Suppose $N_{t}$ is an adapted Poisson process with parameter $\lambda$ on a filtered probability space $(\Omega, \mathcal{F}, \mathbb{P})$. One defines the compensated Poisson process by $M_{t} := N_{t} - \lambda t$. I am trying to show that the compensated process is a martingale. I need to show that
- $M_{t}$ is adapted to $\mathcal{F}_{t}$,
- $M_{t}$ is integrable,
- for $s < t$, $\mathbb{E}(M_{t} | \mathcal{F}_{s}) = M_{s}$.
I am having trouble with $1$ and $3$. For question $1$, I know that $N_{t}$ is adapted by assumption, but why is $N_{t} - \lambda t$ adapted? The definition of adapted says that $N_{t} - \lambda t$ must be $\mathcal{F}_{t}$ measurable; I don't immediately see why this is the case. Perhaps it is something to do with the fact that translations of measurable random variables are measurable?
For question $3$, we calculate
\begin{aligned}\mathbb{E}(N_{t} - \lambda t ~|~ \mathcal{F}_{s}) &= \mathbb{E}(N_{t} - N_{s}~ |~\mathcal{F}_{s}) + \mathbb{E}(N_{s} - \lambda t ~|~ \mathcal{F}_{s}) \\[1ex] &= \mathbb{E}(N_{t} - N_{s}) + N_{s} - \lambda t,\end{aligned}
but I don't understand why $\mathbb{E}(N_{s} - \lambda t ~|~ \mathcal{F}_{s}) = N_{s} - \lambda t$. I assume that $N_{s} - \lambda t$ must be $\mathcal{F}_{s}$ -measurable as then this follows from properties of conditional expectation. But why is this the case? I get that $N_{s}$ is $\mathcal{F}_{s}$ -measurable by hypothesis but I don't see why $N_{s} - \lambda t$ is.
I guess my basic probability theory is rusty so I would appreciate if someone could spell out the argument for me, thank you!