Problem let $\{X(t)\}_{t\geq 0}$ be a Poisson process having rate $\lambda = 2$
Determine $E[X(1)\cdot X(2)]$
Having searched the site, it seems the solution lies in the fact that one can split the values inside the expectation function so they are independent and thus use linearity. However, I have not been able to get the correct answer.
Any assistance is appreciated thank you
Poisson processes have independent increments, so $X_1$ is independent of $X_2-X_1$. Therefore
$$ \mathbb E X_1X_2=\mathbb E X_1^2 + \mathbb E X_1\ \mathbb E (X_2-X_1). $$
Since the process has rate 2, both $X_1$ and $X_2-X_1$ are Poisson with mean 2. The variance of a Poisson random variable is equal to its mean, so $\mathbb EX_1^2=\text{Var}(X_1^2)+(\mathbb E X_1)^2=2+4=6$. Thus $$ \mathbb EX_1X_2=6+2\cdot 2=10. $$