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I am a statistical physicist, and I've come across a problem that I don't know how to solve. I believe my issue lies with how to formulate it mathematically. I'd be very grateful for any assistance, as I've really struggled with this.

Suppose we have a system with two states, say $+$ and $-$. The waiting time distribution for switching from either state to the other is an exponential, $\psi_{\pm}(t) = \kappa_{\pm} e^{-\kappa_{\pm} t}$. So $\psi_{\pm}(t)$ is the probability distribution for the time spent in the $\mp$ state.

Let $X$ be the random variable that we are interested in. If we enter state $+$ at $t = 0$, then its probability distribution at a later time $t$ is $p_{+}(x, t)$, assuming it does not switch to $-$. Define $p_{-}(x, t)$ likewise. These distributions depend on the time spent in the state.

I would like the full probability distribution $p(x, t)$, assuming the initial state is known, say $+$. I simply don't know how to write down a general equation for $p(x, t)$...

Again, any help would be wonderful. Thanks!

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  • $\begingroup$ I didn't understand what $X$ and $x$ are. Real numbers? $\pm$ states? I also didn't understand what is $p_+$ (or $p_-$). Is it the distribution of $X$ and does $X$ depend on time $t$? $\endgroup$ May 7, 2013 at 19:10
  • $\begingroup$ 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 $p_{\pm}(x, t)$. $\endgroup$
    – ionlet
    May 7, 2013 at 19:29
  • $\begingroup$ Somehow, $p(x, t)$ needs to take into account switching between these states. $\endgroup$
    – ionlet
    May 7, 2013 at 19:30
  • $\begingroup$ for example, we are in state 2 at time 5, the process last entered state 2 at time 3.7, $X \sim p(x,1.3)$ ? $\endgroup$
    – mike
    May 7, 2013 at 20:56
  • $\begingroup$ 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 $0$. $\endgroup$
    – ionlet
    May 7, 2013 at 21:17

2 Answers 2

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first simple case: $p_+=p_-\equiv p_0$ and $\kappa_+=\kappa_-\equiv \kappa$; then all you need to know is the time $\delta t$ since the last switching event, which has an exponential distribution, hence:

$$p(x,t)=(1-e^{-\kappa t})^{-1}\int_0^t d\delta t\; \kappa e^{-\kappa \delta t}p_0(x,\delta t)$$

now the general case; you'll need to distinguish even from odd number of switching events, and find the distribution $P_{\rm even}(\delta t)$ of the time $\delta t$ since the last switching event, given that there have been an even number of switches in a time $t$, and similarly for an odd number; the a priori probability that there have been an even or odd number of switches is just given by the Poisson distribution (summing over $n$ even or $n$ odd). The integral over $P_{\rm even}(\delta t)p_{+}(x,\delta t)$ and $P_{\rm odd}(\delta t)p_{-}(x,\delta t)$ then gives the full answer.

a bit more explicit, still assuming $\kappa_+=\kappa_-\equiv \kappa$ for simplicity; it is convenient to set the first switching event at time $0$, so that the number of switching events $m=n+1$ in a time $t>0$ is $\geq 1$; the probability $P_{m,t}(\delta t)$ that there have been $m=n+1\geq 1$ switching events in a time $t$, while the last switching event was a time $\delta t\in[0,t)$ ago is given by a slight modification of the Poisson distribution,

$$ P_{m,t}(\delta t)=\frac{1}{(m-1)!}[\kappa(t-\delta t)]^{m-1}\kappa e^{-\kappa t}.$$

summing over all $m=1,2,3,\ldots$ we recover the exponential probability $\sum_{m}P_{m,t}(\delta t)=\kappa\exp(-\kappa\delta t)$ we had before, but now we have to distinguish between even $m$ (= odd $n$) and odd $m$ (= even $n$):

$$P_{\rm even}(\delta t)=\sum_{m=1,3,5}^{\infty}P_{m,t}(\delta t)=\cosh[\kappa(t-\delta t)]\kappa e^{-\kappa t}$$

$$P_{\rm odd}(\delta t)=\sum_{m=2,4,6}^{\infty}P_{m,t}(\delta t)=\sinh[\kappa(t-\delta t)]\kappa e^{-\kappa t}$$

and we're done:

$$p(x,t)=(1-e^{-\kappa t})^{-1}\int_0^t d\delta t\; [P_{\rm even}(\delta t)p_+(x,\delta t) +P_{\rm odd}(\delta t)p_-(x,\delta t)]$$

$$\quad\quad=(1-e^{-\kappa t})^{-1}\int_0^t d\delta t\; \kappa e^{-\kappa t}\left[\cosh[\kappa(t-\delta t)]p_+(x,\delta t) +\sinh[\kappa(t-\delta t)]p_-(x,\delta t)\right]$$

if we take $p_+=p_-\equiv p_0$ we recover the earlier result.

