Ok so I'm pretty stuck on this, even after searching through a bunch of other questions on here that seem very similar.
Let $\{X(t), t\geqslant0\}$ be a poisson process with rate $\lambda$.
For $0 < s < t, \text{find E}[X(s)X(t)]$.
I have two trains of thought though, one of them being to calculate the joint distribution using the double integral. The other is to use the independent increment ideas, but I'm not sure how to go about it.
I figure that $X(s) \sim \text{Poi}(λs)$ and $X(t-s) \sim \text{Poi}(λ(t-s))$, and obviously $X(t) = X(s) + X(t-s)$.
I then understand that $X(s)$ and $X(t-s)$ are independent. But how can I go from there, figuring out what the distribution of $X(t)$ looks like?
Provided that a Poisson process observes independent increment, i.e., $X_t-X_s$ is independent from $X_s$ for all $t\ge s\ge 0$, the desired expectation is thus \begin{align} \mathbb{E}\left(X_tX_s\right)&=\mathbb{E}\left[\left(X_t-X_s\right)X_s+X_s^2\right]\\ &=\mathbb{E}\left[\left(X_t-X_s\right)X_s\right]+\mathbb{E}\left(X_s^2\right)\\ &=\mathbb{E}\left(X_t-X_s\right)\mathbb{E}\left(X_s\right)+\mathbb{E}\left(X_s^2\right)\\ &=\lambda\left(t-s\right)\lambda s+\lambda s\left(1+\lambda s\right)\\ &=\lambda s\left(\lambda t+1\right). \end{align}