Intermittent Waiting-Time Noises Through Embedding Processes

Abstract

An intermittent two-state noise can be modelled through a renewal process characterized by two different time scales. A (four state) Markovian embedding of this non-Markovian process is presented. The equivalence between the renewal approach and the enlarged master equation is shown. Analytical results for n-time moments of the intermittent dichotomic noise are obtained. The Monte Carlo simulations supports our analytical results. The advantage of using the enlarged master equation for calculating higher order moments is established.

Appendices

Appendix A: 2-Times Moment for a Non-stationary Case

For a non-stationary renewal model, with $\left\langle t\right\rangle =\infty$, even in the asymptotic limit (with $\{t,t^{\prime }\}\rightarrow \infty$ and $t-t^{\prime }<\infty$) the 2-times moment $\left\langle \xi \left( t\right) \xi \left( t^{\prime }\right) \right\rangle$ explicitly depends on both times t and $t^{\prime }$

$$\begin{aligned} \left\langle \xi \left( t\right) \xi \left( t^{\prime }\right) \right\rangle =\theta ^{2}\left( 1-2\mathcal {L}_{\omega }^{-1} \left[ \frac{\rho (\omega , t^{\prime }\rightarrow \infty )}{\omega \left( 1+\psi (\omega )\right) }, t-t^{\prime }\right] \right) . \end{aligned}$$

(A1)

In this asymptotic limit it makes no difference to use $\rho \left( \Delta , t_{i}\rightarrow \infty \right)$ or $\rho ^{0} \left( \Delta ,t_{i}\rightarrow \infty \right)$. To proceed to make an explicit calculation we need to give a particular model of (long-tail) waiting-time function. In this case from (9) and using Tauberian’s theorem we get the asymptotic behavior $\rho (\Delta ,t_{i})\simeq \phi \left( \Delta \right) f\left( t_{i}\right)$, for $t_{i}\rightarrow \infty$. Thus, the expression (A1) coincides with the one given for broad waiting-time distributions [9].

Appendix B: Correspondence Between $P\left( \xi (t_{1})\left| \xi (0)\right. \right)$ and $P\left( \sigma (t_{1})\left| \sigma (0)\right. \right)$

1.1 1. Time Correspondence

In order to demonstrate this correspondence we use the Cayley-Hamilton theorem (if $\mathcal {P}\left( z\right)$ is the characteristic polynomial of the matrix $\mathbf {A}$, then $\mathcal {P}\left( \mathbf {A}\right) =0$). Thus, we can rewrite the matrix $e^{\mathbf {H}t}$ as a polynomial in a closed form. Due to the fact that $\mathbf {H}$ has different eigenvalues $\lambda _{j}$, see (35), we can find a similarity transformation matrix $\mathbf {C}$ and its inverse $\mathbf {C}^{-1}$ such that $\mathbf {H}_{D} =\mathbf {C}^{-1}\mathbf {HC}$. Then $e^{\mathbf {H}t}$ can be written in its diagonal form $e^{\mathbf {H}_{D}t}$. Now using the characteristic polynomial of a matrix $\mathbf {A} \in \mathcal {R}^{N\times N}$ we get:

$$\begin{aligned} \mathcal {P}\left( z\right) =\det \left| z\mathbf {I}-\mathbf {A} \right| =\sum _{n=0}^{N}z^{N-n}c_{n}, \end{aligned}$$

(B1)

where $c_{n}$ are the algebra invariants of matrix $\mathbf {A}$. Therefore, using Cayley-Hamilton theorem for matrix $\mathbf {A}\equiv e^{\mathbf {H} _{D}t}\in \mathcal {R}_{e}^{4\times 4}$, we can write $e^{\mathbf {H}t}$ as a polynomial. From (B1) we get $-c_{3}e^{\mathbf {H}_{D}t} =c_{4}+\sum _{n=0}^{2}e^{\left( 4-n\right) \mathbf {H}_{D}t}c_{n}$. Now applying the similarity transformation we arrive to:

$$\begin{aligned} e^{\mathbf {H}t}=\frac{-1}{c_{3}}\left( c_{4}+\sum _{n=0}^{2} \mathbf {C}e^{\left( 4-n\right) \mathbf {H}_{D}t}\mathbf {C}^{-1}c_{n}\right) . \end{aligned}$$

(B2)

This expression can straightforwardly be handled to calculate the time-dependent marginal probability of the dichotomic process $P\left( \xi (t_{1});\xi (t_{0})\right)$; that is,

$$\begin{aligned} P\left( \xi (t_{1});\xi (t_{0})\right) =\mathcal {M}\oplus \left\langle \sigma ^{\pm }\left| e^{\mathbf {H}\left( t_{1}-t_{0}\right) }\right| \sigma ^{\pm }\right\rangle P\left( \sigma ^{\pm }\left( t_{0}\right) \right) . \end{aligned}$$

(B3)

