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

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.

• The expectation will be determined by the probability that the sign flips exactly an even number of times. To obtain this, you need to calculate the sum of the infinite series whose terms are $\mathbb{P}(N = 2k)$ for all nonnegative $k$. – fblundun Nov 29 '20 at 21:40

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.