 2013 at 0:29 | history | asked | AmberWave | CC BY-SA 3.0 |