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May
8 |
awarded | Scholar |
May
8 |
comment |
Probability distribution for two-state system that depends on residence time
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 |
comment |
Probability distribution for two-state system that depends on residence time
One last question for you: In the case of two rates, the derivation will have to change because we can no longer use the Poisson distribution, right? (But perhaps it can be replaced by another appropriate distribution...) |
May
8 |
comment |
Probability distribution for two-state system that depends on residence time
This was beautiful, thank you! Accepted. I just used it as an intermediate step in a larger calculation, checked the answer against some stochastic simulations, and got a beautiful match. Thanks again! |
May
8 |
accepted | Probability distribution for two-state system that depends on residence time |
May
8 |
awarded | Commentator |
May
8 |
comment |
Probability distribution for two-state system that depends on residence time
Even in the case of one rate, it is not clear to me that $P_{\mathrm{even}}(t')$ and $P_{\mathrm{odd}}(t')$ can be derived by summing the Poission distribution - that gives a probability, which I don't think is the probability distribution of the time since the last switch. Thanks for your help though, it certainly is a step in the right direction. |
May
8 |
comment |
Probability distribution for two-state system that depends on residence time
I've uncovered some other issues with this. In your simple case, you are integrating over the waiting time distribution $\psi(t') = \kappa_{0} e^{-\kappa_{0} t'}$. This means it is very probable to draw a time $t'$ near 0, and then decays it exponentially. Should we not integrate over $\psi(t - t')$, again with proper normalization? This means it is very probable to draw a time $t'$ near $t$, rather than near $0$. |
May
8 |
comment |
Probability distribution for two-state system that depends on residence time
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
7 |
comment |
Probability distribution for two-state system that depends on residence time
Also, in your answer, shouldn't we also account for the fact that the integration interval is from $0$ to $t$. In your simple case, we will not normalize to $1$ if we integrate over $x$. There I think we have to divide by $1 - e^{-\kappa t}$. Likewise for $P_{\mathrm{even}}(t)$ and $P_{\mathrm{odd}}(t)$? |
May
7 |
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Probability distribution for two-state system that depends on residence time
$p_{\pm}(x, t)$ 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 |
comment |
Probability distribution for two-state system that depends on residence time
Thanks! This is along the lines of what I was hoping for, but I did not think of splitting it into even and odd switches. To be clear, does your general case still assume $\kappa_{+} = \kappa_{-}$? I think it does, otherwise would we still have a Poisson distribution for the number of switches? (FYI, I want to play around with the math a bit before I accept the answer.) |
May
7 |
comment |
Probability distribution for two-state system that depends on residence time
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 |
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Probability distribution for two-state system that depends on residence time
Using my notation, we are in state $+$ at time $5$, knowing that the process entered state $+$ at time $3.7$, we have $X \sim p_{+}(x, 1.3)$. If we switch to state $−$ at time $5$ and wait there until time $7$, $X \sim p_{-}(x, 2)$. Whenever we enter either state, we start the clock again from $0$. |
May
7 |
comment |
Probability distribution for two-state system that depends on residence time
Somehow, $p(x, t)$ needs to take into account switching between these states. |
May
7 |
comment |
Probability distribution for two-state system that depends on residence time
X (or x) is some random variable, it is a real number. $p_{\pm}(x, t)$ are the probability distributions of $x$ given that the system enters state $\pm$ at $t = 0$ and remains there until time $t$. To be more clear, $p_{\pm}(x, t = 0)$ is the probability distribution of $x$ right when the system enters state $\pm$. $p_{\pm}(x, t)$ is the probability distribution of $x$ after it waits in state $\pm$ for a time $t$. These are normalized such that $\int_{-\infty}^{\infty} p_{\pm}(x, t) = 1$. For my particular application, I know the exact forms of $p_{\pm}(x, t)$. |
May
7 |
awarded | Student |
May
7 |
awarded | Editor |
May
7 |
revised |
Probability distribution for two-state system that depends on residence time
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May
7 |
asked | Probability distribution for two-state system that depends on residence time |