To prove that $X(t) = N(t+L) - N(t) , L > 0$ is Covariance stationary given $\{N(t) | t \geq 0\}$ is a Poisson Process.

Question

Let $\{N(t) | t \geq 0\}$ be a Poisson process.

I have to prove that $X(t) = N(t+L) - N(t) , L > 0$ is Covariance stationary.

That is I have to prove the following points -

$1) E(X(t)) , E(X(t)^2)$ are finite.

$2) m(t) = E(X(t)) $ is independent of $t$.

$3) Cov(X(t),X(t+h))$ depends only on $h \forall t \in T$.

Well I got the first two bits but I am stuck with $3)$ bit showing the covariance.

In the solution it follows like this -

Let $s < t$.

$cov(X(t), X(s)) = E(X(t)X(s)) − E(X(t))E(X(s))$

$= E((X(t) − X(s) + X(s))X(s)) − ( \lambda L)^2 = E(X(t) − X(s))E(X(s)) + E(X^2(s)) − (λL)^2$

$= 0 ∗ E(X(s)) + λL$

$= λL$ Now which could be taken as the function of $t-s$.

But I am facing trouble in understanding how $E(X(t) -X(s)) = 0 ?$, I think this shoukd be $\lambda (t-s)$.

Answer 1

\begin{align} E(X(t)-X(s))&=E(N(t+L)-N(t)-N(s+L)+N(s)), \text{by definition of } X \\ &=E(N(t+L)-N(t)) -E(N(s+L)-N(s))\\ &=E(N(L))-E(N(L)) , \text{by stationary increment of } N\\ &=0 \end{align}

Alternatively, in part $2$, you ahve proven that $m(t)$ is independent of $t$. Hence $$E(X(t)-X(s))=m(t)-m(s)=0.$$