I have to find the following expectation of a Poisson process with parameter $\lambda$ "in terms of independence":
$E(X(t)X(t+s))$
I supposed that "in terms of independence" means that both random variables are independent, so I tried the following:
$E(X(t)X(t+s)) = E(X=t)E(X=t+s)= \lambda E(X=t+s)$
Here is where I start struggling. I do not know how to work with $E(X(t+s))= \sum_{k=0}^{\infty} k P(X=t+s)$. I tried using conditional probability to see if I could convert the formula into something like $ \sum_{n=s}^{\infty}kP(....) $ but that left me confused.
Also, how would you compute the original expectation if the random variables were dependent?
Poisson process has independent increments property. In terms of independence means, use independent increment property to obtain the required expectation. \begin{align*} E\left\{X(t)\cdot X(t+s)\right\}& = E\left[X(t)\left\{\overbrace{X(t+s)-X(t)}+X(t)\right\}\right]\\ & = E\left[X(t)\left\{X(t+s)-X(t)\right\}\right] + E\left[X(t)\cdot N(t)\right]\\ & = E\left[X(t)\right]\cdot E\left[X(t+s) - X(t)\right] + E\left[X^{2}(t)\right]\\ &\qquad\mbox{(because of independent increment property}\\ & \qquad\mbox{of the Poisson Process.)}\\ & = \lambda t\cdot E\left[X(s)\right] + \left[\lambda t + (\lambda t)^{2}\right]\\ & \qquad \mbox{(on using the stationarity property) }\\ & = \lambda t\cdot \lambda s + \lambda t + (\lambda t)^{2}\\ & = \lambda t + \lambda t\cdot(\lambda t + \lambda s). \end{align*}