Expected value of stochastic process given probability of number of sign changes on interval

Question

We have a stochastic process $X_t$, which at a given time $t$ have a value of $-1$ or $1$. Number of sign changes on an interval $(t; t + \Delta)$ have a Poisson distribution $P(N = k) = e^{-\lambda\Delta} \cdot \frac{(\lambda\Delta)^k}{k!}, \lambda > 0, k \geq 0$. Compute the expected value $E[X_t]$, covariance $K(t_1, t_2)$ and variance $V[X_t]$.

I sort of can't wrap my head around how to take into account all the possible ways the sign change can occur in a sequence (which is probably not the correct way of thinking about it). Any hint would be appreciated.

Answer 1

In the communications literature, this stochastic process is referred to as a random telegraph wave or semi-random telegraph wave depending on whether $X_0$ is modeled as a random variable equally likely to take on values $+1$ and $-1$ or as a deterministic value of $+1$ or $-1$ with no randomness involved. Now consider a Poisson process with arrival rate $\lambda$ and let $N(t,t+s)$ denote the number of arrivals in the interval $(t,t+s]$ so that $N(t,t+s)$ is a Poisson random variable with parameter $\lambda s$. With all this as background material, the process $X_t$ in question has the property that $$X_{t+s} = \begin{cases}\,X_t, & \text{if}~N(t,t+s)~ \text{is an even number,}\\ -X_t, & \text{if}~N(t,t+s)~ \text{is an odd number.}\end{cases}$$ Since \begin{align}P(N(t,t+s)~ \text{is an even number}) &= \exp(-\lambda s)\cosh (s),\\ P(N(t,t+s)~ \text{is an odd number}) &= \exp(-\lambda s)\sinh (s)\end{align} we get that \begin{align}P(X_{t+s}=X_t) &= \exp(-\lambda s)\cosh (s),\\ P(X_{t+s} = -X_t) &= \exp(-\lambda s)\sinh (s)\end{align} and \begin{align}P(X_{s}=X_0) &= \exp(-\lambda s)\cosh (s),\\ P(X_{s} = -X_0) &= \exp(-\lambda s)\sinh (s).\end{align} For the random telegraph wave, $X_0$ is equally likely to have value $+1$ and $-1$, and so \begin{align}P(X_s=+1) &= P(X_0=+1) \exp(-\lambda s)\cosh (s) + P(X_0=-1) \exp(-\lambda s)\sinh (s)\\ &= \frac 12 \exp(-\lambda s)\cosh (s) + \frac 12 \exp(-\lambda s)\sinh (s)\\ &= \frac 12.\end{align} Thus, $E[X_t] = 0$ for all $t\geq 0$ and $K(t,t+s) = E[X_tX_{t+s}]$ can be figured out from the information provided above.

For the semirandom telegraph wave with $X_0=+1$ (say),

$$P(X_t=+1) = \exp(-\lambda t)\cosh (t)$$ which has value $1$ at $t=0$ and then decays away to approach $\frac 12$ as $t\to \infty$ while $$P(X_t=-1) = \exp(-\lambda t)\sinh (t)$$ which has value $0$ at $t=0$ and then increases to approach $\frac 12$ as $t\to \infty$. I will leave it to the OP to determine whether $E[X_t]=0$ for all $t\geq 0$ in this case also, and to figure out what $K(t,t+s)$ is from this information. As $s \to \infty$ for fixed $t$, the semi-random telegraph wave looks asymptotically like the random telegraph wave.