Using the explicit expression of the invariants $c_{n}$ (of the matrix $e^{\mathbf {H}_{D}t}$), introducing the elements of the $4\times 4$ matrix $\mathbf {C}e^{\left( 4-n\right) \mathbf {H}_{D}t}\mathbf {C}^{-1}$ and using a stationary probability $P\left( \sigma ^{\pm }\left( t_{0}\right) \right)$, we can check the Markovian ($2\times 2$) limit. That is, for example for $a\rightarrow b,\forall p\in \left( 0,1\right)$ we get:

$$\begin{aligned} \begin{array}{rcl} c_{1}&{}= &{} -1-2e^{-bt}-e^{-2bt} \\ c_{2}&{}= &{} 2(e^{-bt}+e^{-2bt}+e^{-3bt}) \\ c_{3}&{}= &{} -e^{-2bt}-2e^{-3bt}-e^{-4bt} \\ c_{4}&{}= &{} e^{-4bt} \end{array}, \end{aligned}$$

(B4)

and for $p=(1-q)\rightarrow 0$ and $\forall b$ and $a=0$ we get:

$$\begin{aligned} \begin{array}{rcl} c_{1}&{}= &{} -3-e^{-2bt} \\ c_{2}&{}= &{} 3+3e^{-2bt} \\ c_{3}&{}= &{} -1-3e^{-2bt} \\ c_{4}&{}= &{} e^{-2bt} \end{array}. \end{aligned}$$

(B5)

Introducing any result from (B4) or (B5), in (B2) and (B3) we get

$$\begin{aligned} P\left( \xi (t)\left| \xi (0)\right. \right) =\frac{1}{2}\left( \begin{array}{cc} 1+e^{-2bt} &{} 1-e^{-2bt} \\ 1-e^{-2bt} &{} 1+e^{-2bt} \end{array}\right) , \end{aligned}$$

which is the expected Markovian limit.

1.2 2. Laplace Correspondence

A general one-to-one correspondence is simpler to show when considering the conditional probabilities $P\left( \xi (t)\left| \xi (0)\right. \right) \in \mathcal {R}_{e}^{2\times 2}$ and $P\left( \sigma (t)\left| \sigma (0)\right. \right) \in \mathcal {R}_{e}^{4\times 4}$ in the Laplace representation.

For a stationary renewal process the probability to have n-events in the time interval $\left[ 0,t\right]$ can be red from (19) or (20). Then, using (6) and Feller’s proposition, in the Laplace representation, we get:

$$\begin{aligned} \mathcal {P}_{n}\left( u\right) =\frac{1-\psi \left( u\right) }{u} \psi _{1}\left( u\right) \psi \left( u\right) ^{n-1},\ \forall n\ge 1. \end{aligned}$$

(B6)

Given an arbitrary discrete-time recurrence relation of the form:

$$\begin{aligned} \mathbf {W}_{n}=\mathbf {T}\cdot \mathbf {W}_{n-1},\ n=0,1,2,\ldots , \ \mathbf {W}_{0}=\mathbf {1}\text { (identity),} \end{aligned}$$

(B7)

where matrix $\mathbf {W}\in \mathcal {R}_{e}^{2\times 2}$ and $\mathbf {T}$ is a $\left( 2\times 2\right)$ Markov transition matrix. The Green solution of (B7) is $\mathbf {W}_{n}=\mathbf {T}^{n},\ \forall n\ge 0$. Then, its continuous-time representation can be written using a general renewal process as

$$\begin{aligned} \mathbf {W}\left( t\right) =\sum _{n=1}^{\infty }\mathcal {P}_{n} \left( t\right) \mathbf {T}^{n}+\left( 1-\int _{0}^{t}\psi _{1} \left( \tau \right) d\tau \right) \mathbf {W}_{0},\quad t\ge 0. \end{aligned}$$

(B8)

So in the Laplace representation we get

$$\begin{aligned} \mathbf {W}\left( u\right) =\frac{1-\psi \left( u\right) }{u}\psi _{1} \left( u\right) \left( 1-\psi \left( u\right) \mathbf {T}\right) ^{-1}\mathbf {T} +\frac{1-\psi _{1}\left( u\right) }{u}\mathbf {1}. \end{aligned}$$

(B9)

This formula can easily be applied to our intermittent dichotomic model adopting a symmetric Markov matrix $\mathbf {T}\in \mathcal {R}_{e}^{2\times 2}$, and using $\psi \left( t\right)$ and $\psi _{1}\left( t\right)$. Then, we get an exact expression, in the Laplace representation, for the conditional probability of the intermittent dichotomic noise.

On the other hand the propagator of the 4-dimensional Markov process (29) is $\mathbf {P}\left( u\right) =\left( u-\mathbf {H}\right) ^{-1}\in \mathcal {R}_{e}^{4\times 4}$, where $\mathbf {H}$ is the Master Hamiltonian (30). Then, we can show, using symbolic algebraic computation that the following equality is maintain:

$$\begin{aligned} \mathbf {W}\left( u\right) =\mathcal {M}\oplus \left\langle \sigma \left| \left( u-\mathbf {H}\right) ^{-1}\right| \sigma \right\rangle P\left( \sigma \left( 0\right) \right) . \end{aligned}$$

(B10)

This result concludes our proof.

A particular form of the 4-dimensional initial condition $P\left( \sigma \left( 0\right) \right)$ can be put in a one-to-one correspondence with the ordinary or ongoing renewal process associated to the intermittent waiting-time (12). Explicit calculations are shown Sect. 5.1.1