Expected value Poisson process

Question

Let $\{N(t)\mid t\geq 0\}$ be a Poisson process with rate $\lambda$. Find $\mathbb{E}[N(t) N(2t)]$. Attempt: $\mathbb{E}[N(t) N(2t)]=\mathbb{E}[\mathbb{E}[N(t) N(2t)\mid N(t)]]=\mathbb{E}[N(t) \mathbb{E}[N(2t)\mid N(t)]]=\mathbb{E}[N(t) [N(t)+\lambda t]]=\mathbb{E}[N(t)^2]+\lambda t \mathbb{E}[N(t)]=\lambda t +(\lambda t)^2 + \lambda t \mathbb{E}[N(t)]=\lambda t + 2(\lambda t)^2$. My question is very simple: are my calculations correct, because I have no solution manual and my intuition does not really help.

Answer 1

Your result is indeed correct, although I'd say that the third equality needs some explanation. A slightly different approach would be the following:

We want to calculate $$ \mathbf E (N(2t)N(t)) $$ but this equals $$ \mathbf E (N(2t)N(t))=\mathbf E ((N(2t)-N(t))N(t)+N(t)^2)=\mathbf E ((N(2t)-N(t))N(t))+\mathbf E (N(t)^2) \tag1 $$ and now we just apply the independent (and stationary) increments property of the Poisson process and so we get \begin{align} \mathbf E ((N(2t)-N(t))N(t))+\mathbf E (N(t)^2)=&\mathbf E (N(2t)-N(t)) \mathbf E(N(t))+\mathbf E (N(t)^2)\\ =&\mathbf E^2 (N(t))+\mathbf E (N(t)^2)=(t\lambda)^2+t\lambda+(t\lambda)^2\\ =&2(t\lambda)^2+t\lambda \end{align}