Timeline for How does changing the transition probabilities affect the concentration of a position-dependent random walk?
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Feb 28, 2013 at 8:47 | vote | accept | yves | ||
| Feb 28, 2013 at 8:47 | vote | accept | yves | ||
| Feb 28, 2013 at 8:47 | |||||
| Feb 28, 2013 at 6:32 | answer | added | Ori Gurel-Gurevich | timeline score: 4 | |
| Feb 28, 2013 at 3:21 | history | edited | Douglas Zare | CC BY-SA 3.0 |
changed title
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| Feb 28, 2013 at 0:37 | comment | added | yves | However,I would be very interested to see any result which translate a concentration of $X(t)$ into a concentration of $Y(t)$, so feel free to dispense/modify the technical assumptions. | |
| Feb 28, 2013 at 0:36 | comment | added | yves | I threw in the words "position dependent" in there, and specified that the assumption should hold for all $c$ and for all $t$ large enough. | |
| Feb 28, 2013 at 0:35 | history | edited | yves | CC BY-SA 3.0 |
added 113 characters in body
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| Feb 27, 2013 at 17:04 | comment | added | Günter Rote | Should the assumption hold for all $c$? Is it essential that the walk moves to the right on average (like $(1/3)t$ in your example) our would a movement to the left or a stationary concentration around $0$ be permitted? (In the last cases there would probably be easy counterexamples when some $p_n$ is changed from 0 to $\epsilon$.) | |
| Feb 27, 2013 at 13:36 | comment | added | Lee Mosher | Stick the words "position dependent" in there somewhere? | |
| Feb 27, 2013 at 2:22 | comment | added | yves | I added a sentence emphasizing that all $p_n$ are not the same. About the title, I'm not sure what else captures my question in a pithy way. | |
| Feb 27, 2013 at 2:21 | history | edited | yves | CC BY-SA 3.0 |
added 60 characters in body
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| Feb 27, 2013 at 2:19 | comment | added | David White | I recommend a different title. I only clicked on it to vote to close because I assumed you were asking ``what happens if at each $n$, the probability of going right is $p>1/2$'' which is of course well understood. The fact that your probability changes based on which $n$ you're at should probably be highlighted. | |
| Feb 27, 2013 at 2:05 | history | asked | yves | CC BY-SA 3.0 |