Timeline for Probability distribution for two-state system that depends on residence time
Current License: CC BY-SA 3.0
13 events
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| May 10, 2013 at 10:18 | comment | added | Did | I find a tad surprising that an answer restricted to the symmetric case $\kappa_+=\kappa_-$ is "what you were looking for" since the modifications needed to solve the general case are non trivial. For this reason, to ask whether my answer and the accepted one are the same seems rather moot and I do not feel much motivation to answer your last query. But since you seem happy with what you got, everything is perfect. (Unrelated: one cannot accept two answers.) | |
| May 8, 2013 at 20:49 | comment | added | ionlet | Your answer may be right, but I don't understand it that well. The above answer by Carlo Beenakker is what I was looking for and, I think, correct. Is yours the same as his? If yes, I'll accept yours too. (I can't see if it is or is not.) | |
| May 8, 2013 at 11:51 | history | edited | Did | CC BY-SA 3.0 |
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| May 8, 2013 at 11:39 | history | edited | Did | CC BY-SA 3.0 |
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| May 8, 2013 at 11:16 | comment | added | Did | If $p_+(\ ,t)$ and $p_-(\ ,t)$ are probability distributions, so is $p(\ ,t)$ as a barycenter of these. (But the question is about fixing $x$ once and for all and working on $p(x,\ )$ from $p_-(x,\ )$ and $p_+(x,\ )$, actually.) | |
| May 8, 2013 at 10:08 | comment | added | ionlet | Something like this is what I'm looking for. The way you've written $t - T_{t}$ is what I meant originally. As written, is your $p(x, t)$ a probability distribution? It is not normalized, is it? | |
| May 8, 2013 at 8:07 | history | edited | Did | CC BY-SA 3.0 |
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| May 8, 2013 at 7:54 | comment | added | Did | Could you confirm or infirm the Edit? | |
| May 8, 2013 at 7:53 | history | edited | Did | CC BY-SA 3.0 |
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| May 7, 2013 at 22:53 | comment | added | ionlet | $p_{\pm}(x, t)$ is the probability distribution for $x$ given that it enters state $\pm$ at $t = 0$. This is the key point, I mentioned it in the question. See also the comments above. What is a better way to describe this? | |
| May 7, 2013 at 22:46 | comment | added | Did | The answer is referring very precisely to the model you described. If you are interested in a different dynamics, please explain clearly what it is. Alternatively, describe what you think the problem with the derivation above is, avoiding vague terms such as "the fact that p±(x,t) evolves with time" (of course it "evolves with time", otherwise what would the argument $t$ be there for?). | |
| May 7, 2013 at 21:57 | comment | added | ionlet | Thanks for the answer. Unfortunately, I don't think it is as simple as that, though I'd be glad to be shown I'm wrong. Your $q(t)$ is perfectly fine, but there is an issue with your $p(x, t)$. It doesn't take into account the fact that $p_{\pm}(x, t)$ evolves with time as it waits in either state. In $p_{\pm}(x, t)$ is this time, not the global system time. | |
| May 7, 2013 at 21:48 | history | answered | Did | CC BY-SA 3.0 |