Covariance of Poisson Counts

Split the later count at time $5$ :

N(15) = N(5) + \bigl(N(15)-N(5)\bigr).

By independent increments, $N(15)-N(5)$ is independent of $N(5)$ , so its covariance with $N(5)$ is zero. Bilinearity of covariance then gives

\operatorname{Cov}\bigl(N(5),N(15)\bigr) = \operatorname{Var}\bigl(N(5)\bigr) + \operatorname{Cov}\bigl(N(5),\,N(15)-N(5)\bigr) = \operatorname{Var}\bigl(N(5)\bigr).

Since $N(5)\sim\mathrm{Poisson}(5\cdot 5)=\mathrm{Poisson}(25)$ , its variance is $25$ . Hence

\boxed{\operatorname{Cov}\bigl(N(5),N(15)\bigr)=25}

Generalisation. The same argument shows that for a Poisson process of rate $\lambda$ , $\operatorname{Cov}\bigl(N(s),N(t)\bigr)=\lambda\min(s,t)$ . The correlation is therefore $\sqrt{\min(s,t)/\max(s,t)}$ , which here equals $1/\sqrt3\approx 0.577$ .