Stochastic Processes – Problem 45: Prove that is both strictly stationary and wide-sense stationary

Let $X = X_{t}, t\geq0$ be a continuous-time stochastic process. Give reasonable definitions for $X$ to be strictly stationary and wide-sense stationary. Let $\{N_{t},t\geq0\}$ be a Poisson process with intensity $\lambda$, and $X_{t}=N_{t+1}-N_{t}, t\geq0$. Prove that $X$ is both strictly stationary and wide-sense stationary.

Step 1. Give the definitions of strict stationarity and wide-sense stationarity. A stochastic process $\{X_t, t \ge 0\}$ is called strictly stationary if for any positive integer $n$, any time points $t_1, t_2, \dots, t_n \ge 0$, and any time shift $h > 0$, the random vectors $(X_{t_1}, X_{t_2}, \dots, X_{t_n})$ and $(X_{t_1+h}, X_{t_2+h}, \dots, X_{t_n+h})$ have the same joint distribution. A stochastic process $\{X_t, t \ge 0\}$ is called wide-sense stationary (or second-order stationary) if it satisfies three conditions: 1. Finite second moments: $E[X_t^2] < \infty$ for all $t$; 2. Constant mean function: $E[X_t] = \mu$ for all $t$; 3. The autocovariance function $Cov(X_t, X_{t+s})$ depends only on the time lag $s$, not on the starting time $t$.

Step 2. Prove that $X_t = N_{t+1} - N_t$ is strictly stationary. The Poisson process $\{N_t\}$ has stationary increments, meaning for any $t \ge 0$ and $h > 0$, the increment $N_{t+h} - N_t$ depends only on the time interval length $h$, following a Poisson distribution with parameter $\lambda h$. For $X_t = N_{t+1} - N_t$, this is exactly an increment over an interval of length 1. The shift $t \to t+h$ preserves all interval lengths (all equal to 1) and the relative positions (overlap lengths) of the intervals $[t_i, t_i+1]$. Since the Poisson process has stationary independent increments, the joint distribution is invariant under the shift. Hence $X_t$ is strictly stationary.

Step 3. Prove that $X_t$ is wide-sense stationary. First verify finite second moments: $X_t \sim \text{Poisson}(\lambda)$, so $E[X_t^2] = \lambda + \lambda^2 < \infty$. The mean is constant: $E[X_t] = \lambda$. For the autocovariance, let $s \ge 0$. If $s \ge 1$, the intervals $[t, t+1]$ and $[t+s, t+s+1]$ are disjoint, so $X_t$ and $X_{t+s}$ are independent and $Cov(X_t, X_{t+s}) = 0$. If $0 \le s < 1$, decompose into independent increments: $X_t = A+B$, $X_{t+s} = B+C$ where $A = N_{t+s} - N_t$, $B = N_{t+1} - N_{t+s}$, $C = N_{t+s+1} - N_{t+1}$ are mutually independent. Then $Cov(X_t, X_{t+s}) = Var(B) = \lambda(1-s)$. The covariance $R(s) = \lambda(1-|s|)$ for $|s|<1$ and $0$ otherwise depends only on $s$, not on $t$. Hence $X_t$ is wide-sense stationary.

(1) Strict stationarity: for any $n, t_{1}, \dots, t_{n}, h$, $(X_{t_{1}}, \dots, X_{t_{n}})$ and $(X_{t_{1}+h}, \dots, X_{t_{n}+h})$ have the same distribution. Wide-sense stationarity: finite second moments, constant mean, autocovariance depends only on the time lag. (2) QED.