The standard Poisson process is just the process \(N(t)\) where \(N(t)\) takes integer values, the increments are independent ( \(N(t_2)-N(t_1)\) is independent of \(N(t_4)-N(t_3)\) for \(t_1 < t_2 \leq t_3 < t_4\)) and for \(t> s \geq 0\) and \(n \in \mathbf{N}\)
\[\mathbf{P}\big(N(t)-N(s)=n\big)=e^{-(t-s)}\frac{(t-s)^n}{n!}\]
Here “standard” just means rate one Poisson process.
Define the process \[X(t)=\xi \cdot (-1)^{N(t)}\, ,\] where \(\xi\) is a random variable independent of the standard Poisson process \(N(t)\) that take values \(\pm 1\) with probability \(\frac12\). Clearly \(X(t)\) takes only two values, \(\pm 1\). Show that \(X(t)\) is stationary and that its covariance is \(e^{-2|t-s|}\).
The stationary Ornstein-Uhlenbeck process is a Gaussian process with mean zero and covariance \(R(t,s)=\frac{1}{2}e^{-|t-s|}\). Thus the OU process and \[Y(t)=\frac{1}{\sqrt 2}X(\frac{t}{2})\] are both stationary and have the same covariance but are very different processes. Does \(X(t)\) satisfy the Kolmogorov condition for path continuity? Does the OU process?