$ N=N(t) $ Poisson process with intensity ( \lambda>0 ).
(a) $ X=\{X(t), t \geqslant 0\} $ given by $ X(t)=N(t)-t \lambda , t \geqslant 0 $ Calculate $EX(t)$ and $ {Cov}(X(s), X(t)) \quad \forall s, t \geqslant 0 $.
(b) Show $ \mathbb{E}[X(t) \mid X(s)]=X(s) \quad \forall 0<s<t $.
For a.): Expected value should be 0 since $E(X(t)) = E(N(t))-E(t \lambda)= t \lambda - t \lambda = 0$.
For the covariance I would have done the following : $Cov(X(s), X(t)) = E((X(s)-E(X(s))(X(t)-E(X(t)))) = E(X(s)X(t)) = ...$ , but then I get stuck at the point E(N(s)N(t)), can you just separate that by the property of independent increments?
With b.) I ask myself how one could proceed there at the best, thus linearity brings already much, but then I remain stuck again in the middle.