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  • $\begingroup$ 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.) $\endgroup$
    – ionlet
    May 7, 2013 at 22:18
  • $\begingroup$ 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 $P_{\mathrm{odd}}(t)$? $\endgroup$
    – ionlet
    May 7, 2013 at 23:05
  • $\begingroup$ 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 $0$. $\endgroup$
    – ionlet
    May 8, 2013 at 10:12
  • $\begingroup$ 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. $\endgroup$
    – ionlet
    May 8, 2013 at 10:14
  • $\begingroup$ @ionlet: I expanded the answer, hopefully answering all your queries; is it clear now? $\endgroup$ May 8, 2013 at 14:13
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You seem to be after $p(x,t)=P[S_t=+]p_+(x,t)+P[S_t=-]p_-(x,t)$ where $S_t$ is the state at time $t$. One knows that $S_0=+$ and that $S_t$ switches from $\pm$ to $\mp$ at rate $\kappa_\pm$.

Thus, $p(x,t)=q(t)p_+(x,t)+(1-q(t))p_-(x,t)$ where $q(t)=P[S_t=+]$ and the task is to compute $q(t)$. Note that $q(0)=1$ and that, when $s\to0$, $$ q(t+s)=P[S_{t+s}=+]=q(t)(1-\kappa_+s+o(s))+(1-q(t))\kappa_-s+o(s), $$ that is, $$ q'(t)=-\kappa_+q(t)+\kappa_-(1-q(t)). $$ This ODE yields $q(t)$ for every $t$, hence $p(x,t)$.

Edit: Second try, the model the OP is interested in might (or might not) be the following. Some quantities $p_+(x,t)$ and $p_-(x,t)$ are given for every $t\geqslant0$ and a state process $(S_t)$ is switching back and forth between states $+$ and $-$ at rates $\kappa_+$ and $\kappa_-$ and starting from $S_0=+$. Call $T_t$ the time of the last switch before $t$ (if no switch occurred before $t$, $T_t=0$). One is interested in $$ p(x,t)=E[p_{S_t}(x,t-T_t)]= E[p_+(x,t-T_t);S_t=+]+E[p_-(x,t-T_t);S_t=-]. $$ From now on, let us choose $x$ and shorten $p(x,t)$, $p_+(x,t)$ and $p_-(x,t)$, $q(t)$ and $1-q(t)$ into $p(t)$, $p_+(t)$, $p_-(t)$, $q_+(t)$ and $q_-(t)$ respectively. When $r\to0$, during the time interval $(t,t+r)$ and assuming that $S_t=s$ with $s\in\{\pm\}$, either no switch happen and then $S_{t+r}=s$ and $t+r-T_{t+r}=t-T_t+r$, or a switch happen and then $S_{t+r}=-s$ and $t+r-T_{t+r}=o(1)$. Hence, $$ p(t+r)=\sum_s\kappa_s rq_s(t)p_{-s}(0)+(1-\kappa_sr)E((p_s(t-T_t)+rp'_s(t-T_t));S_t=s)+o(r), $$ which implies that $$ p'(t)=\sum_s\kappa_s q_s(t)p_{-s}(0)+E(p'_s(t-T_t);S_t=s)-\kappa_sE(p_s(t-T_t);S_t=s). $$ Conditionally on $S_t=s$, $t-T_t$ is distributed like $\min\{t,U_s\}$, where $U_s$ is exponential with parameter $\kappa_s$, hence, for every function $v$, $$ E(v(t-T_t);S_t=s)=\mathrm e^{-\kappa_st}q_s(t)v(0)+q_s(t)\int_0^tv(u)\kappa_s\mathrm e^{-\kappa u}\mathrm du. $$ Applying this to $v=p_s$ and to $v=p'_s$ shows that $t\mapsto p(t)$ solves an explicit integro-differential equation.

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  • $\begingroup$ 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. $\endgroup$
    – ionlet
    May 7, 2013 at 21:57
  • $\begingroup$ 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?). $\endgroup$
    – Did
    May 7, 2013 at 22:46
  • $\begingroup$ $p_{\pm}(x, t)$ 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? $\endgroup$
    – ionlet
    May 7, 2013 at 22:53
  • $\begingroup$ Could you confirm or infirm the Edit? $\endgroup$
    – Did
    May 8, 2013 at 7:54
  • $\begingroup$ 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? $\endgroup$
    – ionlet
    May 8, 2013 at 10:08

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