Telegraph vs Bernoulli processes

Question

I am fairly new to stochastic processes, but I have been interested in studying telegraph processes to model attachment and detachment dynamics (between cells, in a biological context). However, I would like to understand a specific detail on these processes.

In the properties of a telegraph process, it is written that "knowledge of an initial state decays exponentially", and so, for a time $t\gg (\lambda_1+\lambda_2)^{-1}$, the stationary values have mean $$\tag{*} \langle X\rangle_s =\frac{\lambda_2}{\lambda_1+\lambda_2} $$ where we took $c_1=1$ and $c_2=0$ (in the Wikipedia definition). Does this imply that, if I am simulating a discretisation of such a process, as long as my time increment $\Delta t$ satisfies $\Delta t(\lambda_1+\lambda_2)\gg 1$, I can approximate this process by a Bernoulli process with probability given by (*)? In other words, if the timescale at which we make the observation is longer than the inverse of these rates, can we expect the process to be memoryless every time we observe, i.e. a Bernoulli process?

If that is the case, do you know of any reference for this? Apologies if this is standard knowledge, but I would like to understand this limit better.

Answer 1

Yes, you can make this approximation. The Wikipedia article you link to is terse, but it gives the basic formulae for computing probabilities for the telegraph process. If the telegraph process is, say, $(X_t)_{t\ge 0}$ and has possible states $c_1$ and $c_2$, then, since it is a time-homogeneous continuous-time Markov process, its behavior can be expressed by the transition probabilities $$p_{ij}(t):={\bf P}(X_{s+t}=c_j\mid X_s=c_i), \qquad s, t\ge 0,\qquad i,j=1,2,$$ which are independent of $s$. We have $$p_{ij}(0)=\delta_{ij}, \qquad \frac{d}{dt} p_{ij}(t)=-\lambda_j p_{ij}(t)+\lambda_{3-j} p_{i(3-j)}(t),$$ and solving these first-order differential equations gives \begin{eqnarray*} p_{ij}(t)&=&\frac{\lambda_{3-j}}{\lambda_1+\lambda_2}+\frac{\lambda_{j}}{\lambda_1+\lambda_2}e^{-(\lambda_1+\lambda_2)t},\qquad i=j, \\ &=&\frac{\lambda_{3-j}}{\lambda_1+\lambda_2}-\frac{\lambda_{3-j}}{\lambda_1+\lambda_2}e^{-(\lambda_1+\lambda_2)t},\qquad i\ne j. \end{eqnarray*} So, if we assume that ${\bf P}(X_0=c_j)$ equals the limiting probability of $c_j$, $\lambda_{3-j}/(\lambda_1+\lambda_2)$, then the probability of a given sequence of observations is $$ {\bf P}(X_{t_1}=c_{i_1}, \dots, X_{t_n}=c_{i_n})= \prod_j \frac{\lambda_{3-i_j}}{\lambda_1+\lambda_2} \prod_{i_j=i_{j+1}} (1+e^{-(\lambda_1+\lambda_2)(t_{j+1}-t_j)}\frac{\lambda_{i_{j+1}}}{\lambda_{3-i_{j+1}}}) \prod_{i_j\ne i_{j+1}} (1-e^{-(\lambda_1+\lambda_2)(t_{j+1}-t_{j})}), $$ where we assume $$i_1, \dots, i_n\in\{1,2\}, \qquad t_n\ge t_{n-1}\ge \cdots \ge t_1\ge 0.$$ The first part of the right-hand side of this formula is the probability that you would observe the given states in a Bernoulli process $(Y_k)_{k=1,...,n}$ with ${\bf P}(Y_k=c_j)=\lambda_{3-j}/(\lambda_1+\lambda_2)$: $$ {\bf P}(Y_{1}=c_{i_1}, \dots, Y_{n}=c_{i_n})= \prod_j \frac{\lambda_{3-i_j}}{\lambda_1+\lambda_2}, $$ and the second and third parts will approach 1 as the smallest time interval between successive observations, $\min_j \ t_{j+1}-t_j$, becomes large.