Poisson process and expected value

Question

Let $N(1);t\leq0$ be a Poisson process with intensity $\lambda >1$. I have

$E((N(2)−N(1))(N(4)−N(2))) + E((N(2)−N(1))^2) + E(N(1)(N(4)−N(1)))$

which somehow ends up as $2\lambda^2+\lambda+\lambda^2+3\lambda^2$. Somehow I can't figure out where the the single $\lambda$ comes from, what am I missing?

Answer 1

Lemma1: $E N(t) = \lambda t$.

Lemma2: $Var N(t) = \lambda t$.

Lemma3: $N(t) - N(s)$ is independent of $N(u) - N(v)$ where the intervals $(s, t)$ and $(u, v)$ is disjoint.

Lemma4: the CDF of $N(t+s) - N(t)$ is equal to $N(s)$.

Part 1: $$ \begin{align} E((N(2)−N(1))(N(4)−N(2))) &= E((N(2)−N(1))E(N(4)−N(2))) \\ &=EN(1)EN(2) \\ &=2\lambda^2 \end{align} $$

Part 2: $$ \begin{align} E((N(2)−N(1))^2 &= EN(1)^2 \\ &= Var N(1) + (EN(1))^2 \\ &= \lambda + \lambda^2 \end{align} $$

Part 3: \begin{align} E(N(1)(N(4)−N(1))) &= EN(1) E[N(4) - N(1)] \\ &= \lambda (4\lambda - \lambda) \\ &= 3\lambda^2 \end{align}

Answer 2

Beware that $X:=N(2)-N(1)\sim\text{Poisson}(\lambda)$ so that:$$\mathbb EX^2=\mathsf{Var}(X)+(\mathbb EX)^2=\lambda+\lambda^2$$