Covariance of a Poisson process at two times

Question

Let $\{N(t)\}$ be a Poisson process with intensity $5$ per unit time. Find $\operatorname{Cov}\big(N(5),N(15)\big)$ .

Short answer

$25$ . Split $N(15)=N(5)+\big(N(15)-N(5)\big)$ ; the increment is independent of $N(5)$ , so the covariance reduces to $\operatorname{Var}(N(5))=5\cdot5=25$ . In general $\operatorname{Cov}(N(s),N(t))=\lambda\min(s,t)$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Defining the Poisson Process: Three Equivalent Constructions
Expectation, Mean, and Variance
Conditional Expectation and Variance Revisited

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